tag:blogger.com,1999:blog-27838424792600037502024-03-12T21:45:43.518-07:00Python for Signal ProcessingUsing Python to investigate signal processing concepts in the IPython notebook format. Source notebooks available at github.com/unpingco/Python-for-Signal-Processing.Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.comBlogger17125tag:blogger.com,1999:blog-2783842479260003750.post-65205588985321510672014-04-30T16:36:00.000-07:002014-05-01T05:56:38.604-07:00Random walks and stumbles<div class="border-box-sizing" id="notebook" tabindex="-1">
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<h2 id="Random-walks-and-random-stumbles">
Random walks and random stumbles</h2>
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Random walks are a gold mine for a wide variety of stochastic theory and practice. They are easy to explain, easy to code, and easy to misunderstand. In this section, we start out with the simplest imaginable random walk and then show how things can go wrong. Note that this post is also available in <a href="http://nbviewer.ipython.org/github/unpingco/Python-for-Signal-Processing/blob/master/random_walk_random_stumble.ipynb">nbviewer</a> where the math below may render better than here in blogger.<br />
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Let's examine the problem of the one-dimensional random walk. Consider a particle at the origin ($x=0$), with $p$ probability of moving to the right and $q=1-p$ probability of moving to the left. When the particle reaches $x=1$, the experiment terminates. On average, how many steps are required for this to terminate if $p=1/2$? This seems like a reasonable question, doesn't it?<br />
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The first thing we have to do is establish the conditions under which termination occurs. Thus, we need the probability that the particle ultimately reaches $x=1$. We'll call this probability $P$. On average, how many steps does it make to terminate?<br />
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This experiment is easy enough to code as shown below:</div>
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In [168]:
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<pre><span class="k">def</span> <span class="nf">walk</span><span class="p">():</span>
<span class="s">'starting at x=0, step left/right with probability 1/2 until x=1'</span>
<span class="n">x</span><span class="o">=</span><span class="mi">0</span>
<span class="k">while</span> <span class="n">x</span><span class="o">!=</span><span class="mi">1</span><span class="p">:</span>
<span class="n">x</span><span class="o">+=</span><span class="n">random</span><span class="o">.</span><span class="n">choice</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span> <span class="c"># equi-probable left-right move</span>
<span class="k">yield</span> <span class="n">x</span>
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Some classical deft reasoning leads to:<br />
$$ P = p + q P^2 $$<br />
where $p$ is the probability it moves to $x=1$ from $x=0$ on the first try, and $q$ is the probability it does not, but then ultimately makes it back to $x=0$ (with probability $P$) and then again makes it from there to $x=1$, again with probability $P$ (thus, $P^2$). <br />
There are two solutions to $P$, $P=1$ and $P=p/(1-p)$. The crossover point is $p=1/2$. This is drawn below.</div>
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In [169]:
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<pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span> <span class="c"># want floating point division</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">p</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.5</span><span class="p">,</span><span class="mi">20</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">p</span><span class="p">,</span><span class="n">p</span><span class="o">/</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">p</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">r'$p<1/2$'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">3.</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">([</span><span class="mf">0.5</span><span class="p">,</span><span class="mi">1</span><span class="p">],[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">label</span><span class="o">=</span><span class="s">r'$p > 1/2$'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">3.</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mf">1.1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'$p$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$P$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'Probability of reaching $x=1$ from $x=0$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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<h2 id="Average-number-of-steps-to-termination">
Average number of steps to termination</h2>
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A straight-forward question to ask is what is the average number of steps it takes to ultimately terminate this random walk? The quick analysis above says that the particle will <em>ultimately</em> terminate, but, on average, how many steps are required for this to happen?<br />
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Unfortunately, the analysis above is not very helpful here because the statement is about the probability of ultimate termination, not the probability of termination <em>given</em> a particular point somewhere on the left of the origin.<br />
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Fortunately, we have all the Python-based tools to experimentally get at this. The following generator describes the $p=1/2$ random walker.<br />
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In [170]:
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<pre><span class="k">def</span> <span class="nf">walk</span><span class="p">():</span>
<span class="s">'starting at x=0, step left/right with probability 1/2 until x=1'</span>
<span class="n">x</span><span class="o">=</span><span class="mi">0</span>
<span class="k">while</span> <span class="n">x</span><span class="o">!=</span><span class="mi">1</span><span class="p">:</span>
<span class="n">x</span><span class="o">+=</span><span class="n">random</span><span class="o">.</span><span class="n">choice</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span> <span class="c"># equi-probable left-right move</span>
<span class="k">yield</span> <span class="n">x</span>
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Let's try generating a realization of this walk.</div>
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In [171]:
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<pre><span class="n">random</span><span class="o">.</span><span class="n">seed</span><span class="p">(</span><span class="mi">123</span><span class="p">)</span> <span class="c"># set seed for reproducibility</span>
<span class="n">s</span> <span class="o">=</span> <span class="nb">list</span><span class="p">(</span><span class="n">walk</span><span class="p">())</span> <span class="c"># generate the random steps</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">s</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">"particle's x-position"</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'step index k'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Example of random walk'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
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Now, that we're set up, we can generate a whole list of these walks and then average their lengths to estimate the mean of these walks.<br />
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<pre><span class="n">s</span> <span class="o">=</span> <span class="p">[</span><span class="nb">list</span><span class="p">(</span><span class="n">walk</span><span class="p">())</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">50</span><span class="p">)]</span> <span class="c"># generate 50 random walks</span>
<span class="n">len_walk</span><span class="o">=</span><span class="nb">map</span><span class="p">(</span><span class="nb">len</span><span class="p">,</span><span class="n">s</span><span class="p">)</span> <span class="c"># lengths of each walk</span>
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The following plots out a few of these random walks so we can get a feel of what's going on with the average length.<br />
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">s</span><span class="p">:</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">i</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">"particle's x-position"</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'step index k'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'average length=</span><span class="si">%3.2f</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="n">mean</span><span class="p">(</span><span class="n">len_walk</span><span class="p">)))</span>
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Now, here's where things get interesting! Fifty samples of the random walk is really not that many, so we'd like to hopefully get a better average by generating a lot more. If you try to generate, say, 1000 realizations, you'd be in for a long wait! This is because some of the random walks are really, really long! Furthermore, this is a <strong>persistent</strong> phenomenon; it's not just a bad draw from the random deck. Even if there is only one really long walk, it seriously distorts the average. It's tempting to conclude that this is just some outlier and get on with it, but <strong>not</strong> doing so will lead us to a very powerful theorem in stochastic processes.<br />
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To get a better picture of what's going on here, let's re-define our random walker function so we can limit how far it can go and thereby how long we'd have to wait.<br />
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<pre><span class="k">def</span> <span class="nf">walk</span><span class="p">(</span><span class="n">limit</span><span class="o">=</span><span class="mi">50</span><span class="p">):</span>
<span class="s">'limited version of random walker'</span>
<span class="n">x</span><span class="o">=</span><span class="mi">0</span>
<span class="k">while</span> <span class="n">x</span><span class="o">!=</span><span class="mi">1</span> <span class="ow">and</span> <span class="nb">abs</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o"><</span><span class="n">limit</span><span class="p">:</span>
<span class="n">x</span><span class="o">+=</span><span class="n">random</span><span class="o">.</span><span class="n">choice</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span>
<span class="k">yield</span> <span class="n">x</span>
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Because we really just want to count the walks, we can save memory by not collecting the individual steps and just reporting the length of the walk.<br />
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<pre><span class="k">def</span> <span class="nf">nwalk</span><span class="p">(</span><span class="n">limit</span><span class="o">=</span><span class="mi">500</span><span class="p">):</span>
<span class="s">'limited version of random walker. Only returns length of path, not path itself'</span>
<span class="n">n</span><span class="o">=</span><span class="n">x</span><span class="o">=</span><span class="mi">0</span>
<span class="k">while</span> <span class="n">x</span><span class="o">!=</span><span class="mi">1</span> <span class="ow">and</span> <span class="n">n</span><span class="o"><</span><span class="n">limit</span><span class="p">:</span>
<span class="n">x</span><span class="o">+=</span><span class="n">random</span><span class="o">.</span><span class="n">choice</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span>
<span class="n">n</span><span class="o">+=</span><span class="mi">1</span>
<span class="k">return</span> <span class="n">n</span> <span class="c"># return length of walk</span>
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<pre><span class="n">len_walk</span> <span class="o">=</span> <span class="p">[</span><span class="n">nwalk</span><span class="p">()</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">500</span><span class="p">)]</span> <span class="c"># generate 500 limited random walks</span>
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The usual practice would be to draw a histogram (i.e. an approximation of the probability <em>density</em>) here, but sometimes the automatic binning makes things hard to see. Instead, a <em>cumulative</em> distribution plot is helpful here. The excellent <code>pandas</code> module and some very useful data structures for this kind of work.<br />
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<pre><span class="kn">import</span> <span class="nn">pandas</span> <span class="kn">as</span> <span class="nn">pd</span>
<span class="kn">from</span> <span class="nn">collections</span> <span class="kn">import</span> <span class="n">Counter</span><span class="p">,</span> <span class="n">OrderedDict</span>
<span class="n">lw</span> <span class="o">=</span> <span class="n">pd</span><span class="o">.</span><span class="n">Series</span><span class="p">(</span><span class="n">Counter</span><span class="p">(</span><span class="n">len_walk</span><span class="p">))</span><span class="o">/</span><span class="nb">len</span><span class="p">(</span><span class="n">len_walk</span><span class="p">)</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">lw</span><span class="o">.</span><span class="n">index</span><span class="p">,</span><span class="n">lw</span><span class="o">.</span><span class="n">cumsum</span><span class="p">(),</span><span class="s">'-o'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">xmin</span><span class="o">=-</span><span class="mi">10</span><span class="p">,</span><span class="n">ymin</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Estimated Cumulative Distribution'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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What's interesting about the above plot is how steep the slope is. For a path length of one (terminating at the first step), we already have 50% of the probability accounted for. By a path-length of 100, we already have about 90% of the probability. The problem is squeezing out the remaining probability mass means computing random walks longer than 500 (our arbitrary limit). We can do all the above steps for higher limits far above 500, but this observation still holds. Thus, the problem with averaging this is that getting more probability further out competes with the lengths of those further paths. For the average to converge, we want to asymptotically get more improbable paths relatively faster than those paths grow!<br />
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Let's examine the standard deviation of our averages.</div>
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<pre><span class="k">def</span> <span class="nf">estimate_std</span><span class="p">(</span><span class="n">limit</span><span class="o">=</span><span class="mi">10</span><span class="p">,</span><span class="n">ncount</span><span class="o">=</span><span class="mi">50</span><span class="p">):</span>
<span class="s">'quick estimate of the standard deviation of the averages'</span>
<span class="n">ws</span><span class="o">=</span> <span class="n">array</span><span class="p">([[</span> <span class="n">nwalk</span><span class="p">(</span><span class="n">limit</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">ncount</span><span class="p">)]</span> <span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">ncount</span><span class="p">)])</span>
<span class="k">return</span> <span class="p">(</span><span class="n">limit</span><span class="p">,</span><span class="n">ws</span><span class="o">.</span><span class="n">mean</span><span class="p">(),</span> <span class="n">ws</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span><span class="o">.</span><span class="n">std</span><span class="p">())</span>
<span class="k">for</span> <span class="n">limit</span> <span class="ow">in</span> <span class="p">[</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">,</span><span class="mi">50</span><span class="p">,</span><span class="mi">100</span><span class="p">,</span><span class="mi">200</span><span class="p">,</span><span class="mi">300</span><span class="p">,</span><span class="mi">500</span><span class="p">,</span><span class="mi">1000</span><span class="p">]:</span>
<span class="k">print</span> <span class="s">'limit=</span><span class="si">%d</span><span class="s">,</span><span class="se">\t</span><span class="s"> average = </span><span class="si">%3.2f</span><span class="s">,</span><span class="se">\t</span><span class="s"> std=</span><span class="si">%3.2f</span><span class="s">'</span><span class="o">%</span> <span class="n">estimate_std</span><span class="p">(</span><span class="n">limit</span><span class="p">)</span>
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<pre>limit=10, average = 4.15, std=0.55
limit=20, average = 6.25, std=1.08
limit=50, average = 10.00, std=2.36
limit=100, average = 15.55, std=3.52
limit=200, average = 21.59, std=7.20
limit=300, average = 25.60, std=8.70
limit=500, average = 32.20, std=12.71
limit=1000, average = 47.38, std=21.40
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If this were converging, then the standard deviation of the mean (and the mean itself) should start converging as the walk limit increased. This is obviously not happening here.</div>
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<h2 id="Graph-based-combinatorial-approach">
Graph-based combinatorial approach</h2>
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So far, we have been looking at this using samples and started suspecting that there is something going on with the convergence of the average. However, this is not uncovering what's going on under the sheets with the convergence of the average. Yes, we've tinkered with varying the sample size, but it could still be the case that there is some much larger sample size out there that would cure all the problems we have so far experienced.<br />
<br />
The next code-block assembles some drawing utilities on the excellent <code>networkx</code> package so we can analyze this problem using graph combinatoric algorithms.<br />
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<pre><span class="kn">import</span> <span class="nn">networkx</span> <span class="kn">as</span> <span class="nn">nx</span>
<span class="kn">import</span> <span class="nn">itertools</span> <span class="kn">as</span> <span class="nn">it</span>
<span class="k">class</span> <span class="nc">Graph</span><span class="p">(</span><span class="n">nx</span><span class="o">.</span><span class="n">DiGraph</span><span class="p">):</span>
<span class="sd">'''</span>
<span class="sd"> operations assuming `pos` attribute in nodes to support drawing and</span>
<span class="sd"> manipulating path lattice.</span>
<span class="sd"> '''</span>
<span class="k">def</span> <span class="nf">draw</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">ax</span><span class="o">=</span><span class="bp">None</span><span class="p">,</span><span class="o">**</span><span class="n">kwds</span><span class="p">):</span>
<span class="sd">'''</span>
<span class="sd"> Draw based on `pos` attribute and pass kwds to nx.draw</span>
<span class="sd"> :param: axes(optional, default is None)</span>
<span class="sd"> '''</span>
<span class="n">pos</span> <span class="o">=</span> <span class="bp">self</span><span class="o">.</span><span class="n">get_pos</span><span class="p">()</span>
<span class="n">node_size</span><span class="o">=</span><span class="n">kwds</span><span class="o">.</span><span class="n">pop</span><span class="p">(</span><span class="s">'node_size'</span><span class="p">,</span><span class="mi">200</span><span class="p">)</span>
<span class="n">alpha</span><span class="o">=</span><span class="n">kwds</span><span class="o">.</span><span class="n">pop</span><span class="p">(</span><span class="s">'alpha'</span><span class="p">,</span><span class="mf">0.3</span><span class="p">)</span>
<span class="k">if</span> <span class="n">ax</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span> <span class="n">nx</span><span class="o">.</span><span class="n">draw</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span><span class="n">pos</span><span class="o">=</span><span class="n">pos</span><span class="p">,</span>
<span class="n">node_size</span><span class="o">=</span><span class="n">node_size</span><span class="p">,</span>
<span class="n">alpha</span><span class="o">=</span><span class="n">alpha</span><span class="p">,</span>
<span class="o">**</span><span class="n">kwds</span><span class="p">)</span>
<span class="k">else</span><span class="p">:</span> <span class="n">nx</span><span class="o">.</span><span class="n">draw</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span><span class="n">pos</span><span class="o">=</span><span class="n">pos</span><span class="p">,</span>
<span class="n">node_size</span><span class="o">=</span><span class="n">node_size</span><span class="p">,</span>
<span class="n">alpha</span><span class="o">=</span><span class="n">alpha</span><span class="p">,</span>
<span class="n">ax</span><span class="o">=</span><span class="n">ax</span><span class="p">,</span><span class="o">**</span><span class="n">kwds</span><span class="p">)</span>
<span class="k">def</span> <span class="nf">get_pos</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span><span class="n">n</span><span class="o">=</span><span class="bp">None</span><span class="p">):</span>
<span class="sd">'''</span>
<span class="sd"> n := str name of node</span>
<span class="sd"> get positions as returned dictionary</span>
<span class="sd"> '''</span>
<span class="n">pos</span><span class="o">=</span><span class="nb">dict</span><span class="p">([(</span> <span class="n">i</span><span class="p">,</span><span class="n">j</span><span class="p">[</span><span class="s">'pos'</span><span class="p">]</span> <span class="p">)</span> <span class="k">for</span> <span class="n">i</span><span class="p">,</span><span class="n">j</span> <span class="ow">in</span> <span class="bp">self</span><span class="o">.</span><span class="n">nodes</span><span class="p">(</span><span class="n">data</span><span class="o">=</span><span class="bp">True</span><span class="p">)])</span>
<span class="k">if</span> <span class="n">n</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span>
<span class="k">return</span> <span class="n">pos</span>
<span class="k">else</span><span class="p">:</span>
<span class="k">return</span> <span class="n">pos</span><span class="p">[</span><span class="n">n</span><span class="p">]</span>
<span class="k">def</span> <span class="nf">getx</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span><span class="n">x</span><span class="p">):</span>
<span class="k">return</span> <span class="nb">sorted</span><span class="p">([(</span><span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="n">i</span><span class="p">[</span><span class="mi">1</span><span class="p">][</span><span class="s">'val'</span><span class="p">]</span> <span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="bp">self</span><span class="o">.</span><span class="n">nodes</span><span class="p">(</span><span class="bp">True</span><span class="p">)</span> <span class="k">if</span> <span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">][</span><span class="mi">0</span><span class="p">]</span><span class="o">==</span><span class="n">x</span><span class="p">])</span>
<span class="k">def</span> <span class="nf">gety</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span><span class="n">y</span><span class="p">):</span>
<span class="k">return</span> <span class="nb">sorted</span><span class="p">([(</span><span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="n">i</span><span class="p">[</span><span class="mi">1</span><span class="p">][</span><span class="s">'val'</span><span class="p">]</span> <span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="bp">self</span><span class="o">.</span><span class="n">nodes</span><span class="p">(</span><span class="bp">True</span><span class="p">)</span> <span class="k">if</span> <span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">][</span><span class="mi">1</span><span class="p">]</span><span class="o">==</span><span class="n">y</span><span class="p">])</span>
<span class="c"># functions to allow diagonal lattice walking</span>
<span class="k">def</span> <span class="nf">diagwalk</span><span class="p">(</span><span class="n">level</span><span class="p">,</span><span class="n">n</span><span class="p">):</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">level</span>
<span class="n">y</span> <span class="o">=</span> <span class="o">-</span><span class="n">level</span>
<span class="k">while</span> <span class="n">y</span><span class="o"><=</span><span class="mi">1</span> <span class="ow">and</span> <span class="n">x</span><span class="o"><</span><span class="n">n</span><span class="o">+</span><span class="mi">1</span><span class="p">:</span>
<span class="k">yield</span> <span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">y</span><span class="p">)</span>
<span class="n">x</span><span class="o">+=</span><span class="mi">1</span>
<span class="n">y</span><span class="o">+=</span><span class="mi">1</span>
<span class="k">def</span> <span class="nf">diagwalker</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<span class="s">'daisy-chains the individual diagonal walkers'</span>
<span class="k">assert</span> <span class="n">n</span><span class="o">%</span><span class="k">2</span> <span class="c"># odd only</span>
<span class="k">return</span> <span class="n">it</span><span class="o">.</span><span class="n">chain</span><span class="p">(</span><span class="o">*</span><span class="p">(</span><span class="n">diagwalk</span><span class="p">(</span><span class="n">i</span><span class="p">,</span><span class="n">n</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="o">+</span><span class="mi">1</span><span class="p">)))</span>
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This next code-block constructs the lattice graph that we'll explain below.<br />
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<pre><span class="k">def</span> <span class="nf">construct_graph</span><span class="p">(</span><span class="n">level</span><span class="o">=</span><span class="mi">5</span><span class="p">):</span>
<span class="n">g</span><span class="o">=</span><span class="n">Graph</span><span class="p">()</span>
<span class="n">g</span><span class="o">.</span><span class="n">level</span> <span class="o">=</span> <span class="n">level</span>
<span class="n">g</span><span class="o">.</span><span class="n">add_nodes_from</span><span class="p">([(</span><span class="n">i</span><span class="p">,</span><span class="nb">dict</span><span class="p">(</span><span class="n">pos</span><span class="o">=</span><span class="n">i</span><span class="p">))</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">diagwalker</span><span class="p">(</span><span class="n">g</span><span class="o">.</span><span class="n">level</span><span class="p">)])</span>
<span class="k">for</span> <span class="n">x</span><span class="p">,</span><span class="n">y</span> <span class="ow">in</span> <span class="n">g</span><span class="o">.</span><span class="n">nodes</span><span class="p">():</span>
<span class="k">if</span> <span class="n">y</span><span class="o">!=</span><span class="mi">1</span> <span class="ow">and</span> <span class="n">x</span><span class="o"><</span> <span class="n">g</span><span class="o">.</span><span class="n">level</span><span class="p">:</span>
<span class="n">g</span><span class="o">.</span><span class="n">add_edge</span><span class="p">((</span><span class="n">x</span><span class="p">,</span><span class="n">y</span><span class="p">),</span> <span class="p">(</span><span class="n">x</span><span class="o">+</span><span class="mi">1</span><span class="p">,</span><span class="n">y</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span> <span class="p">)</span>
<span class="n">g</span><span class="o">.</span><span class="n">add_edge</span><span class="p">((</span><span class="n">x</span><span class="p">,</span><span class="n">y</span><span class="p">),</span> <span class="p">(</span><span class="n">x</span><span class="o">+</span><span class="mi">1</span><span class="p">,</span><span class="n">y</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span> <span class="p">)</span>
<span class="n">g</span><span class="o">.</span><span class="n">node</span><span class="p">[(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">)][</span><span class="s">'val'</span><span class="p">]</span><span class="o">=</span><span class="il">1L</span> <span class="c"># long int</span>
<span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="n">diagwalker</span><span class="p">(</span><span class="n">g</span><span class="o">.</span><span class="n">level</span><span class="p">):</span>
<span class="k">if</span> <span class="n">j</span><span class="o">==</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">):</span> <span class="k">continue</span>
<span class="n">x</span><span class="p">,</span><span class="n">y</span><span class="o">=</span><span class="n">j</span>
<span class="n">g</span><span class="o">.</span><span class="n">node</span><span class="p">[</span><span class="n">j</span><span class="p">][</span><span class="s">'val'</span><span class="p">]</span><span class="o">=</span><span class="nb">sum</span><span class="p">(</span><span class="n">g</span><span class="o">.</span><span class="n">node</span><span class="p">[</span><span class="n">k</span><span class="p">][</span><span class="s">'val'</span><span class="p">]</span> <span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="n">g</span><span class="o">.</span><span class="n">predecessors</span><span class="p">(</span><span class="n">j</span><span class="p">))</span>
<span class="k">return</span> <span class="n">g</span>
<span class="n">g</span><span class="o">=</span><span class="n">construct_graph</span><span class="p">()</span>
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The figure below shows a directed graph with the individual pathways the particle can take from $(0,0)$ to a particular end. The notation $(k,j)$ means that at iteration $k$, the particle is at $x=-j$ on the line. For example, all paths start at $(0,0)$ and there is only one path from $(0,0)$ to $(1,1)$. Likewise, the next terminus is at $(3,1)$, which is the pathway corresponding to reaching $x=1$ in three steps. There is only one path on the digraph that leads here. However, in this case, there are many pathways that are three-steps long, but that do not reach the terminus. For example, $(3,-1)$ is a path that leads to $x=-1$ in three steps.<br />
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Let's use some of the powerful algorithms in <code>networkx</code> to pursue these ideas.<br />
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Directed Path Lattice'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">g</span><span class="o">.</span><span class="n">draw</span><span class="p">(</span><span class="n">ax</span><span class="o">=</span><span class="n">ax</span><span class="p">)</span>
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The lattice diagram above shows the potential paths to termination for the random walk. For example, the all paths that lead to the node labeled (5,1) are those paths that take exactly five steps to terminate. Note that this is a directed graph so the graph can only be tranversed in the direction of the arrows (arrowheads denoted by thick ends as shown). Fortunately, <code>networkx</code> has powerful tools for computing these paths as shown in the following.<br />
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In [182]:
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<pre><span class="n">l5</span><span class="o">=</span><span class="nb">list</span><span class="p">(</span><span class="n">nx</span><span class="o">.</span><span class="n">all_simple_paths</span><span class="p">(</span><span class="n">g</span><span class="p">,(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">),(</span><span class="mi">5</span><span class="p">,</span><span class="mi">1</span><span class="p">)))</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">l5</span><span class="p">:</span>
<span class="k">print</span> <span class="n">i</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">))</span>
<span class="n">g</span><span class="o">.</span><span class="n">draw</span><span class="p">(</span><span class="n">ax</span><span class="p">,</span><span class="n">with_labels</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">g</span><span class="o">.</span><span class="n">subgraph</span><span class="p">(</span><span class="n">l5</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span><span class="o">.</span><span class="n">draw</span><span class="p">(</span><span class="n">ax</span><span class="o">=</span><span class="n">ax</span><span class="p">,</span><span class="n">with_labels</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">node_color</span><span class="o">=</span><span class="s">'b'</span><span class="p">,</span><span class="n">node_size</span><span class="o">=</span><span class="mi">700</span><span class="p">)</span>
<span class="n">g</span><span class="o">.</span><span class="n">subgraph</span><span class="p">(</span><span class="n">l5</span><span class="p">[</span><span class="mi">1</span><span class="p">])</span><span class="o">.</span><span class="n">draw</span><span class="p">(</span><span class="n">ax</span><span class="o">=</span><span class="n">ax</span><span class="p">,</span><span class="n">with_labels</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">node_color</span><span class="o">=</span><span class="s">'b'</span><span class="p">,</span><span class="n">node_size</span><span class="o">=</span><span class="mi">800</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Pathways that terminate in five steps'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
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<pre>[(0, 0), (1, -1), (2, 0), (3, -1), (4, 0), (5, 1)]
[(0, 0), (1, -1), (2, -2), (3, -1), (4, 0), (5, 1)]
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The figure shows the two pathways that terminate in five steps. The other fact we need is how many pathways do <em>not</em> terminate in five steps, then we'll be on our way to computing a probability. Fortunately, we already have this built into our initial graph construction.<br />
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<pre><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">g</span><span class="o">.</span><span class="n">getx</span><span class="p">(</span><span class="mi">5</span><span class="p">):</span>
<span class="k">print</span> <span class="s">'node (</span><span class="si">%d</span><span class="s">,</span><span class="si">%d</span><span class="s">)</span><span class="se">\t</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">]),</span>
<span class="k">print</span> <span class="s">'number of paths = </span><span class="si">%d</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="n">i</span><span class="p">[</span><span class="mi">1</span><span class="p">])</span>
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<pre>node (5,-5) number of paths = 1
node (5,-3) number of paths = 4
node (5,-1) number of paths = 5
node (5,1) number of paths = 2
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Now, we can compute the conditional probability of terminating in five steps as simply:<br />
$$ \mathbb{P}(5|\hat{3},\hat{1}) = \frac{\text{no. paths to (5,1)}}{\text{total no. of paths of length 5}}$$<br />
This is a conditional probability because the only way to terminate in five steps is to not have terminated at step 1 or 3, which is emphasized by the hat on top of those digits in the above equation. To compute the unconditional probability, we need to account for <br />
$$ \mathbb{P}(\hat{3},\hat{1}) = \mathbb{P}(\hat{3}|\hat{1}) \mathbb{P}(\hat{1})$$<br />
by computing this recursively,<br />
$$ \mathbb{P}(\hat{3}|\hat{1}) = 1- \mathbb{P}({3}|\hat{1}) $$<br />
where $\mathbb{P}(\hat{1})=1/2$. Plugging and chugging with the above result gives<br />
$$ \mathbb{P}(5) = \frac{1}{16} $$<br />
Because this is tedious to do by hand, we can automate this entire process as shown below.</div>
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<pre><span class="kn">import</span> <span class="nn">operator</span> <span class="kn">as</span> <span class="nn">op</span>
<span class="k">def</span> <span class="nf">get_cond_prob</span><span class="p">(</span><span class="n">g</span><span class="p">):</span>
<span class="s">'compute conditional probability for later'</span>
<span class="n">cp</span> <span class="o">=</span> <span class="n">OrderedDict</span><span class="p">()</span>
<span class="k">for</span> <span class="n">i</span><span class="p">,</span><span class="n">j</span> <span class="ow">in</span> <span class="n">g</span><span class="o">.</span><span class="n">gety</span><span class="p">(</span><span class="mi">1</span><span class="p">):</span>
<span class="k">if</span> <span class="p">(</span><span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="o">-</span><span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="p">)</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">g</span><span class="p">:</span> <span class="k">continue</span>
<span class="n">cp</span><span class="p">[</span><span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">]]</span> <span class="o">=</span> <span class="n">g</span><span class="o">.</span><span class="n">node</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="s">'val'</span><span class="p">]</span><span class="o">/</span><span class="nb">sum</span><span class="p">(</span><span class="nb">map</span><span class="p">(</span><span class="n">op</span><span class="o">.</span><span class="n">itemgetter</span><span class="p">(</span><span class="mi">1</span><span class="p">),</span><span class="n">g</span><span class="o">.</span><span class="n">getx</span><span class="p">(</span><span class="n">i</span><span class="p">[</span><span class="mi">0</span><span class="p">])))</span>
<span class="k">return</span> <span class="n">cp</span>
<span class="k">def</span> <span class="nf">get_prob</span><span class="p">(</span><span class="n">g</span><span class="p">,</span><span class="n">cp</span><span class="o">=</span><span class="bp">None</span><span class="p">):</span>
<span class="s">'get unconditional probability of termination '</span>
<span class="n">prob</span> <span class="o">=</span> <span class="n">OrderedDict</span><span class="p">()</span>
<span class="n">cq</span> <span class="o">=</span> <span class="n">OrderedDict</span><span class="p">()</span>
<span class="k">if</span> <span class="n">cp</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span> <span class="n">cp</span> <span class="o">=</span> <span class="n">get_cond_prob</span><span class="p">(</span><span class="n">g</span><span class="p">)</span>
<span class="k">for</span> <span class="n">i</span><span class="p">,</span><span class="n">j</span> <span class="ow">in</span> <span class="n">cp</span><span class="o">.</span><span class="n">iteritems</span><span class="p">():</span> <span class="n">cq</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">=</span><span class="mi">1</span><span class="o">-</span><span class="n">j</span>
<span class="n">prob</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">=</span> <span class="mf">0.5</span> <span class="c"># initial condition</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">cp</span><span class="o">.</span><span class="n">keys</span><span class="p">():</span>
<span class="n">prob</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">=</span> <span class="n">cp</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">*</span><span class="n">prod</span><span class="p">(</span><span class="n">cq</span><span class="o">.</span><span class="n">values</span><span class="p">()[:(</span><span class="n">i</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="o">//</span><span class="mi">2</span><span class="p">][::</span><span class="o">-</span><span class="mi">1</span><span class="p">])</span>
<span class="k">return</span> <span class="n">prob</span>
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The following quick check matches the result we computed earlier for termination in exactly five steps.</div>
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<pre><span class="n">g</span><span class="o">=</span><span class="n">construct_graph</span><span class="p">(</span><span class="mi">15</span><span class="p">)</span>
<span class="n">p</span><span class="o">=</span><span class="n">get_prob</span><span class="p">(</span><span class="n">g</span><span class="p">)</span>
<span class="k">print</span> <span class="s">'prob of termination in 5 steps = '</span><span class="p">,</span><span class="n">p</span><span class="p">[</span><span class="mi">5</span><span class="p">]</span>
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<pre>prob of termination in 5 steps = 0.0625
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The following plot shows the probability of termination for each number of steps.</div>
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">p</span><span class="o">.</span><span class="n">keys</span><span class="p">(),</span><span class="n">p</span><span class="o">.</span><span class="n">values</span><span class="p">(),</span><span class="s">'-o'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'exact no. of steps to terminate'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'probability'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=.</span><span class="mi">6</span><span class="p">)</span>
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<pre>(0.0, 16.0, 0.0, 0.6)
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Thus, we have a computational method for computing the probability of terminating in a given number of steps. What we need now is a separate analytical result that confirms our work so far. This is where the book by Feller comes in (Feller, William. <em>An Introduction to Probability Theory and Its Applications: Volume One</em>. John Wiley & Sons, 1950.).</div>
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<h2 id="Using-Generating-Functions-(Feller,-Vol-I,-p.-271)">
Using Generating Functions (Feller, Vol I, p. 271)</h2>
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In random walk terminology, the probability that first visit to $x=1$ takes place at the nth step is denoted as $\phi_n$. Furthermore, the generating function is defined as:<br />
$$ \Phi(s) = \sum_{n=0}^\infty \phi_n s^n$$<br />
we will need two other random variables. $N$ is the first time that the particle reaches $x=1$. This is a random variable with probability $\phi_n$. Likewise, $N_1$ is the number of trials required to move the particle from anywhere to the left of $x=-1$ to $x=0$. Also, $N_2$ is the number of trials required to move the particle from $x=0$ to $x=1$. Now, here comes the key step: these three random variables are independent with the same probability distribution. With these definitions, we have<br />
$$ N = 1 + N_1 + N_2$$<br />
In other words, you need $N_1$ trials to make it from wherever the particle was on the left of the origin back to the origin; and then you need $N_2$ trials to make it from there to $x=1$. The first $1$ accounts for the first step to the left from the origin. Thus we have,<br />
$$ \mathbb{E}(s^N|X_1=-1) = \mathbb{E}(s^{1 + N_1 + N_2}|X_1=-1) = s\Phi(s)^2$$<br />
where we can multiply through because of the mutual independence of the random variables. Additionally,<br />
$$ \mathbb{E}(s^N|X_1=1) = s $$<br />
because $N_1 = N_2 = 0$ in this case. Unwinding the conditional expectation yields,<br />
$$\mathbb{E}(s^N) = \mathbb{E}(s^N|X_1=1) p + \mathbb{E}(s^N|X_1=-1) q $$<br />
Writing this out yields,<br />
$$ \Phi(s) = p s + s \Phi(s)^2 (1-p) $$<br />
This is a quadratic equation in $\Phi(s)$, so we have two solutions:<br />
$$ \Phi_1(s) = \frac{-1-\sqrt{1+4 (p-1) p s^2}}{2 s(p -1 )} $$<br />
and<br />
$$ \Phi_2(s) = \frac{-1+\sqrt{1+4 (p-1) p s^2}}{2 s(p -1 )} $$<br />
How do we pick between them? The limit of $ \Phi_2(s) $ is unbounded as $s \rightarrow 0$. Thus, $\Phi(s=1) = \Phi_1(s=1)$.<br />
One of the properties of the $\Phi$ function is that $\Phi(s=1)$ should sum to one. In this case, we have<br />
$$ \Phi(s=1) = \Phi_1(s=1)= \frac{1-|p-q|}{2 q}$$<br />
Thus, if $p < q$, we have <br />
$$ \sum \phi_n = p/q$$<br />
which doesn't equal one. The problem here is that the definition of the random variable $N$ only considered the conclusion that the particle would ultimately reach $x=1$. The other possibility, which accounts for the missing probability here, is that the particle <em>never</em> terminates. Similarly, when $ p \ge q $, we have<br />
$$ \sum \phi_n = 1$$<br />
This should be no surprise because we saw this exact same result when we first started investigating this! Namely, $p \ge q \implies p \ge 1/2$, which is what we concluded from our first plot. This is the situation where the particle <em>always</em> terminates.<br />
The first derivative of $\Phi(s)$ function evaluated at $s=1$ is the mean. Thus,<br />
$$\bar{N} = \mathbb{E}(N) = \Phi'(s=1)= -\left( \frac{-1 + 4\,p\,q + {\sqrt{1 - 4\,p\,q}}} {-2\,q + 8\,p\,q^2} \right) $$<br />
Note that this is unbounded when $p=1/2$ which is exactly what we have been observing experimentally. <br />
To get at the individual probabilities for $p=1/2$, we can use a power series expansion on<br />
$$ \Phi(s)\bigg|_{p=1/2}=-\left( \frac{-1 + {\sqrt{1 - s^2}}}{s} \right)$$<br />
Using the binomial expansion theorem, this gives<br />
$$\phi_{2k-1} = (-1)^{k-1} \binom{1/2}{k} $$</div>
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Now, we can try this formula out against our graph construction and see if it matches. The downside is that <code>scipy.misc.comb</code> cannot handle fractional terms, so we need to use <code>sympy</code> instead, as shown below.</div>
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<pre><span class="kn">import</span> <span class="nn">sympy</span>
<span class="c"># analytical result from Feller</span>
<span class="n">pfeller</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">([(</span><span class="mi">2</span><span class="o">*</span><span class="n">k</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="n">sympy</span><span class="o">.</span><span class="n">binomial</span><span class="p">(</span><span class="mi">1</span><span class="o">/</span><span class="mi">2</span><span class="p">,</span><span class="n">k</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="o">**</span><span class="p">(</span><span class="n">k</span><span class="o">-</span><span class="mi">1</span><span class="p">))</span> <span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">10</span><span class="p">)])</span>
<span class="k">for</span> <span class="n">k</span><span class="p">,</span><span class="n">v</span> <span class="ow">in</span> <span class="n">p</span><span class="o">.</span><span class="n">iteritems</span><span class="p">():</span>
<span class="k">assert</span> <span class="n">v</span><span class="o">==</span><span class="n">pfeller</span><span class="p">[</span><span class="n">k</span><span class="p">]</span> <span class="c"># matches every item!</span>
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Now that we know exactly why the mean does not converge for $p=1/2$, let's check the calculation for when we know it does converge, say, for $p=2/3$. In this case the average number of steps to terminate is<br />
$$ \mathbb{E}(N)\bigg|_{p=2/3} = 3 $$<br />
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The code below redefines our earlier code for this case.</div>
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<pre><span class="k">def</span> <span class="nf">nwalk</span><span class="p">(</span><span class="n">limit</span><span class="o">=</span><span class="mi">500</span><span class="p">):</span>
<span class="s">'p=2/3 version of random walker. Only returns length of path'</span>
<span class="n">n</span><span class="o">=</span><span class="n">x</span><span class="o">=</span><span class="mi">0</span>
<span class="k">while</span> <span class="n">x</span><span class="o">!=</span><span class="mi">1</span> <span class="ow">and</span> <span class="n">n</span><span class="o"><</span><span class="n">limit</span><span class="p">:</span>
<span class="n">x</span><span class="o">+=</span><span class="n">random</span><span class="o">.</span><span class="n">choice</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span> <span class="c"># twice as many 1's as before</span>
<span class="n">n</span><span class="o">+=</span><span class="mi">1</span>
<span class="k">return</span> <span class="n">n</span> <span class="c"># return length of walk</span>
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<pre><span class="n">mean</span><span class="p">([</span><span class="n">nwalk</span><span class="p">()</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">100</span><span class="p">)])</span>
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<pre>3.2999999999999998
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<pre><span class="k">for</span> <span class="n">limit</span> <span class="ow">in</span> <span class="p">[</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">,</span><span class="mi">50</span><span class="p">,</span><span class="mi">100</span><span class="p">,</span><span class="mi">200</span><span class="p">,</span><span class="mi">300</span><span class="p">,</span><span class="mi">500</span><span class="p">,</span><span class="mi">1000</span><span class="p">]:</span>
<span class="k">print</span> <span class="s">'limit=</span><span class="si">%d</span><span class="s">,</span><span class="se">\t</span><span class="s"> average = </span><span class="si">%3.2f</span><span class="s">,</span><span class="se">\t</span><span class="s"> std=</span><span class="si">%3.2f</span><span class="s">'</span><span class="o">%</span> <span class="n">estimate_std</span><span class="p">(</span><span class="n">limit</span><span class="p">,</span><span class="mi">100</span><span class="p">)</span>
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<pre>limit=10, average = 2.50, std=0.26
limit=20, average = 2.83, std=0.39
limit=50, average = 2.95, std=0.52
limit=100, average = 2.96, std=0.43
limit=200, average = 3.02, std=0.55
limit=300, average = 2.96, std=0.48
limit=500, average = 3.03, std=0.49
limit=1000, average = 3.02, std=0.52
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Notice that the standard deviation of the average did not change much as the limit increased. This did not happen earlier with $p=1/2$. Now we know why.</div>
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<h2 id="Summary">
Summary</h2>
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In this long post, we thoroughly investigated the random walk and the lack of convergence in the average we noted when equiprobable steps are used. We then pursued this using both computational graph methods as well as analytical results. It's important to reflect on what would have happened if we had not noticed the strange convergence of the equiprobable case. Most likely, we would have just ignored it as some kind of sampling problem that is cured asymptotically. However, that did not happen here, and this kind of thing is easy to miss in real problems that have not been so heavily studied as the random walk. Thus, the moral of the story is that it pays to have a wide variety of analytical and computational tools (e.g. from the scientific Python stack) available, and to use both of them in tandem to chase down strange results, however mildly unexpected. Furthermore, concepts that sit at the core of more elaborate methods should be understood, or at least characterized as carefully as possible, because once these bleed into complicated meta-models, it may be impossible to track the resulting errors down to the source.<br />
<br />
As usual, the source notebook for this post is available <a href="https://github.com/unpingco/Python-for-Signal-Processing">here</a></div>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com3tag:blogger.com,1999:blog-2783842479260003750.post-41835267156702278102014-02-06T19:33:00.001-08:002014-02-06T19:36:29.364-08:00Methods of Random Sampling Using Rejection<div class="cell border-box-sizing code_cell">
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<h2 id="Introduction">
Introduction</h2>
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We want to generate samples of a given density, $f(x)$. In this case, we can assume we already have a reliable way to generate samples from a uniform distribution, $\mathcal{U}[0,1]$. How do we know a random sample ($v$) comes from the $f(x)$ distribution? One way to think about it is that a histogram of samples must approximate $f(x)$. This means that <br />
$$ \mathbb{P}( v \in N_{\Delta}(x) ) = f(x) \Delta x$$<br />
which says that the probability that a sample is in some $\Delta$ neighborhood of x is approximately $f(x)\Delta x$. <br />
Let's consider how to create these samples for both discrete and continuous random variables.</div>
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<h2 id="Inverse-CDF-Method-for-Discrete-Variables">
Inverse CDF Method for Discrete Variables</h2>
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Suppose we have a probability mass function,<br />
$$ f(x) = \sum p_i \delta(x-x_i) $$<br />
where $p_i$ is the probability <em>mass</em> at the point $x_i$. For example, for a fair six-sided die, we have<br />
$$ f(x) = \frac{1}{6} \sum_{i=1}^6 \delta(x-i) $$<br />
This generates the corresponding cumulative mass function,<br />
$$ F(x) = \frac{1}{6} \sum_{i=1}^6 U(x-i) $$<br />
where $U$ is the unit-step function. The next block of code develops this using <code>sympy</code>.</div>
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<pre><span class="kn">import</span> <span class="nn">sympy</span> <span class="kn">as</span> <span class="nn">S</span>
<span class="n">x</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">symbols</span><span class="p">(</span><span class="s">'x'</span><span class="p">)</span>
<span class="n">F</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="n">S</span><span class="o">.</span><span class="n">Heaviside</span><span class="p">(</span><span class="n">x</span><span class="o">-</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span><span class="o">/</span><span class="mi">6</span> <span class="c"># offset to satisfy sympy definition</span>
<span class="n">S</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">F</span><span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">),</span><span class="n">ylabel</span><span class="o">=</span><span class="s">'CDF(x)'</span><span class="p">,</span><span class="n">xlabel</span><span class="o">=</span><span class="s">'x'</span><span class="p">);</span>
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Now, we want a random number generator that outputs an element of $\left \{ 1,2,3,..,6\right \} $ with equal probability. We can generate a uniform random variable and think of it as picking a point on the y-axis of the plot above. Then, all we do is pick the corresponding x-axis value as the output. Let's do this in the next code block</div>
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<pre><span class="n">invF</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Piecewise</span><span class="p">((</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="o"><</span><span class="n">x</span><span class="o"><=</span><span class="mi">1</span><span class="o">/</span><span class="mi">6</span><span class="p">),</span> <span class="c"># if uniform sample between 0 and 1/6, choose die-side labeled 1</span>
<span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="o">/</span><span class="mi">6</span><span class="o"><</span><span class="n">x</span><span class="o"><=</span><span class="mi">2</span><span class="o">/</span><span class="mi">6</span><span class="p">),</span>
<span class="p">(</span><span class="mi">3</span><span class="p">,</span><span class="mi">2</span><span class="o">/</span><span class="mi">6</span><span class="o"><</span><span class="n">x</span><span class="o"><=</span><span class="mi">3</span><span class="o">/</span><span class="mi">6</span><span class="p">),</span>
<span class="p">(</span><span class="mi">4</span><span class="p">,</span><span class="mi">3</span><span class="o">/</span><span class="mi">6</span><span class="o"><</span><span class="n">x</span><span class="o"><=</span><span class="mi">4</span><span class="o">/</span><span class="mi">6</span><span class="p">),</span>
<span class="p">(</span><span class="mi">5</span><span class="p">,</span><span class="mi">4</span><span class="o">/</span><span class="mi">6</span><span class="o"><</span><span class="n">x</span><span class="o"><=</span><span class="mi">5</span><span class="o">/</span><span class="mi">6</span><span class="p">),</span>
<span class="p">(</span><span class="mi">6</span><span class="p">,</span><span class="mi">5</span><span class="o">/</span><span class="mi">6</span><span class="o"><</span><span class="n">x</span><span class="o"><=</span><span class="mi">1</span><span class="p">))</span>
<span class="n">samples</span><span class="o">=</span><span class="n">array</span><span class="p">([</span><span class="n">invF</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">rand</span><span class="p">(</span><span class="mi">500</span><span class="p">)])</span>
<span class="n">hist</span><span class="p">(</span><span class="n">samples</span><span class="p">,</span><span class="n">bins</span><span class="o">=</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">6</span><span class="p">,</span><span class="mi">7</span><span class="p">],</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">align</span><span class="o">=</span><span class="s">'left'</span><span class="p">);</span>
<span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mi">2</span><span class="o">/</span><span class="mf">6.</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'Estimated PMF of Fair Six-Sided Die'</span><span class="p">);</span>
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" />
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For comparison, here is the estimated CDF compared with $F$. You can trying using more or fewer samples in <code>rand()</code> to see how this changes.</div>
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In [7]:
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<pre><span class="n">bar</span><span class="p">(</span><span class="o">-.</span><span class="mi">5</span><span class="o">+</span><span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">),</span><span class="n">cumsum</span><span class="p">([</span><span class="n">mean</span><span class="p">(</span><span class="n">samples</span><span class="o">==</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)]),</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'ECDF'</span><span class="p">)</span> <span class="c"># estimated CDF</span>
<span class="n">bar</span><span class="p">(</span><span class="o">-.</span><span class="mi">25</span><span class="o">+</span><span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">),[</span><span class="n">F</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">)</span><span class="o">.</span><span class="n">evalf</span><span class="p">()</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)],</span><span class="n">fc</span><span class="o">=</span><span class="s">'g'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">,</span><span class="n">width</span><span class="o">=.</span><span class="mi">5</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'CDF'</span><span class="p">)</span> <span class="c"># target CDF</span>
<span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'Estimated CDF for fair six-sided die'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ylabel</span><span class="p">(</span><span class="s">'Frequency'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">xlabel</span><span class="p">(</span><span class="s">'Die face index'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mf">1.1</span><span class="p">)</span>
</pre>
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<pre>(0.0, 7.0, 0.0, 1.1)
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By repeating the same argument, we can construct an <em>unfair</em> six-sided die as</div>
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In [8]:
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<pre><span class="n">p</span> <span class="o">=</span> <span class="n">array</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">9</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span> <span class="c"># the faces 2,4 and more probable than 1,5,6 and 3 is the most probable</span>
<span class="n">p</span> <span class="o">=</span> <span class="n">p</span><span class="o">/</span><span class="nb">sum</span><span class="p">(</span><span class="n">p</span><span class="p">)</span> <span class="c"># normalize to 1</span>
<span class="n">Fu</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="n">S</span><span class="o">.</span><span class="n">Heaviside</span><span class="p">(</span><span class="n">x</span><span class="o">-</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span><span class="o">*</span><span class="n">p</span><span class="p">[</span><span class="n">i</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="n">S</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">Fu</span><span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">6</span><span class="p">),</span><span class="n">ylabel</span><span class="o">=</span><span class="s">'CDF(x)'</span><span class="p">,</span><span class="n">xlabel</span><span class="o">=</span><span class="s">'x'</span><span class="p">,</span><span class="n">title</span><span class="o">=</span><span class="s">'Unfair 6-sided Die'</span><span class="p">);</span>
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In the next block, we automate computing the inverse.</div>
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In [9]:
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<pre><span class="n">cp</span><span class="o">=</span><span class="n">cumsum</span><span class="p">([</span><span class="mi">0</span><span class="p">]</span><span class="o">+</span><span class="n">p</span><span class="o">.</span><span class="n">tolist</span><span class="p">())</span> <span class="c"># need to find edges on vertical axis, add [0] to get left edge</span>
<span class="n">invFu</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Piecewise</span><span class="p">(</span><span class="o">*</span><span class="p">[(</span><span class="n">i</span><span class="p">,</span><span class="n">j</span><span class="o"><</span><span class="n">x</span><span class="o"><=</span><span class="n">k</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span><span class="p">,</span><span class="n">j</span><span class="p">,</span><span class="n">k</span> <span class="ow">in</span> <span class="nb">zip</span><span class="p">(</span><span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">),</span><span class="n">cp</span><span class="p">[:</span><span class="o">-</span><span class="mi">1</span><span class="p">],</span><span class="n">cp</span><span class="p">[</span><span class="mi">1</span><span class="p">:])])</span>
<span class="n">hist</span><span class="p">([</span><span class="n">invFu</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">rand</span><span class="p">(</span><span class="mi">1000</span><span class="p">)],</span><span class="n">bins</span><span class="o">=</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">6</span><span class="p">,</span><span class="mi">7</span><span class="p">],</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">align</span><span class="o">=</span><span class="s">'left'</span><span class="p">);</span>
<span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mf">1.</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'Estimated PMF of Unfair Six-Sided Die'</span><span class="p">);</span>
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<h2 id="Inverse-CDF-Method-for-Continuous-Variables">
Inverse CDF Method for Continuous Variables</h2>
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The same idea applies to continuous random variables, but now we have to use squeeze the intervals down to individual points. In the example above, our inverse function was a piecewise function that operated on uniform random samples. In this case, the piecewise function collapses to a continuous inverse function. <br />
We want to generate random samples for a CDF $F$ that is invertible. The criterion for generating an appropriate sample is the following,<br />
$$ \mathbb{P}(F(x) < v < F(x+\Delta x)) = F(x+\Delta x) - F(x) = \int_x^{x+\Delta x} f(u) du \approx f(x) \Delta x$$<br />
which is saying that the probability that the sample $v$ is contained in a $\Delta x$ interval is approximately equal to the density function, $f(x) \Delta x$ at that point. The trick is to use a uniform random sample ($u$) and an invertible CDF $F(x)$ to construct these samples.<br />
Note that for a uniform random variable $u \sim \mathcal{U}[0,1]$, we have,<br />
$$ \mathbb{P}(x < F^{-1}(u) < x+\Delta x) =\mathbb{P}(F(x) < u < F(x+\Delta x)) = F(x+\Delta x) - F(x) = \int_x^{x+\Delta x} f(p) dp \approx f(x) \Delta x$$<br />
This means that $ v=F^{-1}(u) \sim f $, which is what we were after. Let's try this with the exponential distribution,<br />
$$ f_{\alpha}(x) = \alpha\exp(-\alpha x) $$<br />
with the following CDF,<br />
$$ F(x) = 1-\exp(-\alpha x )$$<br />
and corresponding inverse,<br />
$$ F^{-1}(u) = \frac{1}{\alpha}\ln \frac{1}{(1-u)}$$ </div>
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<pre><span class="kn">import</span> <span class="nn">scipy.stats</span>
<span class="n">alpha</span> <span class="o">=</span> <span class="mf">1.</span>
<span class="n">nsamp</span> <span class="o">=</span> <span class="mi">1000</span>
<span class="n">u</span><span class="o">=</span><span class="n">scipy</span><span class="o">.</span><span class="n">stats</span><span class="o">.</span><span class="n">uniform</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span>
<span class="n">x</span><span class="o">=</span><span class="n">scipy</span><span class="o">.</span><span class="n">stats</span><span class="o">.</span><span class="n">expon</span><span class="p">(</span><span class="n">alpha</span><span class="p">)</span>
<span class="n">Finv</span><span class="o">=</span><span class="k">lambda</span> <span class="n">u</span><span class="p">:</span> <span class="mi">1</span><span class="o">/</span><span class="n">alpha</span><span class="o">*</span><span class="n">log</span><span class="p">(</span><span class="mi">1</span><span class="o">/</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">u</span><span class="p">))</span>
<span class="c">#Finv=lambda u: 1+1/alpha*log(1/(1-u)) # shift over to correct</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">xhat</span> <span class="o">=</span> <span class="n">array</span><span class="p">(</span><span class="nb">map</span><span class="p">(</span><span class="n">Finv</span><span class="p">,</span><span class="n">u</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="n">nsamp</span><span class="p">)))</span>
<span class="c"># exponential distrib samples by inverse method</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="n">xhat</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">bins</span><span class="o">=</span><span class="mi">30</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'inverse method'</span><span class="p">)</span>
<span class="c"># exponential distrib samples by scipy.stats</span>
<span class="n">xrvs</span> <span class="o">=</span> <span class="n">x</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="n">nsamp</span><span class="p">)</span>
<span class="n">xe</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">xrvs</span><span class="o">.</span><span class="n">max</span><span class="p">(),</span><span class="mi">100</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="n">xrvs</span><span class="p">,</span><span class="n">bins</span><span class="o">=</span><span class="mi">30</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'scipy.stats'</span><span class="p">)</span>
<span class="c"># exponential theoretical density function</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">xe</span><span class="p">,</span><span class="n">x</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">xe</span><span class="p">),</span><span class="s">'r-'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'theoretical'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">(</span><span class="nb">range</span><span class="p">(</span><span class="nb">int</span><span class="p">(</span><span class="n">xrvs</span><span class="o">.</span><span class="n">max</span><span class="p">())))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mf">1.05</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Inverse CDF for Exponential Distribution'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'pdf(x)'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'x'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">();</span>
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The reason the histogram is shifted over is because we lose knowledge of a constant shift in $v$ because $1-v$ is also uniformly distributed between $\mathcal{U}[0,1]$. This means we have to change the inverse function to the following:<br />
$$ F^{-1}(u) = 1+\frac{1}{\alpha}\ln \frac{1}{(1-u)}$$ </div>
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<h2 id="Rejection-Method">
Rejection Method</h2>
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In some cases, you may not be able to invert for the CDF. The <code>rejection</code> method can handle this situation. The idea is to pick a uniform ($\mathcal{U}[a,b]$) random variable $u_1$ and $u_2$ so that<br />
$$ \mathbb{P}\left( u_1 \in N_{\Delta}(x) \bigwedge u_2 < \frac{f(u_1)}{M} \right) \hspace{0.5em} \approx \frac{\Delta x}{b-a} \frac{f(u_1)}{M} $$<br />
where we take $u_1=x$ and $f(x) < M $. The only job of the $M$ variable is to scale down the $f(x)$ so that the $u_2$ variable can span the range. The <em>efficiency</em> of this method is the probability of not rejecting $u_1$ which comes from integrating out the above approximation.<br />
$$ \int dx \frac{f(x)}{M} \hspace{0.5em} = \frac{1}{M(b-a)}$$<br />
This means that we don't want an unecessarily large $M$ because that makes it more likely that samples will be discarded. Let's use a density that does not have a continuous inverse. <br />
$$ f(x) = \exp\left(-\frac{(x-1)^2}{2x} \right) \hspace{1em} (x+1)/12 $$ <br />
where $x>0$. Note that this does not <em>exactly</em> integrate out to one like a good probability density function should, but I probably need to think of a better function because the normalization constant for this is pretty convoluted.<br />
Nevertheless, here it is with its corresponding CDF.</div>
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In [12]:
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<pre><span class="n">x</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="mf">0.001</span><span class="p">,</span><span class="mi">15</span><span class="p">,</span><span class="mi">100</span><span class="p">)</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">f</span><span class="o">=</span> <span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">exp</span><span class="p">(</span><span class="o">-</span><span class="p">(</span><span class="n">x</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span><span class="o">/</span><span class="mf">2.</span><span class="o">/</span><span class="n">x</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">x</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span><span class="o">/</span><span class="mf">12.</span>
<span class="n">fx</span> <span class="o">=</span> <span class="n">f</span><span class="p">(</span><span class="n">x</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">fx</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'$f(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">cumsum</span><span class="p">(</span><span class="n">fx</span><span class="p">)</span><span class="o">*</span><span class="n">diff</span><span class="p">(</span><span class="n">x</span><span class="p">)[</span><span class="mi">0</span><span class="p">],</span><span class="s">'g'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'$F(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">);</span>
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And, following our rejection plan, the following are the simulated random samples of $f$.</div>
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In [13]:
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<pre><span class="n">M</span><span class="o">=.</span><span class="mi">3</span> <span class="c"># scale factor</span>
<span class="n">u1</span> <span class="o">=</span> <span class="n">rand</span><span class="p">(</span><span class="mi">10000</span><span class="p">)</span><span class="o">*</span><span class="mi">15</span> <span class="c"># uniform random samples scaled out</span>
<span class="n">u2</span> <span class="o">=</span> <span class="n">rand</span><span class="p">(</span><span class="mi">10000</span><span class="p">)</span> <span class="c"># uniform random samples</span>
<span class="n">idx</span><span class="o">=</span><span class="n">where</span><span class="p">(</span><span class="n">u2</span><span class="o"><=</span><span class="n">f</span><span class="p">(</span><span class="n">u1</span><span class="p">)</span><span class="o">/</span><span class="n">M</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span> <span class="c"># rejection criterion</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">u1</span><span class="p">[</span><span class="n">idx</span><span class="p">]</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="n">v</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">bins</span><span class="o">=</span><span class="mi">40</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">fx</span><span class="p">,</span><span class="s">'r'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">3.</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'$f(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Estimated Efficency=</span><span class="si">%3.1f%%</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="mi">100</span><span class="o">*</span><span class="nb">len</span><span class="p">(</span><span class="n">v</span><span class="p">)</span><span class="o">/</span><span class="nb">len</span><span class="p">(</span><span class="n">u1</span><span class="p">)))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
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Out[13]:</div>
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<pre><matplotlib.legend.Legend at 0x9015690>
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The plot above shows a pretty good match between our target density function, $f$, and our histogram of the samples we just generated using the rejection method.<br />
Conceptually, there's nothing wrong with this result. The problem is the efficiency is low -- we are throwing away too many samples, as shown in the figure below. We need to somehow do better.</div>
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In [14]:
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">u1</span><span class="p">,</span><span class="n">u2</span><span class="p">,</span><span class="s">'.'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'rejected'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">u1</span><span class="p">[</span><span class="n">idx</span><span class="p">],</span><span class="n">u2</span><span class="p">[</span><span class="n">idx</span><span class="p">],</span><span class="s">'g.'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'accepted'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
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Out[14]:</div>
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<pre><matplotlib.legend.Legend at 0x7f357d0>
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The above shows the samples that were accepted and rejected by this method. The fact that we threw out so many is very inefficient, and we want to do better.</div>
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The above argument uses $u_1$ to select along the domain of $f(x)$ and the other $u_2$ uniform random variable decides whether to accept or not. One idea would be to choose $u_1$ so that $x$ values are coincidentally those that are near the peak of $f(x)$, instead of uniformly anywhere in the domain, especially near the tails, which are low probability anyway. Now, the trick is to find a new density function $g(x)$ to sample from that has a similiar concentration of probability density. One way it to familiarize oneself with the popular density functions that have adjustable parameters and fast random sample generators already. There are lots of places to look and, chances are, there is likely already such a generator for your problem. Otherwise, the <a href="http://en.wikipedia.org/wiki/Beta_distribution">family of $\beta$ densities</a> is a good place to start looking. </div>
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To be explicit, what we want is $u_1 \sim g(x) $ so that, returning to our earlier argument,<br />
$$ \mathbb{P}\left( u_1 \in N_{\Delta}(x) \bigwedge u_2 < \frac{f(u_1)}{M} \right) \approx g(x) \Delta x \frac{f(u_1)}{M} $$<br />
but this is <em>not</em> what we need here. The problem is with the second part of the $\bigwedge$ clause. We need to put something there that will give us something proportional to $f(x)$. Define the following:<br />
$$ h(x) = \frac{f(x)}{g(x)}$$ <br />
with corresponding maximum on the domain as $h_{\max}$ and then go back and construct the second part of the clause as<br />
$$ \mathbb{P}\left( u_1 \in N_{\Delta}(x) \bigwedge u_2 < \frac{h(u_1)}{h_{\max}} \right) \approx g(x) \Delta x \frac{h(u_1)}{h_{\max}} = f(x)/h_{\max} $$<br />
Recall that satisfying this criterion means that $u_1=x$. As before, we can estimate the probability of acceptance of the $u_1$ as $ 1/h_{\max}$.<br />
Now, to construct such a $g(x)$ function. Here,we choose the chi-squared distribution. The following plots the $g(x)$ and $f(x)$ (left plot) and the corresponding $h(x)=f(x)/g(x)$ (right plot). Note that $g(x)$ and $f(x)$ have peaks that almost coincide, which is what we are looking for.</div>
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<pre><span class="n">ch</span><span class="o">=</span><span class="n">scipy</span><span class="o">.</span><span class="n">stats</span><span class="o">.</span><span class="n">chi2</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span> <span class="c"># chi-squared</span>
<span class="n">h</span> <span class="o">=</span> <span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">f</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">/</span><span class="n">ch</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">x</span><span class="p">)</span> <span class="c"># h-function</span>
<span class="n">fig</span><span class="p">,</span><span class="n">axs</span><span class="o">=</span><span class="n">subplots</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">12</span><span class="p">,</span><span class="mi">4</span><span class="p">)</span>
<span class="n">axs</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">fx</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'$f(x)$'</span><span class="p">)</span>
<span class="n">axs</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">ch</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">x</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">'$g(x)$'</span><span class="p">)</span>
<span class="n">axs</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">axs</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">h</span><span class="p">(</span><span class="n">x</span><span class="p">))</span>
<span class="n">axs</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'$h(x)=f(x)/g(x)$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
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Now, let's generate some samples from this $\chi^2$ distribution with the rejection method.</div>
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In [16]:
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<pre><span class="n">hmax</span><span class="o">=</span><span class="n">h</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">.</span><span class="n">max</span><span class="p">()</span>
<span class="n">u1</span> <span class="o">=</span> <span class="n">ch</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="mi">5000</span><span class="p">)</span> <span class="c"># samples from chi-square distribution</span>
<span class="n">u2</span> <span class="o">=</span> <span class="n">rand</span><span class="p">(</span><span class="mi">5000</span><span class="p">)</span> <span class="c"># uniform random samples</span>
<span class="n">idx</span> <span class="o">=</span> <span class="p">(</span><span class="n">u2</span> <span class="o"><=</span> <span class="n">h</span><span class="p">(</span><span class="n">u1</span><span class="p">)</span><span class="o">/</span><span class="n">hmax</span><span class="p">)</span> <span class="c"># Rejection criterion</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">u1</span><span class="p">[</span><span class="n">idx</span><span class="p">]</span> <span class="c"># keep these only</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="n">v</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">bins</span><span class="o">=</span><span class="mi">40</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">fx</span><span class="p">,</span><span class="s">'r'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">3.</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'$f(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Estimated Efficency=</span><span class="si">%3.1f%%</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="mi">100</span><span class="o">*</span><span class="nb">len</span><span class="p">(</span><span class="n">v</span><span class="p">)</span><span class="o">/</span><span class="nb">len</span><span class="p">(</span><span class="n">u1</span><span class="p">)))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
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<pre><matplotlib.legend.Legend at 0x9344c30>
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Using the $\chi^2$ distribution with the rejection method results in throwing away less than 10% of the generated samples compared with our prior example where we threw out at least 80% of the generated samples. Obviously, this way is much more computationally efficient and this is the kind of thing we are always looking for.<br />
For completeness, here is the corresponding plot that shows the samples with the corresponding threshold $h(x)/h_{\max}$ that was used to select them.</div>
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In [17]:
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">u1</span><span class="p">,</span><span class="n">u2</span><span class="p">,</span><span class="s">'.'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'rejected'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">u1</span><span class="p">[</span><span class="n">idx</span><span class="p">],</span><span class="n">u2</span><span class="p">[</span><span class="n">idx</span><span class="p">],</span><span class="s">'g.'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'accepted'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">h</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">/</span><span class="n">hmax</span><span class="p">,</span><span class="s">'r'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">3.</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'$h(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
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Out[17]:</div>
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<pre><matplotlib.legend.Legend at 0x938fd30>
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Summary</h2>
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In this section, we investigated how to generate random samples from a given distribution, beit discrete or continuous. For the continuouse case, the key issue was whether or not the cumulative density function had a continuous inverse. If not, we had to turn to the rejection method, and find an appropriate related density that we could easily sample from to use as part of a rejection threshold. Finding such a function is an art, but many families of probability densities have been studied over the years that already have fast sample-generators.<br />
<br />
It is possible to go much deeper with the rejection method, but these involve careful partitioning of the domains and lots of special methods for separate domains and corner cases. Nonetheless, all of these advanced techniques are still variations on the same fundamental theme we illustrated here.</div>
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References</h2>
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<a href="http://books.google.com/books?id=ACTrrQk1UgoC&dq=Exploring+Monte+Carlo+Methods&source=gbs_navlinks_s">Exploring Monte Carlo Methods by Dunn, 2011</a><br />
<a href="http://books.google.com/books?id=0QzvAAAAMAAJ&dq=continuous+balakrishnan&hl=en&sa=X&ei=vzcuUtqvOYOKjAKr14BI&ved=0CC8Q6AEwAA">Continuous univariate distributions by Balakrishnan, 1995</a><br />
<a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Sampling_Monte_Carlo.ipynb">IPython Source Notebook</a></div>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-67545746193021487412014-02-06T13:46:00.000-08:002014-02-06T13:47:38.279-08:00Exploring Buffon's Needle Using Python Tools<div class="cell border-box-sizing code_cell">
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<h2 id="Buffon's-Needle">
Buffon's Needle</h2>
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The problem is to randomly drop a needle on the unit square and count the number of times that the needle touches (or "cuts") the edge of the square. This is a well-studied problem, but instead of the usual tack of going directly for the analytical result (see appendix), we instead write a quick simulation that we'll later refine to obtain more precise results more efficiently. This methodology is common in Monte Carlo methods and we will work through this example in careful detail.<br />
Buffon's needle is a classic problem that is easy to understand, and complex enough to be characteristic of much more difficult problems encountered in practice. The overall idea is that when the problem is too complex to analyze analytically (at least in the general case), we write representative computer models that can drive a numerical solution and motivate analytical solutions, even if only on special cases. </div>
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Setting up the Simulation</h2>
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The following code-block sets up the random needle position and orientation and creates the corresponding matplotlib primitives to be drawn in the figure. As shown, the simulation chooses a random center point for the needle in the unit square and then chooses a random angle $\theta \in (0,\pi)$ as the needle's orientation. Because the needle pivots at its center, we only need $\theta$ in this range.</div>
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<pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span>
<span class="kn">from</span> <span class="nn">matplotlib.lines</span> <span class="kn">import</span> <span class="n">Line2D</span>
<span class="kn">from</span> <span class="nn">matplotlib.patches</span> <span class="kn">import</span> <span class="n">Rectangle</span>
<span class="k">def</span> <span class="nf">needle_gen</span><span class="p">(</span><span class="n">L</span><span class="o">=</span><span class="mf">0.5</span><span class="p">):</span>
<span class="sd">'''Drops needle on unit square. Return dictionary describing needle position and </span>
<span class="sd"> orientation. The `cut` key in the dictionary is True when the needle intersects </span>
<span class="sd"> an edge of the unit square and is False otherwise.</span>
<span class="sd"> </span>
<span class="sd"> L := length of needle (default = 0.5)</span>
<span class="sd"> '''</span>
<span class="k">assert</span> <span class="mi">0</span><span class="o"><</span><span class="n">L</span><span class="o"><</span><span class="mi">1</span>
<span class="n">xc</span><span class="p">,</span><span class="n">yc</span> <span class="o">=</span> <span class="n">rand</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span> <span class="c"># uniform random in [0,1] for each coordinate dimension</span>
<span class="n">ang</span> <span class="o">=</span> <span class="n">rand</span><span class="p">()</span><span class="o">*</span><span class="n">pi</span> <span class="c"># uniform random angle in [0,pi]</span>
<span class="n">x0</span><span class="p">,</span><span class="n">y0</span> <span class="o">=</span> <span class="n">xc</span><span class="o">-</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">cos</span><span class="p">(</span><span class="n">ang</span><span class="p">),</span><span class="n">yc</span><span class="o">-</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">sin</span><span class="p">(</span><span class="n">ang</span><span class="p">)</span> <span class="c"># coordinates for one end of needle </span>
<span class="n">xe</span><span class="p">,</span><span class="n">ye</span> <span class="o">=</span> <span class="n">xc</span><span class="o">+</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">cos</span><span class="p">(</span><span class="n">ang</span><span class="p">),</span><span class="n">yc</span><span class="o">+</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">sin</span><span class="p">(</span><span class="n">ang</span><span class="p">)</span> <span class="c"># coordinates for the other end</span>
<span class="n">not_cut</span> <span class="o">=</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">x0</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">y0</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">xe</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">ye</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span>
<span class="k">return</span> <span class="nb">dict</span><span class="p">(</span><span class="n">x0</span><span class="o">=</span><span class="n">x0</span><span class="p">,</span><span class="n">y0</span><span class="o">=</span><span class="n">y0</span><span class="p">,</span><span class="n">xc</span><span class="o">=</span><span class="n">xc</span><span class="p">,</span><span class="n">yc</span><span class="o">=</span><span class="n">yc</span><span class="p">,</span><span class="n">xe</span><span class="o">=</span><span class="n">xe</span><span class="p">,</span><span class="n">ye</span><span class="o">=</span><span class="n">ye</span><span class="p">,</span><span class="n">ang</span><span class="o">=</span><span class="n">ang</span><span class="p">,</span><span class="n">cut</span><span class="o">=</span><span class="p">(</span> <span class="ow">not</span> <span class="n">not_cut</span> <span class="p">))</span>
<span class="k">def</span> <span class="nf">draw_needle</span><span class="p">(</span><span class="n">ax</span><span class="p">,</span><span class="n">d</span><span class="p">):</span>
<span class="sd">'''</span>
<span class="sd"> Draw needle symbol on given axis.</span>
<span class="sd"> </span>
<span class="sd"> If red, then the line cuts an edge of the unit square. If blue, then it does not. </span>
<span class="sd"> One end of the needle is marked with a `o` symbol and the other end is unmarked. </span>
<span class="sd"> </span>
<span class="sd"> ax := matplotlib axes to draw needle symbol</span>
<span class="sd"> d := dictionary output from needle_gen</span>
<span class="sd"> '''</span>
<span class="k">for</span> <span class="n">i</span><span class="p">,</span><span class="n">j</span> <span class="ow">in</span> <span class="n">d</span><span class="o">.</span><span class="n">iteritems</span><span class="p">():</span> <span class="k">exec</span><span class="p">(</span><span class="s">'</span><span class="si">%s</span><span class="s">=</span><span class="si">%r</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="n">i</span><span class="p">,</span><span class="n">j</span><span class="p">))</span> <span class="c"># dump dictionary in local namespace</span>
<span class="k">if</span> <span class="n">cut</span><span class="p">:</span>
<span class="n">line</span><span class="o">=</span> <span class="p">[</span><span class="n">Line2D</span><span class="p">([</span><span class="n">x0</span><span class="p">,</span><span class="n">xe</span><span class="p">],[</span><span class="n">y0</span><span class="p">,</span><span class="n">ye</span><span class="p">],</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">),</span>
<span class="n">Line2D</span><span class="p">([</span><span class="n">xc</span><span class="p">],[</span><span class="n">yc</span><span class="p">],</span><span class="n">marker</span><span class="o">=</span><span class="s">'o'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">)]</span>
<span class="k">else</span><span class="p">:</span>
<span class="n">line</span><span class="o">=</span> <span class="p">[</span><span class="n">Line2D</span><span class="p">([</span><span class="n">x0</span><span class="p">,</span><span class="n">xe</span><span class="p">],[</span><span class="n">y0</span><span class="p">,</span><span class="n">ye</span><span class="p">],</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">),</span>
<span class="n">Line2D</span><span class="p">([</span><span class="n">xc</span><span class="p">],[</span><span class="n">yc</span><span class="p">],</span><span class="n">marker</span><span class="o">=</span><span class="s">'o'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">)]</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">line</span><span class="p">:</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_line</span><span class="p">(</span><span class="n">i</span><span class="p">)</span>
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Now, we can simulate the dropping event on the square (yellow background ). The red needles cut the edge of the square and the blue ones are completely contained therein. The circle marker shows the needle position.</div>
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<pre><span class="n">L</span><span class="o">=</span><span class="mf">0.5</span>
<span class="n">samples</span> <span class="o">=</span> <span class="p">[</span><span class="n">needle_gen</span><span class="p">()</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">500</span><span class="p">)]</span>
<span class="k">def</span> <span class="nf">draw_sim</span><span class="p">(</span><span class="n">samples</span><span class="p">,</span><span class="n">L</span><span class="o">=</span><span class="mf">0.5</span><span class="p">):</span>
<span class="s">'Draw simulation results. Package this plot for reuse later'</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">5</span><span class="p">,</span><span class="mi">5</span><span class="p">)</span>
<span class="n">red_count</span><span class="o">=</span><span class="mi">0</span>
<span class="n">n</span><span class="o">=</span><span class="nb">len</span><span class="p">(</span><span class="n">samples</span><span class="p">)</span>
<span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="n">samples</span><span class="p">:</span>
<span class="n">red_count</span><span class="o">+=</span><span class="n">k</span><span class="p">[</span><span class="s">'cut'</span><span class="p">]</span>
<span class="n">draw_needle</span><span class="p">(</span><span class="n">ax</span><span class="p">,</span><span class="n">k</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span> <span class="c"># set aspect ratio to 1</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">Rectangle</span><span class="p">((</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">),</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'y'</span><span class="p">))</span> <span class="c"># unit square background is light yellow</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'x-direction'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'y-direction'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">([</span><span class="o">-.</span><span class="mi">1</span><span class="p">,</span><span class="mf">1.1</span><span class="p">,</span><span class="o">-.</span><span class="mi">1</span><span class="p">,</span><span class="mf">1.1</span><span class="p">])</span> <span class="c"># add some space around the unit square</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$\mathbb{P}$ (cut)=</span><span class="si">%3.2f</span><span class="s">'</span><span class="o">%</span> <span class="p">(</span><span class="n">red_count</span><span class="o">/</span><span class="n">n</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">draw_sim</span><span class="p">(</span><span class="n">samples</span><span class="p">)</span>
</pre>
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The figure above is a good visual check of our simulation and helps motivate our reasoning. Now, we want to concentrate on counting the number of cuts and then use that to estimate the probability of a cut.</div>
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In [6]:
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">axs</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">12</span><span class="p">,</span><span class="mi">2</span><span class="p">))</span>
<span class="n">ax</span><span class="o">=</span><span class="n">axs</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
<span class="n">samples</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([[</span><span class="n">needle_gen</span><span class="p">(</span><span class="n">L</span><span class="p">)[</span><span class="s">'cut'</span><span class="p">]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">300</span><span class="p">)]</span> <span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">100</span><span class="p">)])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="n">samples</span><span class="p">,</span><span class="n">axis</span><span class="o">=</span><span class="mi">1</span><span class="p">),</span><span class="s">'o-'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">xmax</span><span class="o">=</span><span class="n">samples</span><span class="o">.</span><span class="n">shape</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">r'$trial$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$\mathbb{\hat{P}}(cut)$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'$\overline{\mathbb{\hat{P}}}(cut) = </span><span class="si">%3.3f</span><span class="s">,\hat{\sigma}=</span><span class="si">%3.3f</span><span class="s">$'</span><span class="o">%</span>
<span class="p">(</span><span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(),(</span><span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">))</span><span class="o">.</span><span class="n">std</span><span class="p">()))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">axs</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span> <span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">));</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Distribution of Mean Estimates'</span><span class="p">);</span>
</pre>
</div>
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The figure above shows some statistical history of our Monte Carlo simulation. The <code>samples</code> array in the prior code-block is a 100x300 matrix of True/False values indicating cut or no-cut. The overall average of this array is our estimate of the probability of a cut. In other words, we count the number of cuts and divide by the total number of needle drops to obtain the estimated probability (approximately 55% in this case of $L=1/2$). The plot on the left shows the average across each of the 300 columns. Thus, there are 100 such column-wise averages (recall the <code>samples</code> matrix is 100x300). The title shows the estimated standard deviation across these 100 columnwise averages. <br />
The plot on the right shows a histogram of the column-wise averages. Why did we partition the simulated samples like this and compute column-wise instead of just using all the data at once? Because we want to get a sense of the spread of the estimates by computing the standard error. If we used all our samples at once, we would not be able to do this even though we'll still use our average of these averages to get our ultimate probability of a cut!<br />
Naturally, if we wanted to improve the standard deviation of our ensemble of columnwise means, we could simply run a larger simulation, but remember that the standard error only improves at the rate of $\sqrt{N}$ (where $N$ is the number of samples) so we'd need a sample run <em>four</em> times as large to improve the standard error by a factor of <em>two</em>. Furthermore, we need to understand the mechanics of this problem better per unit of computing time.</div>
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<h2 id="Relative-Errors-and-Using-Weighting-for-Better-Efficiency">
Relative Errors and Using Weighting for Better Efficiency</h2>
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There's nothing wrong with the way we have been using the simulation so far, but in the first figure all of those blue needles did not help us compute the probability we were after because they did not touch the edges of the square. Yes, we could reverse the reasoning and use the blue instead of the red for our estimation, but that does not escape the basic problem. For example, if the length of the needle were really short in comparison, then there would have been a lot of blue in our initial plot (try this using this IPython notebook!). In the extreme case, we could find ourselves in a situation where we run the simulation for a long time and not have a single cut! We want to use our valuable computer time to create samples that will improve our estimated solution.<br />
To recap, we computed a matrix of (100x300) cases of needle drops. Each of element of <code>samples</code> in the code above is a <code>True/False</code> value as to whether or not the needle touched the edge of the square. We averaged over the 300 columns to compute the estimated probability of a cut (the mean value of the boolean array in this case). The estimated standard deviation is taken over the ensemble of these estimated means. <br />
A common way to evaluate the quality of the simulated result is to compute the <em>relative error</em>, which is the ratio of the estimated standard deviation of the mean to the estimated mean using <em>all</em> of the samples ($R_e$). This is computed in the title of the figure below.</div>
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">12</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$\hat{\sigma}=</span><span class="si">%2.3f</span><span class="s">,R_e=</span><span class="si">%2.3f</span><span class="s">$'</span><span class="o">%</span> <span class="p">(</span><span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span><span class="o">.</span><span class="n">std</span><span class="p">(),</span><span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span><span class="o">.</span><span class="n">std</span><span class="p">()</span><span class="o">/</span><span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">()),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'sample history index'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$\mathbb{\hat{P}}(cut)$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">);</span>
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So far, our relative error is pretty good (the rule of thumb is below 5%), but let's see how we do when the needle length is smaller? The code-block below draws the corresponding figure for the simulation for this case.</div>
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In [8]:
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<pre><span class="n">L</span><span class="o">=</span><span class="mf">0.1</span>
<span class="n">samples_d</span><span class="o">=</span><span class="p">[</span><span class="n">needle_gen</span><span class="p">(</span><span class="n">L</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">300</span><span class="o">*</span><span class="mi">100</span><span class="p">)]</span>
<span class="n">samples</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="n">k</span><span class="p">[</span><span class="s">'cut'</span><span class="p">]</span> <span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="n">samples_d</span><span class="p">])</span><span class="o">.</span><span class="n">reshape</span><span class="p">(</span><span class="mi">100</span><span class="p">,</span><span class="mi">300</span><span class="p">)</span>
<span class="k">def</span> <span class="nf">draw_sample_history</span><span class="p">(</span><span class="n">samples</span><span class="p">):</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">12</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span>
<span class="n">mn</span> <span class="o">=</span> <span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">mn</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$\hat{\sigma}=</span><span class="si">%2.3f</span><span class="s">,R_e=</span><span class="si">%2.3f</span><span class="s">$'</span><span class="o">%</span> <span class="p">(</span><span class="n">mn</span><span class="o">.</span><span class="n">std</span><span class="p">(),</span>
<span class="n">mn</span><span class="o">.</span><span class="n">std</span><span class="p">()</span><span class="o">/</span><span class="n">mn</span><span class="o">.</span><span class="n">mean</span><span class="p">()),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'sample history index'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$\mathbb{\hat{P}}(cut)$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">);</span>
<span class="n">draw_sample_history</span><span class="p">(</span><span class="n">samples</span><span class="p">)</span>
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" />
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The figure above shows that for shorter needles, we did <em>not</em> do so well for relative error. This situation is illustrated in the figure below.</div>
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In [9]:
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<pre><span class="n">draw_sim</span><span class="p">(</span><span class="n">samples_d</span><span class="p">[:</span><span class="mi">500</span><span class="p">])</span>
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The figure shows there are very few cuts in our simulation with the short needle (red needles). To remedy this, we can alter the code to bias for randomly generating x-coordinates that are within needle length of an edge instead of uniformly random in the unit interval. The code-block below makes this change for the x-coordinate.</div>
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<pre><span class="k">def</span> <span class="nf">needle_gen_b</span><span class="p">(</span><span class="n">L</span><span class="o">=</span><span class="mf">0.5</span><span class="p">):</span>
<span class="sd">'''Drops needle on unit square using a biased strategy. Return dictionary describing </span>
<span class="sd"> needle position and orientation. The `cut` key in the dictionary is True when the</span>
<span class="sd"> needle intersects an edge of the unit square and is False otherwise.</span>
<span class="sd"> </span>
<span class="sd"> L := length of needle (default = 0.5)</span>
<span class="sd"> '''</span>
<span class="k">assert</span> <span class="mi">0</span><span class="o"><</span><span class="n">L</span><span class="o"><</span><span class="mi">1</span>
<span class="n">yc</span> <span class="o">=</span> <span class="n">rand</span><span class="p">()</span> <span class="c"># uniform on unit interval</span>
<span class="n">r</span> <span class="o">=</span> <span class="n">L</span><span class="o">*</span><span class="n">rand</span><span class="p">()</span><span class="o">/</span><span class="mf">2.</span>
<span class="n">xc</span> <span class="o">=</span> <span class="n">r</span> <span class="k">if</span> <span class="n">rand</span><span class="p">()</span><span class="o"><</span><span class="mf">0.5</span> <span class="k">else</span> <span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">r</span><span class="p">)</span> <span class="c"># pick either edge 50/50</span>
<span class="n">ang</span> <span class="o">=</span> <span class="n">rand</span><span class="p">()</span><span class="o">*</span><span class="n">pi</span> <span class="c"># uniform random angle in [0,pi]</span>
<span class="n">x0</span><span class="p">,</span><span class="n">y0</span> <span class="o">=</span> <span class="n">xc</span><span class="o">-</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">cos</span><span class="p">(</span><span class="n">ang</span><span class="p">),</span><span class="n">yc</span><span class="o">-</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">sin</span><span class="p">(</span><span class="n">ang</span><span class="p">)</span> <span class="c"># coordinates for one end of needle </span>
<span class="n">xe</span><span class="p">,</span><span class="n">ye</span> <span class="o">=</span> <span class="n">xc</span><span class="o">+</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">cos</span><span class="p">(</span><span class="n">ang</span><span class="p">),</span><span class="n">yc</span><span class="o">+</span><span class="n">L</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">sin</span><span class="p">(</span><span class="n">ang</span><span class="p">)</span> <span class="c"># coordinates for the other end</span>
<span class="n">not_cut</span> <span class="o">=</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">x0</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">y0</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">xe</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="p">(</span><span class="mi">0</span><span class="o"><</span><span class="n">ye</span><span class="o"><</span><span class="mi">1</span><span class="p">)</span>
<span class="k">return</span> <span class="nb">dict</span><span class="p">(</span><span class="n">x0</span><span class="o">=</span><span class="n">x0</span><span class="p">,</span><span class="n">y0</span><span class="o">=</span><span class="n">y0</span><span class="p">,</span><span class="n">xc</span><span class="o">=</span><span class="n">xc</span><span class="p">,</span><span class="n">yc</span><span class="o">=</span><span class="n">yc</span><span class="p">,</span><span class="n">xe</span><span class="o">=</span><span class="n">xe</span><span class="p">,</span><span class="n">ye</span><span class="o">=</span><span class="n">ye</span><span class="p">,</span><span class="n">ang</span><span class="o">=</span><span class="n">ang</span><span class="p">,</span><span class="n">cut</span><span class="o">=</span><span class="p">(</span> <span class="ow">not</span> <span class="n">not_cut</span> <span class="p">))</span>
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Now, we re-run this simulation using this biased method.</div>
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<pre><span class="n">samples_b</span><span class="o">=</span><span class="p">[</span><span class="n">needle_gen_b</span><span class="p">(</span><span class="n">L</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">300</span><span class="o">*</span><span class="mi">100</span><span class="p">)</span> <span class="p">]</span> <span class="c"># create biased samples</span>
<span class="n">draw_sim</span><span class="p">(</span><span class="n">samples_b</span><span class="p">[:</span><span class="mi">500</span><span class="p">])</span>
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The figure above shows that the samples are now clustered at both vertical edges, and, compared with the unbiased simulation, there are more red needles, indicating that we now have more relevent samples.<br />
The next figure shows the sample history and relative error which is now <strong>much</strong> better than before.</div>
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In [12]:
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<pre><span class="n">no_bias_samples</span> <span class="o">=</span> <span class="n">samples</span> <span class="c"># save for later</span>
<span class="n">samples</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="n">k</span><span class="p">[</span><span class="s">'cut'</span><span class="p">]</span> <span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="n">samples_b</span><span class="p">])</span><span class="o">.</span><span class="n">reshape</span><span class="p">(</span><span class="mi">100</span><span class="p">,</span><span class="mi">300</span><span class="p">)</span>
<span class="n">draw_sample_history</span><span class="p">(</span><span class="n">samples</span><span class="p">);</span>
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Now, that we have a tighter result, we need to make sure we have not changed the result by correcting for the bias we introduced. The probability of being near the edges in the x-direction is $2 L$. We have to multiply our estimated probability by this <em>weighting</em> factor.</div>
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In [13]:
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<pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">10</span><span class="p">,</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="n">no_bias_samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">),</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'unbiased'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="n">samples</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span><span class="o">*</span><span class="mi">2</span><span class="o">*</span><span class="n">L</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'biased'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Biased samples more efficiently produce useful samples'</span><span class="p">)</span>
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Out[13]:</div>
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<pre><matplotlib.text.Text at 0x7d0a850>
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The histogram above shows the spread of the biased samples is much tighter than before, which is what the relative error we just computed is getting at. Still, it's good to back this up with a quick plot.</div>
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<h2 id="Summary">
Summary</h2>
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In this section, we illustrated some basic concepts of Monte Carlo methods by examing the famous problem of Buffon's needle. We showed how to code up a quick simulation that visually reveals the key issues and then how to boil this down into a simulation to compute the estimated solution. We also showed how to create biased samples to improve the overall efficiency and precision of our result.<br />
I carefully used the word <a href="https://en.wikipedia.org/wiki/Accuracy_and_precision"><em>precision</em></a> as opposed to <em>accuracy</em> because we have created a tighter estimate (as measured by relative error), but how do we know that this is the <em>correct</em> (i.e. accurate) answer? We don't! In this case, we can solve for the analytical solution (shown in the appendix), but in general, this is <em>way</em> too hard to do, otherwise we wouldn't bother with the simulation in the first place.<br />
The reality is that we have to do what we can using Monte Carlo methods and then carefully consider their computed results using as much domain knowledge as we can muster. In this section, we made the extra effort to generate some visual results along the way as a sanity-check and doing so is <em>always</em> worth the extra effort, when feasible. Furthermore, this kind of intermediate result can sometimes suggest a narrow special case for which we <em>can</em> generate a related analytical solution that we can use to further develop our simulation.<br />
As usual, the IPython Notebook corresponding to this section is available for download <a href="http://github.com/unpingco/Pig-in-the-Python/blob/master/Buffons_Needle_Sim.ipynb">here</a>. Play around with the various parameters as suggested above on your own to see what else you can discover about this famouse problem.</div>
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<h2 id="Appendix:-Analytical-Results">
Appendix: Analytical Results</h2>
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In this appendix, we carefully derive the analytical solution for our problem using excessive detail. Note that $x_c$ is the needle's center. The needle is entirely contained in the square (at least in the x-direction) when the following two conditions hold:<br />
$$ 0 < x_c - L/2 \cos(\theta) < 1 $$<br />
$$ 0 < x_c + L/2 \cos(\theta) < 1 $$<br />
combining these two into one big inequality gives:<br />
$$ \max(-L/2 \cos(\theta) , L/2 \cos(\theta) ) < x_c < \min(1-L/2 \cos(\theta) , 1+L/2 \cos(\theta) )$$<br />
In the case where $0< \theta < \pi/2 $, we additionally have the following:<br />
$$ L/2 \cos(\theta) < x_c < 1-L/2 \cos(\theta) $$<br />
Because $\theta$ is uniformly distributed between $0$ and $\pi$, for this domain of $\theta$ and $x_c$, we have the probability of the needle being contained completely in the square (in the x-direction) as the following:<br />
$$ \frac{1}{\pi}\int_0^{\pi/2} \int_{L/2 \cos(\theta)}^{1-L/2 \cos(\theta)} dx_c d\theta = \frac{1}{\pi}\int_0^{\pi/2} 1-L \cos(\theta) d\theta = \frac{1}{2} - \frac{L}{\pi} $$ <br />
Now, we can consider the other half of the domain of $\theta$, $ \pi/2 < \theta < \pi $, where we have<br />
$$ -L/2 \cos(\theta) < x_c < 1+L/2 \cos(\theta)$$<br />
Then, following the same reasoning as before we obtain:<br />
$$ \frac{1}{\pi}\int_{\pi/2}^{\pi} \int_{-L/2 \cos(\theta)}^{1+L/2 \cos(\theta)} dx_c d\theta = \frac{1}{\pi}\int_{\pi/2}^{\pi} 1+L \cos(\theta) d\theta = \frac{1}{2} - \frac{L}{\pi} $$ <br />
Thus, combining these two results gives the probability of the needle <strong>not</strong> cutting the edge in the x-direction as the following:<br />
$$ \mathbb{P}(\text{no x-cut}) = 1 - \frac{2 L}{\pi}$$
Now, we can pursue exactly the same line of reasoning for the y-direction, or, just recognize that by symmetry it would be exactly the same as for the x-direction. Nonetheless, let's be thorough and reproduce the reasoning for the y-direction, just to be on the safe side.<br />
As before, the condition in the y-direction for the needle being entirely contained in the square in the y-direction is the following:<br />
$$ \max(-L/2 \sin(\theta) , L/2 \sin(\theta) ) < y_c < \min(1-L/2 \sin(\theta) , 1+L/2 \sin(\theta) )$$<br />
which can be combined into one big inequality, given the restrictions on $\theta \in (0,\pi)$ as shown:<br />
$$ L/2 \sin(\theta) < y_c < 1-L/2 \sin(\theta) $$<br />
Then, integrating this out as before, we obtain<br />
$$ \frac{1}{\pi}\int_0^{\pi} \int_{L/2 \sin(\theta)}^{1-L/2 \sin(\theta)} dy_c d\theta = \frac{1}{\pi}\int_0^{\pi} 1-L \sin(\theta) d\theta = 1 - \frac{2 L}{\pi} $$ <br />
$$ \mathbb{P}(\text{no y-cut}) = 1 - \frac{2 L}{\pi}$$<br />
which is exactly what we got for the x-direction. It's usually worth the extra work to check things like this instead of trying to be too clever too early.<br />
Then, combining the x-direction and y-direction results, we obtain,<br />
$$ \mathbb{P}(\text{no cut}) = \left(1 - \frac{2 L}{\pi}\right)^2$$<br />
so that<br />
$$ \mathbb{P}(\text{cut}) =1-\mathbb{P}(\text{no cut}) = 1-\left(1 - \frac{2 L}{\pi}\right)^2$$<br />
which we code next.</div>
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<pre><span class="n">prob_cut</span><span class="o">=</span> <span class="k">lambda</span> <span class="n">l</span><span class="p">:(</span><span class="mi">1</span><span class="o">-</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="mi">2</span><span class="o">*</span><span class="n">l</span><span class="o">/</span><span class="n">pi</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span>
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We can use this result later to check or Monte Carlo estimates. Try it!</div>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-35783906700329406062013-03-11T19:32:00.000-07:002013-03-11T19:32:00.601-07:00Spectral Estimation Using the Discrete Fourier Transform
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<h2>Introduction</h2>
<p>In this section, we consider the very important problem of resolving two nearby frequencies using the DFT. This spectral analysis problem is one of the cornerstone problems in signal processing and we therefore highlight some nuances. We also investigate the circular convolution as a tool to uncover the mechanics of frequency resolution as the uncertainty principle emerges again.</p>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span>
<span class="n">Nf</span> <span class="o">=</span> <span class="mi">64</span> <span class="c"># N- DFT size</span>
<span class="n">fs</span> <span class="o">=</span> <span class="mi">64</span> <span class="c"># sampling frequency</span>
<span class="n">f</span> <span class="o">=</span> <span class="mi">10</span> <span class="c"># one signal</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="p">)</span> <span class="c"># time-domain samples</span>
<span class="n">deltaf</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="mf">2.</span> <span class="c"># second nearby frequency</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">sharex</span><span class="o">=</span><span class="bp">True</span><span class="p">,</span><span class="n">sharey</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">8</span><span class="p">,</span><span class="mi">3</span><span class="p">))</span>
<span class="n">x</span><span class="o">=</span><span class="n">cos</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">t</span><span class="p">)</span> <span class="o">+</span> <span class="n">cos</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="p">(</span><span class="n">f</span><span class="o">+</span><span class="mi">2</span><span class="p">)</span><span class="o">*</span><span class="n">t</span><span class="p">)</span> <span class="c"># 2 Hz frequency difference</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">fft</span><span class="o">.</span><span class="n">fft</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">Nf</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">fs</span><span class="p">,</span><span class="n">Nf</span><span class="p">),</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'-o'</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$\delta f = 2$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(k)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">x</span><span class="o">=</span><span class="n">cos</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">t</span><span class="p">)</span> <span class="o">+</span> <span class="n">cos</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="p">(</span><span class="n">f</span><span class="o">+</span><span class="n">deltaf</span><span class="p">)</span><span class="o">*</span><span class="n">t</span><span class="p">)</span> <span class="c"># delta_f frequency difference</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">fft</span><span class="o">.</span><span class="n">fft</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">Nf</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">fs</span><span class="p">,</span><span class="n">Nf</span><span class="p">),</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'-o'</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$\delta f = 1/2$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(k)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'Frequency (Hz)'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_xlim</span><span class="p">(</span><span class="n">xmax</span> <span class="o">=</span> <span class="n">fs</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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"></img>
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>Seeking Better Frequency Resolution with Longer DFT</h2>
<p>The top plot above shows the magnitude of the DFT for an input that is the sum of two frequencies separated by 2 Hz. Using the parameters we have chosen for the DFT, we can easily see there are two distinct frequencies in the input signal. The bottom plot shows the same thing except that here the frequencies are only separated by 0.5 Hz and, in this case, the two frequencies are not so easy to separate. From this figure, it would be difficult to conclude how many frequencies are present and at what magnitude.</p>
<p>At this point, the usual next step is to increase the size of the DFT since the frequency resolution is $f_s/N$. Thus, the idea is to increase this resolution until the two frequencies separate. This is shown in the next figure.</p>
</div>
<div class="cell border-box-sizing code_cell vbox">
<div class="input hbox">
<div class="prompt input_prompt">In [45]:</div>
<div class="input_area box-flex1">
<div class="highlight"><pre><span class="n">Nf</span> <span class="o">=</span> <span class="mi">64</span><span class="o">*</span><span class="mi">2</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">sharex</span><span class="o">=</span><span class="bp">True</span><span class="p">,</span><span class="n">sharey</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">8</span><span class="p">,</span><span class="mi">4</span><span class="p">))</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">fft</span><span class="o">.</span><span class="n">fft</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">Nf</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">fs</span><span class="p">,</span><span class="nb">len</span><span class="p">(</span><span class="n">X</span><span class="p">)),</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'-o'</span><span class="p">,</span><span class="n">ms</span><span class="o">=</span><span class="mf">3.</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$N=</span><span class="si">%d</span><span class="s">$'</span><span class="o">%</span><span class="k">Nf</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(k)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">Nf</span> <span class="o">=</span> <span class="mi">64</span><span class="o">*</span><span class="mi">4</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">fft</span><span class="o">.</span><span class="n">fft</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">Nf</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">fs</span><span class="p">,</span><span class="nb">len</span><span class="p">(</span><span class="n">X</span><span class="p">)),</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'-o'</span><span class="p">,</span><span class="n">ms</span><span class="o">=</span><span class="mf">3.</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$N=</span><span class="si">%d</span><span class="s">$'</span><span class="o">%</span><span class="k">Nf</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(k)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'Frequency (Hz)'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_xlim</span><span class="p">(</span><span class="n">xmax</span> <span class="o">=</span> <span class="n">fs</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
</pre></div>
</div>
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<p>As the figure above shows, increasing the size of the DFT did not help matters much. Why is this? Didn't we increase the frequency resolution using a larger DFT? Why can't we separate frequencies now? </p>
<h2>The Uncertainty Principle Strikes Back</h2>
<p>The problem here is a manifestation the uncertainty principle we <a href="http://python-for-signal-processing.blogspot.com/2012/09/investigating-sampling-theorem-in-this.html">previously discussed</a>. Remember that taking a larger DFT doesn't add anything new; it just picks off more discrete frequencies on the unit circle. Note that we want to analyze a particular signal $x(t)$, but we have only a <em>finite section</em> of that signal. In other words, what we really have are samples of the product of $x(t),t\in \mathbb{R}$ and a rectangular time-window, $r(t)$, that is zero except $r(t)=1 \Leftrightarrow t\in[0,1]$. This means that the DFT is structured according to the rectangular window, which explains the <code>sinc</code> shapes we have seen here.</p>
<p>The following figure shows the updated DFT using a longer duration rectangular window.</p>
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<div class="prompt input_prompt">In [46]:</div>
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<div class="highlight"><pre><span class="n">t</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="p">)</span>
<span class="n">x</span><span class="o">=</span><span class="n">cos</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">t</span><span class="p">)</span> <span class="o">+</span> <span class="n">cos</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="p">(</span><span class="n">f</span><span class="o">+</span><span class="n">deltaf</span><span class="p">)</span><span class="o">*</span><span class="n">t</span><span class="p">)</span>
<span class="n">Nf</span> <span class="o">=</span> <span class="mi">64</span><span class="o">*</span><span class="mi">2</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">sharex</span><span class="o">=</span><span class="bp">True</span><span class="p">,</span><span class="n">sharey</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">8</span><span class="p">,</span><span class="mi">4</span><span class="p">))</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">fft</span><span class="o">.</span><span class="n">fft</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">Nf</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">fs</span><span class="p">,</span><span class="nb">len</span><span class="p">(</span><span class="n">X</span><span class="p">)),</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'-o'</span><span class="p">,</span><span class="n">ms</span><span class="o">=</span><span class="mf">3.</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$N=</span><span class="si">%d</span><span class="s">$'</span><span class="o">%</span><span class="k">Nf</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(k)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">Nf</span> <span class="o">=</span> <span class="mi">64</span><span class="o">*</span><span class="mi">8</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">fft</span><span class="o">.</span><span class="n">fft</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">Nf</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">fs</span><span class="p">,</span><span class="nb">len</span><span class="p">(</span><span class="n">X</span><span class="p">)),</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'-o'</span><span class="p">,</span><span class="n">ms</span><span class="o">=</span><span class="mf">3.</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'$N=</span><span class="si">%d</span><span class="s">$'</span><span class="o">%</span><span class="k">Nf</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(k)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'Frequency (Hz)'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_xlim</span><span class="p">(</span><span class="n">xmax</span> <span class="o">=</span> <span class="n">fs</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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<p>The top plot in the figure above shows the DFT of the longer duration signal with $N=128$. The bottom plot shows the same signal with larger DFT length of $N=512$ and a clear separation between the two frequencies. Thus, as opposed to the previous case, a longer DFT <em>did</em> resolve the nearby frequencies, but it needed a longer duration signal to do it. Why is this? Consider the DFT of the rectangular windows of length $N_s$,</p>
<p>$$ X[k] = \frac{1}{\sqrt N}\sum_{n=0}^{N_s-1} \exp\left( \frac{2\pi}{N} k n \right) $$</p>
<p>after some re-arrangement, this reduces to</p>
<p>$$ |X[k]|=\frac{ 1}{\sqrt N}\left|\frac{\sin \left( N_s \frac{2\pi}{N} k\right)}{\sin \left( \frac{2\pi}{N} k \right)}\right|$$</p>
<p>which bears a strong resemblence to our <a href="http://python-for-signal-processing.blogspot.com/2012/09/investigating-sampling-theorem-in-this.html">original</a> <code>sinc</code> function. The following figure is a plot of this function</p>
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<div class="prompt input_prompt">In [47]:</div>
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<div class="highlight"><pre><span class="k">def</span> <span class="nf">abs_sinc</span><span class="p">(</span><span class="n">k</span><span class="o">=</span><span class="bp">None</span><span class="p">,</span><span class="n">N</span><span class="o">=</span><span class="mi">64</span><span class="p">,</span><span class="n">Ns</span><span class="o">=</span><span class="mi">32</span><span class="p">):</span>
<span class="k">if</span> <span class="n">k</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span> <span class="n">k</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">N</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span>
<span class="n">y</span> <span class="o">=</span> <span class="n">where</span><span class="p">(</span><span class="n">k</span> <span class="o">==</span> <span class="mi">0</span><span class="p">,</span> <span class="mf">1.0e-20</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
<span class="k">return</span> <span class="nb">abs</span><span class="p">(</span><span class="n">sin</span><span class="p">(</span> <span class="n">Ns</span><span class="o">*</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="n">N</span><span class="o">*</span><span class="n">y</span><span class="p">)</span><span class="o">/</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">y</span><span class="o">/</span><span class="n">N</span><span class="p">))</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">N</span><span class="p">)</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">8</span><span class="p">,</span><span class="mi">3</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">abs_sinc</span><span class="p">(</span><span class="n">N</span><span class="o">=</span><span class="mi">512</span><span class="p">,</span><span class="n">Ns</span><span class="o">=</span><span class="mi">10</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">'duration=10'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">abs_sinc</span><span class="p">(</span><span class="n">N</span><span class="o">=</span><span class="mi">512</span><span class="p">,</span><span class="n">Ns</span><span class="o">=</span><span class="mi">20</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">'duration=20'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'DFT Index'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(\Omega_k)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Rectangular Windows DFTs'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">);</span>
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<p>Note that the DFT grows taller and narrower as the sampling duration increases (i.e. longer rectangular window). The amplitude growth occurs because the longer window accumulates more "energy" than the shorter window. The length of the DFT is the same for both lines shown so only the length of the rectangular window varies. The point is that taking a longer duration rectangular window improves the frequency resolution! This fact is just the uncertainty principle at work. Looking at the <code>sinc</code> formula, the null-to-null width of the main lobe in frequency terms is the following</p>
<p>$$ \delta f = 2\frac{N}{2 N_s} \frac{f_s}{N} =\frac{f_s}{N_s} $$</p>
<p>Thus, two frequencies that differ by at least this amount should be resolvable in these plots. </p>
<p>Thus, in our last example, we had $f_s= 64,N_s = 128 \Rightarrow \delta f = 1/2$ Hz and we were trying to separate two frequencies 0.5 Hz apart so we were right on the edge in this case. I invite you to download this IPython notebook and try longer or shorter signal durations to see show these plots change. Incidentally, this where some define the notion of <em>frequency bin</em> as the DFT resolution ($ f_s/N $) divided by this minimal resolution, $ f_s/N_s $ which gives $ N_s/N $. In other words, the DFT measures frequency in discrete <em>bins</em> of minimal resolution, $ N_s/N $.</p>
<p>However, sampling over a longer duration only helps when the signal frequencies are <em>stable</em> over the longer duration. If these frequencies drift during the longer sampling interval or otherwise become contaminated with other signals, then advanced techniques become necessary.</p>
<p>Let's consider in detail how the DFT of the rectangular window affects resolution by considering the circular convolution.</p>
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<h2>Circular Convolution</h2>
<p>Suppose we want to compute the DFT of a product $z_n=x_n y_n$ as shown below,</p>
<p>$$ Z_k = \frac{1}{\sqrt N}\sum_{n=0}^{N-1} (x_n y_n) W_N^{n k} $$</p>
<p>in terms of the respective DFTs of $x_n$ and $y_n$, $X_k$ and $Y_k$, respectively, where</p>
<p>$$ x_n = \frac{1}{\sqrt N}\sum_{p=0}^{N-1} X_p W_N^{-n p} $$</p>
<p>and</p>
<p>$$ y_n = \frac{1}{\sqrt N}\sum_{m=0}^{N-1} Y_m W_N^{-n m} $$</p>
<p>Then, substituting back in gives,</p>
<p>$$Z_k = \frac{1}{\sqrt N} \frac{1}{N} \sum_{p=0}^{N-1} X_p \sum_{m=0}^{N-1} Y_m \sum_{n=0}^{N-1} W_N^{n k -n p - n m}$$</p>
<p>The last term evaluates to</p>
<p>$$ \sum_{n=0}^{N-1} W_N^{n k -n p - n m} = \frac{1-W_N^{N(k-p-m)}}{1-W_N^{k-p-m}} \hspace{2em} = \frac{1-e^{j2\pi(k-p-m)}}{1-e^{j 2\pi (k-p-m)/N}}$$ </p>
<p>This is zero everywhere except where $k-p-m= q N$ ($q\in \mathbb{Z}$) in which case it is $N$. Substituting all this back into our expression gives the <em>circular convolution</em> usually denoted as</p>
<p>$$ Z_k = \frac{1}{\sqrt N} \sum_{p=0}^{N-1} X_p Y_{((k-p))_N} = X_k \otimes_N Y_k $$</p>
<p>where the double subscripted parenthesis emphasizes the periodic nature of the index. The circular convolution tells us to compute the DFT $Z_k$ directly from the corresponding DFTs $X_k$ and $Y_k$.</p>
<p>Let's work through an example to see this in action. <br />
</p>
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<div class="prompt input_prompt">In [48]:</div>
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<div class="highlight"><pre><span class="k">def</span> <span class="nf">dftmatrix</span><span class="p">(</span><span class="n">Nfft</span><span class="o">=</span><span class="mi">32</span><span class="p">,</span><span class="n">N</span><span class="o">=</span><span class="bp">None</span><span class="p">):</span>
<span class="s">'construct DFT matrix'</span>
<span class="n">k</span><span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="n">Nfft</span><span class="p">)</span>
<span class="k">if</span> <span class="n">N</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span> <span class="n">N</span> <span class="o">=</span> <span class="n">Nfft</span>
<span class="n">n</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="n">N</span><span class="p">)</span>
<span class="n">U</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">(</span><span class="n">exp</span><span class="p">(</span><span class="mi">1j</span><span class="o">*</span> <span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="n">Nfft</span> <span class="o">*</span><span class="n">k</span><span class="o">*</span><span class="n">n</span><span class="p">[:,</span><span class="bp">None</span><span class="p">]))</span> <span class="c"># use numpy broadcasting to create matrix</span>
<span class="k">return</span> <span class="n">U</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nfft</span><span class="p">)</span>
<span class="n">Nf</span> <span class="o">=</span> <span class="mi">32</span> <span class="c"># DFT size</span>
<span class="n">U</span> <span class="o">=</span> <span class="n">dftmatrix</span><span class="p">(</span><span class="n">Nf</span><span class="p">,</span><span class="n">Nf</span><span class="p">)</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">U</span><span class="p">[:,</span><span class="mi">12</span><span class="p">]</span><span class="o">.</span><span class="n">real</span> <span class="c"># input signal</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">U</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">x</span> <span class="c"># DFT of input</span>
<span class="n">rect</span> <span class="o">=</span> <span class="n">ones</span><span class="p">((</span><span class="n">Nf</span><span class="o">/</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span> <span class="c"># short rectangular window</span>
<span class="n">z</span> <span class="o">=</span> <span class="n">x</span><span class="p">[:</span><span class="n">Nf</span><span class="o">/</span><span class="mi">2</span><span class="p">]</span> <span class="c"># product of rectangular window and x (i.e. chopped version of x) </span>
<span class="n">R</span> <span class="o">=</span> <span class="n">dftmatrix</span><span class="p">(</span><span class="n">Nf</span><span class="p">,</span><span class="n">Nf</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">rect</span> <span class="c"># DFT of rectangular window</span>
<span class="n">Z</span> <span class="o">=</span> <span class="n">dftmatrix</span><span class="p">(</span><span class="n">Nf</span><span class="p">,</span><span class="n">Nf</span><span class="o">/</span><span class="mi">2</span><span class="p">)</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">z</span> <span class="c"># DFT of product of x_n and r_n</span>
</pre></div>
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<div class="prompt input_prompt">In [93]:</div>
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<div class="highlight"><pre><span class="n">idx</span><span class="o">=</span><span class="n">arange</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span><span class="o">-</span><span class="n">arange</span><span class="p">(</span><span class="n">Nf</span><span class="p">)[:,</span><span class="bp">None</span><span class="p">]</span> <span class="c"># use numpy broadcasting to setup summand's indices</span>
<span class="n">idx</span><span class="p">[</span><span class="n">idx</span><span class="o"><</span><span class="mi">0</span><span class="p">]</span><span class="o">+=</span><span class="n">Nf</span> <span class="c"># add periodic Nf to negative indices for wraparound</span>
<span class="n">a</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span> <span class="c"># k^th frequency index</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">(</span><span class="mi">4</span><span class="p">,</span><span class="mi">8</span><span class="p">,</span><span class="n">sharex</span><span class="o">=</span><span class="bp">True</span><span class="p">,</span><span class="n">sharey</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">12</span><span class="p">,</span><span class="mi">5</span><span class="p">))</span>
<span class="k">for</span> <span class="n">i</span><span class="p">,</span><span class="n">j</span> <span class="ow">in</span> <span class="nb">enumerate</span><span class="p">(</span><span class="n">ax</span><span class="o">.</span><span class="n">flat</span><span class="p">):</span>
<span class="c">#markerline, stemlines, baseline = j.stem(arange(Nf),abs(R[idx[:,i],0])/sqrt(Nf))</span>
<span class="c">#setp(markerline, 'markersize', 3.)</span>
<span class="n">j</span><span class="o">.</span><span class="n">fill_between</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="n">Nf</span><span class="p">),</span><span class="mi">1</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">)</span><span class="o">*</span><span class="nb">abs</span><span class="p">(</span><span class="n">R</span><span class="p">[</span><span class="n">idx</span><span class="p">[:,</span><span class="n">i</span><span class="p">],</span><span class="mi">0</span><span class="p">])</span><span class="o">.</span><span class="n">flat</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">)</span>
<span class="n">markerline</span><span class="p">,</span> <span class="n">stemlines</span><span class="p">,</span> <span class="n">baseline</span> <span class="o">=</span><span class="n">j</span><span class="o">.</span><span class="n">stem</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="n">Nf</span><span class="p">),</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">))</span>
<span class="n">setp</span><span class="p">(</span><span class="n">markerline</span><span class="p">,</span> <span class="s">'markersize'</span><span class="p">,</span> <span class="mf">4.</span><span class="p">)</span>
<span class="n">setp</span><span class="p">(</span><span class="n">markerline</span><span class="p">,</span><span class="s">'markerfacecolor'</span><span class="p">,</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">setp</span><span class="p">(</span><span class="n">stemlines</span><span class="p">,</span><span class="s">'color'</span><span class="p">,</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">j</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="s">'off'</span><span class="p">)</span>
<span class="n">j</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'k=</span><span class="si">%d</span><span class="s">'</span><span class="o">%</span><span class="k">i</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">8</span><span class="p">)</span>
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<p>The figure above shows the <font color="Blue">rectangular window DFT in blue, $R_k$ </font> against the sinusoid <FONT color="red"> input signal in red, $X_k$, </font> for each value of $k$ as the two terms slide past each other from left to right, top to bottom. In other words, the $k^{th}$ term in $Z_k$, the DFT of the product $x_n r_n $, can be thought of as the inner-product of the red and blue lines. This is not exactly true because we are just plotting magnitudes and not the real/imaginary parts, but it's enough to understand the mechanics of the circular convolution.</p>
<p>A good way to think about the rectangular window's <code>sinc</code> shape as it slides past the input signal is as a <em>probe</em> with a resolution defined by its mainlobe width. For example, in frame $k=12$, we see that the peak of the rectangular window coincides with the peak of the input frequency so we should expect a large value for $Z_{k=12}$ which is shown below. However, if the rectangular window were shorter, corresponding to a wider mainlobe width, then two nearby frequencies could be draped in the same mainlobe and would then be indistinguishable in the resulting DFT because the DFT for that value of $k$ is the inner-product (i.e. a complex number) of the two overlapping graphs.</p>
<p>The figure below shows the the direct computation of the DFT of $Z_k$ matches the circular convolution method using $X_k$ and $R_k$.</p>
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<div class="prompt input_prompt">In [69]:</div>
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<div class="highlight"><pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">7</span><span class="p">,</span><span class="mi">3</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">a</span><span class="p">,</span><span class="nb">abs</span><span class="p">(</span><span class="n">R</span><span class="p">[</span><span class="n">idx</span><span class="p">,</span><span class="mi">0</span><span class="p">]</span><span class="o">*</span><span class="n">X</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nf</span><span class="p">),</span> <span class="n">label</span><span class="o">=</span><span class="s">r'$|Z_k|$ = $X_k\otimes_N R_k$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="nb">abs</span><span class="p">(</span><span class="n">Z</span><span class="p">),</span><span class="s">'o'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$|Z_k|$ by DFT'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'DFT index,k'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|Z_k|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="n">ax</span><span class="o">.</span><span class="n">get_xticks</span><span class="p">()</span><span class="o">.</span><span class="n">max</span><span class="p">()))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">tick_params</span><span class="p">(</span><span class="n">labelsize</span><span class="o">=</span><span class="mi">8</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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"></img>
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Summary
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>In this section, we unpacked the issues involved in resolving two nearby frequencies using the DFT and once again confronted the uncertainty principle in action. We realized that longer DFTs cannot distinguish nearby frequencies unless the signal is sampled over a sufficient duration. Additionally, we developed the circular convolution as a tool to visualize the exactly how a longer sampling duration helps resolve frequencies.</p>
<p>As usual, the corresponding IPython notebook for this post is available for download <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Frequency_Resolution.ipynb">here</a>. </p>
<p>Comments and corrections welcome!</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>References</h2>
<ul>
<li>Oppenheim, A. V., and A. S. Willsky. "Signals and Systems." Prentice-Hall, (1997).</li>
<li>Proakis, John G. "Digital signal processing: principles algorithms and applications". Pearson Education India, 2001.</li>
</ul>
</div>
Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-26734198950616387962013-02-25T19:00:00.000-08:002013-02-25T19:00:03.687-08:00Discrete Fourier Transform<div class="text_cell_render border-box-sizing rendered_html">
<h2>Introduction</h2>
<p>The Fourier Transform is ubiquitous, but it has singular standing in signal processing because of the way sampling imposes a bandwidth-centric view of the world. The Discrete Fourier Transform (DFT) is the primary analysis tool for exploring this perspective. Our development unconventionally starts with a matrix/vector representation of the DFT because that facilitates our visual approach which in turn is designed to develop intuition about the operation and usage of the DFT in practice.</p>
<p>Let us start with the following DFT matrix</p>
<p>$$ \mathbf{U} = \frac{1}{\sqrt N} \left[ \exp \left( j \frac{2\pi}{N} n k \right) \right]_{n\in{0,N_s},k\in{0,N-1}} $$</p>
<p>where $n$ counts rows through the number of samples and $k$ indexes the discrete frequencies as columns. </p>
<p>The following figure shows the discrete frequencies on the unit circle and their corresponding real and imaginary parts that are the columns of $\mathbf{U}$.</p>
</div>
<div class="cell border-box-sizing code_cell vbox">
<div class="input hbox">
<div class="prompt input_prompt">In [1]:</div>
<div class="input_area box-flex1">
<div class="highlight"><pre><span class="c"># must start notebook with --pylab flag</span>
<span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span>
<span class="kn">from</span> <span class="nn">matplotlib.patches</span> <span class="kn">import</span> <span class="n">FancyArrow</span>
<span class="kn">import</span> <span class="nn">mpl_toolkits.mplot3d.art3d</span> <span class="kn">as</span> <span class="nn">art3d</span>
<span class="kn">from</span> <span class="nn">mpl_toolkits.mplot3d.art3d</span> <span class="kn">import</span> <span class="n">Poly3DCollection</span>
<span class="kn">import</span> <span class="nn">matplotlib.gridspec</span> <span class="kn">as</span> <span class="nn">gridspec</span>
<span class="k">def</span> <span class="nf">dftmatrix</span><span class="p">(</span><span class="n">Nfft</span><span class="o">=</span><span class="mi">32</span><span class="p">,</span><span class="n">N</span><span class="o">=</span><span class="bp">None</span><span class="p">):</span>
<span class="s">'construct DFT matrix'</span>
<span class="n">k</span><span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="n">Nfft</span><span class="p">)</span>
<span class="k">if</span> <span class="n">N</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span> <span class="n">N</span> <span class="o">=</span> <span class="n">Nfft</span>
<span class="n">n</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="n">N</span><span class="p">)</span>
<span class="n">U</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">(</span><span class="n">exp</span><span class="p">(</span><span class="mi">1j</span><span class="o">*</span> <span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="n">Nfft</span> <span class="o">*</span><span class="n">k</span><span class="o">*</span><span class="n">n</span><span class="p">[:,</span><span class="bp">None</span><span class="p">]))</span> <span class="c"># use numpy broadcasting to create matrix</span>
<span class="k">return</span> <span class="n">U</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">Nfft</span><span class="p">)</span>
<span class="n">Nfft</span><span class="o">=</span><span class="mi">16</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">ones</span><span class="p">((</span><span class="mi">16</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">U</span> <span class="o">=</span> <span class="n">dftmatrix</span><span class="p">(</span><span class="n">Nfft</span><span class="o">=</span><span class="n">Nfft</span><span class="p">,</span><span class="n">N</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="c"># --- </span>
<span class="c"># hardcoded constants to format complicated figure</span>
<span class="n">gs</span> <span class="o">=</span> <span class="n">gridspec</span><span class="o">.</span><span class="n">GridSpec</span><span class="p">(</span><span class="mi">8</span><span class="p">,</span><span class="mi">12</span><span class="p">)</span>
<span class="n">gs</span><span class="o">.</span><span class="n">update</span><span class="p">(</span> <span class="n">wspace</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="mf">0.01</span><span class="p">)</span>
<span class="n">fig</span> <span class="o">=</span><span class="n">figure</span><span class="p">(</span><span class="n">figsize</span><span class="o">=</span><span class="p">(</span><span class="mi">10</span><span class="p">,</span><span class="mi">5</span><span class="p">))</span>
<span class="n">ax0</span> <span class="o">=</span> <span class="n">subplot</span><span class="p">(</span><span class="n">gs</span><span class="p">[:,:</span><span class="mi">3</span><span class="p">])</span>
<span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="n">ax0</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">a</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="n">Nfft</span><span class="o">*</span><span class="n">arange</span><span class="p">(</span><span class="n">Nfft</span><span class="p">)</span>
<span class="n">colors</span> <span class="o">=</span> <span class="p">[</span><span class="s">'k'</span><span class="p">,</span><span class="s">'b'</span><span class="p">,</span><span class="s">'r'</span><span class="p">,</span><span class="s">'m'</span><span class="p">,</span><span class="s">'g'</span><span class="p">,</span><span class="s">'Brown'</span><span class="p">,</span><span class="s">'DarkBlue'</span><span class="p">,</span><span class="s">'Tomato'</span><span class="p">,</span><span class="s">'Violet'</span><span class="p">,</span> <span class="s">'Tan'</span><span class="p">,</span><span class="s">'Salmon'</span><span class="p">,</span><span class="s">'Pink'</span><span class="p">,</span>
<span class="s">'SaddleBrown'</span><span class="p">,</span> <span class="s">'SpringGreen'</span><span class="p">,</span> <span class="s">'RosyBrown'</span><span class="p">,</span><span class="s">'Silver'</span><span class="p">,]</span>
<span class="k">for</span> <span class="n">j</span><span class="p">,</span><span class="n">i</span> <span class="ow">in</span> <span class="nb">enumerate</span><span class="p">(</span><span class="n">a</span><span class="p">):</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span> <span class="n">FancyArrow</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="n">cos</span><span class="p">(</span><span class="n">i</span><span class="p">),</span><span class="n">sin</span><span class="p">(</span><span class="n">i</span><span class="p">),</span><span class="n">width</span><span class="o">=</span><span class="mf">0.02</span><span class="p">,</span>
<span class="n">length_includes_head</span><span class="o">=</span><span class="bp">True</span><span class="p">,</span><span class="n">edgecolor</span><span class="o">=</span><span class="n">colors</span><span class="p">[</span><span class="n">j</span><span class="p">]))</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.1</span><span class="p">,</span><span class="s">'0'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="mf">0.1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="s">r'$\frac{\pi}{2}$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">22</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.1</span><span class="p">,</span><span class="s">r'$\pi$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="mf">0.1</span><span class="p">,</span><span class="o">-</span><span class="mf">1.2</span><span class="p">,</span><span class="s">r'$\frac{3\pi}{2}$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">22</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">array</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span><span class="o">*</span><span class="mf">1.45</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Radial Frequency'</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'Real'</span><span class="p">)</span>
<span class="n">ax0</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'Imaginary'</span><span class="p">)</span>
<span class="c"># plots in the middle column</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">8</span><span class="p">):</span>
<span class="n">ax</span><span class="o">=</span><span class="n">subplot</span><span class="p">(</span><span class="n">gs</span><span class="p">[</span><span class="n">i</span><span class="p">,</span><span class="mi">4</span><span class="p">:</span><span class="mi">8</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">([]);</span> <span class="n">ax</span><span class="o">.</span><span class="n">set_yticks</span><span class="p">([])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$\Omega_{</span><span class="si">%d</span><span class="s">}=</span><span class="si">%d</span><span class="s">\times\frac{2\pi}{16}$'</span><span class="o">%</span><span class="p">(</span><span class="n">i</span><span class="p">,</span><span class="n">i</span><span class="p">),</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">,</span>
<span class="n">rotation</span><span class="o">=</span><span class="s">'horizontal'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">U</span><span class="o">.</span><span class="n">real</span><span class="p">[:,</span><span class="n">i</span><span class="p">],</span><span class="s">'-o'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="n">colors</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">U</span><span class="o">.</span><span class="n">imag</span><span class="p">[:,</span><span class="n">i</span><span class="p">],</span><span class="s">'--o'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="n">colors</span><span class="p">[</span><span class="n">i</span><span class="p">],</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.2</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mi">4</span><span class="o">/</span><span class="n">Nfft</span><span class="o">*</span><span class="mf">1.1</span><span class="p">,</span><span class="n">ymin</span><span class="o">=-</span><span class="mi">4</span><span class="o">/</span><span class="n">Nfft</span><span class="o">*</span><span class="mf">1.1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">16</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'n'</span><span class="p">)</span>
<span class="c"># plots in the far right column</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">8</span><span class="p">):</span>
<span class="n">ax</span><span class="o">=</span><span class="n">subplot</span><span class="p">(</span><span class="n">gs</span><span class="p">[</span><span class="n">i</span><span class="p">,</span><span class="mi">8</span><span class="p">:])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">([]);</span> <span class="n">ax</span><span class="o">.</span><span class="n">set_yticks</span><span class="p">([])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$\Omega_{</span><span class="si">%d</span><span class="s">}=</span><span class="si">%d</span><span class="s">\times\frac{2\pi}{16}$'</span><span class="o">%</span><span class="p">(</span><span class="n">i</span><span class="o">+</span><span class="mi">8</span><span class="p">,</span><span class="n">i</span><span class="o">+</span><span class="mi">8</span><span class="p">),</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">,</span>
<span class="n">rotation</span><span class="o">=</span><span class="s">'horizontal'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">U</span><span class="o">.</span><span class="n">real</span><span class="p">[:,</span><span class="n">i</span><span class="o">+</span><span class="mi">8</span><span class="p">],</span><span class="s">'-o'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="n">colors</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">8</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">U</span><span class="o">.</span><span class="n">imag</span><span class="p">[:,</span><span class="n">i</span><span class="o">+</span><span class="mi">8</span><span class="p">],</span><span class="s">'--o'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="n">colors</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">8</span><span class="p">],</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.2</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mi">4</span><span class="o">/</span><span class="n">Nfft</span><span class="o">*</span><span class="mf">1.1</span><span class="p">,</span><span class="n">ymin</span><span class="o">=-</span><span class="mi">4</span><span class="o">/</span><span class="n">Nfft</span><span class="o">*</span><span class="mf">1.1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">yaxis</span><span class="o">.</span><span class="n">set_label_position</span><span class="p">(</span><span class="s">'right'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">16</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'n'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
</div>
</div>
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<p>On the left, the figure shows the discrete frequencies corresponding to each of the columns of the $\mathbf{U}$ matrix. These are color coded corresponding to the graphs on the right. For example, the $k=1$ column of the $\mathbf{U}$ matrix (i.e. $ \mathbf{u}_1 $) corresponds to discrete frequency $\Omega_1=\frac{2\pi}{16}$ marked on the y-axis label which is shown in the second row down the middle column in the figure. The real part of $ \mathbf{u}_1 $ is plotted in bold and the corresponding imaginary part is plotted semi-transparent because it is just an out-of-phase version of the real part. These real/imaginary parts shown in the graphs correspond to the conjugacy relationships on the leftmost radial plot. For example, $\Omega_1$ and $\Omega_{15}$ are complex conjugates and their corresponding imaginary parts are inverted as shown in the plots on the right. </p>
<p>The rows of the matrix correspond to the sample index given a particular sampling frequency, $f_s$. This means that if we have $N_s$ samples, then we have sampled a time duration over $N_s/f_s$. However, if we are only given a set of samples without the sampling frequency, then we can say nothing about time. For this reason, you will find discussions based on discrete frequency (i.e. between zero and $2\pi$) that do not reference sample rates. Thus, $N$ frequencies either divide the unit circle in discrete frequencies between 0 and $2\pi$ or divide the sample rate into sampled frequencies between zero and $f_s$. There is a one-to-one relationship between discrete and sampling frequency. In particular, we have for discrete frequency,</p>
<p>$$ \Omega_k = \frac{2\pi}{N} k $$</p>
<p>and for sampled frequency,</p>
<p>$$ f_k = \frac{f_s}{N} k $$</p>
<p>for the same value of $k$. Note that $\Omega_k$ is periodic with period $N$ (one full turn around the circle). One immediate consequence of the one-to-one correspondence between $\Omega_k$ and $f_k$ is that when $k=N/2$, we have $\Omega_{N/2}=\pi$ (halfway around the circle) and $f_{N/2}=f_s/2$ which is another way of saying that the Nyquist rate (the highest frequency we can unambiguously sample) occurs when $ \Omega_{N/2} = \pi$. We can see this by noting that as the discrete frequency rotates counter-clockwise away from zero and towards $\pi$, the plots on the right get more and more jagged. These also get smoother as the discrete frequency continues to rotate counter-clockwise towards zero again. This is because the higher frequencies are those close to $\pi$ and the lower frequencies are those close to zero on the complex plane. We will explore these crucial relationships further later, but for now, let's consider computing the DFT using this matrix.</p>
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<h2>Computing the DFT</h2>
<p>To compute the DFT using the matrix, we calculate the following,</p>
<p>$$ \mathbf{\hat{x}} = \mathbf{U}^H \mathbf{x}$$</p>
<p>which individually takes each of the columns of $\mathbf{U}$ and computes the inner product as the $i^{th}$ entry,</p>
<p>$$ \mathbf{\hat{x}}_i = \mathbf{u}_i^H \mathbf{x}$$</p>
<p>That is, we are measuring the <em>degree of similarity</em> between each column of $\mathbf{U}$ and the input vector. We can think of this as the coefficient of the projection of $\mathbf{x}$ onto $\mathbf{u}_i$.</p>
<p>We can retrieve the original input from the DFT by calculating</p>
<p>$$ \mathbf{x} = \mathbf{U} \mathbf{U}^H \mathbf{x} $$</p>
<p>because the columns of $\mathbf{U}$ are orthonormal (i.e. $\mathbf{u}_i^H \mathbf{u}_j = 0$). An important consequence of this is that $||\mathbf{x}||=||\mathbf{\hat{x}}||$ for any $\mathbf{x}$. This is Parseval's theorem and it means that the DFT is not <em>stretching</em> or <em>distorting</em> the input which makes it an ideal analysis tool.</p>
<h2>Zero-Padding and Frequency Sampling</h2>
<p>The only relationship between $N$, the size of the DFT, and the number of samples $N_s$ is that $N \ge N_s$. For implementation reasons, we will always choose $N$ as a power of 2. In the code below, let's now turn to the consquences of choose $N$ much larger that $N_s$. </p>
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<div class="prompt input_prompt">In [2]:</div>
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<div class="highlight"><pre><span class="n">U</span> <span class="o">=</span> <span class="n">dftmatrix</span><span class="p">(</span><span class="mi">64</span><span class="p">,</span><span class="mi">16</span><span class="p">)</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">ones</span><span class="p">((</span><span class="mi">16</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">U</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">x</span>
<span class="n">f</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">f</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">8</span><span class="p">,</span><span class="mi">4</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mf">0.8</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">64</span><span class="p">)</span><span class="o">*</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="mf">64.</span><span class="p">,</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'o-'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(\Omega)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span> <span class="n">pi</span><span class="o">/</span><span class="mf">2.</span><span class="p">,</span> <span class="n">pi</span><span class="p">,</span> <span class="mi">3</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="mi">2</span><span class="p">,</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">r'$\Omega$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mf">2.1</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticklabels</span><span class="p">([</span><span class="s">'0'</span><span class="p">,</span><span class="s">r'$\frac{\pi}{2}$'</span><span class="p">,</span> <span class="s">r'$\pi$'</span><span class="p">,</span><span class="s">r'$\frac{3\pi}{2}$'</span><span class="p">,</span> <span class="s">r'$2\pi$'</span><span class="p">],</span>
<span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">);</span>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>As you may recall from our <a href="http://python-for-signal-processing.blogspot.com/2012/09/investigating-sampling-theorem-in-this.html">earlier discussion</a>, this plot looks suspiciously like the <code>sinc</code> function. If you've been following closely, you may realize that for the above example we had $\mathbf{x}=\mathbf{1}$. But isn't this one of the columns of the $\mathbf{U}$ matrix? If all the columns of that matrix are orthonormal, then why is there is more than one non-zero point on this graph? The subtle point here is that the DFT matrix has dimensions $64 \times 16$. This means that computationally,</p>
<p>$$ \mathbf{U}_{16\times64}^H \mathbf{x} = \mathbf{U}_{64\times64}^H \left[\mathbf{x},\mathbf{0}\right]^T$$</p>
<p>In other words, filling the original $16\times 1$ vector $\mathbf{x}$ with zeros and using a larger compatible $\mathbf{U}_{64\times64}$ matrix has the same effect as using the $\mathbf{U}_{16\times64}$ matrix. The answer to the question is therefore that $\mathbf{x} = \mathbf{1}_{16\times1} \ne \left[ \mathbf{1}_{16\times1},\mathbf{0}\right]^T$ and the zero-augmented ones vector is <em>not</em> orthnormal to any columns in $\mathbf{U}_{64\times64}$. This explains why there are so many non-zero points on the graph at different discrete frequencies.</p>
<p>Let's drive this point home in the next figure.</p>
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<div class="prompt input_prompt">In [3]:</div>
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<div class="highlight"><pre><span class="n">U</span> <span class="o">=</span> <span class="n">dftmatrix</span><span class="p">(</span><span class="mi">64</span><span class="p">,</span><span class="mi">16</span><span class="p">)</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">ones</span><span class="p">((</span><span class="mi">16</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">X</span> <span class="o">=</span> <span class="n">U</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">x</span>
<span class="n">f</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="n">subplots</span><span class="p">()</span>
<span class="n">f</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">8</span><span class="p">,</span><span class="mi">4</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mf">0.8</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">64</span><span class="p">)</span><span class="o">*</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="mf">64.</span><span class="p">,</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">),</span><span class="s">'o-'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'zero padded'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">stem</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">16</span><span class="p">)</span><span class="o">*</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="mf">16.</span><span class="p">,</span><span class="nb">abs</span><span class="p">(</span><span class="n">dftmatrix</span><span class="p">(</span><span class="mi">16</span><span class="p">)</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">x</span><span class="p">),</span>
<span class="n">markerfmt</span><span class="o">=</span><span class="s">'gs'</span><span class="p">,</span> <span class="n">basefmt</span><span class="o">=</span><span class="s">'g-'</span><span class="p">,</span><span class="n">linefmt</span><span class="o">=</span><span class="s">'g-'</span><span class="p">,</span>
<span class="n">label</span><span class="o">=</span><span class="s">'no padding'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">r'$|X(\Omega)|$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span> <span class="n">pi</span><span class="o">/</span><span class="mf">2.</span><span class="p">,</span> <span class="n">pi</span><span class="p">,</span> <span class="mi">3</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="mi">2</span><span class="p">,</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">([</span><span class="o">-.</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="p">,</span><span class="o">-.</span><span class="mi">1</span><span class="p">,</span><span class="mf">4.1</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticklabels</span><span class="p">([</span><span class="s">'0'</span><span class="p">,</span><span class="s">r'$\frac{\pi}{2}$'</span><span class="p">,</span> <span class="s">r'$\pi$'</span><span class="p">,</span><span class="s">r'$\frac{3\pi}{2}$'</span><span class="p">,</span> <span class="s">r'$2\pi$'</span><span class="p">],</span>
<span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">);</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Zero padding samples more frequencies'</span><span class="p">);</span>
</pre></div>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>In the figure above, without zero-padding, $\mathbf{x}$ is the $ 0^{th} $ column of the 16-point DFT matrix and so all the coefficients except for the $0^{th}$ column are zero due to orthonormality (shown by the green squares). But, the zero-padded 64-element-long $\mathbf{x}$ vector is definitely <em>not</em> a column of the 64-point DFT matrix so we would <em>not</em> expect all the other terms to be zero. In fact, the other terms account for the 64 discrete frequencies that are plotted above. This means that zero-padding $\mathbf{x}$ and using the 64-point DFT matrix analyzes the signal across more frequencies. </p>
<p>Notice that for the $ 0^{th}$ frequency the height of the DFT magnitude is different for the zero-padded constant signal compared to the unpadded version. Recall from Parseval's theorem that $ ||\mathbf{x}||=||\mathbf{\hat{x}}|| $ but this does not account for how the signal may be spread across frequency. In the unpadded case, <em>all</em> of the signal energy is concentrated in the $ \mathbf{u}_0 $ column of the DFT matrix because our constant signal is just a scalar multiple of $ \mathbf{u}_0 $. In the padded case, the signal's energy is spread out across more frequencies with smaller signal magnitudes per frequency, thus satisfying Parseval's theorem. In other words, the single non-zero term in the unpadded DFT is smeared out over all the other frequencies in the padded case.</p>
<p>The problem with the figure shown is that it does not emphasize that the discrete frequencies are periodic with period $N$. The following figure below plots the 64-point DFT on the face of a cylinder to emphasize the periodicity of the discrete frequencies.</p>
</div>
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<div class="prompt input_prompt">In [4]:</div>
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<div class="highlight"><pre><span class="n">a</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="mf">64.</span><span class="o">*</span><span class="n">arange</span><span class="p">(</span><span class="mi">64</span><span class="p">)</span>
<span class="n">d</span><span class="o">=</span><span class="n">vstack</span><span class="p">([</span><span class="n">cos</span><span class="p">(</span><span class="n">a</span><span class="p">),</span><span class="n">sin</span><span class="p">(</span><span class="n">a</span><span class="p">),</span><span class="n">array</span><span class="p">(</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">))</span><span class="o">.</span><span class="n">flatten</span><span class="p">()])</span><span class="o">.</span><span class="n">T</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="n">d</span><span class="p">[:,</span><span class="mi">2</span><span class="p">]</span><span class="o">.</span><span class="n">max</span><span class="p">()])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">view_init</span><span class="p">(</span><span class="n">azim</span><span class="o">=-</span><span class="mi">30</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'real'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'imag'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlabel</span><span class="p">(</span><span class="s">'Abs'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'64-Point DFT Magnitudes'</span><span class="p">)</span>
<span class="k">def</span> <span class="nf">facet_filled</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.5</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'b'</span><span class="p">):</span>
<span class="s">'construct 3D facet from adjacent points filled to zero'</span>
<span class="n">a</span><span class="p">,</span><span class="n">b</span><span class="o">=</span><span class="n">x</span>
<span class="n">a0</span><span class="o">=</span> <span class="n">a</span><span class="o">*</span><span class="n">array</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">])</span>
<span class="n">b0</span><span class="o">=</span> <span class="n">b</span><span class="o">*</span><span class="n">array</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">])</span>
<span class="n">ve</span> <span class="o">=</span> <span class="n">vstack</span><span class="p">([</span><span class="n">a</span><span class="p">,</span><span class="n">a0</span><span class="p">,</span><span class="n">b0</span><span class="p">,</span><span class="n">b</span><span class="p">])</span> <span class="c"># create closed polygon facet</span>
<span class="n">poly</span> <span class="o">=</span> <span class="n">Poly3DCollection</span><span class="p">([</span><span class="n">ve</span><span class="p">])</span> <span class="c"># create facet</span>
<span class="n">poly</span><span class="o">.</span><span class="n">set_alpha</span><span class="p">(</span><span class="n">alpha</span><span class="p">)</span>
<span class="n">poly</span><span class="o">.</span><span class="n">set_color</span><span class="p">(</span><span class="n">color</span><span class="p">)</span>
<span class="k">return</span> <span class="n">poly</span>
<span class="n">sl</span><span class="o">=</span><span class="p">[</span><span class="nb">slice</span><span class="p">(</span><span class="n">i</span><span class="p">,</span><span class="n">i</span><span class="o">+</span><span class="mi">2</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">d</span><span class="o">.</span><span class="n">shape</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">-</span><span class="mi">2</span><span class="p">)]</span> <span class="c"># collect neighboring points</span>
<span class="k">for</span> <span class="n">s</span> <span class="ow">in</span> <span class="n">sl</span><span class="p">:</span>
<span class="n">poly</span><span class="o">=</span><span class="n">facet_filled</span><span class="p">(</span><span class="n">d</span><span class="p">[</span><span class="n">s</span><span class="p">,:])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_collection3d</span><span class="p">(</span><span class="n">poly</span><span class="p">)</span>
<span class="c"># edge polygons </span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_collection3d</span><span class="p">(</span><span class="n">facet_filled</span><span class="p">(</span><span class="n">d</span><span class="p">[[</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],:]))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_collection3d</span><span class="p">(</span><span class="n">facet_filled</span><span class="p">(</span><span class="n">d</span><span class="p">[[</span><span class="o">-</span><span class="mi">2</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">],:]))</span>
<span class="c"># add 0 and pi/2 arrows for reference</span>
<span class="n">a</span><span class="o">=</span><span class="n">FancyArrow</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="n">width</span><span class="o">=</span><span class="mf">0.02</span><span class="p">,</span><span class="n">length_includes_head</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>
<span class="n">b</span><span class="o">=</span><span class="n">FancyArrow</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">width</span><span class="o">=</span><span class="mf">0.02</span><span class="p">,</span><span class="n">length_includes_head</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
<span class="n">art3d</span><span class="o">.</span><span class="n">patch_2d_to_3d</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>
<span class="n">art3d</span><span class="o">.</span><span class="n">patch_2d_to_3d</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
</div>
</div>
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<p>The figure above is very important and we will be using it as a glyph in the following. It shows the same magnitude of the $X(\Omega)$ 64-point DFT as before, but now that it is plotted on a cylinder, we can really see the periodic discrete frequencies. The two arrows in the xy-plane show the discrete frequencies zero and $\pi$ for reference.<br />
</p>
<p>We will need the following setup code below.</p>
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<div class="prompt input_prompt">In [5]:</div>
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<div class="highlight"><pre><span class="k">def</span> <span class="nf">drawDFTView</span><span class="p">(</span><span class="n">X</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span><span class="bp">None</span><span class="p">,</span><span class="n">fig</span><span class="o">=</span><span class="bp">None</span><span class="p">):</span>
<span class="s">'above code as a function. Draws 3D diagram given DFT matrix'</span>
<span class="n">a</span><span class="o">=</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">/</span><span class="nb">len</span><span class="p">(</span><span class="n">X</span><span class="p">)</span><span class="o">*</span><span class="n">arange</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">X</span><span class="p">))</span>
<span class="n">d</span><span class="o">=</span><span class="n">vstack</span><span class="p">([</span><span class="n">cos</span><span class="p">(</span><span class="n">a</span><span class="p">),</span><span class="n">sin</span><span class="p">(</span><span class="n">a</span><span class="p">),</span><span class="n">array</span><span class="p">(</span><span class="nb">abs</span><span class="p">(</span><span class="n">X</span><span class="p">))</span><span class="o">.</span><span class="n">flatten</span><span class="p">()])</span><span class="o">.</span><span class="n">T</span>
<span class="k">if</span> <span class="n">ax</span> <span class="ow">is</span> <span class="bp">None</span> <span class="ow">and</span> <span class="n">fig</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span>
<span class="k">if</span> <span class="n">ax</span> <span class="ow">is</span> <span class="bp">None</span><span class="p">:</span> <span class="c"># add ax to existing figure</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">([</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="n">d</span><span class="p">[:,</span><span class="mi">2</span><span class="p">]</span><span class="o">.</span><span class="n">max</span><span class="p">()])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">view_init</span><span class="p">(</span><span class="n">azim</span><span class="o">=-</span><span class="mi">30</span><span class="p">)</span>
<span class="n">a</span><span class="o">=</span><span class="n">FancyArrow</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="n">width</span><span class="o">=</span><span class="mf">0.02</span><span class="p">,</span><span class="n">length_includes_head</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>
<span class="n">b</span><span class="o">=</span><span class="n">FancyArrow</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">width</span><span class="o">=</span><span class="mf">0.02</span><span class="p">,</span><span class="n">length_includes_head</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
<span class="n">art3d</span><span class="o">.</span><span class="n">patch_2d_to_3d</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>
<span class="n">art3d</span><span class="o">.</span><span class="n">patch_2d_to_3d</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
<span class="c">#ax.set_xticks([])</span>
<span class="c">#ax.set_yticks([])</span>
<span class="c">#ax.set_zticks([])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="s">'off'</span><span class="p">)</span>
<span class="n">sl</span><span class="o">=</span><span class="p">[</span><span class="nb">slice</span><span class="p">(</span><span class="n">i</span><span class="p">,</span><span class="n">i</span><span class="o">+</span><span class="mi">2</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">d</span><span class="o">.</span><span class="n">shape</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">-</span><span class="mi">2</span><span class="p">)]</span> <span class="c"># collect neighboring points</span>
<span class="k">for</span> <span class="n">s</span> <span class="ow">in</span> <span class="n">sl</span><span class="p">:</span>
<span class="n">poly</span><span class="o">=</span><span class="n">facet_filled</span><span class="p">(</span><span class="n">d</span><span class="p">[</span><span class="n">s</span><span class="p">,:])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_collection3d</span><span class="p">(</span><span class="n">poly</span><span class="p">)</span>
<span class="c"># edge polygons </span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_collection3d</span><span class="p">(</span><span class="n">facet_filled</span><span class="p">(</span><span class="n">d</span><span class="p">[[</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],:]))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_collection3d</span><span class="p">(</span><span class="n">facet_filled</span><span class="p">(</span><span class="n">d</span><span class="p">[[</span><span class="o">-</span><span class="mi">2</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">],:]))</span>
</pre></div>
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<div class="prompt input_prompt">In [6]:</div>
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<div class="highlight"><pre><span class="k">def</span> <span class="nf">drawInOut</span><span class="p">(</span><span class="n">X</span><span class="p">,</span><span class="n">v</span><span class="p">,</span><span class="n">return_axes</span><span class="o">=</span><span class="bp">False</span><span class="p">):</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">8</span><span class="p">,</span><span class="mi">8</span><span class="p">)</span>
<span class="n">gs</span> <span class="o">=</span> <span class="n">gridspec</span><span class="o">.</span><span class="n">GridSpec</span><span class="p">(</span><span class="mi">8</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax1</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">subplot</span><span class="p">(</span><span class="n">gs</span><span class="p">[</span><span class="mi">3</span><span class="p">:</span><span class="mi">5</span><span class="p">,:</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax2</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">subplot</span><span class="p">(</span><span class="n">gs</span><span class="p">[:,</span><span class="mi">2</span><span class="p">:],</span><span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">stem</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">v</span><span class="p">)),</span><span class="n">v</span><span class="p">)</span>
<span class="n">ymin</span><span class="p">,</span><span class="n">ymax</span><span class="o">=</span> <span class="n">ax1</span><span class="o">.</span><span class="n">get_ylim</span><span class="p">()</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">(</span><span class="n">ymax</span> <span class="o">=</span> <span class="n">ymax</span><span class="o">*</span><span class="mf">1.2</span><span class="p">,</span> <span class="n">ymin</span> <span class="o">=</span> <span class="n">ymin</span><span class="o">*</span><span class="mf">1.2</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'input signal'</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'time sample index'</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">tick_params</span><span class="p">(</span><span class="n">labelsize</span><span class="o">=</span><span class="mi">8</span><span class="p">)</span>
<span class="n">drawDFTView</span><span class="p">(</span><span class="n">X</span><span class="p">,</span><span class="n">ax2</span><span class="p">)</span>
<span class="k">if</span> <span class="n">return_axes</span><span class="p">:</span>
<span class="k">return</span> <span class="n">ax1</span><span class="p">,</span><span class="n">ax2</span>
</pre></div>
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<h2>Real Signals and DFT Symmetry</h2>
<p>Note the symmetric lobes in the following figure showing the DFT of a real signal. The plot on the left is the signal in the sampled time-domain and the plot on the right is its DFT-magnitude glyph.</p>
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<div class="highlight"><pre><span class="n">v</span> <span class="o">=</span> <span class="n">U</span><span class="p">[:,</span><span class="mi">6</span><span class="p">]</span><span class="o">.</span><span class="n">real</span>
<span class="n">drawInOut</span><span class="p">(</span><span class="n">U</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">v</span><span class="p">,</span><span class="n">v</span><span class="p">)</span>
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<p>Because the input signal is real, the DFT is symmetric. To see this, recall that in our first figure we observed for every $\mathbf{u}_i$, we had its complex conjugate $\mathbf{u}_{N-i}$ and since the real parts of complex conjugates are the same and there is no imaginary part in the input signal, the resulting corresponding inner products are complex conjugates and thus have the same magnitudes.</p>
<p>The next block of code illustrates this.</p>
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<div class="highlight"><pre><span class="k">print</span> <span class="nb">abs</span><span class="p">(</span><span class="n">U</span><span class="p">[:,[</span><span class="mi">6</span><span class="p">,</span><span class="mi">64</span><span class="o">-</span><span class="mi">6</span><span class="p">]]</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">v</span><span class="p">)</span> <span class="c"># real signal has same abs() inner product for conjugate columns</span>
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<pre>[[ 0.125]
[ 0.125]]
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<p>This fact has extremely important computational consequences for the fast implemention of the DFT (i.e. FFT), which we will take up another time. For now, it's enough to recognize the symmetry of the DFT of real signals. </p>
<h2>High/Low-Frequency Signals and Their DFTs</h2>
<p>Now that we have all the vocabulary defined, we can ask one more intuitive question: what does the highest frequency signal (i.e. $\Omega_{N/2}=\pi$) look like in the sampled time-domain? This is shown in the top figure below that shows a signal toggling back and forth positive and negative. Note that the amplitudes of this toggling are not important, it's just the <em>rate</em> of toggling that defines the high frequency signal. Also, what does the lowest frequency signal (i.e. $\Omega_0=0$) look like as a sampled signal? This is shown in the bottom figure. Note that it is the mirror image of the high frequency signal.</p>
<p>I invite you to please download the <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Fourier_Transform.ipynb">IPython notebook corresponding to this post </a> and play with these plots to develop an intuition for where the various input signals appear.</p>
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<div class="highlight"><pre><span class="n">v</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">(</span><span class="n">cos</span><span class="p">(</span><span class="n">pi</span><span class="o">*</span><span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">16</span><span class="p">)))</span><span class="o">.</span><span class="n">T</span>
<span class="n">ax1</span><span class="p">,</span><span class="n">ax2</span><span class="o">=</span><span class="n">drawInOut</span><span class="p">(</span><span class="n">U</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">v</span><span class="p">,</span><span class="n">v</span><span class="p">,</span><span class="n">return_axes</span><span class="o">=</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Highest Frequency'</span><span class="p">)</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">ones</span><span class="p">((</span><span class="mi">16</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">ax1</span><span class="p">,</span><span class="n">ax2</span><span class="o">=</span><span class="n">drawInOut</span><span class="p">(</span><span class="n">U</span><span class="o">.</span><span class="n">H</span><span class="o">*</span><span class="n">v</span><span class="p">,</span><span class="n">v</span><span class="p">,</span><span class="n">return_axes</span><span class="o">=</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Lowest Frequency'</span><span class="p">)</span>
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<pre><matplotlib.text.Text at 0x6bab1f0></pre>
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</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Summary
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>In this section, we considered the Discrete Fourier Transform (DFT) using a matrix/vector approach. We used this approach to develop an intuitive visual vocabulary for the DFT with respect to high/low frequency and real-valued signals. We used zero-padding to enhance frequency domain signal analysis.</p>
<p>As usual, the corresponding IPython notebook for this post is available for download <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Fourier_Transform.ipynb">here</a>. </p>
<p>Comments and corrections welcome!</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>References</h2>
<ul>
<li>Oppenheim, A. V., and A. S. Willsky. "Signals and Systems." Prentice-Hall, (1997).</li>
</ul>
</div>
Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com1tag:blogger.com,1999:blog-2783842479260003750.post-25333912487065388082013-02-11T19:00:00.000-08:002013-02-11T19:00:03.001-08:00Gauss Markov<div class="text_cell_render border-box-sizing rendered_html">
<h2>Introduction</h2>
<p>In this section, we consider the famous Gauss-Markov problem which will give us an opportunity to use all the material we have so far developed. The Gauss-Markov is the fundamental model for noisy parameter estimation because it estimates unobservable parameters given a noisy indirect measurement. Incarnations of the same model appear in all studies of Gaussian models. This case is an excellent opportunity to use everything we have so far learned about projection and conditional expectation.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>Following Luenberger (1997), let's consider the following problem:</p>
<p>$$ \mathbf{y} = \mathbf{W} \boldsymbol{\beta} + \boldsymbol{\epsilon} $$</p>
<p>where $\mathbf{W}$ is a $ n \times m $ matrix, and $\mathbf{y}$ is a $n \times 1$ vector. Also, $\boldsymbol{\epsilon}$ is a $n$-dimensional random vector with zero-mean and covariance</p>
<p>$$ \mathbb{E}( \boldsymbol{\epsilon} \boldsymbol{\epsilon}^T) = \mathbf{Q}$$</p>
<p>Note that real systems usually provide a <em>calibration mode</em> where you can estimate $\mathbf{Q}$ so it's not fantastical to assume you have some knowledge of the noise statistics. The problem is to find a matrix $\mathbf{K}$ so that $ \boldsymbol{\hat{\beta}} = \mathbf{K} \mathbf{y}$ approximates $ \boldsymbol{\beta}$. Note that we only have knowledge of $\boldsymbol{\beta}$ via $ \mathbf{y}$ so we can't measure it directly. Further, note that $\mathbf{K} $ is a matrix, not a vector, so there are $m \times n$ entries to compute. </p>
<p>We can approach this problem the usual way by trying to solve the MMSE problem:</p>
<p>$$ \min_K \mathbb{E}(|| \boldsymbol{\hat{\beta}}- \boldsymbol{\beta} ||^2)$$</p>
<p>which we can write out as</p>
<p>$$ \min_K \mathbb{E}(|| \boldsymbol{\hat{\beta}}- \boldsymbol{\beta} ||^2)
= \min_K\mathbb{E}(|| \mathbf{K}\mathbf{y}- \boldsymbol{\beta} ||^2)
= \min_K\mathbb{E}(|| \mathbf{K}\mathbf{W}\mathbf{\boldsymbol{\beta}}+\mathbf{K}\boldsymbol{\epsilon}- \boldsymbol{\beta} ||^2)$$</p>
<p>and since $\boldsymbol{\epsilon}$ is the only random variable here, this simplifies to</p>
<p>$$\min_K || \mathbf{K}\mathbf{W}\mathbf{\boldsymbol{\beta}}- \boldsymbol{\beta} ||^2 + \mathbb{E}(||\mathbf{K}\boldsymbol{\epsilon} ||^2)
$$</p>
<p>The next step is to compute</p>
<p>$\DeclareMathOperator{\Tr}{Trace}$
$$ \mathbb{E}(||\mathbf{K}\boldsymbol{\epsilon} ||^2) = \mathbb{E}(\boldsymbol{\epsilon}^T \mathbf{K}^T \mathbf{K}^T \boldsymbol{\epsilon})=\Tr(\mathbf{K \mathbb{E}(\boldsymbol{\epsilon}\boldsymbol{\epsilon}^T) K}^T)=\Tr(\mathbf{K Q K}^T)$$</p>
<p>using the properties of the trace of a matrix. We can assemble everything as</p>
<p>$$ \min_K || \mathbf{K W} \boldsymbol{\beta} - \boldsymbol{\beta}||^2 + \Tr(\mathbf{K Q K}^T) $$</p>
<p>Now, if we were to solve this for $\mathbf{K}$, it would be a function of $ \boldsymbol{\beta}$, which is the same thing as saying that the estimator, $ \boldsymbol{\hat{\beta}}$, is a function of what we are trying to estimate, $ \boldsymbol{\beta}$, which makes no sense. However, writing this out tells us that if we had $\mathbf{K W}= \mathbf{I}$, then the first term vanishes and the problem simplifies to</p>
<p>$$ \min_K \Tr(\mathbf{K Q K}^T) $$</p>
<p>with</p>
<p>$$ \mathbf{KW} = \mathbf{I}$$</p>
<p>This requirement is the same as asserting that the estimator is unbiased,</p>
<p>$$ \mathbb{E}( \boldsymbol{\hat{\beta}}) = \mathbf{KW} \boldsymbol{\beta} = \boldsymbol{\beta} $$ </p>
<p>To line this problem up with our earlier work, let's consider the $i^{th}$ column of $\mathbf{K}$, $\mathbf{k}_i$. Now, we can re-write the problem as</p>
<p>$$ \min_k (\mathbf{k}_i^T \mathbf{Q} \mathbf{k}_i) $$</p>
<p>with</p>
<p>$$ \mathbf{k}_i^T \mathbf{W} = \mathbf{e}_i$$</p>
<p>and from our previous work on contrained optimization, we know the solution to this:</p>
<p>$$ \mathbf{k}_i = \mathbf{Q}^{-1} \mathbf{W}(\mathbf{W}^T \mathbf{Q^{-1} W})^{-1}\mathbf{e}_i$$</p>
<p>Now all we have to do is stack these together for the general solution:</p>
<p>$$ \mathbf{K} = (\mathbf{W}^T \mathbf{Q^{-1} W})^{-1} \mathbf{W}^T\mathbf{Q}^{-1} $$</p>
<p>It's easy when you have all of the concepts lined up! For completeness, the covariance of the error is</p>
<p>$$ \mathbb{E}(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta}) (\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})^T
= \mathbb{E}(\mathbf{K} \boldsymbol{\epsilon} \boldsymbol{\epsilon}^T \mathbf{K}^T)=\mathbf{K}\mathbf{Q}\mathbf{K}^T =(\mathbf{W}^T \mathbf{Q}^{-1} \mathbf{W})^{-1}$$</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>The following simulation illustrates these results.</p>
</div>
<div class="cell border-box-sizing code_cell vbox">
<div class="input hbox">
<div class="prompt input_prompt">In [18]:</div>
<div class="input_area box-flex1">
<div class="highlight"><pre><span class="kn">from</span> <span class="nn">mpl_toolkits.mplot3d</span> <span class="kn">import</span> <span class="n">proj3d</span>
<span class="kn">from</span> <span class="nn">numpy.linalg</span> <span class="kn">import</span> <span class="n">inv</span>
<span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">as</span> <span class="nn">plt</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="kn">from</span> <span class="nn">numpy</span> <span class="kn">import</span> <span class="n">matrix</span><span class="p">,</span> <span class="n">linalg</span><span class="p">,</span> <span class="n">ones</span><span class="p">,</span> <span class="n">array</span>
<span class="n">Q</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">eye</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span><span class="o">*.</span><span class="mi">1</span> <span class="c"># error covariance matrix</span>
<span class="c">#Q[0,0]=1</span>
<span class="n">beta</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">(</span><span class="n">ones</span><span class="p">((</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">)))</span> <span class="c"># this is what we are trying estimate</span>
<span class="n">W</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">],</span>
<span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">],</span>
<span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">]])</span>
<span class="n">ntrials</span> <span class="o">=</span> <span class="mi">50</span>
<span class="n">epsilon</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">multivariate_normal</span><span class="p">((</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">),</span><span class="n">Q</span><span class="p">,</span><span class="n">ntrials</span><span class="p">)</span><span class="o">.</span><span class="n">T</span>
<span class="n">y</span><span class="o">=</span><span class="n">W</span><span class="o">*</span><span class="n">beta</span><span class="o">+</span><span class="n">epsilon</span>
<span class="n">K</span><span class="o">=</span><span class="n">inv</span><span class="p">(</span><span class="n">W</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">inv</span><span class="p">(</span><span class="n">Q</span><span class="p">)</span><span class="o">*</span><span class="n">W</span><span class="p">)</span><span class="o">*</span><span class="n">matrix</span><span class="p">(</span><span class="n">W</span><span class="o">.</span><span class="n">T</span><span class="p">)</span><span class="o">*</span><span class="n">inv</span><span class="p">(</span><span class="n">Q</span><span class="p">)</span>
<span class="n">b</span><span class="o">=</span><span class="n">K</span><span class="o">*</span><span class="n">y</span> <span class="c">#estimated beta from data</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">([</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">])</span>
<span class="c"># some convenience definitions for plotting</span>
<span class="n">bb</span> <span class="o">=</span> <span class="n">array</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
<span class="n">bm</span> <span class="o">=</span> <span class="n">bb</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">yy</span> <span class="o">=</span> <span class="n">array</span><span class="p">(</span><span class="n">y</span><span class="p">)</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">,</span> <span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">(</span><span class="n">yy</span><span class="p">[</span><span class="mi">0</span><span class="p">,:],</span><span class="n">yy</span><span class="p">[</span><span class="mi">1</span><span class="p">,:],</span><span class="n">yy</span><span class="p">[</span><span class="mi">2</span><span class="p">,:],</span><span class="s">'mo'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'y'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">beta</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">beta</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="s">'r-'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\beta$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">bm</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">bm</span><span class="p">[</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="s">'g-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\hat{\beta}_m$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">(</span><span class="n">bb</span><span class="p">[</span><span class="mi">0</span><span class="p">,:],</span><span class="n">bb</span><span class="p">[</span><span class="mi">1</span><span class="p">,:],</span><span class="mi">0</span><span class="o">*</span><span class="n">bb</span><span class="p">[</span><span class="mi">1</span><span class="p">,:],</span><span class="s">'.g'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.5</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\hat{\beta}$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
</div>
</div>
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"></img>
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>The figure above show the simulated $\mathbf{y}$ data as magenta circles. The green dots show the corresponding estimates, $\boldsymbol{\hat{\beta}}$ for each sample. The red and green lines show the true value of $\boldsymbol{\beta}$ versus the average of the estimated $\boldsymbol{\beta}$-values, $\boldsymbol{\hat{\beta_m}}$. The matrix $\mathbf{K}$ maps the magenta circles in the corresponding green dots. Note there are many possible ways to map the magenta circles to the plane, but the $\mathbf{K}$ is the ones that minimizes the MSE for $\boldsymbol{\beta}$. </p>
<p>The figure below shows more detail in the horizontal <em>xy</em>-plane above.</p>
</div>
<div class="cell border-box-sizing code_cell vbox">
<div class="input hbox">
<div class="prompt input_prompt">In [19]:</div>
<div class="input_area box-flex1">
<div class="highlight"><pre><span class="kn">from</span> <span class="nn">matplotlib.patches</span> <span class="kn">import</span> <span class="n">Ellipse</span>
<span class="n">fig</span><span class="p">,</span> <span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">((</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">bb</span><span class="p">[</span><span class="mi">0</span><span class="p">,:],</span><span class="n">bb</span><span class="p">[</span><span class="mi">1</span><span class="p">,:],</span><span class="s">'g.'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">([</span><span class="n">beta</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">beta</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'r--'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\beta$'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">4.</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">([</span><span class="n">bm</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">bm</span><span class="p">[</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'g-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\hat{\beta}_m$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">bm_cov</span> <span class="o">=</span> <span class="n">inv</span><span class="p">(</span><span class="n">W</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Q</span><span class="o">*</span><span class="n">W</span><span class="p">)</span>
<span class="n">U</span><span class="p">,</span><span class="n">S</span><span class="p">,</span><span class="n">V</span> <span class="o">=</span> <span class="n">linalg</span><span class="o">.</span><span class="n">svd</span><span class="p">(</span><span class="n">bm_cov</span><span class="p">)</span>
<span class="n">err</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">sqrt</span><span class="p">((</span><span class="n">matrix</span><span class="p">(</span><span class="n">bm</span><span class="p">))</span><span class="o">*</span><span class="p">(</span><span class="n">bm_cov</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">matrix</span><span class="p">(</span><span class="n">bm</span><span class="p">)</span><span class="o">.</span><span class="n">T</span><span class="p">))</span>
<span class="n">theta</span> <span class="o">=</span> <span class="n">arccos</span><span class="p">(</span><span class="n">U</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span><span class="o">/</span><span class="n">pi</span><span class="o">*</span><span class="mi">180</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">Ellipse</span><span class="p">(</span><span class="n">bm</span><span class="p">,</span><span class="n">err</span><span class="o">*</span><span class="mi">2</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">S</span><span class="p">[</span><span class="mi">0</span><span class="p">]),</span><span class="n">err</span><span class="o">*</span><span class="mi">2</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">S</span><span class="p">[</span><span class="mi">1</span><span class="p">])</span>
<span class="p">,</span><span class="n">angle</span><span class="o">=</span><span class="n">theta</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'pink'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.5</span><span class="p">))</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
</div>
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"></img>
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<p>The figure above shows the green dots, which are individual estimates of $\boldsymbol{\hat{\beta}}$ from the corresponding simulated $\mathbf{y}$ data. The red dashed line is the true value for $\boldsymbol{\beta}$ and the green line ($\boldsymbol{\hat{\beta_m}}$ ) is the average of all the green dots. Note there is hardly a visual difference between them. The pink ellipse provides some scale as to the covariance of the estimated $\boldsymbol{\beta}$ values. I invite you to download this <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Gauss_Markov.ipynb">IPython notebook</a> and tweak the values for the error covariance or any other parameter. All of the plots should update and scale correctly.</p>
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<h2>
Summary
</h2>
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<p>The Gauss-Markov problem is the cornerstone of all Gaussian modeling and is thus one of the most powerful models used in signal processing. We showed how to estimate the unobservable parameters given noisy measurements using our previous work on projection. We also coded up a short example to illustrate how this works in a simulation. For a much more detailed approach, see Luenberger (1997).</p>
<p>As usual, the corresponding IPython notebook for this post is available for download <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Gauss_Markov.ipynb">here</a>. </p>
<p>Comments and corrections welcome!</p>
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<h2>References</h2>
<ul>
<li>Luenberger, David G. <em>Optimization by vector space methods</em>. Wiley-Interscience, 1997.</li>
</ul>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-77567314941650072762013-01-28T20:00:00.000-08:002013-01-28T20:00:01.337-08:00Inverse Projection as Constrained Optimization<body>
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<h2>Inverse Projection</h2>
<p>In our <a href="http://python-for-signal-processing.blogspot.com/2012/11/projection-in-multiple-dimensions-in.html">previous discussion</a>, we developed a powerful intuitive sense for projections and their relationships to minimum mean squared error problems. Here, we continue to extract yet another powerful result from the same concept. Recall that we learned to minimize</p>
<p>$$ J = || \mathbf{y} - \mathbf{V}\boldsymbol{\alpha} ||^2 $$</p>
<p>by projecting $\mathbf{y}$ onto the space characterized by $\mathbf{V}$ as follows:</p>
<p>$$ \mathbf{\hat{y}} = \mathbf{P}_V \mathbf{y}$$</p>
<p>where</p>
<p>$$ \mathbf{P}_V = \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T$$</p>
<p>where the corresponding error ($\boldsymbol{\epsilon}$) comes directly from the Pythagorean theorem:</p>
<p>$$ ||\mathbf{y}||^2 = ||\mathbf{\hat{y}}||^2 + ||\boldsymbol{\epsilon}||^2 $$</p>
<p>Now, let's consider the inverse problem: given $\mathbf{\hat{y}}$, what is the corresponding $\mathbf{y}$? The first impulse in this situation is to re-visit </p>
<p>$$ \mathbf{\hat{y}} = \mathbf{P}_V \mathbf{y}$$</p>
<p>and see if you can somehow compute the inverse of $\mathbf{P}_V$. This will not work because the projection matrix does not possess a unique inverse. In the following figure, the vertical sheet represents all vectors in the space that have exactly the same projection, $\mathbf{\hat{y}}$. Thus, there is no unique solution to the inverse problem which is to say that it is "ill-posed".</p>
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<div class="highlight"><pre><span class="c">#http://stackoverflow.com/questions/10374930/matplotlib-annotating-a-3d-scatter-plot</span>
<span class="kn">from</span> <span class="nn">mpl_toolkits.mplot3d</span> <span class="kn">import</span> <span class="n">proj3d</span>
<span class="kn">import</span> <span class="nn">matplotlib.patches</span> <span class="kn">as</span> <span class="nn">patches</span>
<span class="kn">import</span> <span class="nn">mpl_toolkits.mplot3d.art3d</span> <span class="kn">as</span> <span class="nn">art3d</span>
<span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">as</span> <span class="nn">plt</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">([</span><span class="mi">8</span><span class="p">,</span><span class="mi">8</span><span class="p">])</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">,</span> <span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'x-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'y-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlabel</span><span class="p">(</span><span class="s">'z-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">y</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span><span class="o">.</span><span class="n">T</span>
<span class="n">V</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.25</span><span class="p">],</span> <span class="c"># columns are v_1, v_2</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.50</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.00</span><span class="p">]])</span>
<span class="n">alpha</span><span class="o">=</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">y</span> <span class="c"># optimal coefficients</span>
<span class="n">P</span> <span class="o">=</span> <span class="n">V</span><span class="o">*</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span>
<span class="n">yhat</span> <span class="o">=</span> <span class="n">P</span><span class="o">*</span><span class="n">y</span> <span class="c"># approximant</span>
<span class="n">u</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">xx</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">yy</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">zz</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">size</span><span class="p">(</span><span class="n">u</span><span class="p">)),</span> <span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">sphere</span><span class="o">=</span><span class="n">ax</span><span class="o">.</span><span class="n">plot_surface</span><span class="p">(</span><span class="n">xx</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">yy</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">zz</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="n">rstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">cstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s">'gray'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">0.25</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'r-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'ro'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'g--'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'go'</span><span class="p">)</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{y}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">40</span><span class="p">,</span> <span class="mi">20</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\hat{\mathbf{y}}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">40</span><span class="p">,</span> <span class="mi">10</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_1$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">120</span><span class="p">,</span> <span class="mi">10</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_2$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">30</span><span class="p">,</span> <span class="mi">30</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">xx</span> <span class="o">=</span> <span class="n">array</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">]])</span>
<span class="n">yy</span> <span class="o">=</span> <span class="n">array</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">]])</span>
<span class="n">zz</span> <span class="o">=</span> <span class="n">array</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_collection3d</span><span class="p">(</span> <span class="n">art3d</span><span class="o">.</span><span class="n">Poly3DCollection</span><span class="p">([</span><span class="nb">zip</span><span class="p">(</span><span class="n">xx</span><span class="p">,</span><span class="n">yy</span><span class="p">,</span><span class="n">zz</span><span class="p">)],</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.15</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'m'</span><span class="p">)</span> <span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'The magenta sheet contains vectors with the same projection, $\mathbf{\hat{y}}$'</span><span class="p">)</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
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<p>Thus, since the unique inverse does not exist, we can impose constraints on the solution to enforce a unique solution. For example, since there are many $\mathbf{y}$ that correspond to the same $\mathbf{\hat{y}}$, we can pick the one that has the shortest length, $||\mathbf{y} ||$. Thus, we can solve this inverse projection problem by enforcing additional constraints. </p>
<p>Consider the following constrained minimization problem:</p>
<p>$$ \min_y \mathbf{y}^T \mathbf{y}$$</p>
<p>subject to:</p>
<p>$$ \mathbf{v_1}^T \mathbf{y} = c_1$$
$$ \mathbf{v_2}^T \mathbf{y} = c_2$$</p>
<p>Here, we have the same setup as before: we want to minimize something with a set of constraints. We can re-write the constraints in a more familiar form as</p>
<p>$$ \mathbf{V}^T \mathbf{y} = \mathbf{c}$$</p>
<p>We can multiply both sides to obtain the even more familiar form:</p>
<p>$$ \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T \mathbf{y} = \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1}\mathbf{c} $$</p>
<p>which by cleaning up the notation gives,</p>
<p>$$ \mathbf{P}_V \mathbf{y} = \mathbf{\hat{y}} $$</p>
<p>where</p>
<p>$$ \mathbf{\hat{y}} = \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1}\mathbf{c}$$</p>
<p>and</p>
<p>$$ \mathbf{P}_V=\mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T $$</p>
<p>So far, nothing has really happened yet. We've just rewritten the constraints as a projection, but we still don't know how to find the $\mathbf{y}$ of minimum length. To do that, we turn to the Pythagorean relationship we pointed out earlier,</p>
<p>$$ ||\mathbf{y}||^2 = ||\mathbf{\hat{y}}||^2 + ||\boldsymbol{\epsilon}||^2 $$</p>
<p>where $\boldsymbol{\epsilon} = \mathbf{y} - \mathbf{\hat{y}}$. So, we have</p>
<p>$$ ||\mathbf{y}||^2 = ||\mathbf{\hat{y}}||^2 + || \mathbf{y} - \mathbf{\hat{y}}||^2 \ge 0$$</p>
<p>and since this is always non-negative, the only way to minimize $||\mathbf{y}||^2$ is to set </p>
<p>$$ \mathbf{y} = \mathbf{\hat{y}} = \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1}\mathbf{c} $$</p>
<p>which is the solution to our constrained minimization problem.</p>
<p>If that seems shockingly easy, remember that we have already done all the heavy lifting by solving the projection problem. Here, all we have done is re-phrased the same problem.</p>
<p>Let's see this in action using <code>scipy.optimize</code> for comparison. Here's the problem</p>
<p>$$ \min_y \mathbf{y}^T \mathbf{y}$$</p>
<p>subject to:</p>
<p>$$ \mathbf{e_1}^T \mathbf{y} = 1$$
$$ \mathbf{e_2}^T \mathbf{y} = 1$$</p>
<p>where $\mathbf{e}_i$ is the coordinate vector that is zero everywhere except for the $i^{th}$ entry.</p>
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<div class="highlight"><pre><span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="kn">from</span> <span class="nn">scipy.optimize</span> <span class="kn">import</span> <span class="n">minimize</span>
<span class="c"># constraints formatted for scipy.optimize.minimize</span>
<span class="n">cons</span> <span class="o">=</span> <span class="p">[{</span><span class="s">'type'</span><span class="p">:</span><span class="s">'eq'</span><span class="p">,</span><span class="s">'fun'</span><span class="p">:</span><span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">x</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="s">'jac'</span><span class="p">:</span><span class="bp">None</span><span class="p">},</span>
<span class="p">{</span><span class="s">'type'</span><span class="p">:</span><span class="s">'eq'</span><span class="p">,</span><span class="s">'fun'</span><span class="p">:</span><span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">x</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="s">'jac'</span><span class="p">:</span><span class="bp">None</span><span class="p">},</span>
<span class="p">]</span>
<span class="n">init_point</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">0</span><span class="p">])</span> <span class="c"># initial guess</span>
<span class="n">ysol</span><span class="o">=</span> <span class="n">minimize</span><span class="p">(</span><span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">x</span><span class="p">),</span><span class="n">init_point</span><span class="p">,</span><span class="n">constraints</span><span class="o">=</span><span class="n">cons</span><span class="p">,</span><span class="n">method</span><span class="o">=</span><span class="s">'SLSQP'</span><span class="p">)</span>
<span class="c"># using projection method</span>
<span class="n">c</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]])</span><span class="o">.</span><span class="n">T</span> <span class="c"># RHS constraint vector</span>
<span class="n">V</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">matrix</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">eye</span><span class="p">(</span><span class="mi">4</span><span class="p">)[:,</span><span class="mi">0</span><span class="p">:</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ysol_p</span> <span class="o">=</span> <span class="n">V</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">linalg</span><span class="o">.</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">c</span>
<span class="k">print</span> <span class="s">'scipy optimize solution:'</span><span class="p">,</span>
<span class="k">print</span> <span class="n">ysol</span><span class="p">[</span><span class="s">'x'</span><span class="p">]</span>
<span class="k">print</span> <span class="s">'projection solution:'</span><span class="p">,</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">(</span><span class="n">ysol_p</span><span class="p">)</span><span class="o">.</span><span class="n">flatten</span><span class="p">()</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">allclose</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">(</span><span class="n">ysol_p</span><span class="p">)</span><span class="o">.</span><span class="n">flat</span><span class="p">,</span><span class="n">ysol</span><span class="p">[</span><span class="s">'x'</span><span class="p">],</span><span class="n">atol</span><span class="o">=</span><span class="mf">1e-6</span><span class="p">)</span>
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<pre>scipy optimize solution: [ 1.00000000e+00 1.00000000e+00 -2.17476398e-08 -2.93995881e-08]
projection solution: [ 1. 1. 0. 0.]
True
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<h2>
Weighted Constrained Minimization
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<p>We can again pursue a more general problem using the same technique:</p>
<p>$$ \min_y \mathbf{y}^T \mathbf{Q} \mathbf{y} $$</p>
<p>subject to:</p>
<p>$$ \mathbf{V}^T \mathbf{y} = \mathbf{c}$$</p>
<p>where $\mathbf{Q}$ is a positive definite matrix. In this case, we can define $\eta$ such that:</p>
<p>$$ \mathbf{y} = \mathbf{Q}^{-1} \boldsymbol{\eta}$$</p>
<p>and re-write the constraint as</p>
<p>$$ \mathbf{V}^T \mathbf{Q}^{-1} \boldsymbol{\eta} = \mathbf{c}$$ </p>
<p>and multiply both sides,</p>
<p>$$\mathbf{P}_V \boldsymbol{\eta}=\mathbf{V} ( \mathbf{V}^T \mathbf{Q^{-1} V})^{-1} \mathbf{c} $$</p>
<p>where</p>
<p>$$\mathbf{P}_V=\mathbf{V} ( \mathbf{V}^T \mathbf{Q^{-1} V})^{-1} \mathbf{V}^T \mathbf{Q}^{-1}$$</p>
<p>To sum up, the minimal $\mathbf{y}$ that solves this constrained minimization problem is</p>
<p>$$ \mathbf{y}_o = \mathbf{V} ( \mathbf{V}^T \mathbf{Q^{-1} V})^{-1} \mathbf{c}$$</p>
<p>Once again, let's illustrate this using <code>scipy.optimize</code> in the following</p>
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<div class="prompt input_prompt">In [31]:</div>
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<div class="highlight"><pre><span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="kn">from</span> <span class="nn">scipy.optimize</span> <span class="kn">import</span> <span class="n">minimize</span>
<span class="c"># constraints formatted for scipy.optimize.minimize</span>
<span class="n">cons</span> <span class="o">=</span> <span class="p">[{</span><span class="s">'type'</span><span class="p">:</span><span class="s">'eq'</span><span class="p">,</span><span class="s">'fun'</span><span class="p">:</span><span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">x</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="s">'jac'</span><span class="p">:</span><span class="bp">None</span><span class="p">},</span>
<span class="p">{</span><span class="s">'type'</span><span class="p">:</span><span class="s">'eq'</span><span class="p">,</span><span class="s">'fun'</span><span class="p">:</span><span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">x</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="s">'jac'</span><span class="p">:</span><span class="bp">None</span><span class="p">},</span>
<span class="p">]</span>
<span class="n">Q</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">diag</span> <span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">])</span>
<span class="n">init_point</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">0</span><span class="p">])</span> <span class="c"># initial guess</span>
<span class="n">ysol</span><span class="o">=</span> <span class="n">minimize</span><span class="p">(</span><span class="k">lambda</span> <span class="n">x</span><span class="p">:</span> <span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">Q</span><span class="p">,</span><span class="n">x</span><span class="p">)),</span><span class="n">init_point</span><span class="p">,</span><span class="n">constraints</span><span class="o">=</span><span class="n">cons</span><span class="p">,</span><span class="n">method</span><span class="o">=</span><span class="s">'SLSQP'</span><span class="p">)</span>
<span class="c"># using projection method</span>
<span class="n">Qinv</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linalg</span><span class="o">.</span><span class="n">inv</span><span class="p">(</span><span class="n">Q</span><span class="p">)</span>
<span class="n">c</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]])</span><span class="o">.</span><span class="n">T</span> <span class="c"># RHS constraint vector</span>
<span class="n">V</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">matrix</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">eye</span><span class="p">(</span><span class="mi">4</span><span class="p">)[:,</span><span class="mi">0</span><span class="p">:</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ysol_p</span> <span class="o">=</span> <span class="n">V</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">linalg</span><span class="o">.</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Qinv</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Qinv</span><span class="o">*</span><span class="n">c</span>
<span class="k">print</span> <span class="s">'scipy optimize solution:'</span><span class="p">,</span>
<span class="k">print</span> <span class="n">ysol</span><span class="p">[</span><span class="s">'x'</span><span class="p">]</span>
<span class="k">print</span> <span class="s">'projection solution:'</span><span class="p">,</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">(</span><span class="n">ysol_p</span><span class="p">)</span><span class="o">.</span><span class="n">flatten</span><span class="p">()</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">allclose</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">array</span><span class="p">(</span><span class="n">ysol_p</span><span class="p">)</span><span class="o">.</span><span class="n">flat</span><span class="p">,</span><span class="n">ysol</span><span class="p">[</span><span class="s">'x'</span><span class="p">],</span><span class="n">atol</span><span class="o">=</span><span class="mf">1e-5</span><span class="p">)</span>
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<pre>scipy optimize solution: [ 1.00000000e+00 1.00000000e+00 -9.93437889e-09 -2.31626259e-06]
projection solution: [ 1. 1. 0. 0.]
True
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<h2>
Summary
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<p>In this section, we pulled yet another powerful result from the projection concept we developed previously. We showed how "inverse projection" can lead to the solution of the classic constrained minimization problem. Although there are many approaches to the same problem, by once again appealing to the power projection method, we can maintain our intuitive geometric intuition. </p>
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References
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<p>This post was created using the <a href="https://github.com/ipython/nbconvert">nbconvert</a> utility from the source <a href="www.ipython.org">IPython Notebook</a> which is available for <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Inverse_Projection_Constrained_Optimization.ipynb">download</a> from the main github <a href="https://github.com/unpingco/Python-for-Signal-Processing">site</a> for this blog. The projection concept is masterfully discussed in the classic Strang, G. (2003). <em>Introduction to linear algebra</em>. Wellesley Cambridge Pr. Also, some of Dr. Strang's excellent lectures are available on <a href="http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/">MIT Courseware</a>. I highly recommend these as well as the book.</p>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-30622138698724134872013-01-14T21:00:00.000-08:002013-01-14T21:00:09.834-08:00Conditional Expectation for Gaussian Variables<body>
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<h2>Introduction</h2>
<p>By this point, we have developed many tools to deal with computing the conditional expectation. In this section, we discuss a bizarre and amazing coincidence regarding Gaussian random variables and linear projection, a coincidence that is the basis for most of statistical signal processing.</p>
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<h3>
Conditional Expectation by Optimization
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<p>Now, let's consider the important case of the zero-mean bivariate Gaussian and try to find a function $h$ that minimizes the mean squared error (MSE). Again, trying to solve for the conditional expectation by minimizing the error over all possible functions $h$ is generally very, very hard. One alternative is to use parameters for the $h$ function and then just optimize over those. For example, we could assume that $h(Y)= \alpha Y$ and then use calculus to find the $\alpha$ parameter.</p>
<p>Let's try this with the zero-mean bivariate Gaussian density,</p>
<p>$$\mathbb{E}((X-\alpha Y )^2) = \mathbb{E}(\alpha^2 Y^2 - 2 \alpha X Y + X^2 )$$</p>
<p>and then differentiate this with respect to $\alpha$ to obtain</p>
<p>$$\mathbb{E}(2 \alpha Y^2 - 2 X Y ) = 2 \alpha \sigma_y^2 - 2 \mathbb{E}(XY) = 0$$</p>
<p>Then, solving for $\alpha$ gives</p>
<p>$$ \alpha = \frac{ \mathbb{E}(X Y)}{ \sigma_y^2 } $$</p>
<p>which means we that</p>
<p>\begin{equation}
\mathbb{ E}(X|Y) \approx \alpha Y = \frac{ \mathbb{E}(X Y )}{ \sigma_Y^2 } Y =\frac{\sigma_{X Y}}{ \sigma_Y^2 } Y
\end{equation}</p>
<p>where that last equality is just notation. Remember here we assumed a special linear form for $h=\alpha Y$, but we did that for convenience. We still don't know whether or not this is the one true $h_{opt}$ that minimizes the MSE for all such functions.</p>
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<h3>
Conditional Expectation Using Direct Integration
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<p>Now, let's try this again by computing $ \mathbb{E}(X|Y)$ in the case of the bivariate Gaussian distribution straight from the definition.</p>
<p>\begin{equation}
\mathbb{E}(X|Y) = \int_{\mathbb{ R}} x \frac{f_{X,Y}(x,y)}{f_Y(y)} dx
\end{equation}</p>
<p>where </p>
<p>$$ f_{X,Y}(x,y) = \frac{1}{2\pi |\mathbf{R}|^{\frac{1}{2}}} e^{-\frac{1}{2} \mathbf{v}^T \mathbf{R}^{-1} \mathbf{v} } $$ </p>
<p>and where</p>
<p>$$ \mathbf{v}= \left[ x,y \right]^T$$ </p>
<p>$$ \mathbf{R} = \left[ \begin{array}{cc}
\sigma_{x}^2 & \sigma_{xy} \\
\sigma_{xy} & \sigma_{y}^2 \\
\end{array} \right] $$ </p>
<p>and with</p>
<p>\begin{eqnarray}
\sigma_{xy} &=& \mathbb{E}(xy) \nonumber \\
\sigma_{x}^2 &=& \mathbb{E}(x^2) \nonumber \\
\sigma_{y}^2 &=& \mathbb{E}(y^2) \nonumber <br />
\end{eqnarray}</p>
<p>This conditional expectation (Eq. 4 above) is a tough integral to evaluate, so we'll do it with <code>sympy</code>.</p>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">Matrix</span><span class="p">,</span> <span class="n">Symbol</span><span class="p">,</span> <span class="n">exp</span><span class="p">,</span> <span class="n">pi</span><span class="p">,</span> <span class="n">simplify</span><span class="p">,</span> <span class="n">integrate</span>
<span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">stats</span>
<span class="n">sigma_x</span> <span class="o">=</span> <span class="n">Symbol</span><span class="p">(</span><span class="s">'sigma_x'</span><span class="p">,</span><span class="n">positive</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">sigma_y</span> <span class="o">=</span> <span class="n">Symbol</span><span class="p">(</span><span class="s">'sigma_y'</span><span class="p">,</span><span class="n">positive</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">sigma_xy</span> <span class="o">=</span> <span class="n">Symbol</span><span class="p">(</span><span class="s">'sigma_xy'</span><span class="p">,</span><span class="n">real</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">fyy</span> <span class="o">=</span> <span class="n">stats</span><span class="o">.</span><span class="n">density</span><span class="p">(</span><span class="n">stats</span><span class="o">.</span><span class="n">Normal</span><span class="p">(</span><span class="s">'y'</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="n">sigma_y</span><span class="p">))(</span><span class="n">y</span><span class="p">)</span>
<span class="n">R</span> <span class="o">=</span> <span class="n">Matrix</span><span class="p">([[</span><span class="n">sigma_x</span><span class="o">**</span><span class="mi">2</span><span class="p">,</span> <span class="n">sigma_xy</span><span class="p">],</span>
<span class="p">[</span><span class="n">sigma_xy</span><span class="p">,</span><span class="n">sigma_y</span><span class="o">**</span><span class="mi">2</span><span class="p">]])</span>
<span class="n">fxy</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="p">)</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">R</span><span class="o">.</span><span class="n">det</span><span class="p">())</span> <span class="o">*</span> <span class="n">exp</span><span class="p">((</span><span class="o">-</span><span class="n">Matrix</span><span class="p">([[</span><span class="n">x</span><span class="p">,</span><span class="n">y</span><span class="p">]])</span><span class="o">*</span><span class="n">R</span><span class="o">.</span><span class="n">inv</span><span class="p">()</span><span class="o">*</span> <span class="n">Matrix</span><span class="p">([[</span><span class="n">x</span><span class="p">],[</span><span class="n">y</span><span class="p">]]))[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]</span><span class="o">/</span><span class="mi">2</span> <span class="p">)</span>
<span class="n">fcond</span> <span class="o">=</span> <span class="n">simplify</span><span class="p">(</span><span class="n">fxy</span><span class="o">/</span><span class="n">fyy</span><span class="p">)</span>
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<p>Unfortunately, <code>sympy</code> cannot immediately integrate this without some hints. So, we need to define a positive variable ($u$) and substitute it into the integration</p>
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<div class="highlight"><pre><span class="n">u</span><span class="o">=</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'u'</span><span class="p">,</span><span class="n">positive</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span> <span class="c"># define positive variable</span>
<span class="n">fcond2</span><span class="o">=</span><span class="n">fcond</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sigma_x</span><span class="o">**</span><span class="mi">2</span><span class="o">*</span><span class="n">sigma_y</span><span class="o">**</span><span class="mi">2</span><span class="o">-</span><span class="n">sigma_xy</span><span class="o">**</span><span class="mi">2</span><span class="p">,</span><span class="n">u</span><span class="p">)</span> <span class="c"># substitute as hint to integrate</span>
<span class="n">g</span><span class="o">=</span><span class="n">simplify</span><span class="p">(</span><span class="n">integrate</span><span class="p">(</span><span class="n">fcond2</span><span class="o">*</span><span class="n">x</span><span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="o">-</span><span class="n">oo</span><span class="p">,</span><span class="n">oo</span><span class="p">)))</span> <span class="c"># evaluate integral</span>
<span class="n">gg</span><span class="o">=</span><span class="n">g</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span> <span class="n">u</span><span class="p">,</span><span class="n">sigma_x</span><span class="o">**</span><span class="mi">2</span> <span class="o">*</span><span class="n">sigma_y</span><span class="o">**</span><span class="mi">2</span> <span class="o">-</span> <span class="n">sigma_xy</span><span class="o">**</span><span class="mi">2</span> <span class="p">)</span> <span class="c"># substitute back in</span>
<span class="n">use</span><span class="p">(</span> <span class="n">gg</span><span class="p">,</span> <span class="n">simplify</span><span class="p">,</span><span class="n">level</span><span class="o">=</span><span class="mi">2</span><span class="p">)</span> <span class="c"># simplify exponent term</span>
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$$\frac{\sigma_{xy} y}{\sigma_{y}^{2}}$$
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<p>Thus, by direct integration using <code>sympy</code>, we found</p>
<p>$$ \mathbb{ E}(X|Y) = Y \frac{\sigma_{xy}}{\sigma_{y}^{2}} $$ </p>
<p>and this matches the prior result we obtained by direct minimization by assuming that $\mathbb{E}(X|Y) = \alpha Y$ and then solving for the optimal $\alpha$!</p>
<p>The importance of this result cannot be understated: the one true and optimal $h_{opt}$ <em>is a linear function</em> of $Y$. </p>
<p>In other words, assuming a linear function, which made the direct search for an optimal $h(Y)$ merely convenient yields the optimal result! This is a general result that extends for <em>all</em> Gaussian problems. The link between linear functions and optimal estimation of Gaussian random variables is the most fundamental result in statistical signal processing! This fact is exploited in everything from optimal filter design to adaptive signal processing.</p>
<p>We can easily extend this result to non-zero mean problems by inserting the means in the right places as follows:</p>
<p>$$ \mathbb{ E}(X|Y) = \bar{X} + (Y-\bar{Y}) \frac{\sigma_{xy}}{\sigma_{y}^{2}} $$</p>
<p>where $\bar{X}$ is the mean of $X$ (same for $Y$).</p>
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<h2>Summary</h2>
<p>In this section, we showed that the conditional expectation for Gaussian random variables is a linear function, which, by a bizarre coincidence, is also the easiest one to work with. This result is fundamental to all optimal linear filtering problems (e.g. Kalman filter) and is the basis of most of the theory of stochastic processes used in signal processing. Up to this point, we have worked hard to illustrate all of the concepts we will need to unify our understanding of this entire field and figured out multiple approaches to these kinds of problems, most of which are far more difficult to compute. Thus, it is indeed just plain lucky that the most powerful distribution is the easiest to compute as a conditional expectation because it is a linear function. We will come back to this same result again and again as we work our way through these greater concepts.</p>
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<h3>References</h3>
<p>This post was created using the <a href="https://github.com/ipython/nbconvert">nbconvert</a> utility from the source <a href="www.ipython.org">IPython Notebook</a> which is available for <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Conditional_Expectation_Gaussian.ipynb">download</a> from the main github <a href="https://github.com/unpingco/Python-for-Signal-Processing">site</a> for this blog. </p>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-90703218049322872042012-12-31T19:30:00.000-08:002012-12-31T19:30:00.252-08:00Worked Examples for Conditional Expectation
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<h2>Introduction</h2>
<p>Brzezniak (2000) is a great book because it approaches conditional expectation through a sequence of exercises, which is what we are trying to do here. The main difference is that Brzezniak takes a more abstract measure-theoretic approach to the same problems. Note that you <em>do</em> need to grasp the measure-theoretic to move into more advanced areas in stochastic processes, but for what we have covered so far, working the same problems in his text using our methods is illuminating. It always helps to have more than one way to solve <em>any</em> problem. I urge you to get a copy of his book or at least look at some pages on Google Books. I have numbered the examples corresponding to the book. </p>
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<h2>Examples</h2>
<p>This is Example 2.1 from Brzezniak:</p>
<blockquote>
<p>Three coins, 10p, 20p and 50p are tossed. The values of those coins that land heads up are added to work out the total amount. What is the expected total amount given that two coins have landed heads up?</p>
</blockquote>
<p>In this case we have we want to compute $\mathbb{E}(\xi|\eta)$ where</p>
<p>$$ \xi = 10 X_{10} + 20 X_{20} +50 X_{50} $$</p>
<p>where $X_i \in { 0,1} $. This represents the sum total value of the heads-up coins. The $\eta$ represents the fact that only two of the three coins are heads-up. Note </p>
<p>$$\eta = X_{10} X_{20} (1-X_{50})+ (1-X_{10}) X_{20} X_{50}+ X_{10} (1-X_{20}) X_{50} $$</p>
<p>is a function that is only non-zero when two of the three coins is heads. Each triple term catches each of these three possibilities. For example, the first term is when the 10p and 20p are heads up and the 50p is heads down.</p>
<p>To compute the conditional expectation, we want to find a function $h$ of $\eta$ that minimizes the MSE</p>
<p>$$ \sum_{X\in{0,1}^3} \frac{1}{8} (\xi - h( \eta ))^2 $$</p>
<p>where the sum is taken over all possible triples of outcomes for $ {X_{10} , X_{20} ,X_{50}}$ and the $\frac{1}{8} = \frac{1}{2^3} $ since each coin has a $\frac{1}{2}$ chance of coming up heads.</p>
<p>Now, the question boils down to what function $h(\eta)$ should we try? Note that $\eta \in {0,1}$ so $h$ takes on only two values. Thus, we only have to try $h(\eta)=\alpha \eta$ and find $\alpha$. Writing this out gives,</p>
<p>$$ \sum_{X\in{0,1}^3} \frac{1}{8} (\xi - \alpha( \eta ))^2 $$</p>
<p>which boils down to solving for $\alpha$,</p>
<p>$$\langle \xi , \eta \rangle = \alpha \langle \eta,\eta \rangle$$</p>
<p>where </p>
<p>$$ \langle \xi , \eta \rangle =\sum_{X\in{0,1}^3} \frac{1}{8} (\xi \eta ) $$</p>
<p>This is tedious and a perfect job for <code>sympy</code>.</p>
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<div class="highlight"><pre><span class="kn">import</span> <span class="nn">sympy</span> <span class="kn">as</span> <span class="nn">S</span>
<span class="n">eta</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'eta'</span><span class="p">)</span>
<span class="n">xi</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'xi'</span><span class="p">)</span>
<span class="n">X10</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'X10'</span><span class="p">)</span>
<span class="n">X20</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'X20'</span><span class="p">)</span>
<span class="n">X50</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'X50'</span><span class="p">)</span>
<span class="n">eta</span> <span class="o">=</span> <span class="n">X10</span> <span class="o">*</span> <span class="n">X20</span> <span class="o">*</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">X50</span> <span class="p">)</span><span class="o">+</span> <span class="n">X10</span> <span class="o">*</span> <span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">X20</span><span class="p">)</span> <span class="o">*</span><span class="p">(</span><span class="n">X50</span> <span class="p">)</span><span class="o">+</span> <span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">X10</span><span class="p">)</span> <span class="o">*</span> <span class="n">X20</span> <span class="o">*</span><span class="p">(</span><span class="n">X50</span> <span class="p">)</span>
<span class="n">xi</span> <span class="o">=</span> <span class="mi">10</span><span class="o">*</span><span class="n">X10</span> <span class="o">+</span><span class="mi">20</span><span class="o">*</span> <span class="n">X20</span><span class="o">+</span> <span class="mi">50</span><span class="o">*</span><span class="n">X50</span>
<span class="n">num</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">summation</span><span class="p">(</span><span class="n">xi</span><span class="o">*</span><span class="n">eta</span><span class="p">,(</span><span class="n">X10</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="n">X20</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="n">X50</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">den</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">summation</span><span class="p">(</span><span class="n">eta</span><span class="o">*</span><span class="n">eta</span><span class="p">,(</span><span class="n">X10</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="n">X20</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="n">X50</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">alpha</span><span class="o">=</span><span class="n">num</span><span class="o">/</span><span class="n">den</span>
<span class="k">print</span> <span class="n">alpha</span>
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<pre>160/3
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<p>This means that</p>
<p>$$ \mathbb{E}(\xi|\eta) = \frac{160}{3} \eta $$</p>
<p>which we can check with a quick simulation</p>
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<div class="highlight"><pre><span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="kn">from</span> <span class="nn">numpy</span> <span class="kn">import</span> <span class="n">array</span>
<span class="n">x</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">randint</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">,(</span><span class="mi">3</span><span class="p">,</span><span class="mi">5000</span><span class="p">))</span>
<span class="k">print</span> <span class="p">(</span><span class="mi">160</span><span class="o">/</span><span class="mf">3.</span><span class="p">,</span><span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">[:,</span><span class="n">x</span><span class="o">.</span><span class="n">sum</span><span class="p">(</span><span class="n">axis</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span><span class="o">==</span><span class="mi">2</span><span class="p">]</span><span class="o">.</span><span class="n">T</span><span class="p">,</span><span class="n">array</span><span class="p">([</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">,</span><span class="mi">50</span><span class="p">]))</span><span class="o">.</span><span class="n">mean</span><span class="p">())</span>
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<pre>(53.333333333333336, 53.177920685959272)
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<h2>Example</h2>
<p>This is example 2.2:</p>
<blockquote>
<p>Three coins, 10p, 20p and 50p are tossed as before. What is the conditional expectation of the total amount shown by the three coins given the total amount shown by the 10p and 20p coins only?</p>
</blockquote>
<p>For this problem,</p>
<p>$$\eta = 30 X_{10} X_{20} + 20 (1-X_{10}) X_{20} + 10 X_{10} (1-X_{20}) $$</p>
<p>which takes on three values (10,20,30) and only considers the 10p and 20p coins. Here, we'll look for affine functions, $h(\eta) = a \eta + b $.</p>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">sympy.abc</span> <span class="kn">import</span> <span class="n">a</span><span class="p">,</span><span class="n">b</span>
<span class="n">eta</span> <span class="o">=</span> <span class="n">X10</span> <span class="o">*</span> <span class="n">X20</span> <span class="o">*</span> <span class="mi">30</span> <span class="o">+</span> <span class="n">X10</span> <span class="o">*</span> <span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">X20</span><span class="p">)</span> <span class="o">*</span><span class="p">(</span><span class="mi">10</span> <span class="p">)</span><span class="o">+</span> <span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">X10</span><span class="p">)</span> <span class="o">*</span> <span class="n">X20</span> <span class="o">*</span><span class="p">(</span><span class="mi">20</span> <span class="p">)</span>
<span class="n">h</span> <span class="o">=</span> <span class="n">a</span><span class="o">*</span><span class="n">eta</span> <span class="o">+</span> <span class="n">b</span>
<span class="n">J</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">summation</span><span class="p">((</span><span class="n">xi</span> <span class="o">-</span> <span class="n">h</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="o">*</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">8</span><span class="p">),(</span><span class="n">X10</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="n">X20</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="n">X50</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">sol</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">solve</span><span class="p">(</span> <span class="p">[</span><span class="n">S</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">J</span><span class="p">,</span><span class="n">a</span><span class="p">),</span> <span class="n">S</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">J</span><span class="p">,</span><span class="n">b</span><span class="p">)],(</span><span class="n">a</span><span class="p">,</span><span class="n">b</span><span class="p">)</span> <span class="p">)</span>
<span class="k">print</span> <span class="n">sol</span>
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<pre>{b: 25, a: 1}
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<p>This means that</p>
<p>$$ \mathbb{E}(\xi|\eta) = 25+ \eta $$</p>
<p>since $\eta$ takes on only four values, ${0,10,20,30}$, we can write this out as</p>
<p>$$ \mathbb{E}(\xi|\eta=0) = 25 $$
$$ \mathbb{E}(\xi|\eta=10) = 35 $$
$$ \mathbb{E}(\xi|\eta=20) = 45 $$
$$ \mathbb{E}(\xi|\eta=30) = 55 $$</p>
<p>The following is a quick simulation to demonstrate this.</p>
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<div class="highlight"><pre><span class="n">x</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">randint</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">,(</span><span class="mi">3</span><span class="p">,</span><span class="mi">5000</span><span class="p">))</span> <span class="c"># random samples for 3 coins tossed</span>
<span class="n">eta</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">[:</span><span class="mi">2</span><span class="p">,:]</span><span class="o">.</span><span class="n">T</span><span class="p">,</span><span class="n">array</span><span class="p">([</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">]))</span> <span class="c"># sum of 10p and 20p</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">[:,</span><span class="n">eta</span><span class="o">==</span><span class="mi">0</span><span class="p">]</span><span class="o">.</span><span class="n">T</span><span class="p">,</span><span class="n">array</span><span class="p">([</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">,</span><span class="mi">50</span><span class="p">]))</span><span class="o">.</span><span class="n">mean</span><span class="p">()</span> <span class="c"># E(xi|eta=0)</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">[:,</span><span class="n">eta</span><span class="o">==</span><span class="mi">10</span><span class="p">]</span><span class="o">.</span><span class="n">T</span><span class="p">,</span><span class="n">array</span><span class="p">([</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">,</span><span class="mi">50</span><span class="p">]))</span><span class="o">.</span><span class="n">mean</span><span class="p">()</span><span class="c"># E(xi|eta=10)</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">[:,</span><span class="n">eta</span><span class="o">==</span><span class="mi">20</span><span class="p">]</span><span class="o">.</span><span class="n">T</span><span class="p">,</span><span class="n">array</span><span class="p">([</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">,</span><span class="mi">50</span><span class="p">]))</span><span class="o">.</span><span class="n">mean</span><span class="p">()</span><span class="c"># E(xi|eta=20)</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">dot</span><span class="p">(</span><span class="n">x</span><span class="p">[:,</span><span class="n">eta</span><span class="o">==</span><span class="mi">30</span><span class="p">]</span><span class="o">.</span><span class="n">T</span><span class="p">,</span><span class="n">array</span><span class="p">([</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">,</span><span class="mi">50</span><span class="p">]))</span><span class="o">.</span><span class="n">mean</span><span class="p">()</span><span class="c"># E(xi|eta=30)</span>
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<pre>26.4120922832
34.5614035088
45.8196721311
55.3743104807
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<h2>Example</h2>
<p>This is Example 2.3</p>
<p><img src="data:image/jpeg;base64,/9j/4AAQSkZJRgABAQECWAJYAAD/2wBDAAEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEB%0AAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQH/2wBDAQEBAQEBAQEBAQEBAQEBAQEBAQEB%0AAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQH/wAARCACZAp8DASIA%0AAhEBAxEB/8QAHwAAAQUBAQEBAQEAAAAAAAAAAAECAwQFBgcICQoL/8QAtRAAAgEDAwIEAwUFBAQA%0AAAF9AQIDAAQRBRIhMUEGE1FhByJxFDKBkaEII0KxwRVS0fAkM2JyggkKFhcYGRolJicoKSo0NTY3%0AODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqDhIWGh4iJipKTlJWWl5iZmqKjpKWm%0Ap6ipqrKztLW2t7i5usLDxMXGx8jJytLT1NXW19jZ2uHi4%2BTl5ufo6erx8vP09fb3%2BPn6/8QAHwEA%0AAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoL/8QAtREAAgECBAQDBAcFBAQAAQJ3AAECAxEEBSEx%0ABhJBUQdhcRMiMoEIFEKRobHBCSMzUvAVYnLRChYkNOEl8RcYGRomJygpKjU2Nzg5OkNERUZHSElK%0AU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6goOEhYaHiImKkpOUlZaXmJmaoqOkpaanqKmqsrO0tba3%0AuLm6wsPExcbHyMnK0tPU1dbX2Nna4uPk5ebn6Onq8vP09fb3%2BPn6/9oADAMBAAIRAxEAPwD%2B/ivD%0A9W/ab/Zt0DVdT0LXf2hPgfout6LqF7pOs6Nq3xY8Badquk6rp1zJZ6hpmp6fea/Dd2GoWF3DNa3t%0AldQxXNrcxSQTxxyxso9wr%2BH3/gux/wAEh/2f/wBjr/ghD488S%2BGfhZ8BfHP7Sfwv%2BKvw48ffGr9r%0A6y%2BCHhLwB8ZPiLd/Fj49jTvF%2Bqw6/eah458eaTpWreLfijoemQeBYfiPq2h6N4btotP01LfRtKtd%0APtwD%2BzrwP8ZPhD8Tbu%2BsPht8Vfhv8Qr7S7eO71Kz8D%2BOfDHiy7060mkMMNzfW%2Bg6pfzWlvLMDFHN%0AcJHG8gKKxbiug8aeOPBfw38L6x44%2BIni/wAL%2BAvBfh63S71/xf401/SfC3hfQ7SW4htI7nWNf1y7%0AsdJ0y3kuri3tkmvbuCN7ieGFWMkqK34Z%2BMf2VvgT%2BwxdaJ/wVI/4Jqfsh%2BFV8ReK/wBlrw9%2Bz3cf%0Asy/BX4dXPwr8FfEnSP2iPj1%2BzHrvw8/aB%2BJVt8J/DHi3VvDmn/s4%2BGtK8deI/ilqvhn4KfEbxxr3%0AgXxJq2o6lqWiab4Du7678y8a/thWX/BTn9jz/gr3/wAE9vjV8HvhFaftT/s/fsu%2BObLxDpHwr8ea%0AL%2B1F%2Bzl8QNX8U/CbxJrvwc%2BMnwb8c6t4X8F3Nj4j8I/GTwrDq9l4H8b%2BHfD3j74S%2BN/DXha6h1%2B8%0A8SaZeal4fAP6IfCnizwr478N6L4y8D%2BJvD/jLwh4k0%2B31bw74q8Kazp3iLw3r%2BlXaeZa6nouuaRc%0A3mmapp9yhD297Y3U9tMnzRyMOa6Cv5C/%2BCLf/BUfxzJ%2Bzh/wSo/Yq/ZW/ZHh/al8MD9mTwlrH7Wv%0Axp8H/HOy%2BHtn%2Bxlolj8dPHXwe8S678U/B/j/AMD3mmapqkjeEPE3xA8PeD7b4oaB4y%2BJmjaN4mvv%0Ahh4Lv/DNlot9qH1Z8Ef%2BDhvRfi/4N8FftPal%2By/YfDv9gv4i/tPx/s0%2BEvjDr/7RWkah%2B03LDrOq%0A6b4I8H/GvU/2P9L%2BFckf/CpNZ%2BLGpn4faxN4X%2BPHi3x34b/sTxP4iPgjVdO0q0h1QA/pIrz34o/F%0Az4UfA/wdqHxE%2BNXxO%2BHnwg%2BH%2BkvFHqvjr4o%2BNPDfw/8AB2mST7/ITUPE3izUtJ0SyebY/lLc30TS%0AbH2Btpx6FX8p/wCxF8Pj/wAFB/8Agt5/wVt8b/t1eGLb4hR/8E8PHnwe%2BEX7H37PnjWefxh8HfhZ%0A4J%2BKWg/EGGL40W3gLxRoDaBdePvin4Q%2BGHhrxzpmsG5v38Nan4%2B8ex21o1v/AMIRrFgAf0t/B/47%0AfBD9oXwovjz4BfGT4VfHHwM93c2CeM/g/wDEPwj8S/Cj31nNLbXdkviLwXrGtaQ13a3EE1vc24vD%0ANBNDLFKiyRuo9Vr%2Bar9nXw34asf%2BDlj9p6w/Za8LeGPAPwG%2BG/8AwTJ%2BGfhH9rC1%2BDvhrRNP%2BG2u%0AftSXnxun1j4WaT8To/DVhp2h6N8Y7T4Patfnw8bwXnihfBPgzVdNWSHTLrUbSy/fX49/GTTv2f8A%0A4TeK/i7q3gH4wfE/T/CX9hfaPA3wE%2BF/ir4zfFnXP7e8S6P4Zi/4RT4a%2BCrLUPE3iT%2BzJ9aj1jXf%0A7Ms5v7H8NafrGv3vl6fpd3KgB7BRX4Y6z/wXP0XS7%2BKzsv8Agkz/AMFz/EVvJ4gvtGfVNG/4Jr%2BP%0AoLCDTrSHRZbfxZLH4g8WaFqbeH9UfVL63sbaDTpvFUc3hvWm1Lw1p1vP4dn1/wDW74CfGTTv2gPh%0AN4U%2BLuk%2BAfjB8MNP8W/279n8DfHv4X%2BKvgz8WdD/ALB8S6x4Zl/4Sv4a%2BNbLT/E3hv8AtOfRZNY0%0AL%2B07OH%2B2PDWoaPr9l5mn6paSuAcFcftlfs12/wC1FYfsWr8TLe9/acv/AAlc%2BPH%2BFWkeGPGuuaho%0Avgy301tWXxN4u1/RvDd94Q8E6Pf2qiHRrzxh4h0OLXtTkh0XRDqGszwWEnuui%2BOPBfiXWfEvhzw5%0A4v8AC%2Bv%2BIfBlxZ2njDQdF1/SdV1nwnd6it0%2Bn23iXS7G7nvtCuL5bK9azh1SC1kultLowK4t5Sn8%0Azn7X/jb4u/sz/tp/8F6fi98LjFpvxUk/4IqfBj4//BzxHZ6hpVpfeFbf4ZWn7V3g298bx3mo%2BGdd%0Aih1vwR4j8Nap4u0nwtcyRWPiybw3pmn3trOt/a6p4f8Azkn0f9nz9lX/AIJV/wDBC3/gqL%2Bx98P/%0AAAZ4L%2BM%2BhftPfscwftnfHnwjFZQfGv47%2BEvi/o/jPwd%2B254A%2BLXxAvY/EHif4or8RfjVczxvp3je%0A78QTeCFgs7rwJaeFLfQdN0%2BzAP7pqx9W8Q6BoEmjw67rmj6LN4i1iDw94fi1bU7LTpNd1%2B5tb2/t%0AtD0dLyeFtT1i4sdO1C8g0yyE97La2F7cRwNDazvHsV/Kd/wWy8A%2BD/g9/wAFV/8AghB%2B03pmgald%0Aa38TP299L%2BDfxCvJfiH45vDd6h4zs/hb8M/hjLongHWNQ1L4faLo/huDVPEWq69eeHbTwzqcty7u%0AbPXtQ8T6xqViAfuFe/8ABTH9ijTv249J/wCCbuofGDULD9s7X9Pk1bQvg/e/Cb41WttrWlQ/DLU/%0AjBNqek/E6f4dR/CDUdPi8AaLrGqSXtn4%2Bmto9V0nU/CLSDxjYXWgRfd9fhd/wUrOh6J/wVK/4IH%2B%0ANdUgu7e8sf2h/wBsvwLp%2BuW76Rbx2g%2BJX7IviTSBoc13PbXeur/wketafoERstNiTSdRtdPvINau%0AbO9Hh%2B8g/Yn4v%2BIPiZ4V%2BGfjLxF8G/hxo/xd%2BJ%2Bk6PJd%2BC/hp4g8fw/CzRvGOsrNCsWj6j8Qp/DP%0AjKHwnbyQvNKdVk8L60sbQrF9ibzd6AHpFV7m7tLKNZry5t7SF7i0tEluZo4I3u7%2B7hsLC2V5WRWu%0AL2%2BubeztIQTJc3c8NvCrzSojfzb/AAz/AOC8vxy8S6v%2BwPN8Tf2BfAfwr8FftsftTeLv2M/EF%2Bv7%0AcOi%2BLviL%2Bzz%2B0J4B%2BM3xI%2BEvi7wB43%2BFq/s6%2BHbrXdS0%2B38C6P4stZ7DxJpemX1r498L6T/aFvdX%0A%2BnXGp/Mv7bf/AAUw1P4h/sqt%2B1x%2B2j/wSS0vUvhD%2BwR/wUo0PQLDxXo//BRWDT5/B3xJ%2BD/jvQfh%0A34c/aJ%2BCeufCb4feH9e%2BL%2BreDPiV4k8SeE/EPwysIm03SdT8N6za6zqmpXPh7x1Y%2BAwD%2BuqvJ/h/%0A8evgZ8WPFHxA8EfCz4z/AAn%2BJfjT4TXGg2nxU8IfD/4i%2BD/GXij4Z3fiq31G78L23xA0Dw5rGpat%0A4MuPElro%2BrXOgw%2BI7TTZNYt9L1GbT1uI7K5aL8APiF%2B3J%2B354X/4OHtK/ZN8L/s92/j34K6v%2Bw7o%0Afifw74MP7VLeCfC9l8L9T/aHg0r4gftf%2BIPCUvgXxB4a1T4m6Nd%2BHNU%2BHHg/4Vyw2niSfw9Zw3af%0AEnQ7Pxrq%2BlaZ8ffD74j/ABw/Y4/4OGP%2BCv3w0/Y%2B/YK8H/tIfEH9p74P/ssftLaFoGm/Hjw5%2By34%0AL8M%2BFfB3w80zSvjP4j8S6lL8NPHHhu/%2BIHxl%2BO3xEufEFxearpdnq%2BteJ/7b8TatrjS%2BLdf1EAH9%0AjFFfIH7A37Zvw4/4KE/shfBL9sT4T6T4g0DwT8afD%2Br6jaeHvFMMUOu%2BHNe8KeLPEHgDxr4cvnt2%0Aa2vv7A8beFPEWj2ur2m2z1uzsrfWLOOO2voo1/L3/gov%2B1h/wUm%2BD3/BRz9gH4efsy/sqWvxb%2BDv%0AjG9/aQ0V9Avf2rPCHwY0L9oLxXb/AAM8P%2BKrO48aNdeC/F%2Bp%2BBfD3wRsLb4ia7p1vqfh/wAXL461%0A6ztX0%2B20DUbTw42tAH7mQ/Er4c3HxAufhPB4/wDBU/xTs/DR8aXnw0h8VaFJ8QLTwcL6y0s%2BLLnw%0Aal%2B3iKDw0NS1LTtOOuy6amli%2B1CytDdfaLqCOTta/l9/bZ8Y%2BL/2WP8Ag4L/AGaPif8As9fsgeHf%0A2jPjn%2B11/wAE5/jb8FND0Gw8a%2BEvgkureOPht8VvAfxK1P4i/Ff4q6t4T8UjSNI8JfCDwO/gux8R%0AXOh694hvrbUdC8DaJbXiDTtIk%2Bg/iX/wXO0HSf8AgntpP7Vngj4Oaf4f%2BO/i79srSv8Agnp4f%2BDH%0Axr%2BImn%2BDfhfoP7VT/EW68FeLF8S/HNdPt9H1D4NeENC0Pxb49HxJ0rR7ODU9H0Iadrll4Ivk8Rye%0AFQD9/wCivyL/AGB/%2BCl8/wC0v%2B0h%2B0T%2Bxt8UpP2f9W%2BN/wADfDXhj4ueHviD%2Byd8VH%2BM/wAAPip8%0ADfG9%2BdG0fWbXxDcW1rrngP4oeFPEMf8AYnxC%2BGfieGVrM6jomueEvEHivw/fy6hafR3/AAUM/bt8%0AA/8ABPb9nTU/jb4p8K6/8U/G%2Bu%2BKfCPwr%2BA/wC8D32lW/wASv2gvjl8R9ctPDfgD4W%2BAbTU5hNe3%0A9/qF4%2BteI59I07xDreh%2BCNF8TeIdI8K%2BKtS0u08N6qAfc1FfgN%2Bx9/wVo/bu%2BPn7d%2Bt/sTfH/wD4%0AJNRfsqn4Y%2BBdP%2BJf7QvxPuP29/gt8ZoPgt4H8Y%2BD/G%2Bs/CvXpvC3hP4daFb%2BPIvH/i3whb%2BC5LPw%0Af4vvNR8HJrDeJ/FFnZ6TpsqXP0j/AMFJv2jU%2BIv/AATr/bf039ifx18D/jz8Vz%2By18Wo73T/AAn%2B%0A0d4Y0HWPAHgbxj8O/FGg6l8YtI1HwrpHxAlvdU8B2V5L4y8OeHr%2BLwrZeNJ9El0XT/G2iapPYyTA%0AH1N8JP8AgoL%2BxB8e/jZ4w/Zx%2BCf7VXwN%2BK3xv8BWV5qPin4deAfiFoHijXLC00uSzg1qS2bSbu6s%0ANYfw7d39rp/iiDRL3UbjwxqjvpfiCLTdRguLWL7Cr%2BPP/gnl8TfHngLwZ/wbw67%2B0X/wS%2B%2BGfwo%2B%0AHNr8PtJ/Zx/ZR/a78F/tgt4r8Q6X4g/ab/Y21zxpqnjXVv2WfAfw78C%2BHrTXP2rl%2BEVj4u8W6j45%0A8Q/ES8%2BHk%2Bt%2BJrvUNSi8capdX2qfQPxm/wCDiDVvBPj/APai%2BJ3w3%2BGv7MXjv9if9iv48Q/AH4wx%0Aat%2B1n4L8OftrfGO68OzaDbfF34rfsr/Bdo7r4f8AjvwP8Pv%2BEt0%2B98M6N4k%2BIegah8U4vDeraXpW%0AueHNd1HV9P8Ah8Af0Z%2BAfj18DPit4u%2BJnw/%2BF3xn%2BE/xJ8efBbWLLw98Y/BPgH4i%2BD/GPi74Ta/q%0AU%2BtWunaH8TPDfh3WNR1nwHrF/c%2BG/EVvZaZ4pstKvbqfQdahggkk0u%2BWBvjL4%2B/An4dfED4ffCb4%0Ag/Gr4S%2BBPip8W5by3%2BFXw08ZfEfwd4Y%2BIHxNn09okv4fh94N1vWbHxH4zlsXmhS8j8OabqT2zSxL%0AMqGRAf5wf2Q/j58Av2VP%2BCpH/BwP%2B0X4k%2BweD/gl49/Zb/4J8/t/6Nc%2BFvD9pZHVvhX4Y/Zy%2BId9%0A8TvE02h2y6aD8RvG3j/xHfaumizg69438U%2BIbnULie41nVpnn%2BRP2lP2q/2j/wBrH46f8G9X7f8A%0A8Z/2Afh/%2BzJ8FviP%2B3V8LfCfwg%2BJo/aR0D4x/GTWdC/am0fWLL4R%2BHvF/hTTPgd4I1PwT4f8f%2BHd%0AJT4t6VbaN4w8b6HHBHpVj4ybwlr80JhAP7U6Kz9W1bStA0rU9d13U9P0XRNF0%2B91bWdZ1a9ttO0r%0ASdK062kvNQ1PU9QvJIbSw0%2BwtIZrq9vbqaK2tbaKSeeSOKNmH873gb/guf8AEbXf2brT/gp74v8A%0A2TNH8Mf8EktV8Z654Gt/ixofxguPGH7XXhjQtO%2BOep/ATSv2hPHP7OyeANE8JwfC2/8AH8en6D4k%0A%2BH3gf4reOvi14O0%2BDWPHFhp3jfSIrXSJQD9wfHH7Tf7Nvwy%2BJvgb4KfEj9oP4H/D74y/E/7B/wAK%0A0%2BEnjj4seAvCfxN%2BIf8AauqXGh6X/wAIN4C17X7DxV4t/tLWrS60ew/sDStQ%2B2apbXGn2/mXcMkK%0A%2Bl%2BLfGng7wBok/iXx34s8NeCvDlrLb29zr/i3XdL8N6Jbz3cqwWkM%2Bq6zdWVjDLdTusNvHJOrzSs%0Ascas7BT/ABofs6%2BDtN/a9/4OM/24dM/an/YV%2BE3xs8P69p37Hn7Q3wP%2BM2ofHK%2B8ep%2By38JPgh8H%0A/EA%2BDPxE%2BGK6X8L7C6vbX9qHxnN4F8a%2BKfhR4i1v4RweD9TvprvV9O%2BLP9l22o6v/Tf/AMFM/gHp%0An7T3/BPf9tH4G33h/wCHXiDVPH37Mvxs0vwOPisvh6LwP4e%2BJyfDrxFc/DDxxqus%2BKLe50fwlL4D%0A8ew%2BHvF%2Bm%2BOZhBceCNR0a18VWN5ZXmkwXcIBV0n/AIKmf8Ex9f1XTNC0L/goz%2BwhrWt61qFlpOj6%0APpP7Xn7P2o6rq2q6jcx2en6Zpmn2fxCmu7/UL%2B7mhtbKytYZbm6uZY4II5JZFU/d9f5pHwJX4k/F%0A/wD4NhNV%2BGfws/4JF6h8cf8AhDfB/wAZ2179uGe//ZiNz4Y1jRf2oPGnxL8Q%2BLfCHgyTxfN%2B094l%0Al%2BG/wz1mLQotc0XwnC0Wqadq%2BmLbXPgrRr6/vP6vfBH/AAUC8V%2BCfh1/wRx/Yw/ZM8R/DD9qn46/%0Atb/svaZ4nk%2BP/wATfE3iTS/hbZfCj9nf4C6BcePfi94ug0OyPxG1Txt8TfGkEegeF/B8WhWV9Z%2BI%0A4vGEXj2bwnL4fnhlAP36or%2BbD9pf/gsj%2B2f%2Bzv8AB/8Aao1i/wD2S/grq/xs/wCCdPjv4Zz/ALbn%0Ag%2Bw%2BLfinVPBurfs8fG7SNQ1v4OfHP9nHVrfwvZ69q/8AwlcGj6tpviPwZ8RNJ8P6t4C1Sznl1CfV%0A9Ds9U1jT9bx9/wAFiv2tv2cf2X/hZ44/bD/Zg%2BB/7P8A%2B0N%2B1Z%2B118D/ANn39kzTdU%2BN8Orfs865%0A8MvjbovgTW7j43fFXxxpz3fxE8C6J8G4NW8Y2/xVsvEvgDwfcwXVj4VfQoL7TtX1HUdNAP3J%2BN/7%0ATf7Nv7MulaHrv7SP7QfwP/Z80TxPqFzpPhrWPjf8WPAfwo0rxDqtnbC8u9M0PUPHmv6BaatqFraE%0AXVzZWE1xcwWxE8saxENXuFfxD/8ABWT9pzxZ/wAFEv8Agkj/AMFSPg98aPA37HXiL9o7/gnr4r/Z%0AY%2BKNr8XfgB8R4fi/8E9a%2BGfxR8YaZf6R8TvgP441aCTxD4C%2BJWt6L4R%2BLvwx8X/D/XLg61baM19o%0A51HVLfxhpMs/9qHgfxfo3xC8F%2BEPH3hx7iXw9448L6B4v0GS7ga1u5NG8S6Taa1pb3Nq5Zra4exv%0AYGmgZmaGQtGSSpNAHUUV/Nf8G/8Agtx8dvjF%2B2NrHwK0b4Ofs0S6D4F/b98ffsR/Fn4A%2BH/jX418%0ASftx%2BAPC/hTXbXwFpP7UVr8NH%2BH%2Bi6J4%2B%2BDE3iX%2B1PHfxD13wLaa14a%2BEvwu0q91DxJ4yvLXSfFn%0Ainwv4bf/APByJrVhpvxN/aqk8GfsmN%2BwR8Jv2uNW/Z38SaCP2i78/t3678KdO8WeEvhnD%2B098PPg%0Akvhp/DXjvwRdeLvE8viZ/Cek6uviKLwlo2o2NvPex2mpeMtMAP6wqK/mM/4Jtftrf8FMfiz/AMFn%0AP%2BClv7Nn7SHwe%2BH%2BifDb4Z6f%2BzLqfifw14f/AGlNd8a%2BBv2UvD%2Bp/Bnxt4l8CWHwf029%2BFmmy/GT%0Axh%2B0VdeKvA%2Bu/ETWriD4KaV4XtPCGu39xol5rK6R4bm6/wDY6/4LV/tC/tVftR%2BAPAf/AAob9mrS%0A/hH4q%2BP/AMaf2dfih8J/h/8AtCeK/iV%2B3n%2ByX4g%2BFNr8RtI0T4gftI/A2T4YeEdJ0r4X%2BP8Axp8O%0AkgTxl4PvfEHg/wAF6L4z0uDxH4wOu6FfWeqAH9IlFeWfHDxx4y%2BGXwb%2BKXxF%2BHnww1j41%2BOvAvgL%0AxV4t8J/CDw9q9noOv/EzXvD%2BjXeqab4H0PWNQtb2zsNY8S3NsmlaZPcWd1GL25gXyJSwQ/hn%2BxV/%0AwWm%2BK/xf%2BOPwx/Zg/ae%2BCPwO%2BG/7Q/7T3wJ8bfHD9mv4Y/Cb43%2BIPEviLw54j%2BH/AIc1TxV4s/Zf%0A/a38O%2BO/h74S8YfAH486H4fsBrks/wDwjfiTwxeaZZ%2BJ0aPTtV8OJpmsAH9D9Y/iHxBo3hPQNc8U%0A%2BI9Rt9I8PeGtH1PxBr2rXbMtppejaNZT6jqmo3LIrstvZWNtPczMqswjiYhWOAf5dtX/AOC9H7b%2B%0AifDX9rb4v%2BKf%2BCb3w88D/Df/AIJz/tEaN8Lf21/EP/DVb%2BO7%2BbRLjxnZ6d4l8J/s16JafCPwbp/x%0AF%2BJHhDwf4k8JaxrGueNfFXgHwn9u1LTVsNJv7fxBNH4V%2B2f2k/8Agox%2B1f8AEP8Aaq%2BD/wCxJ/wS%0A/wDAH7M3i74xeLv2TNF/bp%2BJfxH/AGxfEXxV0D4T%2BBvgT4j8baN4I8H%2BCYtH%2BDGjaz4wHxc8b3Wr%0APq1pDqdxDZ%2BG9CtrbVLjQ/EFnqLTacAfqx%2Bzx%2B0p8Bf2s/hfpXxp/Zu%2BKvg/4yfC3WtQ1jSdP8Z%2B%0ACtSGo6VJqvh/UJtM1nTLhXSG7sNQsLuEiWyv7a1uWtpbS%2Bijksb2zuZ%2B4%2BJXxM%2BHnwa8BeK/il8W%0APG3hf4cfDjwNo9zr/jDxx401vT/Dvhfw3o1oF8/UdY1nVJ7axsrdWeOFDNMrT3EsNtAslxNFG/4M%0Af8G0lnr3hf8AYk/aO%2BFPjbRvD/hr4m/BD/go/wDto/Cf4seFfCeoWeqeFfCnxH0LxvpGreIvDnhm%0A/s3H2rw/pP8Ab9pbaNc3Fpp8t3py215FZLZXFpPcfr1%2B3J8VIPgZ%2BxX%2B118aLjTPDWtx/Cf9mX47%0AfEVdC8ZxpP4S1%2Bfwb8MPFHiC10DxJbSaTr63eia5c2EOk6lZDQdde9tbyW1j0bVZJlsLgA%2BWvCX/%0AAAW6/wCCRPjW8isdG/4KL/sl2U0sQmV/Fvxh8LeALMITqA2y6j47vPDenwy50y5zbzXMc4Eunkx4%0A1bSje/p/YX9jqtjZanpl7aajpuo2lvf6fqFhcQ3ljf2N5ClxaXtld27yW91aXVvJHPb3EEkkM8Mi%0ASxOyMrH/ADzP%2BCd%2BjeLdQ/4IKR/syfs7/wDBGX9oz9pf9p39pTwv8YdE0P8AaW%2BK3wR/Zq0L9ndf%0AEfxl8Xa/4V8PfE3wr8T/AIm/EHUvF8nh74UeDpfC%2BqeGb/xB4C0jwlqnjXwpcajc%2BIfDvhu5l1my%0A/uZ/Yg/Z%2B1/9lH9jz9mT9mrxV471D4m%2BJvgZ8D/hv8MPEHjrUJpZv%2BEh1nwh4X07R9SudL%2B0W9td%0A2/hq3u7aWx8JWGoJNqem%2BFrXR9P1O91C/trm/uQD6lr51/Z4/a4/Zn/azsfHupfs2fG34f8Axnsv%0Ahf41vvh38QJvAuuQ6t/wivjHT4Y559H1SMLHNH5kUhex1CKOXStT8i8XTb67axvFg/Iv4e/G/wD4%0AKYt/wXP%2BP3wI17wl%2ByTb/s7Xf7MH7PnxEtrRPip8bdS8XaX%2BzjoHx2/ap8KeDPiD4W04/D628KD9%0Aorx7r3iDxBZ/FrwXqVho3gvTNB8HfDzTdA%2BI/igeHZta8S/kr8Hf2jf2of8AgnP%2B3v8A8HCXhb9m%0AP9lr9nrxr8NP2efHWk/8FCvj3B43%2BLHjb4VRn4Oa38Drb4uT%2BBPg/a%2BA/gz8QvDNr8VfGPhm48R%2B%0AKLKbxlqOk6HbammoWqaPqZ0%2B8yAf20UV/N9%2B0d/wXO1XVtd/YS%2BHH7GTfsr%2BCtW/bi/Zd8Q/tbQf%0AtBft3/EzXfh98A/gp4A0TSp1tPCfivRvCraHrnivxxqXiy01bwxf21h448Nw6BfaBKDb6xZ6pNqf%0Ah79RP%2BCb37cWmft6fs%2B6p8RpdI8OeHfiX8Kvil43/Z6%2BPPh7wN4qs/iF8NdL%2BNXwxGkt4qk%2BFvxM%0A0lptB%2BI3w31/TNe8P%2BK/CXibRry9EGm6%2BnhvX5LTxh4f8SaZYAHp3xe/bz/Ya/Z88ZT/AA5%2BPf7Z%0A37KHwQ%2BIVtp9hq1z4E%2BL37RXwg%2BGvjK30rVY2l0zU5/C/jPxhoutxafqMSPJYXslitteRoz28kiq%0ASPo/wn4s8K%2BPfCvhnx14F8TeH/Gngnxp4f0bxZ4O8Y%2BE9Z07xH4V8WeFfEenW2seHvE3hnxDo9ze%0AaRr3h/XdIvLPVNG1nS7y607VNOura%2Bsbme2nilb%2Ban/gvH8N/hPf/t2/8EEPiN8UPh74M8VaDe/8%0AFAtO%2BCWvXms%2BFtH17U9VuviQnhv/AIVh4Z1JbyzludT8JWvj6zOtXOn3cs2j6beb9SlsZJJpS/7u%0AftMftA/Cn9if9mn4m/GHxMvhrQ/CXwQ%2BDXxD8b%2BHPh/a6pofg9/EunfB/wCHOueNE%2BH3gKwlSO3/%0AALQm0HwvLp2jaTo%2Bm3Zs4Ej%2Bz6c9vb%2BVQB7l428b%2BC/hr4R8R/ED4jeL/C/gDwH4O0e%2B8Q%2BLvG3j%0AbX9J8K%2BEfCugaZA9zqWueI/Emu3dho2h6Pp9tG9xfanqd7a2VpAjy3E8calhj/C34sfCz44%2BBNC%2B%0AKXwU%2BJfw/wDjB8MvFH9p/wDCM/EX4W%2BMvDnxA8CeIv7E1jUPDus/2F4u8J6lq/h/V/7I8QaRquha%0An/Z%2BoXH2DWNM1DTLryr2yuYIv57v2NP%2BCuvir9rXx/8Asv8Awb/bK8C/sEaj8I/%2BCkXwi8Qa98H/%0AAAF8Evjinx18ceAPEyeAIPixZ/Az9rf4aeOdE03SbO48SfC%2Bz8RSza3Y2SWUHxIg0r4U3HhO51K8%0Al1qP8xf%2BCSX7X37f/wDwT9/4I1eEPivdfA79nfxt%2Bxh%2Bwz8Tfj78Nfj14bm8feNYP2n/ABXp1l%2B0%0A1488R/FT4sfB7VF0HR/hFb6P4GvvG8ngrw38KfEzeINW%2BIOu6NrXi2f4ueBNJl0/wVbAH9xdFfy5%0A/tTf8F4fHEf7YvxL/Zd/ZW%2BJv/BO74I6B8H/ANlf4aftEt8Vf%2BChvxVfwFof7QHjH4yeEvBnxK%2BH%0AHwN%2BFiWvxe%2BEdv4F1PVvhx420nXL/wAc%2BJ7vxp/Y9zKDqPgW20SSw1vUuA8Y/wDBwH8VfiTpv7Kn%0Aiv4J%2BM/2KP2O/Cv7Rv7HunftAeEtQ/4KT%2BGPjJ4Y8FfGL4uaT8SviL8L/iv8Gvh18a/BHxO8GeBv%0AhpY/DnxN4H0OMeLPidpepTeKvDPxG8O/ELw7ot74ZtCmqgH9ZdFfzP8A7T3/AAWW/a/8BfEzwd8J%0APgr8Dv2adO8QeEv%2BCUP/AA9i%2BOniL9oj4geL/D%2Bg%2BJ/hjb29r4c1v4X/AAotfAU%2BuDwb428OeNv7%0Ab1ifUfGGueN9H1fwn4avdLsvs13fWXieuY8O/wDBYT/goP4V/ZG/Yg/by%2BMPwe/Yy1/9mP8Aa4/a%0AI%2BBXgrxFdfCzX/jjo3xT%2BEnwj%2BPXiHwd4dtHvPBXjK11LRvFPxC%2BHmuS/EXw5quo6J4tm8O%2BNV0T%0Awn4s03QvC1tr2p%2BHNFAP6h6K/C0/8FcPiH8Kf2sP%2BCpH7OX7S3wM8O6DY/sN/skeI/27fgvr3wy8%0AU3%2Bt/wDC4v2cfB1l4rv9b/4TO78QWmnz%2BFviFdpB4OsLLSfD3hrX9BsdUk8aW17rky6PoMniP5z/%0AAOCfv/BcHxx8aviH%2Bxp4U/aa8dfsFeKLf/goB4a8Uan8JfCf7F/xL1fxz8Zv2cviFH4e8P8Aj3wb%0A8Gf2pfhlqXjvxzrtrLc%2BDD420vxZ8V9BtNA0nwf8T9AsvBPjjwB4Gt7x9dUA/peor5C/bx%2BL/wAf%0AvgB%2Byj8ZPjZ%2BzZ8Pfhl8UPiV8KvCWp/EF/CPxZ8a%2BIPAvha88G%2BDrSfxH45mt9T8OeG/EV7qHiOH%0AwvpmpL4d0S4fw/p17qksEt74htYrb7FqP4naj/wV7/4KURfDr/gnz8UNG/Yx/Zf1LR/%2BCpN34c8H%0Afs0aBqnx68feG9X%2BE/jDx7/Y/jPwB4q/aAvLv4f3jap8OtY%2BBkPir4narH8KtD13xBpl7aWfhCxO%0AtXz6Le%2BMAD%2Bm6iv58/iJ/wAFrvFn7MH7K3/BS3x3%2B1l8EPBmiftNf8E1vHHw7%2BH3ifwD8LfiDfaz%0A8JPjXqv7RWjeEfEv7NOseAPFfiHRLLxfo9l4n0Px5ot38QdE1Tw1qOu%2BDdN0bXNbWO8/f6Lo2F8N%0Af%2BC0PxW1X4h%2BM/2edS8D/softCftE%2BK/2O/E37UX7KWl/siftGad4u%2BHnxT8W%2BB/BuqeIfGXwK%2BI%0AfifX3uYfhp4hsL%2BHSk0DxTe3k%2Bm%2BKNIn1XUY9L0qDS1uLwA/opor%2Bfz9kD/gsR8Q/Hn7Xv7On7H/%0AAO1J4a/Zr0vx1%2B2H8IPGPxJ%2BD1l%2Bzl8X7bx94%2B%2BDPj34VeB7Pxv8T/2f/wBrH4Zw614uk%2BHPjGx0%0AO18X6t4a8a2Xi2Tw3rV94W1bwLFpMniXR9cuNP8AO/2dP%2BCzn7Rv7Tfx58EeAfhZ8Ov2Mtfh8V/G%0Aj9rD4H%2BMv2btR/aB8beGv2yf2d/FnwP8N/Hq7%2BFurftEeCLXwF4vi8LfCzx74m%2BFXgvRfiF8X/A3%0AhL4iaJ8PJviIumaDoHxB1bS4o9RAP6R6K/kx8Z/8F0f29dM/4Jx/En9u%2Bz/Z8/Zf8Da3%2BxH%2B3/4k%0A/ZC/4KG/B7x5qXxZ12/8PaJ4b%2BIPwi%2BHd3/wovWvDWqaZFeeMINc%2BLWjQazrHiO317SNKhae903w%0At4kk0y40y6%2Bqv2uf%2BCjf/BS/wF/wUmsP2FP2TPg9%2BxL8VtG%2BMf7NVh%2B1R%2BzX4q%2BIfjP4u6Nq%2BueD%0APh5Zaynxb8B%2BMNR0Ir4KuvGvxP8AE2mRaF8GNe0jU9H8E%2BA9EvNN8aeP9R8YLqU/hnRwD%2BiSiv5d%0Av2if%2BC237TkP7ZX7Z/7O/wCzTafsC%2BCdB/4J8eGvhlqXxO8OftYfFrXNL%2BJn7WHjHxboq%2BLvF3w7%0A/Zj1TRPG3gLTPCV34T0zTtb8HS%2BIPEXgj4qXFv4sufC15rXhLTbjXrPwbP8A0CfsnftHeFP2vP2b%0APgr%2B014I0DxX4V8K/G34f6F4%2B0fwz4406LSvFvh%2BHWbffNo2v2Vtc3lomoabdpcWkk9ld3VheLEl%0A7YXM9ncQSuAfQtFfzd6R/wAFUP8Agod8aPCn7bP7Yf7O3wD/AGWk/Yl/YT%2BNXx6%2BEfiT4ZfGLXvj%0AB4e/aa/aH0D9lmEa58Yfiz8K/ipYQWvwZ8DaVrHh6eTTfAfhHxH4D8ZxReNPDHifRfE/jyx8pRaY%0A%2Bs/8Fjv2y/i1%2B2d%2Byl8B/wBjH4Dfs2/EP9nf/gol8Ar743/siftGfFHxR8bfDeofD3Qvhz4ag1/4%0A3an%2B0r8NtC8D3mq2974e1XR/EPgjwX4S8NX%2BgeGvFWv%2BLfhZO3xotX1LxHpGjgH9LlFfAH/BOj9t%0Aq8/be%2BEvxV1rxV8O5fhd8YP2bP2lfjL%2Bx5%2B0D4Us9ROu%2BCk%2BN/wE1TTdL8aat8L/ABPLHbXniT4d%0Aa3HrGlalod9qVhY6lp13can4avBqUuhHXNV%2B/wCgAr8T/wDguN%2Bxv%2B3R/wAFDv2SPGv7Ff7LFv8A%0AsnaD8OvjNpvhCX4ofEb9oD4n/GDwt410HV/h98X/AIefFLwxY%2BAvCnw6%2BCHxH0K/027k8BtY69qv%0AiHxFbXUiaysWn6NanSze6n%2B2FeVaN8dvgh4i%2BKviP4E%2BH/jJ8Ktd%2BN/g7w/F4s8XfBvRviH4R1P4%0Aq%2BFvCs13Y2EPibxH8PLLWJ/F2h%2BH5b7U9NsotZ1PR7XTpLvULG2S5M13AkgB%2BKX7Tn/BPz9v39rz%0A/gmv%2Bzx%2Byj8S9U/Zc8FfEn4M/HP4Fax8X/hl4D%2BN/wC0FJ%2Bzl%2B2J%2BzL8FNHuLG%2B%2BBfxK%2BIVr8GvB%0AnxV%2BG9n8ULiXRJ/FVtpXgL4kWGl6h4J03XtLvJ7jWYNN8JfN3hr/AIJYf8FL/h9%2B1N%2B0V%2B0b8EPD%0Af/BMX9nHwH8eP2arX4Fn9lv4X6n8QLrwBpWkeB/hT8Wvhf8ADL4cX/jO6/ZH0Ox/4QQ%2BLPiJ4P8A%0Aj7418W%2BCvhD4E1u/v/g34W%2BAlt4Ln8F%2BJfFPxCn/AKk6KAP44f2Bv%2BCEX/BRn/gnbP8AseeL/gf4%0Am/Yk/wCEw0LwP8bfgV/wUV8E6j8XfjzZ/B/9rf4MeLvi9r/xA%2BEHiiDSdC/Zs0rWL748fDTw18Sv%0AH3gu18Z%2BMm8nTdB8KfDHw1pEsvhiTxbp199t/sSf8El/2qf2F/gnYfsj%2BB9C/wCCffxR%2BHngX41L%0Ar/wg/a%2B%2BLngjVdb/AGjPCnwW8SfErUfiN428P%2BKfgjafBSDw142%2BJWmDUNS8P%2BA/FE37TGk6Zp82%0AvLrtxZW2keEND8CX39IdeP8AgL9oX4BfFTxp8QPhx8MPjj8H/iR8Q/hNqtzoPxU8BeAviX4L8YeN%0APhprllezaZeaN8QPC3h7WtR1zwbqtpqNvPp9zp3iOx028gvYJrSWFJ43jUA9gr8P/wBjD9j3/goz%0A8Jvif%2B2P%2B0t%2B0F4i/Y%2B1L9o/44/sn/s7fBPw7r/wq8S/Fy90r4yfG/8AZr0P41ab4X/aE%2BPban8L%0APh94N%2BHN/wCL1%2BIPhbR/FPhL4XfBvxjomk6VobTeCoPD9vaan4f8bfuBRQB%2BIn/BEr/gnx%2B1b%2BwF%0A8MfjPof7V3jL4EeJPG/xV8aWXjfWJPgFqnjPxLpXjz4k6nrvxA8XfFL9oT4k%2BJfH/wAO/htrcvxZ%0A%2BKeoeOtB8D6h4b0HQYPAHhf4c/BX4Y2/h23i1u98W3mo/qn%2B0D8IdW%2BOvwr8QfDHRvjZ8aP2er7X%0A7jRJl%2BKn7P2s%2BC/D3xU0OPRtasNZktPD%2BtePfAnxH8PWFvrQsBpGttJ4Wubu50W8v7K1ubNrlp19%0AoooA/HiH/gkv8QIJLuVP%2BCw//BYdmvbhbmYTfHb9mO5jSRLW2swtpDcfsfSw2FuYbSJ2tLBLa0e7%0Ae5v3ha%2Bvby4n/S34FfCu%2B%2BCfwp8JfDDU/i18W/jpfeFotWjufit8ddc8OeJPit4ufVde1TXBP4t1%0Arwl4T8DeHLyXTE1NdD0ldK8KaPBa6Dpml2bwzT28t3cetUUAfjH4j%2BGvgb4w/wDBYj9rv4VfEOws%0Atc8I/E7/AIIx/s7fDXxb4YvINLnTxB4G8eftR/ty%2BHPGNhLFe2d1cS2Vzpl6mn3sCltLmGoRLqtn%0AeONP8j5N8K/8EiP2wdU/Zh/YW/4J0fFj4i/s3n9kr9in9ov4b/F7Xfjh4Dv/AIgTfGf9pf4efAzx%0ALrHjL4a/CvXv2d/Fnw5uvBnwXu/E2ua1ZxfEHxnoH7TfxKltF8IWd5oOhXGn%2BKtV8MaT%2BwnjD9ij%0A9lHxJ%2B238Kf29fF/hJZf2t/APwq1f4DfCvxpc%2BPfFemQweBbk%2BPte1XRNP8AANt4is/CHiLVIrLx%0A54/uJb%2B%2B8PapqVtpmqXVy0if2PpV1pn2VQB%2BVH7PHx2/ab8Uf8FWP%2BCgv7PPiv4lfDf4m/sx/CL4%0ARfsteO/Ami%2BEvC%2BlaN4m/Z38ffFbR/EyS/CLx3r0WojXfEnjLxto3g/WPjFq1rqdrrlpofgrxL8K%0AdT0zUfClt4wXwzP88f8ABWv/AIJ0/teft7/G79hvxr8H9c/Zf8L%2BAf2I/wBorwF%2B03pVh8X9X%2BJ2%0Ao638UvGPhHXdE1u58GeI9F8L%2BBLzRdC8FXsegWumTTW2sa3ql9FdXlzIlkpjsh%2B5MVhoehtrurQW%0AWlaO2rXf9veJdTit7PT21O%2Bs9H07Rv7Z128RITe3droGh6RpX9o38kk0Gj6Pp1j5y2Wn20UPNSfF%0AH4Zw6BB4ql%2BIvgSLwvdag2k2viSTxd4fTQLnVVimmbTINYbUBp02oLDbXEzWUdy1yIoJpDHsicqA%0Afix/wUj1vx0/7W//AAQN8MeOF8OWGuaz%2B2l4q8T%2BP9O8HyXF7ocPxA8K/sn/ABMhmi8K%2BKdU0rSP%0AGEnhK11DxL4ktre0uY9Hh8T6XNp83ivQp7mysrew/eGviDxx/wAE4P2L/iT%2B2T8OP%2BCgPjb4Lxa9%0A%2B1z8JPD9t4X%2BHvxZm8e/FG1Tw9olnY%2BLNNtLQfD/AE/xtafC/VJba08b%2BJ1gv9a8FajqMU%2Box3iX%0Aa3mm6VPZfWPib4geA/BctpB4x8beEfCc1/HLLYw%2BJvEmjaDLexQMiTSWkeqXtq9xHC8iLK8IdY2d%0AFcgsAQD%2BZj4uf8G8fifxt8af%2BCoXx08EfG3QvBPjP9o7x54O%2BNP/AAT8lj1Xx7/Z37Iv7Qur%2BMvg%0Ab8bf2h/2gP7Jso4bbwh8Yvir8aP2fvhpaxfEL4ayT%2BIz8PfC0WmeJbrWtM1abwZpvuv/AAUX/wCC%0APnxr%2BPH/AASB%2BD3/AAS3/ZO8ZfB3T5/DmqfCsfFD4sftG6v4su9V1%2BPwRLqXjzxt8QNF1DRvh38R%0Addf4p/Fb40PB4n8UavbzeEFXR/EnjjTotQfStXn8M339EFFAH4I/taf8E9v28viT%2B2Da/tz/ALOX%0AxO%2BBfw0%2BNHxD/wCCYPj/AP4J0fE/SdU8efEjQLT4Saz428d618S9A/aD%2BB3j7Rfg94p1zxZrfww8%0Ae67beKNG8H%2BI/DPw2m8Tf8IfpOiyeNvCX9u3%2BtaVs%2BHP2PP%2BCi/gT/gqT%2B21%2B3x4T0f9inUPA3xx%0A/ZhsPgP8F/AHiH4z/HSHxZbeKPhNeaVf/Brxt8UNT039nNtH0jQPFt1N4sk%2BJ3hDwoviXUfClvee%0AHbTwr4u8WSaVqV/rX7gXevaHp%2BqaRol/rOlWWteIP7Q/sHSLvUbO21TW/wCyrdbvVP7IsJpku9S/%0As20dLrUPsUU32O3dZrjy42DHVoA/K3/gi3%2BxZ8bf%2BCd3/BO/4L/sa/HrWPhZ4m8YfBjVPiSlj4q%2B%0AEWv%2BLde8NeIdF%2BI3xF8S/F2We6/4TLwR4F1TTNU0bxN4/wDEnhSO1h06%2BtNR0Lw9ofiZryw1DxBf%0A%2BGfD3p/7ZfwF/aH8dfGD9jT9oj9ms/CbxH4x/ZZ%2BIfxc1jxH8JvjL458UfCbwl8UfBfxg%2BB/jH4W%0A3drD8WfB/wAIfj5rvg3WfB/iTVPC/jCztrf4T6zF4ot9KvdEudd8NwyvNe/oHUUNxBco0lvNFPGs%0Atxbs8MiSos9rPJa3ULMjMoltrmGa3uIyd8M8UkMirIjKAD8APCP7An/BSDwh%2B2D/AMEyfj94v%2BI3%0A7Ov7SFl%2Bxx%2Byr44/Z4%2BPvxT%2BJPxW%2BKfgL4yfGPW/i7DoEHizxx4f8JaL%2Bz38QPDj6h4ETwvo76bq%0AHiz4kSax8YJTqN14ruPh/rF/NrR%2BRPhD/wAELP2vx%2ByR%2B0L8Jvi94/8A2WPCPxt/4ejab/wVl/Yz%0A8S/DzUfiH8U/Afw/%2BPtrqitP4L%2BPcHiz4T/CWf4n%2BAU8L2Q8MQzReFGv7pfEuqXt5Z/YfDPhvQm/%0Aqvg1bSrrUdQ0i21PT7jVtJisZ9V0uC9tptR0yDUxcNps2oWUcjXNnFqC2d21jJcxRpdi1uDbtIIJ%0Adsr39jFHfTSXtpHFpm/%2B0pXuYUj0/wAu1ivpPtzs4W02WU8N4/2gx7bWaK4bEUiOQD4b/Zg%2BFX7V%0Anhz4keI/Gvxy079kr4U%2BBm8Ff8Iz4e%2BC37LPhnW9ci13xtf%2BIodc8UfGLxx8X/Gvgn4b%2BI0l1p7S%0AZ/D/AMKNB8GLpPhdvE%2BvnxT4%2B%2BKetWWi%2BJ4fKv8Ago1%2Bw78RP2qvHH7DPxt%2BEes/Ds/Eb9hz9pux%0A/aC0b4ffFm61zQvAPxSsP%2BEevdE1Lwtf%2BN/DPhfx3rvw/wBcguG0zVvDfjC18BeOIdIv7Np5/DOo%0AOLcxfo14G8d%2BB/if4Q8OfEL4a%2BMvCnxD8A%2BMNKtde8JeOPA3iHSPFvhDxTod8nm2Os%2BHPEugXmoa%0ALrelXkZElrqOmXt1Z3CHfDM681L4w8aeDvh54b1Xxn4/8WeGvA3g/QooZ9b8V%2BMNd0vwz4b0eC4u%0AYLKCbVdd1q6stL0%2BKe8uba0hku7qFJLm4ggRmlljRgD80P2J/wDgnv4v%2BEfxm/bd/ay/ay8b%2BBPj%0AV%2B0h%2B3drfgnSvH%2BjeDPDF/p/wR%2BHXwL%2BGPgODwP8P/gb4T8P%2BLptT1TxRaWdnda9/wAJv4x16LSo%0A/iGJtIurzwZoeoW%2BtXOvdh8Rv%2BCcH7PHgX4BftNeEP2Gf2V/2Ov2c/jT8ffgr42%2BDP8Awmnhf4M%2B%0AD/hHpkujePNIi8P6jbeJ9a%2BE3giPxPc%2BH9NttviG18OWlvLpmoeJdM0%2B4uYLae4udUi/SSs/VdW0%0ArQtOu9X1zU9P0bSbCIz32qare22nadZQAhTNd3t3JDbW0QZlUyTSogLAFskUAfzl6X/wTR/4KV6X%0A8FP%2BCRPwal%2BJX7HWsad/wTL%2BIHg3x1rl7da/8fLX/haH/Co/h/rfwb%2BFulWUDeE9SgsPI%2BGfjDxL%0A/a19dwJ/ZGu2%2Bm2Xh2zXw/eX9tD2fwy/4JPfF79krxT%2B1j4Q/Zn%2BGX7Cfxb%2BDP7Snxh8Z/HX4XeM%0Av2rbHxXqnxQ/ZG8Z/EXQbS38W%2BH7Xwhb/B74kH9p34eQ%2BL9PtfFXhfw/rfx0%2BAGo6NDPquj32t6z%0ANqUmtJ/QDpWraVrunWmr6Hqen6zpN/EJ7HVNKvbbUdOvYCSomtL20kmtrmIsrKJIZXQlSA2Qa0KA%0AP5yPHn/BIz9rD46ftQ/t5%2BOvjX8VvgPL8G/25f2LYf2KfEHiLwq/jlvjN4W0vwV4c1608AfHD/hE%0Az4N8O/DrVPFuv%2BK7jS/EPjj4bWOv6D4W0i0ur/SPDvie/h022k1bxH4v/wDBL7/grZ8Q/wBlP/gl%0Ax%2BypY6j/AME%2BDp//AATB/aO/Zh%2BMmi/EnUvjT%2B0vo%2BsfHXRP2StB1bwd8JbK88MeGv2cLd/g1fR%2B%0ADNQGk%2BMbHSfGvxKvfE%2Btm18UaD458BNpc2i67/VLRQBxninwnafEb4e%2BI/AvjazSCw8d%2BDNX8J%2BL%0ArDRdUuJ0htPE%2BiXGj6/Z6TrU2n6ZdTJHDfXcNhqkulafcSARXb6faSE20f8ANNof/BIP/god4a/4%0AJlz/APBGLQfin%2ByFpn7LFz4u8Q%2BGZ/2wra4%2BJi/tEy/s3eLPjBqfxf8AEfhJv2V2%2BFbfCb/hbur6%0AlrOqaDd/EYftIDR28NXsyJ4TTxuV%2BIi/1GUUAfhV%2Bzb/AME//wBsT9n3/grR8X/2r7WX9mnU/wBj%0Ar4nfs7fCf9mTRtCPxN%2BKMn7QXgHwJ8A/CT2fw71e18Nv8DrbwJ4o1bxD4j0/T4vF9nqvxOtP7L03%0AW9W1bSdZ1OXRdN8N6p%2Bn37YvhT45ePv2ZvjL8Pv2ctI%2BFGsfFv4g%2BBtc8B%2BHYvjZ4z8X%2BA/hzYW3%0AjGym8O67rWta94F%2BH3xN8TNPouh6jqGpaTpNj4WePWdVtrPTbzVdFs7ifU7b6XooA/lk/ZJ/4Jc/%0A8Fi/2RP%2BCbtr/wAE4Phb8bv%2BCdHhnw7qSfFDQvEHx0uvCPx/8c/EPSfCfxovvEupeM5vD3he%2Bs/C%0AngrWfHGhXfim5i8J694mtpNBm0bS9O0jV/CbXDHWLb6P%2BGH/AARKuv2QLH/glz40/Y%2B%2BIHhLU/i1%0A/wAE5PAvxZ%2BFvibTfjani2y8D/tJ%2BA/2jdUv/FPxvWXXPDMvifV/g14mtviB4o8d%2BOPhbq2k%2BFPH%0A%2Bn6C/iSLwr4r0bxboWkaYbX%2BgyigD%2Bbn9pP/AIJL/tv/ALRHwF/4KQPq/wAZf2bbX9qH/gpvqf7P%0AXw/%2BJMMafGWw%2BAPwK/Zu/Zq8M67pXgrw18KZrMTfEXxX8SNf8Qazrfinxdq3iyPSfBeoXPjfWtMP%0AhWXSNBisfFH1r%2B1d/wAE6Pi5%2B1b%2BzB%2BxRaa34g%2BCXhP9sf8AYZ%2BMXwf%2BPPw21eTSfEXjr9njxX4m%0A%2BFnneHta%2BHPjaw1vR9M8dTfDL4o%2BCXjXX59GttM8WeG/FVnomqaLqF3b6I8WrfsjRQB%2BB/7ev7EH%0A/BTT9v39iX9p39mnxJ47/Y0%2BCF5%2B0VL8B/DOkfD/AMHXvxL8f%2BAvhV4W8B%2BKNW%2BKXxu8fR/FfUvh%0AH4B8afEP4lfHb4if8IroKaDqfwx8MeEfh/8AD3w8b3TL7V/iFqmveJ9b/Yb9m/wT43%2BGn7P3wV%2BH%0AHxIufCt547%2BH/wALvA/gjxTeeCJNXl8JXmreFPDun6DPd6BJr1tZ6w2nXQsFnh/tG1guQZGV41wB%0AXtVFAH8xOv8A/BFP9p/4yeL/ANnc/tIePf2cviZf/s4f8FEbD9q7wV%2B2Tq2s/GLW/wBuOP8AZs8M%0A/FzxP8WfDf7Kw8d2GheBNclQT6rpfhbS/FmsfF3WdD%2BGeh2lm/gbwT9r8G%2BAn0D6Y/Zw/wCCZ/7R%0Av7Dfwy%2BPH7On7Jt1%2ByhbfDrx/wDHj4u/Er9nz43%2BOvD2rWvxU/ZQ8AftDah4Wb4j%2BBtG%2BD8Hwz8Y%0AeAfizqfgnSNGubD4d32p/Ej4f%2BEfER0rwRafEXwNqfhzw/c%2BHtR/d6igD8BfGP8AwSi/abtP%2BCsH%0Axk/bp%2BC/7SOnfDv4NftQeJv2JPFfx48Mad4o%2BIHhn4nov7HV94NkuvAnh%2Bz8KaVb%2BFfF3w7%2BM/hf%0A4d6X8PPHGh%2BMvEVsP%2BEd8fePpZF1PT47fwrrPg/w9/4Iy/tPeKvin%2Bwn8Q/2lb39lyf4yfsc/tJ6%0AV8ePGn/BQbwL4u8beOP23v2pPAHhBPGFv4M/Zz%2BNcnib9n74e6Z4h0NdD1DwL4K8TfE7W/id4hu5%0APCfgxbPwJ8Nvh/8Ab54X/pyooA%2BSf29PgD48/ap/Yx/aa/Zv%2BGPxAHwt8e/G34N%2BNvhv4Y8eSTaj%0Ab2mgaj4o0ifTd%2BpzaRFPqsei6hbzT6TrjabDLf8A9kX96LONrgxg/wAyXh7/AIIDft/eBvFX/BN7%0Ax7%2Bz74u/YF/Yz%2BIf7DPwP8WfAvxV8RPglZ%2BOPGnjD4ra98V/h/4p%2BEfxY/alv5/FH7P3hHRNX%2BLW%0AoeDtVsPFWgeBfGWia3Zw%2BPb7xGL74rto50Y2P9kdFAH82Gs/8Eof25/iD%2ByV/wAFUP2WPGnj39ln%0AwpH/AMFKv2ofEf7Rtt498F%2BIfjBrtz8Gh8Qo/hfp/jvwpdeHdV8AeG1%2BIlqug/B/w5pegzR694HE%0Atz4i8R6hrAuLO0sNDuvpq7/4J3/tT/DD46fs1/tm/s6fEj9nmH9pnwF%2BxfpH7DP7RngDx5oHxR0T%0A9nD4w/DHwprll4y8B%2BMPA0eh674r%2BI3gHx/4K8Y2lyLfUfF2pfE%2BTV/BGpv4PbUNKktJdd1P9taK%0AAPyG/wCCSn7D37U/7Dmn/toaZ%2B0t8UvgN8WZf2nf2xPiZ%2B2LpOv/AAY8D%2BLfAt7a%2BOfjl9luPirY%0A%2BJNL8SajqcEWhrqHh/w1L4F06DUvEOr6JZya1pmreLtc0%2BLw9a6P23/BXD9jr9pH9v39kTx1%2ByH8%0AB/iz8K/gr4e%2BNNjZ6D8WvHfjzwx4w8U%2BKLXwzpXjXwN4nh0rwDZ%2BHNc0XS7dvEFhofiLQ/FU3iJN%0ASjudH1CLTNMg0%2B6vpdd0n9Q6KAPmr9jj4QeN/wBn39lf4BfAb4han8OtY8S/BX4X%2BFfhQNQ%2BE/hr%0AxH4P%2BH8/hv4e6dH4T8EJoHhvxb4n8Z6/pX2fwTpPh621WK%2B8S6mkutw6lc2TWunz2tlbfStFFAH4%0Afft2f8E2f2qf2jP2u/E3xv8AgP8AtC/D34R/Dj45fscfDD9jT42x6np3xG0/4weBvD/w6/ab8QfH%0Ah/il8DvF3gDX9CYeOdQ8OeNPFXgrSbDVdX8LWPh/U5bHxRNqWuiNNJtPmLUP%2BCSn7eg%2BJn/Bb3xj%0ApHxO/Yvt/Dn/AAV4%2BDDfCm10CDwp8WdBv/hhf6H4Y8WfBfwf4tvr7R7dtKvbm6%2BC/wAQ/HviL4gw%0ADQtY1PxX8a9T0DW01%2Bx0TR/EaeOv6XqKAP5qPgf/AMEZv2rf2Y9J/wCCcnxc%2BEHxs/Z31b9qf9gj%0A4J/E79lzV9L8d%2BFPiYfgb8b/AIDfEHWLzV9O%2B1ajo2qjxt8PviF4R1HU9V1m31bS9E13Sdcur620%0AvULKDTdMLaj%2B7/7PXhz9oDQfBur3n7S3xC8JeOvib4q8W6n4ll0n4c%2BH49B%2BGHwv0CbTtH0fRfhp%0A8O577TbPxt4m0LTotGm8Tat4t%2BJF7rHizXPGvivxXNZP4c8ER%2BDvAnhH3iigD8M/%2BCuP/BPb9sb9%0AuX4vfsJ%2BOf2dvGv7Lvgzw/8AsT/tD/Dz9q7Rx8a7D4pXfibxR8V/h14nTWLDwpdS%2BBrW4sovhlqt%0AlpmiDWIbWbT/ABJc6hDM0OoW0NvaPXP/ABQ/4J0/t3/twfHmbxR%2B3X8Zf2Z/Cn7PGi/sm/tY/s5%2B%0ADfhB%2BzBonxS1bXx4l/a7%2BGcPwh8dfE/xB4p%2BJ76TZSeI9A8GTana%2BEr200eQaPp95qfh%2B00%2BBPFX%0AijW7797KKAPy0/Zb/Z//AG%2BPhn4N/ZK%2BCvxO8d/s0%2BG/h/8As26LoelfE34pfBDSr27%2BIH7U2n/D%0Az4eah8Kvh14cv/hb47%2BENt4N/Z4sNR8Pw%2BEdf%2BJviPwH8QfHPiLVL3wxL4K%2BHJ%2BGfg/V3%2Bw/nLon%0A/BGn9tpv2RfjB/wTZ1b9qT4E%2BG/2Q/jt8e/ih8WPiR8WvC3w48Ta7%2B0dL8MPix8TD8SvFfwC8E%2BD%0AfEbWnwz8I3d/qK3VuPjlrvib4gapYW/iG%2Bh074X27aLpdxc/0y0UAfiZ4i/4Je/E/wCAH7Zsf7bv%0A/BO3x38IfAXizxf%2BzH4N/ZU%2BNHwV%2BP8A4Y8U618LPG/hj4VWPg7Q/gx8TdN8QfDe90XxrY/EP4d%2B%0AFvB2m%2BEr221iTW9P8XeE7Ow0K11HwZJBJqsuN4%2B/4J%2Bft96vD4p03VP2j/2Yv2sdA%2BMv7Lvh34Mf%0AG/wr%2B2b8EPFfiHwK3xftvjP%2B0v8AE7UPjF8Nvh94M8VL4V0PwboHhr9pCf4XeG/hFfWj63rXw%2B%2BG%0AXww8PeJPjCl14XXX739zKKAP8/345fsX%2BG/DP7eHw3/Yo%2BJ3x9/Yq%2BAFp%2BwB/wAExf2V/hF8OPiD%0A/wAFO/gN4D%2BJnwU/bI03wddeO/i/8W/j1%2Bz1oHxJn8M%2BE/ANp8MfEniDX/A3xG8Mz/E7xVe614b0%0AG/j1/S/E%2Bi/D3X9e0L9Rvhlpf7RH/Bdr9gH45fBfxX8Qv2crXxZ%2ByB%2B3F8N2/Zd/bI%2BEfwr8Wf8A%0ADL37SY/Z1t/C3iLTfE%2Bm%2BBrzxxNrNlZajc33iHwh431r4aeL7Xw1p82r/YvB%2BnrZ6RqWkXn9Qnjj%0A4Y/Db4m2tjYfEn4e%2BB/iFY6XcSXem2fjjwnoPiy1067mj8ma5sbfXtPv4bS4lhAikmt0jkeMbGYr%0AxXQeH/D2geE9G07w54W0PR/DXh7SLdbTSdB8P6ZZaNo2l2iszrbadpenQW1jZW6s7ssNtBFGGZiF%0AyxJAP57PB/8AwSl/bY%2BIX7cP7ZX7S/7V/wAav2Y9V%2BEn7eH7Hmp/sffGL4SfCLwp8Vl1zwF4AuPA%0AGreC7Cx%2BF3iHxPq2m6Dql62qTReLta8W/Erwx4p1O5bxJ4q8LeHdG8K6LFp8tx9V/sPfsn/8FHv2%0Ad/h/%2Byn%2Bzb8Uv2nv2frj9nn9kjwzoXgeLxP8HfhPqtp8Zv2j/h38OvCGpeAfhR8NfiHY/EhfE3gj%0A4P8AhnR9Ck8Lan42134cy%2BJPiN4u1bwHpNppHjXwfZ6/4pudR/YCigD5x/a%2B%2BGfxK%2BNP7L3x4%2BDn%0Awh1jwN4d%2BIPxZ%2BGPir4Z6J4g%2BJFjr%2Bp%2BDdCtvHmmy%2BFvEOsappvhi6sdb1CfTPDWq6xe6NY2l9Zx%0A3euQabb311b2El1Mn432/wDwTB/bzsPDf/BGHQbT40/soF/%2BCWA0bS/El1F4N%2BLNlJ8T/Dnh/wCH%0ANp%2Bz7pdvo5utW1yGyv5PglbXep6/Nd21gusfETVzFoz%2BENC0iK61L%2Bh6igD%2Bcv4lf8EXfjT%2B1Q3/%0AAAV88FftVfFz4Pp8P/8Agpnd/BXxb4J8X/CTw74rPjn4FeP/ANmbSPDPhX4F3V74T8VJBovi3w/Z%0AaJ4I8J33j6DT/Gvh3WvFIh8R%2BFbHVdDtPEh8SaZ9D/tCfsa/8FNP2wfgL8TP2efiv%2B1f8Bvgf4D8%0Aa/smfE79n/XrP4K/D7xN8Qr741fEX4i6T4c8PP8AFT4k%2BI/iHF4X8S/D3wtpvh/S/FWmQfCz4b6n%0Afy3zfEjxBN4w8eeNbLRdB0mL9raKAP5aPHf/AARv/wCCh/iLxB/wTf8AEfwZ%2BOf7Hn7HL/8ABOD4%0AY/FD4M/CvSPgp4I8X%2BMIYrX9or4V3Hwt%2BO/x00y88ZeCNH0iL4j3dnbaR4h8N%2BA9d8H65pep%2BL59%0Af8VeIviWur6lO9z9L%2BCf%2BCWP7R3in4qfsE/En9pDU/2U9Z%2BKn7Fnxn1f4peLP2xvAdl8S9b/AGov%0A2jPCtn4H8ceCfBnwf8U3nj/R5de8P%2BERaeL9KfxfqPiz43fGO/1S38B%2BFrextrSb%2B0ry7/oAooA/%0AlP8Ajz/wQV/bR%2BJ/7Of/AAUM/Y88F/tq/A3wn8BP29P%2BCg/xF/bu8WazrvwA8S%2BIPi3ct498WeAv%0AHEXwm1vVbX4g6V4Q0nR9E8V/Dvwbr7%2BJfCHhqw1nV9R8KNbqNJ8NeL9T8M6N9seI/wDgnB%2B274s/%0A4Khfsw/8FKNa%2BP8A%2ByMniH4Pfs8%2BEf2b/iV4L0f9nn4uafF4z8E6xrt54m%2BNd14S1K7/AGgNVvtA%0A1q41fxP42X4N3Gv33iTSPBltf6ZJ4s0Hxrcpqz6n%2B7FFAH44Sf8ABOL44/AT9uH9qn9s79hn41/C%0A/wAAx/tw%2BBPB9n%2B0d8H/AI3fDTVvG3hxfjP8LbLUtI%2BG3xp%2BG/iXwn4j8Na/ostvpHiPxQPFXgPW%0AF1PQNf1rXNW1u5uppLzRLTwl%2Bn/wU8E%2BNfhv8I/hx4C%2BJHxY8QfHXx/4S8H6HoXjL4w%2BKdB8MeFt%0Ad%2BI/iSwsYodX8V33hvwbp2meHNC/tW9WWa10nT7e4exszb29/quualHd61f%2BoUUAfz%2Bzf8EqP2yP%0Aht8H/wBu39kP9nH9rH4N%2BHv2WP23/HPxq8X6frPxU%2BDPirxP8dv2abb9p2K6i%2BP2i%2BC9S8D/ABA8%0AA%2BG/inBfPrfiO%2B%2BGF/4rm8Ja74Eu7qxXVNZ8XR2kbRdN4J/4JC%2BOfgZ%2B3N%2Bwh8ef2efix8N/C/7N%0AP7D37Jv/AAyH4f8Ag7468FeMPFnxI8V%2BC9bm1G78c%2BNLvx3o/jTwt4Xs/iH4hvZNLvV1VfBcmk/b%0AoNTu9Q0W%2BjvrCz0j93KKAPyT/wCCTn7Anxz/AGAtG/bC0b4xfFr4T/FeL9qT9rz4j/tj2s3w08A%2B%0AMPA0nhPx38Z7XSrX4j%2BGJ18UeNvF66j4Ss18JeEX8Exx%2BTrOns3iMa7quti70z%2Bzv1soooAK/h3/%0AAGlP2ovGH7En7NX/AAc0/tD/AAak1Dwp8edU/wCChXg34KeBPih4Y0y3j8U%2BCovib4K%2BCXh7U9XT%0AxUltLqPhyLQ/D%2Bq%2BJ9S8M6lDIrab46m0CbTJLDV7q21K2/uIr%2BPTUf2Drv8A4KSfCT/g5K/ZS8P3%0AWj6Z8RPFn/BRLSvFHwj1vXryTTtI0n4pfD/4U/Brxf4UGp6lDpetzado%2Bvy6Vc%2BD/EF7BpN9c2/h%0A3xJq0lnHFeCC5gAPXX/Zm0P/AIJB/wDBTf8A4JF%2BHv2bPE3jCa2/b5tP2gPgT/wUIk8QfEn4ra/p%0A/wC1V8Sfhp8EPh/4g%2BH37RWr%2BCviN4w%2BJ%2Bg%2BE/iVpnj2z8T%2BLdW8Q6Fqlp4nubXxXrnhEa1f6d4x%0A8Zapdf1XV/Nl4c%2BBf7cX/BRf/goD/wAEzv2n/wBqv9krxf8AsVeCv%2BCa3gT42eJfHzeMviN8I/F2%0Aq/HX9p74xeDfh/4M1vwl8KLb4O/E3xhc2nwb8H694NTxhZfEPxTa6XbeOtAifwxb%2BHmXXLq60P8A%0ARb/gmp%2B1j%2B03%2B04f22PDH7VXwo%2BG/wAMvH37Lf7a/wARf2eNBk%2BEXiTSvFXgjxF8PLHwF8Mvid8P%0A7i/1a0%2BIvjy/m%2BJFp4O%2BJmh3HxCjmtfCmn6ffapp2iv4c8OeMtK8c%2BC/CAB%2BTd5pfxX%2BE3/B018G%0AvhrcftOftG/FL4Q/FX9g74zftDWPwn%2BKvxbi8SfD/wCFni7xn8SPib4e1bwx8L/hzolh4d0rwp4C%0A0/SvAfhNPDs2u6PrniS6mtLq0uPG2uQaHaWOg/W/hvwzZ6F/wcu/EC88O29poGl6/wD8ES/Beu%2BL%0A9K0WUaPZeLfGd9%2B3f45s7Hxh4j0Sw0lbHxN4l0/R9IutItvFWr6rFrOk6bezaVaW%2BqWer3Uuj/KH%0Axv8AhD/wUkvf%2BDgz4S/8FAPBH/BOHx14z/Zm%2BE37Neo/sZX3i5f2mf2R/D2r%2BJdGv/iN8V/ErfHb%0Aw54S1n4wReJ18NRW3xC06WPwR4g0rQ/Gc1hp%2BoTyW1tqn2XQ7j6yupriD/g5L8bPo1jqWpeLY/8A%0Agg94Xm8OWF7d3ej/AA9vriH9vP4hkWPi7xHZRa1caHqV3qbaPDod3b%2BC/FF3FojeNL%2B1iR9POla8%0AAfu5HbwQvcSQwxRSXUq3F08caI9zOsENqs1wyqGmlW2tre3WSQs4gghhDeXEir/Jz/wXP/ZR/Zp%2B%0AAX/BQf8A4Jbf8FefH/w78Jav4c0L9rLwH%2Bzz%2B1daaxpOp67D4js/F3hrVbf9nj4z6no58QaLoMVx%0A%2Bz54h8N6je3OruZtT1Cab4dpfaX4r0fwZF4auv6Fv2G/GH7YXjz9mbwH4o/bz%2BE/w6%2BCP7Ueoar8%0ARIfiB8NvhTr8XiXwPoek6b8S/F%2BmfDm50rVrfxf4%2BguJdd%2BGdn4Q1/Uini/VHbUdTupbqy8L3stz%0A4O8PfNH/AAWn/Yl13/goP/wTQ/aj/Zo8E6dcat8Utc8G2vjf4OaVba3pXh6TW/iz8MNZ0/x/4I8L%0ANq3iG8sPDFlb%2BN9U0AeBru88T3tjoen2viSbUrvVdDezh13TAD%2BcKz8P/sD/ALM//B174K8aeGvg%0APpeieB/jNp3xT/Z40f4iadrdzeeAdB/4Kk3vhD4ffGT4meL9C0K1v761tdSuvhF%2B0F8Mvg3rnh60%0AsI9C0L4u/GHUPEUlnpWu6LrN9o/3z%2BzZof7Bv/BPH43ftNf8FKz8O/E/xA/aO/4Kff8ABRH46fsy%0AfsreEPg1qEXif4ifFTRdX%2BLOmeGvEXhXwXpvjH4m%2BGPg7Y%2BGPiJ8ffgr8QP2gfEHxm8Z%2BJfA/gfQ%0AfB/ifwRp%2BoeMtG0ptBsdY4H9r/8A4I4ftXeLP%2BCM/wCyR4I%2BCTeIPEn/AAVU/Z4%2BOHwb/bZ1Dx3r%0A3xQ8Mt451T9rrx74qstf/aO1%2B8%2BKnjDxEPB%2BqHwb4h8Xaj4o07VJ9bkt9W074Q%2BGbbw5HquoyaZo%0A%2Bqe7/wDBQf8A4I7eKPH/AOyj/wAEtdE%2BFHw2svj747/4Jpan4J0bxZ8H4PjJ4g%2BEni349fB/xP4B%0A8NeAP2idA%2BHHx7TxB8MdQ8KfFHxHqHhvQvHfh3xXr/ir4e2N9qOk3d1qWpabePa6NfAH7Bfst/tz%0AfDf9qb4oftQfAzTfAXxV%2BD/xy/Y/8YeBfCfxw%2BEvxhsPAMPibQrb4qeE5PHPwt8Y6Vq3wt%2BInxS8%0ADa74P%2BIfha3vdV8O3th4ufVUisbmHW9G0e4EUMtD9of9vLwF8A/jt8LP2YNJ%2BFXxk%2BPn7Qvxd8Ae%0APfit4Z%2BFPwah%2BE2n6zB8NfhvJZ2niXxbe6/8dPi38E/A14F1XULTSrDwr4b8V%2BIPHl7NJNqKeFF0%0ACx1DWLX5x/4Jt/st2vwb%2BLP7VXxe0v8AYF8O/sO%2BGPi2vwo0fw5D45%2BLOn/Gv9rX4oav4N/4T9vG%0Afir4veJfB3xc%2BOvwp8FfC2HSda%2BGfh34M/DTwJ8Tte1bSr7w38RtW8YGz0vUvAGkaR8Xf8Fcf2Df%0AGX7X/wC2d8GPEPxI/wCCf3jn9sX9mDwL%2BzhrFv4H%2BJX7NHxz%2BDf7O37Vv7O37UcXxNm8Q2%2Bv6T4p%0A%2BKP7QvwHtvHXgTxD4U07wraWGh6rJ4i0r4f6/aax4y0Kzk1/UJrK8APnn9qn48/Cb4zf8FA/%2BDeb%0A/gp78B/2ePj18RfiD8b/AA/%2B2l4G8MfDG0tfD3h/4t6z4LP7M3xIi8P%2BC/FWieKviNY/BvwRqHw/%0A8b%2BP/FWv%2BMPEVz43tBpvhNfF2rXOr%2BKNN8MaZZWX7JfCf/gpZpfxE/Zy/a4%2BMur/ALNvxy0f4o/s%0ATfE/4h/B342fsw%2BBrfwt8avi5eePPA2geFfF9jafDKP4ba3qmn%2BPdK8W%2BEvHPhjVNJ1OE6U9vdjx%0ALpl7ZRt4YvLmf8nfht%2Byj/wVm8CeI/8Agg3dfE34QWv7Tnj39gfwV8avFP7VPxs8XftDfDvwhaXF%0A3%2B0d4I8ZfAfSfhLol1dXXjjx/wDEf4ofs9/CnxFpuueMPGd/4btfA/xU1DwtplhovxIutY8Xa7rP%0AhPA0v9ij/gsh4g%2BGv/BZdPBvg/4U/sf/ABZ/bq%2BKHg745fBPxb4L/aA8PeIvElnb6NY%2BEfAvjz4J%0AXPjTwf8ACvQdW8Ga/wCOfhd4Z1GDRPjJbOl14Y8YeIJ7yyt9H1a%2B1X4hoAfq98Ef%2BChfhn9ob9pb%0AxN/wTy/ab/Zd8S/AT4y%2BN/2UNP8A2mdE%2BGfxS1r4ffEvwP8AGD9nPxtr178M/E2m6lBYSi407xLp%0A2ujV/DnjP4S%2BPvCOnaxNpFprd9JZ3uiWl7IP4wvEH7N/w9tv%2BDW//gpx4l8R%2BAfDV7L8Fv8AgqT4%0A88Sfsw6lcaRpg1L4TaLd/tFfss/AXW7bwfqdrdXGqQWmqaG/jvw7qUWrG3lvbLUEimsLi10rw7ql%0Afr3%2Bz7/wSS/bc/Zw/wCCl3wf/bR/Y5/Ys/Zp/Y2%2BCGgfA6D4IeOvhb4r/aL1T4y/FnxvYabrXhj4%0AkfEPxn8TPENvZapoZ%2BI/xrs11L4I%2BGfG%2BheKfH9/4c17SIvij8QbPU9MW0sdV43X/wDgmF/wVa8Q%0Af8EW/wBrP/gm7N%2BzD8ILb4mftLftdal8YrLxY/7V/he68NeCfAPiH43eDv2i79ktovhjY3Ou6rpf%0AiTwBZfD%2BLSbnULB9Sg8U33jgavp0Gh23ge%2BAP7Tq/g1%2BJXjLwB%2B1j8aP%2BDiX4vftp/8ABOv9p34m%0AWHwx8JeEfgxF8QPh1P8AsVa78Uv2GPgR8LPgx4o1vU4tEk%2BI37VuoxaHrurahpmpfFzxZqH7P1j8%0AXNO1u5m8QHXY4bN7TwlF/dL4L1LxJrXg7wnrHjPwxF4J8Yat4a0LUvFfgyDXbbxTB4R8SX%2Bl2t1r%0AvhiHxNZWlhZ%2BI4tA1SW60qPXbSxsrbV0tF1CC0t4rhIU/l5179jz/gqU2jf8HAdy37Kfwo8Uat/w%0AU80K/wDBXwGsrL9prwJ4YuNH8L3Hwf8AFP7NOmX3iiGD4cWmiRaxofw01TTPiBJBeXlnceKfFFmf%0ACWueJIbvVtV%2BJduAfX/h3/gqp8J/2U/hZ/wSB%2BA3gf8AZp/bX%2BPmhftx/su/CWT9mLxtbj4C3vij%0AxPpnhz4B6H4isfB/xL1Dxh%2B0Do0mnfHgaRefDW%2B8cXvi2%2B0P4RSW/jrWvFen/G3XE8C%2BNrLTPePh%0A9/wWW/Znl/Z9/ax%2BOH7Sug%2BNf2QtV/YX%2BIEPwq/ak%2BGXxRuPCPjPXfBnj/U9K0XVPC%2BieCvEvwf8%0AR%2BPPB3xT/wCE0k16z0XwtH4S1iXWrrXIbu01LQ9JtfsN9ffmr4R/Yw/4KcarqH/Bu7pOvfs3fDrw%0AR4D/AOCafgrw/oX7UVrd/tD%2BCPEGuXGv6T4Ftf2XtN1zwomjeGp4tRi8P/CrwrL8bJtJ0rV7231m%0A8%2BJsPwzN7Hq/gaTxRqngXib/AII6ftv/ALcnww/4LEfCr9oP4Q%2BDP2UU/bh/aX8KftR/s5%2BLrr48%0AeGfiuPDfjT4R6zF4P8OeF/iT4f8Ah94Zv2Xwv498GabceKRrulapcatoNv4ktdNv/DcevaLcWk4B%0A1vx5%2BJFx%2B0P/AMF9/wDggj8cNV/Yn%2BMP7Nuu%2BM/Bf7cFv4e8b/HuL4J6nr/xh%2BDWifsuXnjvwDZz%0A%2BG/hj8bfH154C8S/CjxF46%2BJ/iOXwl43XRfEvhGH4keEvEL2useK77xH8OPBP9d1fy7fE/wn/wAF%0AeP2jf%2BCjf/BJj9qrx7/wTU8P/Dbwh%2Bw/4g/aG8L/ABg1HSf2vvgLrN1rl3%2B0r8IdL%2BFHxC%2BJfgnT%0ABr2qXKfBXQ1sLLxB4P0XU9Mvfjd4njt9R0LWfC3w4lFl4iv/AOlH4leIPGPhXwF4r8R/D/wBcfFT%0Axpo2j3N/4b%2BHVp4l0Lwdd%2BMdTgCtDoVt4o8TyQ%2BHtEuLwblhvtZnt9OjkCi5uII2aZAD%2BHn/AIKX%0AaZ/wTi8X/wDByB458Xftx/DPUdZ/Z6/Zv/YM8Laz8ZdL8KfAj4sfGu1%2BL/7SvifT9Qj%2BHWk/Elfg%0A14R1rW/DcsPwL8RaRqvhXVvGd3D4IsJfgzpyS6rpFxrFzLF%2B0X7Hv7Ff7E37F/hnS/8Ags58BPgd%0Ab/sf%2BDvFn/BMLxv43%2BPH7OPh/wAWaxrXhiLQ9Z0/4P8A7TPhbW7uTU1j0rSvHHw88KeAfGHhHxTf%0AaTpttpeo3HiJZLa3mOm3ep6z85fsk23/AAVJ%2BDnxO/4KM/GH4/8A/BFvx18WPH3/AAUD%2BOsHivX5%0AfBn7ZH7CtrpGi/s4eFPhL4e%2BF/wW%2BBXiO/8AE3xq0zX/ABDrXw38Mv4v0PxDrNuLXwzqd/rF9f8A%0AhTTNKsbyX7T9r/AT9j79sP8AaA/Y2/4KC6R%2B2Z4c8L/s6fGr9t/4G%2BKP2XfhX8AvBXxItviF8Kv2%0AUf2atN%2BBfi74e/BjwDDqnhu0vvDWqeONC8YfFz4reIfiZ428N2eox%2BKUu/Dmn2cU%2Bi%2BFfD9jCAcR%0A/wAEyk8b/s4f8Ekfi5/wUz%2BIvhnUPjJ%2B2J%2B1p8FfiV/wUq%2BN4062hGq/FLWr74Y618RfgV8JvBmm%0A6Je%2BIJdE8H6B8INP8B%2BBPA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iXnhfSdW1CWSHTru6%2By3bQfovX8tHwi8D/HPxj/wckf8A%0ABQHwx4l8Xfs76z4Fm/ZT/Zc8SeJfDXib9mfx540g8V/BHSfHEWq%2BEPBlvqN5%2B0BYeF/BHxj8M620%0AMz/EzWdJ8d%2BHvE0%2Bk2/ibw78GfDEtnf6fo/9D/7TXjj40/Dr4G%2BPPFf7Onwht/jt8b7a30TSfhr8%0AMtQ8Uab4L0LWfEvinxPovhW21jxZ4n1SaGHSPA/giHWp/HXjqezM2uS%2BDvDWuW3hyzv/ABBPpljc%0AAHvFFfzifD39sz/gtP8AFr9rr9p79nX4NaB/wTT%2BLXgr9jL9oH9mX4YfGP4nXXhz45fC7/hMfDPx%0As%2BFemfEv4q2/gDRrX9ob4vvoHjH9nSfUrfwv4ibxJJqaeNbqaWXR/DOj%2BINI1rwZZ/tt%2B1x8cp/2%0AYf2Uf2nP2lbXw3F4yuf2ef2evjR8crfwfcaq%2BhQeK5/hL8N/Evj6Hw3Nrcen6tJo0WuyaAulyaqm%0Al6k%2BnpdNdrp940It5AD6For%2BFf4b/F//AIJ5ftz/AAn%2BGP7Rn/BTP/goJ/wUv8Z/tJ/FLwfoPxD8%0AQfDj9nj4Mf8ABRv4T/sxfAVvGXh7RL1PhZ8DfBPwh/Z78SeCbvSvDuk2uh6N4o%2BJVl4z8by/FrxL%0Ao9147TxPfaZqmmJB/cb4auNMuvDmgXWizahc6Pc6JpVxpNxq0mszarPpk1jBJYTanN4jZvEMuoS2%0ArRPeya6zay9y0jaoxvjOaAPhr4Kf8FJf2fPjTfftnXI0v4o/CX4bfsMfE/x38KPi58fPjl4Nt/ht%0A8BPEPiD4Valruh/FbUfhr8TtQ1u60bxDoHw01/w5qWi%2BMb3WF8NXen3Rs7q30%2B80m%2BtdSl91/Z4/%0Aa6/Zk/ay03xJqv7N3xw%2BHfxktPB02gQ%2BK18EeILXU77w6ni3RYvEng%2B91jSiYtUsNJ8Y%2BHpk13wf%0ArVxZppHivRxJqXh691KzhlmT%2BI/9uTxN4v8AD/8AwTu/aF%2BHgd7DwF%2B0t/wdl/tGfBr4l6TbTxWd%0Av8Rfg9qXxi%2BLHi7WvDevf2JJY3d/pmo%2BNfhnpUF9a%2BIpL%2B5ntNDghXbpkWirZftF%2B034iHwf/wCD%0Amj/glppHgPTLTw9b/tHfsVftN/B74nLos15ott4g8G/Czwr8S/i34BsdW0zSp7TTNai8K%2BIvB8Ee%0AhW2r2l3b6Zb30klpHFcWGlyWYB/SzX5IePf28/2iPBP/AAWG/Z//AOCfl/8ABn4VWn7PHxu/Z6%2BM%0AHxb0b4vxePfEniH4r6zrHw603w9css3g2z8O6XoXw50rRdefXPDN1p2uXPiy18dWWq6J4n0Px34X%0A1fQ9V%2BHmsfrfX8jX7YX/AAUC/ZR8J/8ABbX4M/GLxT%2B3T4X%2BGWpfsh%2BD/jF%2Bzrrv7N/ib/gn/wD8%0AFDviJ8RdatfEmnya98atV8M614M8EaX4H1rUIh4b8Han4S%2BIXhI%2BIPBV38ONRHjaaHxdomk%2BHZvE%0A4B/Rb8ev2zfhn%2Bzl8Xv2ePg/8RvCfxd%2B0ftMePdN%2BGHgP4j%2BH/h3qOsfB7RPH2uDUR4c8KeP/iMt%0Axb6N4R13xRPpstr4a0m78/UtcuZI0061nRLqS2%2BuK/Ij9vzxfonxn8D/APBK/wCJHwt8QvqPg74i%0Af8FF/wBi74m%2BEPESLrXh1td8A654U8f%2BNbS7NnqOmW2uWKeIfCEpSTRNY0vT7uaO/bR9Yi0wzXb2%0A3670AFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABR%0ARRQAUUUUAFFFFABRRRQAUUUUAFFFFABRWVo2vaH4js5tQ8PazpWu2FvquvaDcX2jajZ6pZwa54W1%0AzUfDHifRprmxmnhi1Xw54k0jVvD2vac7reaRrml6jpOoQ29/ZXNvF5Z8NP2iPgt8ZPG/xi%2BHfws%2B%0AIGj%2BPPFP7P8A4o03wN8YYfDcWpajo3gfx1qNjPqTeBNQ8WRWP/CJ3njjRLOBZfGHg7SNb1HxJ4Ga%0A90mHxnpegz61pEV8Ae0UUV5P8Hvjn8Jfj94d1zxT8H/HOj%2BN9J8LeOPF/wAMfF6WAvLLWfBXxJ%2BH%0A%2Bry6F42%2BHvjnwzq9rp3iTwV448K6pF9m1zwn4q0nSNe09ZrS5uLBLS%2Bs55wD1iiiigAoor53%2BK/7%0AXn7J3wG8aeF/hv8AHL9qD9nf4M/ETxxb2N34K8BfFf41/Db4d%2BNPGFpqerTaDpt14X8LeL/Euj65%0A4gt9Q1y3uNGsZtJsbuO71aCbToGkvInhUA%2BiKKK8F/ac%2BKXxG%2BDHwQ8Z/Eb4SfBXxL%2B0N8RdFl8K%0AWHhf4R%2BEr%2Bz0vWPEl34p8aeHfCVzqD6hfJJDa6J4N07Xbzxx4okjhnvf%2BEY8N6wunW9xqDWsEgB7%0A1RX5m/sNft2%2BLf2kP2gP26P2TPif4I8GaX8W/wBgrxx8JvCvjf4lfCHxJqfiT4KfE%2B0%2BOHhPX/H3%0AhRPCQ8Q2Nl4m8L%2BOPAei6Qvhb4weB9Xn16Pwx41jlg03xNqtpciKx/TKgAooooAKKKKACiiigAor%0AyfQ/jl8K/E3xh8d/ATw54tt9d%2BK3wv8AC/hPxf8AEjwzpGm61qNv4D0nx2%2Bof8IZaeLvE1ppsvhP%0Aw/4o8TWel3uuaN4H1TXbXxpfeF0h8WwaA3hm7s9Wn9YoAKK8a%2BGXx/8AhV8XPF/xj%2BHngvxL9o%2BI%0AP7P/AI1tfAfxe8C6tp2paB4s8Gatq%2BjWnibwnql5oms2tleXvg/x74VvrTxJ4B8b6Ul/4T8Xaab%2B%0ALR9XuNU0LxFp2key0AFFFFABX8rn/BtXpeq2Hxi/4LovqOn6raBf%2BCrPxr0uSTVbLS7Sf%2B1dL8U%2B%0ANjqmn3X/AAj2heGfDKarp/22zbVLLw/4f0HS7J7uD7DoWj6fPZWUf9Udfyef8GxWm%2BHPE3j/AP4L%0AL/GCTwx4ag8ZeIv%2BCnPx201fElloFgNVtfDmqa1N4on8Mad4q0XzPCk3hr%2B1bqC/XQPB0kdhBc29%0ArqN79q0258JtagH9YdeAftQ/Df4wfF74G%2BO/hr8CvjF4Z%2BAvxE8YWFto1l8UPFvwih%2BOekaDo1zf%0AWw8T2p%2BG9346%2BHVlrN1rvh4aloNtdXviRLfSH1I6oLC/ntYYD7/WVr2vaH4W0PWfE/ifWdK8OeG/%0ADmlajr3iHxDr2o2ekaHoWh6RZzahq2s6zq2oTW9hpelaXYW9xfajqN9cQWdlZwTXNzNFDE7qAfk7%0A/wAEg/8AgnF8fv8Agl58BbP9lrxn%2B2P4X/af%2BAngu31hvg54ch/Zgtvgh4u%2BG%2Bo%2BKvHfiz4heMmu%0A/Gth8bviQfG%2Bj%2BIfEPjLWL1dO13Q49T0if7NHpviCHSoTpD/AK715/8ADD4s/Cv42%2BDdL%2BI3wY%2BJ%0Afw/%2BLvw91zzv7F8d/DDxl4c8feDdX%2BzyGK4/svxR4U1LVtE1DyJQY5vsl9N5UgKPtYYr0CgAoooo%0AAK/KP/gq1/wTDuv%2BCofgz9nvwBeftJ%2BLf2ffDXwF%2BOVh%2B0LEng/4deCvHGpeJ/iR4U8O6voPw41Y%0A33jB/K0ZPBR8SeJrg6ebPV9E8SRa5Nba5pE8thpF5Yfq5XBeN/ir8L/hnd%2BCNP8AiR8SPAXw/v8A%0A4meM9N%2BHPw3svG/jDw94Uu/iD8QtZgu7rR/Angi217UbCbxX4z1W10%2B/udN8L6Cl/rd9BZXc1rYy%0Ax20zIAflR8Kv%2BCWnxm%2BGf/BSj4rf8FFpf2%2BviB4hvvjPongTwT8QfgjN8C/hPaeENe%2BGfgDw3/Y%2B%0Ah%2BAT4muTq/iDw9p%2Bn62kXi/T9V8Bx%2BDdafWRdL4kvPFFvqWpJd/cX7bf7OHxA/au/Z28X/Bb4X/t%0APfGL9j/xtr974e1HRfjl8DLyGy8c%2BH5dB1m01WXTi/2jTdRn0LW4rd7DWrLRfEPhnUry2cW7a0NO%0Ak1DTtRwv2mv%2BCiX7Ef7G3jDwB8Pf2nP2k/hr8IfHvxR%2ByP4D8FeItSu7nxX4gtb/AMQWfhWy1OLw%0A/oljquq2WiXfiC9Gm2%2Bu6na2WjSy2OtyJfGDw9rsunfZ9AH4t/suf8Ek/iL%2Bzt8TPEXiXxJ/wUI/%0AaK%2BMvw28dftS3X7bnxR%2BHOs%2BG/A3w51v4vftOax8PfCnhXV9X%2BI/xH%2BFUPhTUbv4K6b4s8F%2BG/HX%0Agf8AZ38JaF4O%2BHOiReHPDXgXxfB8QvBukzadqH69eO/A3hD4n%2BB/GXw1%2BIXhzSvGHgH4h%2BFPEPgb%0Axx4S161S%2B0PxT4Q8W6ReaB4l8OazYygx3mla3ouoXumajayApcWd1NC42ua6qigD%2Be7wT%2Bx9/wAF%0Atv2FfD3h79n39gj9pD9hL9pD9knwRcQaV8JND/4KL%2BHvjxofx6%2BCfwo0fTdC0zwv8FND%2BJn7Ntqd%0AA%2BI3hfwza2mqWnhrxL438IW/iHw7pn9laHbx3/hvT9J0LQ/6BrD7d9hsv7U%2Byf2l9kt/7R%2Bwed9h%0A%2B3eSn2v7F9o/0j7J9o8z7P5/77ydnm/Puq3RQB/LD%2Byn%2Bzj%2BzX/wUn/Zp/4KufsR678XtP8ADvxk%0A%2BH//AAWU/bz%2BL%2Bm3fhTWdG1r4tfs6fE3Tf2pfFfiT4IfGW28Ba3cvL/wjV5dafeWAWS00/w1460s%0A/EDwpY67p/iH%2B3NR0j9JP2bv%2BCc3xrtv2urf9vv9uz9ovwV%2B0L%2B034O/Z/f9mn4Iaf8ACD4OT/CH%0A4Q/BbwbdeINY1Hxh8SdG8OeLvHXxUvtZ%2BOHxWguLObxZ4ttZPCmjaHpup%2BIfh3p/h7W/B8WgPpv3%0AVD46/Zi%2BFH7QGjfBDSF%2BH/g79oj9ofw/41%2BK6%2BFPCfg6O18YeP8Aw38OtQtP%2BEs%2BIHjnU/DGhNFb%0A6fp2t%2BOY7Wy8Q%2BPtTsE1zxBr2paf4fudU1dtVt4/mXxt/wAFBdZ8A/8ABTj4H/8ABO7xB8BZrfRv%0A2gPg/wDFL4p%2BAfj2Pi74Mc3k3wp0XR9Y1zRZPg%2Blg/imHT5ZL7VdIh1y71y0v59R0ae60rwzq3h6%0ADXNd8PgHOf8ABIrR/wBrPQv2b/iRpP7WvxJ%2BLvxe1DTf2ovjzpHwG%2BJP7QHw5ufhN8b/ABp%2Bzh4f%0A8RWfh34f%2BJviR4A1C5OraFe%2BINb0jxlrnhL%2B2dF8H6tqXw31HwVq174P0GW/%2BzN%2BVf8AwWA8W/tP%0AfBb9o/Q/it8JP2hv2rfCfhq6/ab/AGGdA8YfGvw58TtD0z9iT9hz4NeNPEmjfC/xt8LPjZ%2Bzx4X%2B%0AMd/4n%2BMfjr40%2BK/Edh8S9T8cfET4AWNpoWieNfhZ4V8H/FHQ9Nt/Ehu/6oa/Fv8A4KDfsKfsjeFd%0AL/aU/wCCgnj34W/tFfGbV/DWn/D/APaY%2BJv7MHwp%2BK/xStvh3%2B0v8XP2TPB01h8CNX8Z/CfwxeXd%0AjrWq%2BGLbSfB9pcIlkfA81n4G8M%2BIvH3hDxRB4ZvBeAHcf8FMviv8N9G%2BK/8AwTT%2BEl5468GL8XPF%0AH/BQ34D%2BLfDvwpfxh4VsviTrvgrSvD3xU0bxF430bwVqetWGv6t4S8LXWpW3/CQ63ptjdWtizpZo%0AZ9UubLT7r0b4SeHf2qLX/gqN%2B2DrniD4m/EfxJ%2Bxrdfs4/sz3Hw48CeMPCs/h/wJ4H%2BPWsal8QdK%0A%2BIPh74Oa03hexg8eabY%2BDvAfhTx58RfEKeJNVGkeK/jHY%2BEEaeTQbrTvDHk3wP8A2qfh9%2B0h/wAF%0ACfGP7Kn7Sf7HPwv8M/tefsmfBrwZ%2B0f8Mvi3oDXXx%2B8LeGPAvxZx4Yu4vCPxh%2BI/7P3wO8ffB/4l%0ASLrqadqPg1vCelXviPQrnxG9jqeoQaB4ks7L9Ovht8Wvh38X7HxXqXw48T2nie08DfEXx78JfFrW%0A9tqNlNoPxE%2BGPiO98KeN/DF/aapZ2N2l3o2t2E8KXKQPp2q2L2WtaLeajouo6fqF0Aei0UUUAFFF%0AFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUU%0AAFFFFABRRRQAV%2BTH/Bcj48fGn9nL/gl3%2B1D8SP2bfibo/wAKP2hl0fwP4Z%2BD3iG/k019d1XxL4p%2B%0AJXg/SNa8HfDjTNQ07V5Nd%2BLHi3wHP4w0b4Z6RpOl3muS%2BMLnSrvSjZXVmup2X6z1%2BRv/AAVm/ZK%2B%0APXx58Ofs2ftC/szfGb4O/Cf4z/sFfGLVv2mvClh%2B0f4Z8TeJv2e/HENh4A8ReGfEGlfE8eC5X8W%2B%0AHLLT/Durard2Xi3wxpWr%2BItEhl1MaAmk6zd2PiHSAD8%2Bv2SvD/7N/wCxv/wTy/bF/wCCiHwR/ZC/%0Aaj%2BFH7bPwj/Yu%2BId98cPi7%2B238LP2hfCvxs/aP8AjR4K%2BDsHxZ8UeNfGVt8WbibW/HmjeMPi3odj%0A4g8dfEbSPDmk%2BHFv31qVLnStJ0rVbfTPuP8AZEg8T/sE/wDBJD9lOb4cfAHxr%2B0N%2B0P42%2BGn7P8A%0Aq3ijwH4NutY1PXfi9%2B1f%2B043g%2B%2B%2BKnxR%2BMXxft/DXi6bwv4KHxF8aeIvH3xh%2BP8A4807Wrbwd8Ot%0AD1HV7lNevrDRPD2qeZ/shftlr/wXA/Yb/bv%2BA3iDwT4Y%2BFfipfAPxC/ZB%2BInjn4Q/F/wz8b/AIIe%0AKPEHxo%2BBd7aXXjT4M/FPw9pNzZ6jo1jpXjSx1K60fVdH1LWfB19dw6BrkWrajY3Xn8t8AvEX7en7%0AR/8AwTS/Z%2B%2BD/wCyJ8fPCv7J/wDwUB/YQ%2BIHwo/Zw/bN8L/tE/DPSvidofi3xV%2Bzn4DTwV8S/h14%0AtaPStUntPh/%2B0FY3vgP9oH4efGD4SyS6j4k%2BHWr%2BH7Dwr4u8P3PiLW9U8PgGN8Af%2BCof/BS/4wft%0AMR/B2y/YE%2BDnxG%2BDvwd/bS8TfsW/tYftEfBr49a5qGjeEvFuh%2BFfDXizxN4w8H%2BC/Engy11fR/CX%0Awg0zxf4dHi%2B58aajJqfj/XtV/wCEQ8L6T4Y1PSddvNN9v8K3118Af%2BC9Xj34Z%2BFruax%2BGH7d37Al%0At%2B0N418H2aXE2mx/tM/sz/FbRPhLN8ShZ%2BUdP0O68c/BHxr4a8L%2BJNVsGtbvxHffDfwvH4jbVZbT%0Aw42lZ/8AwT0/4Jwftofsl%2BOZrf4s/teeBfG/wn/4aU/aQ/a/8ZW/wm8A%2BLfAfi79pL43ftOaHqel%0A6zpnxos/Ffi/xtpvh/4U/CzUfEniTxZ4U8I%2BFtc1y98beOYfh54v8R6voVz8Ojp3ivofgEkn7VX/%0AAAWA%2BP37XvgvUNM1f9nv9kT9l1/%2BCfng3xRZJaappHxG/aF8afFbw/8AG79ovVvCHiK3ieC80z4O%0AWfhP4cfB/X2s7qaGz%2BIEnxA8PG4a/wBH1qysQD9Ev2sPjb8RP2efgl4j%2BKXwt/Zu%2BKH7V3jLRdQ8%0AO2Vh8GvhDf8AhLTvGWsW2ta5Y6Vf6tBceL9Y0m0On6BaXUupX8emxaxqrRxIy6ZHpq6lq%2BlfmnZ/%0A8Faf2pJfClzqt5/wRG/4KZW3iW313QbAaFb2H7OlzpMmj6jY2txq2pprFx8bdP1%2BW90y6g8RW1pY%0A2vgm50m7isfD8ur%2BJPDd34oTTtL/AE0/an/Zi%2BG/7YPwW8TfAf4sal8SNJ8F%2BKbjSL29vvhV8T/H%0AHwl8XW97oOpW%2Br6VPbeKPAetaLqM9vbaja293Jo%2BqtqWgXs9vbS6hpN3JaWrwfAfiD/gh1%2BxTq/w%0Av0n4T6R4y/bk8F%2BHdJ1BdWifwz/wUU/bcnjn1WfRtR0HWtTk8KeLPjp4o%2BG9pqHiXTNX1S11y90n%0AwPptzNa6jfafZyWOmXt1YzAH35%2Byx8a/HP7QfwW8M/FH4jfs6/Fj9ljxZrdxq9rqPwd%2BNE3g6bxv%0AoraRqVxpo1CR/BXiPxHZHR9Wa3e60WXUzoms3dj5d9caDZ2d1YXF5%2BAX/BSj9ijwH8CvgN/wWI%2BN%0Afx68b2H7Wvxh/wCClNhpXwe/Yc%2BBniP4beH/ABB8Rfh98V9V%2BF2tfDH4J/C39nez1K58U%2BOfFPi/%0AT/HPiWL4hapD4Gj02x8M%2BEfhkfiKNA0BNH8ba%2BP6JfgH8FvD/wCzv8IPBHwX8K%2BKvih418P%2BA9Pu%0A9O0zxP8AGb4meMPjB8StUhvdW1DWHbxJ8Q/Hmqax4m1z7JPqUtjo9td332Dw/oFrpfhrQbPTPD%2Bj%0A6Xptp%2BC37Y3/AASp/wCCqfxz/wCCh%2Bqft1fs9f8ABSf4S/s8f8Id4K1D4Vfs5%2BD9f/Zg8KfHOf4P%0AfDzxTovg1PiLcaN/wta18S%2BHND8d/ETxF4auZ/EnjHwzolh4ik8M3kvhG31yLwzqmsaJdAHZa946%0A/b//AOCZ/wDwS3/Z4%2BG3gux%2BEXjj4t/shfsEeJPjd%2B1R8Zv2kdO%2BLviD4O%2BH9B/Z78AaNdSfADwX%0Arfw4vdPm8afFjxfeXmp%2BEvAGup4ze30Twd8KdU8XeIPAt7L4v8F%2BHrzf/ZW/ah/4LF/t8fs8/HbX%0A7v8AZs/Z%2B/Yes/i7%2Bzh8CPHn7D/7QN/8R9U%2BMNhLqPx18Iab428U6r4j8FWf9h%2BL5Nf%2BHfgDxho7%0AaQ2teD/CGgaJ8XNDutGuoPih4bbU30fzv9qj/gjD%2B25%2B0N%2BzT4H/AGQNF/4Ki6x4V%2BAWofsW%2BEP2%0AfP2hPD3ib4HXXjfxF8Uvj14Y%2BJtl8X/EX7R%2Bi%2BMY/i14b8SaC/xW8SafYeB/FPgDV9S8Q6DoPwjj%0A8ReFLH%2B3b7xidX8M/rl%2Bxr8APjX8C/DPjM/Hf432Hxb8U%2BLNU8IweG/DPgfwxrHgH4LfBX4eeAPh%0A74Y8A%2BF/hz8KvA2seK/F91aQ3M2hav418X%2BKLvUYNR8TeJfFNxbPYWOjaBoNnbgHzf8A8Epv2Hv2%0AgP2H/hv4/wDCHx1%2BKHwa8YyeJdV0O60Pw38B/APi3wr4WbWLSfxRrXjv42fErxb8TvFXjr4s/Fn9%0Aor426/4ujt/ir8QfHHjHV21PQvhv8OYbBLa8h1qS7/VqvN/jJNqdt8IfircaJ8Rbf4QazB8N/HM2%0AkfFq78Pad4utPhdqcXhjVHsPiLc%2BE9XePSfE9v4Juli8SzeHtUkj07Wo9MbTb10trmVh8i/8Er/E%0AH7Qfir/gn7%2BzN4j/AGofGd98R/jDrHgzV72%2B%2BImseDdS%2BHniTx/4Bl8aeJ/%2BFK%2BN/GXgbWI49U8K%0AePfFvwUHw98QePtD1Dzbuw8Z6hrsE9zdyK11MAfoFRRRQAUUUUAflZ/wV0/4KDfFD/gnL%2BzXB8a/%0AhP8As66j%2B0BrKarr9/4ol1O98S%2BF/hX8M/h34A8EeI/iJ4v8T/En4g6D4U8VWvhe/wDFUPhyy%2BFv%0Awo07WI9Mt/FfxW8d%2BFdNhvbpoJdI1LzD4D/8FEv2tNf/AGuf2XP2fP2nP2ZfgZ8E9P8A20fgb8Wv%0Ajv8AC/wZ4J/aG8TfE346/BDw18H/AA78MdSuNP8A2h9E1D4O%2BB/CUmu%2BPtT%2BIV7p%2BjWvgXU7nQ/D%0AV54Q8TaQfE/jK70m4uj9gf8ABQz9kG7/AG7P2c1/ZouPF%2BmeEvAXi/4xfAHxP8Y49R8PW/iKXxl8%0AIPhX8ZfBfxV8a/DrS4L3zdMsNT8cW/gy10CPUtY07XNEitbq8tdZ0HWNMu7qxm%2BAv2bfGsH7Rf8A%0AwX4/4KEeONJe6ufDX7Cf7GH7NH7Ftrfxaml/4av/ABv8dvGXib9pz4j3OhLbalfafDrenHQvB/gf%0Ax3HHBp%2Br2eq%2BBNN0nXbRf7N0uacA2P8Agkd8U7Zv%2BCW2u/8ABQI/Dfxn8T/i7%2B1R4j/aa/bM%2BJfh%0Ar4c%2BHm1z4t/FzxXq3xN%2BIFn8NPhl4a0eTVpbC51zwx8K/Cfwz%2BBfgSwTUtD8JWtl4U0vUb1tBtbr%0AWtRXwXS/%2BCvf/BRex/a88YfsW3P/AATJ8GfHH4kfBLxn%2BzPJ%2B0l4x/Zn/aZvvEngX4TfDr9qqz0m%0ATwZa3q%2BLfhBo%2BrWfxC8Gx6tqnirxCfF6%2BE/B2t/D74e%2BJfF51bwfoOuaJq1r6H%2BxT8O/jb4N/wCC%0Ac3xu/wCCZ/7H3xs8N/s9ftr/ALCXxL8TfBfRPHnxB%2BH3/Ce%2BH/DXgbxD8b9U%2BM3wR%2BIU3hLxRZ3s%0APinw38bP2X9estPtfEDWOpWeh%2BPLvxfbWkOqXngaX7Tn/sdf8Euv27PgN%2B038fvi18TP2qfg3feD%0Af2sv2qfBH7YPx/1X4HeC/HPw1%2BKmoat8O7TXLrwj%2By54Vl1DVda0ZP2foPEF34bsPFviXxLqmteM%0AvH/w10XxP4I1PQNL1j4jax420oA9U/al0jVPg/8A8Fvf%2BCWfxi8A6Xp9n/w1z8H/ANsz9jb9ojVp%0AJrdrvXPCHwl%2BFzftY/Ai2tdPntZ2iu9B8Z%2BDfiL9q1rT5rG%2Bk0/WbfSr%2B6udPS1sq/Xr4q%2BJvGvg%0Az4ceNPFfw5%2BGuofGPx14f8P3%2Bq%2BFvhXpPifwx4L1Xx9q9pEZbXwxpnirxrfaZ4R0LUNVKm3stQ8S%0A6npmiQ3LR/2jqNjatJdw/kh4r8Pan%2B1x/wAFrvgb4t0Gwluvgt/wSo%2BCvx2i8R%2BPLe28QHRdQ/bH%0A/a38I/D/AMNS/CGLUrPWbbwzqfiD4bfs1X9n458QW99YX914Vg%2BLei2TafeX/iSLUvBf6z/GD4Se%0AAPj18LfHvwX%2BKuiXHiP4b/E7wvq3g3xroVpr3iPwtdar4d1u2ez1KztvEnhDV9B8U6FcSwOwh1Xw%0A/rWlavZSbZ7G%2BtrhElUA/Mj/AIbs/wCCmP8A0g5%2BP/8A4m7/AME9/wD5/Vfe/wCzL8Uvjd8Xvh5e%0AeKvj7%2BzB4o/ZJ8aQ%2BKNS0iy%2BF/i/4o/Cb4t6ze%2BHbTT9IuLHxc/iX4M%2BJfFng61t9Wv7zVNOg0Vt%0AZl1e1XRjeX0NvHqFrGPyZ/4hh/8Agh5/av8Abv8AwxZqH9t/2h/a39sf8NTftmf2r/av2n7Z/af9%0Aof8ADQ32v%2B0Ptf8ApX2zzvtP2n9/5nm/NX6k/si/sW/s1/sJfDHVvg7%2Byx8PLv4bfDzXfHfiH4ma%0Azo1/49%2BJPxIvtT8deLLfSrXxF4ivfE/xV8YeN/Fc93qsWiaabiF9bNkJoHuYraO5uruWcA%2BNf%2BCv%0An/BUDQv%2BCbnwS8I2XhLwtqHxQ/a7/ae8QXHwg/Y0%2BDFjaXiW/j/4ualqHhnw5Df67r72E%2BgaR4f8%0AHah448Napfabq1/p994tubqw8NaPJbC/1HXdB0/%2BCLH7B/iD/gnn%2BwB8L/g18S47Kf8AaH8c6x4q%0A%2BPP7UniC01I6xN4k%2BPnxY1BNX8SvqmrRahqOl6tqfg/w3aeEfhjPrWgzJouvReBLfXbRZ31Ke9u/%0Ap39u/wDYm%2BCn/BQr9lz4p/ss/HbQNP1Twv8AEHw/qMXh3xHNplvqOvfC74gx6Zf23gz4r%2BCZJpLd%0A7Hxh4G1O9/tPTmjure11eyOpeF9eS/8AC%2Bv67pV/%2BcX/AAbq/tW/Ev8Aac/4J3R%2BFPjTreseM/jD%0A%2Bxp8d/iv%2BxP8RPiZrOoz6rN8UtQ%2BDJ8O6x4X8axX1%2B8mt3jt8O/Hfgzw5q2peI5Zdf13xH4d1rxB%0AqMjSasuAD93K/M//AIKc/sIeJ/2//An7O/w4t/jD4e%2BG/wAL/hd%2B1T8I/j58dfA3jH4a6F8UfBfx%0A8%2BGXwvvL/Vb/AOEni/w34mu08MXmh6pfy2eqPZ%2BLNI8T%2BE5b/TbC%2B1nw9fyaXYmL9MK/OX/gqd%2Bz%0A7%2B2V%2B1P%2ByP4o%2BAf7Evxu%2BHX7P3xA%2BJet2fhr4j/EH4hWXiudz8EdQ0TxFb%2BOvDHgnUvB0F1rHhrx%0Ah4rvpfDmiP4lggW70zwhdeLjoN/ofiqbQNf0sA/HP/gkB8OvB%2Bo/8Fiv%2BCm/x1/YO8F6F8K/%2BCXE%0Afwy%2BG3wRsbD4VaHZ%2BG/2bvjr%2B1n4NvfDEOv/ABL%2BA%2BneHLaL4eXej/DLTNC%2BJPgrxHqPw7%2Bz6Be6%0Aj41s9feO8uPE4vW/qmr8aP8Agmz%2Bxl/wUK%2BAXxI1fxl%2B2h%2B0r8CPFnw88I/A3QvgP%2Bz7%2BzR%2ByJ8P%0AfEvwV/Z3%2BHOiweKo/E2teLrj4YWreGvAZ8Ww2ek%2BHvB/hi%2Bj8GajfeG/CcGo6R4Y1fwxpur%2BJtN8%0AUfsvQAUUUUAFfhl/wWe%2BLurfBb4if8Ejtem/Z2/Zz%2BP3gv4gf8FS/wBnP4H31x8YtK8R6x8UPhP8%0AQviTqFxc/Dj4l/s8R2Os%2BGvCWieOfD8PhXxheyeLPGmtXNnpGq2vhPT7fw3qVnrms634X/c2vyP/%0AAOCtP7A37RH7fHh/9kTTv2f/AI7/AAt%2BA2tfsq/tcfDv9snTdf8AiH8LfEHxMutU%2BKPwXstXg%2BFF%0ArZW2meNvDGlW/hiyu/FXia98Y6Rq2latc%2BIpF8OR6dqugW2m6vba%2BAfkB4Q%2BGnx8%2BPv/AAchft5%2B%0ACfjD8Gv2IfjD8I7X9m/9mzw/4y0H4haz8QfFWr%2BGf2XdM%2BJmmfFD4Xv4X0rV/hpqui3/AMa7/wCJ%0APhTwh8TPGHgjUZvCvw50TWtN8MapovibxLq3h%2Bx8QX/9N/7U/wAZ/FX7Pf7PvxR%2BMHgP4LfED9ov%0Ax74P8Pxv4C%2BB3ww0/Ub3xf8AE7x3rmp6f4a8GeFIbrS9F8RT%2BGPD9/4m1nSj42%2BIF1oWq6P8NfBM%0AfiH4g%2BILKfQvDOorX5qfCP8A4J8/tjfDL/gq18ef%2BChF38d/2Y9d%2BHv7RfhHwJ8KPHvwts/gV8Rt%0AB8a2Pwv%2BGGlfZfBk/hfxnF8Yr7SoPiFJqVvp954m13xVofjDStR0uKbQdH0nw/bJpVzpX29%2B3t8F%0AP2ov2gP2dNc%2BHP7Hv7Ub/sf/ABsuvFHgvWtJ%2BMaeBdN8fi30bw94gtNV1vw1LpGpTQpbW%2Bv29ukN%0A1eQLcNcW8Euh3trPpGs6mlAH40aX/wAFQP8Agsbc/tM/FL9mvwx/wT4/Zq/aFu/2Xv2hf2ZfhR%2B0%0A38QPgd8cfG%2Bj%2BHrTQv2nPhp4U%2BKyXHw50z4l%2BHNGu5bX4F6Le6v4f%2BJ/xA8R38f27U9f8Fa9F8LP%0AC2iW3iiKy/eT9q79oXwr%2ByX%2BzJ8f/wBp3xrbf2j4Z%2BAXwf8AiF8WdT0RNV07RLzxN/wg3hfUtfsv%0ACGkalq0kenweIPGGoWVp4X8OR3Bb7Xr2r6dZxRTz3EcMn5a/sV/8E0f20v2cfiP4w1P4oftveGfG%0A/gH4l/tzfFf9vz4y/wDCq/hn46%2BGHjP4zfEH4n/C3wr4K074Ka5Jr/xc%2BIdr4Q%2BAHw68RaFZavo/%0AhjR9V1mfxL4e8EeCvCmsR2OkXOrWlv8AsV8Wfhh4N%2BNvws%2BJfwY%2BI2l/238Pfi78P/GXww8d6L50%0Alv8A2v4N8feHNS8KeKNL%2B0RESwf2homrX1p50ZEkXm70IZRQB/JX%2BzV%2B058FP2zPh78N/jd%2B25/w%0AcYzfAL47/HKHw/8AFOw/Zf8A2P8A9rD9m/8AZq%2BFH7O3hXxnpEPjDQf2dfES%2BLvBXifxd4n8beGt%0AF1vSvDnxB1zx34ktvGun%2BOtM1LwPpmpzappGo6t4h/sE0k27aVpjWeotq9o2n2RtdWe5hvX1S3Nt%0AGYNRa8tlW3u2vYtty1zAqwzmUyxKEdRX5B/spfBf/gp9%2Bw98DvAn7KXhPwx%2Bxl%2B138J/gZoVj8Ov%0Agl8W/if%2B0r8a/wBln4sf8Kg8MK%2BlfDzwh8SfAXhD9jb9qPwbrHiLwV4TttH8ML4v8MeM/DthrWna%0AXZmbwhY30NzqGpfsLatcvbW73sMFveNBC13Ba3El5bQXLRqZ4be7ltbGW6gjlLpDcS2VnJPGqyva%0A27MYkAPxA/4JH%2BOb/wDaO/aS/wCCx37T3jcXlx498Mf8FHfiN%2BwD4VW%2B1e91i08J/AP9iLwt4U03%0AwD4Y8KxXRisvDmjeIPG/xT%2BJ/wAUNe0bR9Ps47rxj461jUNVvNcvvK1AfHf7TP7Ufhqb/gtZ8Afj%0A/cftzf8ABJTwf%2Bz7%2Bxh4W%2BMv7N3jn4W/Gf8A4KKWHwv/AGhdK1/4rXPgax%2BO/iq4%2BFusfCK40Lwz%0A8QvCGo%2BH9P0Xw94CPizW9D8dQfC97Lxd8UPhfq/iOPTvBP3J/wAE3vhfP%2Bx7%2B2n/AMFSv2Utdv1v%0ALb9oL9pjV/8AgqX8GNbv1EWqeMPBf7V9rpfhH4yaHCttjTtvwR%2BNPwwuPCtzYRb9Xs/C/jP4beI9%0AedT450uKP8%2Bf%2BCpn7Sf/AAUK/Y6/bS%2BGHjTSf2g/jH4a%2BHHxh/aZ/Zm%2BHPwQmm%2BEnw2t/wDgl98L%0Afgv4/wBb0XwN8TfCP7cHxH1DQNS%2BNlv8avFPiux1/wAQ6B4r8KeI9G0Hwr4Q1PSbvTvE3hn7Vdab%0AoIB/UZ4T8WeFfHvhXwz468C%2BJvD/AI08E%2BNPD%2BjeLPB3jHwnrOneI/Cvizwr4j0621jw94m8M%2BId%0AHubzSNe8P67pF5Z6po2s6XeXWnapp11bX1jcz208UreJftcfF8fAn9m/4vfEu1%2BJ37Pvwd8SaL4M%0A1Wy8AfEP9qn4iW3wq/Z90D4ma9AdC%2BHD/FLxvPDcSaX4XuvGmoaJa3ttp9vPq%2BriZdI0iP8AtG%2Bt%0AmX4//wCCmXxe/af%2BFXiH9gGx/Za%2BKHhzw34l%2BKH7ePwS%2BGvxA%2BDV/wCA9P8AGHiT9ob4M6xfXWo/%0AGfwzo2uXeieKb34Z%2BG/hr8ItJ8e/Gb4gfEXRPDr6v4c8O%2BA/tFtrNipk0fxH9%2BfHHw54/wDGPwU%2B%0AMHhH4Ua54Y8MfFLxV8LfiB4c%2BG3iXxtoEHivwZ4e8f654T1fTPB2ueLvC9zb3lt4k8MaT4iutOv9%0Af0C4tLqDWNKt7vTpbeeO5aJgD%2Bb3/ghpq/iPQP2i/jhfftA/t4/8Ea/2m/j/APHrwf4fvtc1/wDY%0AV/aTv/jv%2B1T%2B0b448F2kMGqfED436r8RNQOvad4a%2BHXgPQbPSPAXwv8AgVongv4GeBdH1PXhofwx%0A8H6fp2mA/buj694y%2BDP/AAcAeKfht4ea4/4Ut%2B2n/wAE54fjt438O6TZa3qrQftN/s2/Fvwv8HY/%0Ain4j8mFNC8J6RrnwD17wD8M59fnkubnxDrHg/wADeHbp9Oez0KLV/nz/AIIw/H748eN/2tv29v2e%0A/iv8Vv2hfFfgf9nnT/hDB8KvCX/BQD4Q/Df4Yft43l54sh1y1%2BJ/xamvvhv4V8G2vjL9nDx14h8K%0AaRb/AA08Q68njLXX0DTPA%2Bjy6r4a0zQtKtPEn1H8HdOj/aQ/4LRftHftH%2BHb6DVPhR%2BxB%2ByZ4a/Y%0AQ0vV7bS7HU9B8RftKfF74k237QH7QNloXjCKV1/t34J%2BB/CvwP8AA/i7R7JJ/wCxfEXxC8S%2BHNRv%0ArTXNK8Q6HZAH7LUV%2BYH/AATI1D9qq/t/22Yf2jvjZqvx88EaF%2B3V8afD37L3j/xB8JtM%2BF2qXfwY%0AtNM8HalrWj6Q2lapcx%2BKvh/4C%2BM2r/FH4TeB9d1Cxe%2Bvbf4c6nq%2Bk%2BINQ8A6z4G8OeDf0/oAKKKK%0AACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooA%0AKKKKACiiigAooooAx9B8PaB4V0yLRPDGh6P4c0aC41C7g0jQdMstH0yG71bUbrV9VuYrDT4Le1ju%0ANT1a%2BvtU1CZIhJe6jeXV7ctJc3E0rlp4e0Cw1nWPEdjoej2XiHxDb6Vaa/r1ppllb6zrlpoS3qaH%0AbaxqkMCX2p2%2BjLqWorpUN7PPHpy396LRYRdTiTYooAKx/D/h7QPCejad4c8LaHo/hrw9pFutppOg%0A%2BH9MstG0bS7RWZ1ttO0vToLaxsrdWd2WG2gijDMxC5Yk7FFABRRRQAUUUUAFFFFABRRRQAUUUUAF%0AFFFABWfZaTpWnXOrXmn6Zp9hd6/qEera7dWVlbWtzrWqw6VpmhQ6nq08Eccuo6hFomi6No0d7eNN%0Acx6VpOmaesgtLC1hi0KKAOXXwP4LTxpN8SU8IeF0%2BIlz4XtvA9x4%2BXQNJXxpP4LstWu9es/CE3ik%0AWg1yXwvaa5f3%2BtW2gPfNpMGrXt3qMVol5czTP1FFFAFKx0zTdMF0um6fZaet7e3Wp3q2NpBaC71K%0A9k8291C6EEcYuL27l/eXV3LvnuJPnlkduau0UUAFFFFABX8yn/BuFc3Xg/xT/wAFrf2fmdNU0/4S%0A/wDBYb9p7VrPxRNfy3%2Bra9L4qudP8JzHVHe91HFxDafCvTb6Yz391qS6jq%2BpWuqSG6tDX9Ndfy%2Bf%0A8G80jn9q7/g4RiMEqxp/wV5%2BP8i3TNB5Ezy/Ej4qq8EarM1yJbZYY5Jmlt4oGS6gFvNPIt0lsAf1%0AB0UV%2BGf/AAXn8e%2BJ9J%2BB/wCxF8CNE1W90Pwx%2B3D/AMFQv2O/2JvjDqOkXMmna6/wP%2BOF/wCOB8VN%0AA0nV7PyNX0tPG/hbw1d%2BAvEUmi6lo9/qXg7xR4k0GbUf7L1fUbK8AP3HhuILlGkt5op41luLdnhk%0ASVFntZ5LW6hZkZlEttcwzW9xGTvhnikhkVZEZRLX853wR8NfD79h7/gvzdfsg/s26ToXwp/Zu/ag%0A/wCCbFz%2B0J4i/Zp%2BHlna%2BGvhh4G%2BPPw9%2BP8AqfhOw%2BLPhL4f6Y1t4X8DW/jjwCmu%2BGNYsPCOh6PB%0Ar2saJFqWpNdtp1tHZf0Y0AFFFFABWfqOraVpCWsmranp%2Blx32oWWk2L6je21kl5qupTra6dplq1z%0AJEtxqGoXLpb2VlCXubqdligjkkYKdCvxI/4Lk%2BMPgH4H%2BDX7Jvib9pr9k/U/2pfhXoH7dv7O2r6X%0A/ZXxivvhXP8ACH4uT63feFPht8RLjSdO/e/FHTFHizxJoeq%2BAdUEnhzUNO1K5n1VYLqLStRsAD9Z%0AviR8cfgp8G5dAh%2BL3xg%2BFvwqm8VyX8XhaL4kfEDwn4Gl8Sy6U2npqkegR%2BJ9X0t9Zk019X0pb9NO%0AFy1m2p6etwIze2wl9Rr%2BN742/BzxV%2B3X/wAHGf7RXwl/aj/Yg8G/tGfs0%2BBf2CfhX8B7bwdrXxg8%0AA29x4F%2BF/wAQf2kfCfxAsP2p4jBqeleJPDPiy21TRPiQg8KeDtS0/wCL58ITaBaLfLputpYN/V/%2B%0A0J8b/Cn7NnwR%2BKPx68b6T4z8QeFvhR4M1nxnq/h74d%2BF7/xp488QxaTbNLb%2BH/B/hfTtk%2BseI9cv%0ADbaVpNvPcWGmpe3cVxrGq6RpEN9qdoAexUV/Pj4i/wCC6vxE8N/Gbxz8Ao/%2BCTX7fPxB%2BJ/wd8Sf%0ABb/henh34IWPwr%2BNt38JPh1%2B0FoP/CTfC3xN4jtfh5411fUV8Y6toun%2BJ73VvBFxY2Gg%2BHotBD6r%0A8RoUvXay/eHx/wCOPDXwx8CeNfiV401BNI8HfD3wl4k8ceLNVkUummeGvCejXuv67qDqMFkstL0%2B%0A6uWUHLCIgcmgDraK/lR%2BG/7V37cv/BRPw7pf7Uvh7/gsT%2ByV/wAEmvhF44tPDNz8I/2StB%2BHf7Kv%0A7T/xksvh34nvtLv/AAT44/aW8efHfxNpw8HfGX4paTqdxJovw08B6FD4W8NeG9V%2BHjXcus%2BMoPF%2B%0An6p/UH4RTVNK8F%2BErbxH4o/4TvXLXw/4a0/WvGtjolvpsfjDWjZWFlfeK4tC0I3mn6Paa9qDSaw9%0Anp8kuk6Na3ThbldOtDcKAcb4w1L4C2XxZ%2BDll4%2B1D4RWnx11e3%2BI9p%2Bz9aeMLrwZB8WdUtLfQ9L1%0AH4u23wcg1qRPGF7bweGrPRb/AOI8PglJY4tDtdLuvE6rYQWkifyt/t3v%2Byz4U%2BOv/BRO5%2BJGif8A%0ABQvXv2HPh/8AtTfsT/GT/gpHp3gb4pfA7RP2fNI%2BM3iTTPhGPCHjrw98KT8Mr/8Aa08dfDvwnpGg%0AfA7xJ8bD8NfiRpUF/f6Db3fhrw5rQ%2BHMVzZfrL%2Bw34w8NftLf8FM/wDgqb%2B0HOqarq37NniP4Pf8%0AE7fhZNqGn6G154K8EfDnwmvxn%2BNsOialpct1NCnxM%2BO3xGuD4ogubqS%2B1HS/hF8MU1hLK60C20LQ%0Avzf/AOCxPjv9sr9oT9pTR/gb4T%2BD37HvxT/Ys%2BBl/pnjR/hp8XP28/gX8HF%2BO/7TXh600/XPh5rv%0A7SXgnUZdS8fal%2Bz38EPG7i81H9m/TLn4f6t8Xte0BNW8X/ECDQJNH8MaeAf1eV45%2B0P8DvCH7TPw%0AJ%2BL37PXxA1DxdpPgj40/DzxX8NPFepeAvE%2BoeDvGFloPjDR7vRdSuNA8RacTLZXyWt3IVhvLfUdF%0A1KLzNL8Q6Preg3upaRe%2Bp6Tq2la/pWma7oWp6frWia1p9lq2jazpN7bajpWraVqNtHeafqemahZy%0ATWl/p9/aTQ3Vle2s0ttdW0sc8EkkUisfmX9tn4Q6R8bP2Y/i14O1nx18avhxb2nhTVvFlv4s/Z/%2B%0AMfjj4GfEmzvPCWn3euRWVj468B6jYakuk6wlpLpGu6NqUOq6HqmmXs6Xmlz3EVnPbAHy1bfAT9kf%0A/gnV8Rvjn/wUd/ae/aa8V6z8R/ij4a%2BF3wW8d/tK/tQa78HvCdl4a8B2Ou6Vo3gn4c6DpPwX%2BFHw%0AU%2BHmlaVqvi2503U9V1S%2B8Ian4kubuFLzVPEkHh3SDDY/evwNX4H3Xwv8MeKP2crP4aQ/B/4jxXvx%0AX8Kap8I9O8O6d4E8YD4pahdePdV8f6WfC1tbaPqt34%2B1jxBfeL9Z8QxpJeeItZ1i%2B1rUrm61G%2Bub%0AiT%2BeP/ghdd3V/wD8Gxfw5vr65uL2%2BvfgP/wUEu7y8u5pLm6u7q5/aC/aomuLm5uJmeae4nmd5Zpp%0AXeSWR2d2ZmJP6Wf8ERbz7d/wSJ/4J0Tf23F4g2fsl/B6z%2B3w6NoWgpb/ANneFrPTzojWPh28vtPn%0Al8Mm2/4Ru41m4mTXfEc%2BlSeIfFFnp/iXVNXsLYA/UqiiigAooooAKKKKACiiigAooooAKKKKACii%0AigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKK%0AACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAr%2BQ3/AIJ%2B/EKH/gmb/wAF%0A9f8AgpN%2BxR%2B0Ouk%2BCvDP/BTv4nwftjfsj/EnVJdGtNF%2BIXibxj4n8ea/qvw%2Bi8UXV1p9/wD23qOv%0A%2BM/Fngrw74P1Szeb/hNPAGoad4cWYeN/Dl94w/ryrwj48/swfs8ftQ6HoXh39oT4NfD34vaZ4V17%0ATfFPhL/hNvDdhq2o%2BEfEek6rpWtWWueEdbkiXW/C2pjUNE0t7q80DUNOm1C2tRp%2BoPdadLPaygHu%0A9flz/wAFaf2OfiX%2B17%2Bz/wDCbUvgTJo1x%2B0T%2Bx7%2B1d8BP25f2ffDPinWo/Dng74gfFH9nnXNSvrf%0A4d%2BKtdl0vV49IsfGfhTxD4r0TSNTuLRdN07xbc%2BHL7Xbm20C31SVf1GooA/Ej9lz9lr9o342f8FM%0AfEv/AAVa/aq%2BCNj%2Bylqfh79jjS/2I/gr%2BzTqPxG8EfGb4hf2QPi1q3xf8YfGz4g%2BOPhfqeq/Dbw3%0Ae3eo61feCPBfg/wr4h8az3/hW7vtd8UX/hfVxb6DL%2B29FFABRRRQAV%2BMv/BbT9jb9sP9ub9nz4N/%0AB39kY/s62t94Z/aV%2BE/xr%2BIOo/tAeM/iL4Tj/sn4Ta/b6/4c0jwgvgH4e%2BOhcS6jr8iXniu/1j7K%0A%2BneGdIvLPw/pWu67rVo2lfs1RQB%2BBHwl/Y0/4KXeC/8Agr/8Rf8AgoL4j8LfsETfBr46fAj4O/s%2B%0AfEvwjpPxs%2BPt/wDGPwb4f8C2ei6n4g8WfDnVL79mTS/CurXEvjaPVr2HwXr9zpFl4o8P6f4Ysr3x%0AX4U1a3udWT9SP247L9sG/wD2V/i3B%2BwPq3wv0X9rWPT/AA5e/CK8%2BMkc0vw8mudO8aeG9Q8Y6TrC%0Aw6ZqyjUNf%2BHtr4t0LwtJeWsWlQ%2BLdR0KbWdT0XSY73WbD6vooA/mx/Yf/Y5/4K//AA8/am%2BPvxj%2B%0APWl/sQfC/wAXftbftK/BT4//ALRv7UHwX8WeMPip4wv/ANnz4QeBtC8J6H%2Bwh8NPhZ8TvhbpM2ix%0AX0XhHT/DniD4za/8Qby28F%2BDvGHji%2B%2BHmh6749fwn4l8Jf0GfFz4XeDvjh8KPid8FfiJp76t8P8A%0A4v8Aw88afC7x1pUcvkSan4O8f%2BG9S8J%2BJtPSfa/kve6Jq19bLLsfy2kD7W24PoVFAH8bPw%2B/Yj8R%0Af8E5vA9t%2Bzf8bP8Ag3x/Zi/4Ki%2BHfh9qc3h74O/tqfAj4IfshSeP/iz8NUSSbw8P2jPh58UPD9/8%0ARNG%2BNGgWaJY%2BLvGKXWseEPFs01qLDXNU1ew1PXNf/r78CRrD4H8GxJ4K/wCFbJF4U8Oxp8Ov%2BKdH%0A/CAqmkWar4Kx4QvdS8Jj/hFgBof/ABTGo6h4d/0D/iS3t1pv2aeTq6KAPxs/YC8Lw/AD/goP/wAF%0Aav2ftZEsWpfF/wCNfwt/b4%2BG2oapA8%2BseL/hl8dfhT4b%2BGvi6e31uSGaW40P4f8Axr%2BDvjrwdp3h%0A6bVZV8OWE2mXVlpWlWXiGIXP4hf8Fr/hF4J%2BGf7VXibxd4t/Zh/4Jm%2BIPFfxk%2BLv7KehfsVfA/Wf%0A2QG8SfFf9uzxb8U/il8PPB/7T2p/tX/tMaJ4T0dfghqfhTxJ4jtB8OvF2o%2BOLnSfEfh6LUNN1/R9%0Ab1/xHb2Vx/Ydrnwn%2BGviT4g%2BB/izrfgnw7ffE/4a2fiXSvA3xAfToYfGHh3Q/GVilh4s8M2fiG3E%0AOpyeE/Eq2%2Bm32ueEru5uPDep65oHhbxHeaXLr/hTw3qWlfkL%2B1t/wSb%2BNP7V/ir9qn4feIf229Tt%0Af2Jf22PGfwJ8a/HP9n/xN8KNX8c/EfwYnwTT4eWmq%2BCf2avi1d/GTSfBXwg8L/GG0%2BH%2Bj3fjG91L%0A4H%2BN9b8PazpsWpeH5p7vVLi604A7/wD4KJ/Fr4nf8E2P2Wv2a7/9iDwH%2BzR4J%2BGPg79pf9nX4KeL%0AvhD4s0LXNGtr34N/FHxtbfDlPAv7PPhT4fyaaknxS1Hxb4m8PXNkk1re2VhocfirxZqdjeQaXqMo%0A%2B6/2vJvjen7OfxSs/wBnb4UeF/jX8V9d8OT%2BFtA%2BHvi/4qJ8GdG1K08Vunh3W9Ufx7N4J8f21jce%0AGtG1O98RwaZc6AsevNpZ0ZNT025vYLlfctd8J%2BFfFM3h%2B48TeGfD/iK48JeILbxZ4Vn13RtO1ebw%0Az4qs7HUdMs/E3h%2BXULa4fRfEFrpur6tp9trOmtbajBY6pqNpFcpb3tzHL0FAH83P/BL79nD9tT9i%0AT/gjv8Wv2J/2s/gb8N/h7Y/s%2B/s7/tJN4H%2BKfgj492vxQu/i5d/FXxD8fPivr8N94BsPh5oMPw5t%0A/A0Pi/SdDjubjxz4sk8UyS/2hb2mkqlzZ23df8EsrD9trXf%2BCYv/AARUvf2Z/iB%2Bz7pfwv0n4ZfD%0Axv2t4PjL4a8Q6z4z8QfBOx0%2Bzt7Tw18Dn%2BHmqzaBp3j2xtbLUfDkd/4z1CyjtZZdM8Q%2BILM6to2t%0A%2BBtX/Yr9qT9mj4dfte/A/wAa/s/fFfUPiHpngTx5b2Vtrd18L/iR4x%2BFni0R2F/b6jBHbeKfBOq6%0ATqT2UtxbRpqOjX73%2Bg6zal7PWNKv7VzDXdfBv4T%2BCPgL8IfhX8DPhnptxo/w4%2BDHw38D/Cf4f6Rd%0A6jfaxd6V4I%2BHXhjS/B/hTTbnVtTnutS1S4sdB0ewtZtR1C5uL69kia5u55biWSRgD0iiiigAoooo%0AAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigA%0AooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACi%0AiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKK%0AKACiiigAooooAKKKKACiiigAooooAKKKKACiiigD/9k%3D%0A"></p>
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<p>Note that "Lebesgue measure" on $[0,1]$ just means uniformly distributed on that interval. Also, note the the <code>Piecewise</code> object in <code>sympy</code> is not complete at this point in its development, so we'll have to work around that in the following.</p>
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<div class="prompt input_prompt">In [15]:</div>
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<div class="highlight"><pre><span class="n">x</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'x'</span><span class="p">)</span>
<span class="n">c</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Symbol</span><span class="p">(</span><span class="s">'c'</span><span class="p">)</span>
<span class="n">xi</span> <span class="o">=</span> <span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">**</span><span class="mi">2</span>
<span class="n">eta</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Piecewise</span><span class="p">(</span> <span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span> <span class="o"><</span> <span class="n">x</span><span class="o"><</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">3</span><span class="p">)),</span>
<span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">3</span><span class="p">)</span> <span class="o"><</span> <span class="n">x</span><span class="o"><</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">)),</span>
<span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">)</span> <span class="o"><</span> <span class="n">x</span> <span class="o"><</span> <span class="mi">1</span><span class="p">))</span>
<span class="n">h</span> <span class="o">=</span> <span class="n">a</span> <span class="o">+</span> <span class="n">b</span><span class="o">*</span><span class="n">eta</span> <span class="o">+</span> <span class="n">c</span><span class="o">*</span><span class="n">eta</span><span class="o">**</span><span class="mi">2</span>
<span class="n">J</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">integrate</span><span class="p">((</span><span class="n">xi</span> <span class="o">-</span> <span class="n">h</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">sol</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">solve</span><span class="p">(</span> <span class="p">[</span><span class="n">S</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">J</span><span class="p">,</span><span class="n">a</span><span class="p">),</span>
<span class="n">S</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">J</span><span class="p">,</span><span class="n">b</span><span class="p">),</span>
<span class="n">S</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">J</span><span class="p">,</span><span class="n">c</span><span class="p">),</span>
<span class="p">],</span>
<span class="p">(</span><span class="n">a</span><span class="p">,</span><span class="n">b</span><span class="p">,</span><span class="n">c</span><span class="p">)</span> <span class="p">)</span>
<span class="k">print</span> <span class="n">sol</span>
<span class="k">print</span> <span class="n">S</span><span class="o">.</span><span class="n">piecewise_fold</span><span class="p">(</span><span class="n">h</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sol</span><span class="p">))</span>
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<pre>{c: 8/9, b: -20/9, a: 38/27}
Piecewise((2/27, x < 1/3), (14/27, x < 2/3), (38/27, x < 1))
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<p>Thus, collecting this result gives:</p>
<p>$$ \mathbb{E}(\xi|\eta) = \frac{38}{27} - \frac{20}{9}\eta + \frac{8}{9} \eta^2$$</p>
<p>which can be re-written as a piecewise function as</p>
<p>$$\mathbb{E}(\xi|\eta) =\begin{cases} \frac{2}{27} & \text{for}\: 0 < x < \frac{1}{3} \\ \frac{14}{27} & \text{for}\: \frac{1}{3} < x < \frac{2}{3} \\ \frac{38}{27} & \text{for}\: \frac{2}{3}<x < 1 \end{cases}
$$</p>
<p>The following is a quick simulation to demonstrate this.</p>
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<div class="highlight"><pre><span class="n">x</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">rand</span><span class="p">(</span><span class="mi">1000</span><span class="p">)</span>
<span class="n">f</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">**</span><span class="mi">2</span><span class="p">,</span><span class="n">bins</span><span class="o">=</span><span class="n">array</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="o">/</span><span class="mf">3.</span><span class="p">,</span><span class="mi">2</span><span class="o">/</span><span class="mf">3.</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span><span class="o">**</span><span class="mi">2</span><span class="o">*</span><span class="mi">2</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="bp">True</span><span class="p">,</span><span class="n">alpha</span><span class="o">=.</span><span class="mi">5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">vlines</span><span class="p">([</span><span class="mi">2</span><span class="o">/</span><span class="mf">27.</span><span class="p">,</span><span class="mi">14</span><span class="o">/</span><span class="mf">27.</span><span class="p">,</span><span class="mi">38</span><span class="o">/</span><span class="mf">27.</span><span class="p">],</span><span class="mi">0</span><span class="p">,</span><span class="n">ax</span><span class="o">.</span><span class="n">get_ylim</span><span class="p">()[</span><span class="mi">1</span><span class="p">],</span><span class="n">linestyles</span><span class="o">=</span><span class="s">'--'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">r'$2 x^2$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">);</span>
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IMIdAAxEuAOAgf4fJ7LzjzqgRPwAAAAASUVORK5CYII=
"></img>
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>This plot shows the intervals that correspond to the respective domains of $\eta$ with the vertical dotted lines showing the $\mathbb{E}(\xi|\eta) $ for that piece.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>Example</h2>
<p>This is Example 2.4</p>
<p><img src="data:image/jpeg;base64,/9j/4AAQSkZJRgABAQECWAJYAAD/2wBDAAEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEB%0AAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQH/2wBDAQEBAQEBAQEBAQEBAQEBAQEBAQEB%0AAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQH/wAARCACJAqADASIA%0AAhEBAxEB/8QAHwAAAQUBAQEBAQEAAAAAAAAAAAECAwQFBgcICQoL/8QAtRAAAgEDAwIEAwUFBAQA%0AAAF9AQIDAAQRBRIhMUEGE1FhByJxFDKBkaEII0KxwRVS0fAkM2JyggkKFhcYGRolJicoKSo0NTY3%0AODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqDhIWGh4iJipKTlJWWl5iZmqKjpKWm%0Ap6ipqrKztLW2t7i5usLDxMXGx8jJytLT1NXW19jZ2uHi4%2BTl5ufo6erx8vP09fb3%2BPn6/8QAHwEA%0AAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoL/8QAtREAAgECBAQDBAcFBAQAAQJ3AAECAxEEBSEx%0ABhJBUQdhcRMiMoEIFEKRobHBCSMzUvAVYnLRChYkNOEl8RcYGRomJygpKjU2Nzg5OkNERUZHSElK%0AU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6goOEhYaHiImKkpOUlZaXmJmaoqOkpaanqKmqsrO0tba3%0AuLm6wsPExcbHyMnK0tPU1dbX2Nna4uPk5ebn6Onq8vP09fb3%2BPn6/9oADAMBAAIRAxEAPwD%2B/iii%0Avyo8bf8ABcb/AIJHfDvxd4j8C%2BLv2/f2eNO8U%2BEtYvtA8Q6ba%2BLJ9bi07WdMne11HTm1PQtP1PSb%0Ai4sbqOW0uxZ31wtvdQzW0rLPDLGgB%2Bq9FfGH7K3/AAUS/Yh/be1XxhoX7J37S/wv%2BOet%2BANP0nVv%0AGGjeCtYmn1XRNK1y5vLPTdTuNP1C0sLubT5ruwntZb2zhuLazuWtIL2S3lv7Bbn3f44/HX4P/s0f%0ACvxb8b/j58RPC/wo%2BEngS30258X%2BPvGWoppfh3Qo9Z1rTfDejreXbhibjV/EOs6ToelWkMct3qOr%0A6nY6dZwz3d1DE4B6xRXK%2BBPHPhD4n%2BB/BvxK%2BHviPSvGHgH4h%2BFPD3jnwP4t0G6S%2B0PxT4Q8W6RZ%0A6/4a8R6NfREx3mla3ouoWWp6ddRkpcWd1DMh2uK6qgAoorwL9oz9qj9m79kTwGnxO/af%2BOHwy%2BBH%0AgS41NNE07xF8TfF2keFrbXNeksrzUovD3hu31G5ivvE/iOXTdO1HUYfD/h%2B11PWZdP07UL6Oxa1s%0ArqWIA99or5y%2BAf7Xf7Mv7UTeKYPgD8bPAXxN1XwNepp3jjwzoWsJF408F3cyRy2sfi/wPqken%2BL/%0AAAwl/DNFPps%2BuaJYW%2BpwOs%2Bny3MR319G0AFFFFABRXyF4u/bf%2BB3g39sP4ffsMahH8TdU%2BPXxE%2BG%0AUfxesoPCvwj%2BIfizwD4W8C32qeOdF0DVviH8SPD/AIf1Dwh4DTxPqvwz8e6foTeJNTsbea78NzW9%0A7cWE2qaAmrdr4W/a8/ZO8cfGDXf2efBX7UH7O/jD4/eF7jXbTxL8DvC3xr%2BG3iD4weHbvwu/leJr%0AXXfhnpPiW78aaRceHZP3euw6holvJpD/ACagtu3FAH0RRRXn/jv4s/Cv4XTeD7b4m/Ev4f8Aw6uP%0AiH4w0P4e%2BAIPHfjLw54Rm8c%2BP/E9/b6X4a8D%2BD4vEGpae/ibxh4h1O7tdO0Pwzoq3utatf3NvZ2F%0AlcXE0cbAHoFFfGXjL9un4N/D79sj4Q/sPeM/D3xj0D4sfHvQvFGtfB3xdefCbxY3wR8f3XgjwZ4k%0A%2BIPjDwtofxgt7Wbwg3jPwz4Q8J61rWt%2BGrq7ttQ062XS/tSRt4h8PjVPs2gAoory/wCLXxv%2BC/wD%0A8N2PjH46/F74X/BbwhqfiDTPCem%2BKvi14/8ACnw58N6h4q1pLqTRvDNjrnjHVtG0y78QatHY3r6Z%0Ao1vdSajfpZ3TWttKtvKUAPUKKK8P%2BGf7TH7PPxo8f/Fv4W/B/wCNnwv%2BKfxC%2BAmoaVo3xt8KfDzx%0AroHjLVfhP4h1nWvG3h2y8K/EKPw9fahF4T8YJrfw58babqPhDWpbPxNo1zoF0usaVYrNZtcgHuFF%0AFFABRXzva/tb/swX/wC0bqH7Idh8ffhPf/tQaR4Xl8Zaz8B7DxtoV78TtD8OxWGg6wLzXPCtreS6%0AlotxcaD4n0HxLY6VqkNpq%2Bo%2BFtTt/FFhYXPh8SalH9EUAFFFFABRRRQAUV%2BeHwu/4Kx/8E7PjZ%2B0%0AFH%2By/wDCb9qn4f8AxA%2BMF1qGvaJpFl4as/FuoeAPFHifwtplnrniPwV4L%2BNaeHB8E/HHj/RdDvYt%0Ab1H4f%2BEPiFrfjW20WK%2B1aTQRp%2Bm6jc2v6H0AFFfOvgH9rj9mf4p/HP4sfs0fDn42/D/xn8evgZaa%0AfffFv4W6BrkN/wCKfAlrqYsfs8ut2kSmFfJl1PT7PUktp7iTR9Ru4dM1dLHUW%2By1j/Hj9tf9lD9m%0AHx38Hfhj%2B0D8efh78JvHv7QGut4a%2BDnhjxhq50/UfHetJrPh3w%2B1ppYEMsFug1nxX4f0/wC2anNY%0AWJn1FQLkpb3bW4B9RUUUUAFFfJ/7Q37df7HH7Jnir4c%2BB/2lf2lfg/8ABLxd8WtQh074eeHfiH4z%0A0rw/qviNp9Rt9IXUI7W6mD6f4fi1O6gsbvxRq/8AZ/huyuWaO81WAxy7PoDx/wDELwB8KPBviH4j%0AfFLxx4P%2BGvw98I6e%2BreK/Hfj/wATaL4N8G%2BGNKjkjik1PxD4o8RXunaJounpLLFG97qV9bWyySxo%0A0gZ1BAOwor84Nf8A%2BCw//BKDw1/Yn9o/8FI/2ILn%2B3/EGm%2BGbH%2BwP2nPg/4s8jUdV877Lc63/wAI%0At4t1n/hGvD8XkP8A2l4s8R/2V4V0bdD/AGvrNj9pt/N/Re0u7S/tba/sLm3vbG9t4buzvLSaO5tb%0Au0uY1mt7m2uIWeGe3nhdJYZoneOWN1dGZWBIBYooooAKK8K/aP8A2mvgH%2ByH8KNb%2BOP7SvxU8KfB%0A34VeHrvS9P1Txj4uu5oLFdS1u9j0/SdLsbSzgvNU1fVdQupNttpekWF9qEkMVzdC2%2By2l1ND7LpO%0AraVr%2BlaZruhanp%2BtaJrWn2WraNrOk3ttqOlatpWo20d5p%2Bp6ZqFnJNaX%2Bn39pNDdWV7azS211bSx%0AzwSSRSKxANCiiigAooooAKKKKACiiigAor5N/ZW/bm/ZV/bZT4vSfsv/ABasvimvwH%2BJurfB74rC%0A18MeOPC8nhX4haKC1/o3leOPDPhmXXLLCy/YfE3htNY8K6qbe6XStbvWtbkRfVV3d2lha3N/f3Nv%0AZWNlbzXd5eXc0dta2lpbRtNcXNzcTMkMFvBCjyzTSukcUaM7sqqSACxRX5Af8P8Aj/gjb/0kJ%2BAH%0A/g313/5Q1%2Bp/w%2B%2BIPgf4r%2BB/CXxM%2BGfi3w/48%2BHvjzw/pfivwZ4z8Kapaa34b8T%2BG9btIr/Sda0X%0AVrCWa0v9Pv7SaKe3uIJWRlbBwwZQAdhRRXzf8K/2vv2Zvjf8XPjJ8BvhJ8Z/BXj/AOMH7PeoQ6T8%0AavAXh2%2BmvNc%2BHOp3F9e6ZHYeI1Nslrb3Y1HTb%2BwlghuZ5Le8s7m3nWOWF0AB9IUUUUAFFfFX7VH/%0AAAUa/Ya/Yj1zwn4Y/av/AGnfhV8DvEnjnStQ17wr4e8aa48OuavoemXkWn3Wsx6Tp9tf39rpRv5W%0AsbPUb63trPUby01K206a6m0rU0tPoX4L/Gv4R/tF/C/wf8avgR8R/B/xa%2BE/j7T59T8H/EDwHrlj%0A4h8M67bWeoXmj6lHa6jYSyxR6hout6dqegeINJufI1Xw94h0vVdA1uysNZ0y/sbcA9Qoqvd3dpYW%0Atzf39zb2VjZW813eXl3NHbWtpaW0bTXFzc3EzJDBbwQo8s00rpHFGjO7Kqkjw/8AZ7/ah/Z1/ay8%0AI6/4/wD2ZvjV8N/jv4G8L%2BONe%2BG%2Bu%2BMPhb4q0vxj4asfG/hmDTbzWNA/tvRri6065uINO1rRdXtb%0AizuLix1TQ9a0fXNKur3SNV0%2B9uAD3iiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKK%0AACiiigAr%2BTb/AILm/sjfA79k3/gkP8Gfgp4X0vRLzTYf27/2XtS8S%2BL9Z8O%2BDND8UfEHxf4u%2BLni%0ATxB418XeI5fCXhGysr3xBqy%2BJ/FzT3Vt4d1K/h0O7vrWaHVk%2B3G8/rJr8Mv%2BCxn/AATu/bc/4KT%2B%0AHvBXwU%2BFXx3/AGZvg9%2Bzv4U%2BInwi%2BMVzB40%2BGXxb8T/F/WPiR8N9V8W3k8F/4h8OfE7QvA938OJI%0AdS8N3Fr4Qk8FR69d61ptzqD%2BN9KSCwgoAh/4KP8Aw%2B%2BI37GHj/RP%2BChn7DfgL4LWvxw8Y%2BDfDP7G%0AXxYtfjLrXhP4WfsueC/hN4n%2BJOk/Ei0/an%2BOOo2mufDPWNbk%2BBR8MeIPCFpp0PxA0u8v9D%2BK1wNL%0Af7VoUOka/wCffA/9u3xX/wAFEv2H/wDgq38KvGGu/steOfH37NXgT4pfCZP2gv2Y76b4tfsqfF6f%0Axf8As7XvxB8HfEjwJ4Y8S%2BI/EN0bXwlqmpf2N4j8IXvxE8VW02v%2BF7k2ni%2B3a6e00j1n9sP/AIJ7%0A/te/tjeDf2J/F/xc8c/sZ/ED47fse/tGa38ddS%2BHWqfCH4t6B%2Byf8aludE1bwx4QtdZ8M6x8TPin%0A438Ja58P7DVV8Q6dqF3d/ECy1bxVYKyWWg6TqFzZxfKGuf8ABKD/AIKp%2BLfjL%2B278W9W/be/Zk8O%0AaT%2B3N4A1S2%2BI3wf8JfB74np8PrXxvrfwV039kvTNI1m7uPHNp4n8ceEPhZ%2BzxaN4r8I65Nqfhi78%0AV/Haz0fVdb8EeH/DdtcJcAHnH/BNv/gqJ8T/ABH8If8Agnz%2Bz7%2Byl8Pfg78Wf2Wv2Tv%2BCbv7Hmv/%0AAPBSv9qDxb438W%2BHbL9lrUtJ%2BAVjf%2BJfhzoOl6X4b1JviN8TdN8GeA9R1KLw14PsvFUVj4gu7bRf%0AHN94FtIrrVYuc%2BA3/Bx14r%2BI2u/s6fHXxrH%2BxTZfsnftZftc6R%2Byr4E/Z68B/G%2BDxN/wUU%2BBejeM%0A9Y17wz8Pfj/%2B0D4JtvGOoeDB4Y1DXdH0258e/DTw54asPE3gLwn4u8MeI9L8S%2BPnvl0tvSPhV/wQ%0A2/aq%2BAfjX9j7x/8ABv43fsxeEtS8DfshN%2Bwl%2B3Z4V8O/D34peDvAX7YfwI0S61LwP4D8Y32hWXiz%0AXZdF%2BOXhz4SXOl%2BJLrx1HdWN5qnxt0u5ubi8Pwz1u88HQ/Vv7If/AATy/wCCh37G3wU%2BHH7Dvwu/%0Aa1%2BB%2Bn/skfCP4tw6/wCCfjPD8Ldbk/bFi%2BAy/Ftfite/BK60vVJdT/Z/bxXrX2rWvA%2Bu/GTUfDOu%0A2N94L1bUrWy%2BEeleIrvT/E%2BgAH7v1/Mr%2BwXpeo/tWf8ABej/AIK8/En9p/Rb3xd4o/4J8X/7MPwp%0A/Ys8OeL7bUpfDHwC8DfHDwD8Q9S8Z%2BNPhr4S8QveWugeMvi9pHw%2B8J%2BJNU%2BIulJb6pr2l%2BI9Vj0u%0A8i8Javpml2X9Hvj/AEbxB4j8CeNfD3hLxC/hLxVrvhLxJo3hnxXHCbh/DHiDVNGvbHRvEKW6sjTv%0AoupT22pLCHQytbBAylsj8WP2BP8Agn7/AMFA/wBm7QP20fil8d/2kf2eviB%2B2V%2B0H%2Bzf%2Bzx8Cvhz%0A8WPB/gX4j6joh8W/srfDX4y%2BEfhZ8c/2g7vx94gutX%2BJvxA8Rat8U9E/4WLDoej%2BGdN1TSPh/a3U%0Aa3GreJdQFiAfInwG8L%2BKP2jv%2BDn79qH9qH4L%2BEL/AMLfs/8A7Jv7I2mfsgfHf4u6Xpuo6V4T%2BPf7%0AQV1e6P4m1DwO2t2jpp3jbxN8NY9S0Lw54vs75pZfCVx8E/C1tf6fAZPB2qX376/tc/Hj4ifs4/Bf%0AVfib8Lf2Zvi1%2B1r4ys9b0DSLD4P/AAYufCVl4tvLbV74QX%2Bv3F34v1rSLSDRNDtElmvG02HWtVku%0ApbCFdKj02XUtZ0j4d/4I9/8ABPj4%2Bf8ABP74K694M/aE%2BPPh34s%2BKtbsvBWhaZ4V%2BF9p430f4PeD%0A7HwUfFt/rnj23sPHWv6xrHin47/H/wAb%2BOfFfxR/aJ%2BLF5ZaDfeOPEt74d0CPTIfDXgHwtFD9xft%0Ac/so/Dv9tD4L6r8Cfil4m%2BLXhXwbrOt6Brt/f/Bj4peLfhH4tuLnw5fDUbCyuPEfhC%2Bsru%2B0SS7S%0AK4vND1Jb3Srm6tbDUGtV1LTNNvLQA/LK0/4K%2Bftmva2z3/8AwQo/4KL218%2BsQwXlvaX/AMEb21g0%0ABo1NxqdteTeN7Ca71iKYukOhy2FlZTxqsr%2BIbZnMKfrv%2Bzd8VfGvxu%2BCXgH4p/ET4G/ED9mzxn4v%0A0/ULzXPgl8Ur3wxqHjvwNNZa5qmk21trt34P1fW9Bk/tmw0%2B18SaYsF%2Bt/Do%2Bs6fDrmnaLrsepaN%0AYfjh/wAQ3v7Cf/RYf2//APxNr4w//J9fs7%2Bz/wDBLw5%2Bzl8HfAvwT8I%2BJvib4x8N/D/TLnStI8R/%0AGP4leLvi78RtQtbrVb/VseIfiB461PV/EuspZS6jJp%2Bi2lze/wBn%2BHfD9ppXhnw/Z6Z4e0fS9MtA%0AD87P2ufi5q/wD8Sf8Fafjn4Htnh%2BIfwT/wCCRXwb%2BLnh/UZdQszazav8O7//AIKjeL/CNsNPW%2Bu7%0A6xex1zQ7qe41C60G1s7%2BO%2Bggsr7WZtL1Cz0b8k/j1pngr4Of8Eb/APgg/wDtM%2BH/AIX6de/Fn4Zf%0AG/8A4JR/FvTtYENvbfEnWfFPx2sfh3cftCW1/wCN49GuNbudd%2BPkWt6zZfFHU7rT79/Fus3trr2s%0AaTf6vpGjmy/djXfhP4N%2BMP7Vv7d3wl8axTal4P8AjR%2BwN%2ByZ8OvHWlza/ca0s/hbxr48/wCCingv%0AXLa18GeJLbWvB%2Bh2t7ot9dxfaYdIubLxTeG8j8RaVfR6RH9s/P74W/8ABLv9uSfwv/wT7/Zk/aI/%0AaY/Z%2B8T/ALIP/BOj4qfBf4oeCL34VfDPx94R%2BPHx6X9lrRrS0/Ze8I/E5NV8V6h4C8FaT8PNY0/R%0AdW8ZX3hl/E8/xKPhjRRcWnh28l1O%2BugD%2Bgqv5H/%2BC%2B37PnwT8J/8FDf%2BCGf7RHhv4ZeEtG%2BOPxW/%0A4Kw/sseGPiN8UbHS44vGPjHw74R8d/Cax8M6Jrer5M11pmiWtjaRafafLDB5IdV8xndv2v8A2U/F%0AH7RV5%2B3/AP8ABTLwP8SfjVc/F74F%2BDpP2S9T%2BCOgw/DCPwHofwF8UeN/A3xL1r4mfBC18SOJLn4k%0A69ZeDYPgZ8UfEPiaO%2B1DS4o/iro9rbHSdWOveHNE%2BSv%2BCnn/AATT/bb/AG8f2if2Vvih4A/aF/ZY%0A%2BF/w5/Ym/aH%2BG37T3wG8M%2BMPgZ8W/GXjXWviT4Ek8K63c6d8XfEWi/HHwvoWv%2BB9Q8UeHpDBo3g3%0Aw54I16DQJktZPFcmok30YB6L%2B3itq3/BWT/ghGbia4imT4m/8FDGsUhto547m6P7DHjdHhu5Xu7Z%0ArK3Wye8uFuYYb%2BR7qC2szaJDdy31l%2BuPxDn8d2vgDxzc/C7T/DWrfE238H%2BJp/h1pXjS%2B1DTPB2p%0A%2BO4dFvZPCOn%2BLNS0m3vNV07w1eeIF0%2B212%2B0y0utQtNLkuriyt57mOKJvx4/ahvpLz/gqJ/wQ18K%0A/ELx58Im%2BNnhnR/24PGvjLw1o2q2vh6fxBqN3%2ByPN4P13WPhd4G8Sv4s8Z/2Dd%2BI7zXbzRrSTW7P%0AU7PwjoniM3/jHVxoWtaPr37b0AfyV/s%2Bf8HAv7U3xQ0//gmd4s%2BJfwZ/Y6%2BFnhL9uv8Aa1%2BIv7Kv%0AjvQr/wCMXxRtfi78GPFXwu8d23hzxLpfiDwN4p8EeG9G0PxLc6Xd6JHo2nT%2BMvEcs2veP/h7Z6nY%0A6Vca2%2BlxcR/wUZ/b4%2BLnxH/4Jh/Hn9rj9p/9ib/gnb%2B03%2Bzp%2Bzz%2B3vP4E%2BEvhXxR8QviX8Q/CHxu%0A0P4Y/EzUPgjpfxy%2BE%2Bq6Z4PudEv9St/iLqPiPwrfWWs6loHh/wAS/D/S/i5pOpSXdhc2HhXx57l%2B%0A0F/wbL/DT4qfED/gqb8YfCHxl/4Rn4nftxeIPh/8Sv2btdm8PX1jqX7JfxcsPiR4X%2BNvxm8W6f4i%0A0zVrh9al%2BM3xY8FeG7i51jw94e8KeKvDHgm3u/DkOs6xqVz/AG%2Bn1F/wUC/4I4eO/wBoX/gk38G/%0A%2BCVv7Lvxa%2BG/wi8DfDrTPg74c8U%2BLfiH4L8Sa/ceLfD/AMIYrHU4ZNKg0XxA0vhvXfFXjvT7Xxlr%0A%2Bo3b%2BIg%2B290WBEGqSapaAHjfxT/aG/4KXan/AMHGHww/Zr%2BH3h39l24%2BCfgH9g3xX8Z7fwv4k%2BKX%0Axh0qS8/Z0%2BKn7S3wp%2BH3xR%2BJHiG/0H4dxWl3%2B0lb%2BJfg7o2lfDL4d3fhrxj8MfC/hjVr%2Bebx3/a/%0AiPXNX0jO8TfEr9sr4Y/8FmP%2BCo1p%2BxP%2Bz98IvidLrn7If7EnxQ%2BNXxE%2BNfjnV/AXhPwdrfwr8Aft%0ATW3wx%2BGmkaH4K0bxB4o8d%2BM/jlcavaWWl%2BLr250DRfh/oPw88Ux6xp2uE%2BGVuvd/21v%2BCWn7W37R%0AX7ROkftWfCT9o/4XfCb4t/Fr/gm/r3/BNP8AamgHhjxemi6N8O/HvjmT4k%2BMviv%2Bzdq1tdal4k0f%0AxzpPinWNei8JaT4vuYimmWvh6U%2BLtH1aC%2Bvp/ZLX9kL9v3wR%2B3H/AMFCv2qPhl8Sf2Tv%2BES/at%2BB%0APwr%2BHHwn8I%2BMdB%2BNK614V8Z/s/8AhfxTp/wl1z4j6h4R13Q7240i7134pfEp/GkngLxLoWqjRYvC%0An/CPCz1lLvULQA%2BxP%2BCeX7XVr%2B3n%2BxR%2Bzn%2B13a%2BDrj4fN8cfh7aeKdS8FXGox6wvhvX7PUdR8O%2BJ%0AdLsdXjitjqujweIdF1M6HqVxZ6fe3%2BjNYXV/pmmX0txYW35/f8FGviV/wUr0X9vz/gnl8N/2OdB/%0AZfu/AniOP9o3xndWPxv%2BLvxe8JWfjnxL4M%2BDer6Lq9j8R9K%2BH3wn8RtoXhPwjp/j7SdU8BXWhXvx%0AB1LxN40vZV17w54N0rQ7bxBL9Zf8Emf2SPi9%2Bwh%2BwP8AAr9kf41%2BKfhv428XfBGy8UeG7LxV8LYv%0AE8HhzWvDOo%2BMNd8T6FPdW/iy1tNTi123t9dksdW8qFNPnmtEurNIEna2gw/%2BCgv7LX7Vfxs8bfs5%0AfGL9kH4j/BzwP8Tfgdpn7R/hS%2B0342WvxHh8M654f/aD%2BE6eBY9Z0zxJ8KdQsPG2geIvAfibSPDn%0AinTk0W50m71ZLe4trfxT4cmijnugD8Lv259a%2BMngP/g6F/ZE1D9lf4LeBPHn7S3xF/4JjXGn67Ye%0AN/Fsvgj4N6NbXvxT%2BOOl3/xM%2BIXjTQPCmo%2BN9btPBfhbwvq2mSC08L3HiHxPFpvw68J6Nb6YJ5Dp%0A30LL/wAHDd9B/wAE37L9o/X/AIZ/Br4cftS%2BK/23/Fn/AATa8NeH/HnxVm0/9mHwv%2B0Lp637WXxx%0A%2BIPj3VLfwz4u0z9mDwVpD6P41%2BJ17qFp4Y1/TNL%2B1aGdb0UahpXio9/4C/4JM/ts/Cv/AIKM/sff%0Atk6H8Wf2efiH4N/ZH/4J7fCn/gn7aWfxB1n4tWvxQ%2BKOheCPC2sL4l%2BMHi7UrPwrrulWnjbxD458%0AU%2BINSj0yLUtSs5NEi08X%2BpyaxcXt2Pnv4W/8ECP2uLf9mSL4VfEb9pP4FeEPjT8Cv%2BCkmh/8FMv2%0ANviv8O/DnxE%2BIXhfwl8T7rXbnXvGvwz%2BLvw98dN4Stdf%2BHT6ja6Nq3hubwrqOl6yNce7vNWaWwi1%0APSvFYB%2BnP/BOb9v74gftB/tHftRfsr/FH4s/syftJ6l8GvCHwh%2BL3w7/AGkv2ONB1LRvgn418F/F%0AW31qx8U%2BANVs7n40/Hy00vx38JvH2hahoNuY/iFJd%2BKPBN94d13UdC0nWIdYN19F/wDBR39umT9h%0AT4M%2BE/Efg74R%2BIP2iP2gfjd8UPDXwF/Zm/Z88Kavpmh6z8VfjH4ytdUvdJsb/WdSZ08OeCtA0zRt%0AT17xt4uayvLPw7pNost/9kguhe2/ovwD%2BGX7WmmfE7xz8Uv2m/j54K8X2Or%2BCvB3gbwJ8D/gp4B1%0ALwT8IPB1xotxqOr%2BMfiZqN54z8QeM/H/AIl%2BIHjjVNRtdJtrebxDY%2BGfB/hDw7YafaWGtaxq2qa2%0AvCftofsq/E741/Eb9j79oP4FeNPAfhj42fsa/Fvx94/8JaJ8WtE8Qa58LfHnhv4ufBH4g/Anx/4b%0A8RHwdqWj%2BLdB1y00Px3F4l8H%2BJtMutUsLLVtCk0vWvC%2Br2etrf6IAfmH%2Byz/AMFHv%2BCsvxN/b28e%0A/se/tB/ss/sMfBzSvgJ4H%2BFnxe/aH16z/aJ8fX%2BseHfhB8R5PEzDxN4BuYvDGs6N4p1iwttAJutL%0A1i38N6JpN09lFrvimwtdajvdL%2Bw/2uf27PgF8eP2Jf219L/YU/bI/YL%2BMvxn8Ffsv/FPx3qOnXH7%0AQ3gv4ieDfCXw50jRjb/ELxl8QLT4O%2BL9Y8V6D4f0vwpe6pBp3iGeO00Ky8WXfh2LW7yLTbi4V/SP%0A%2BCfv7DHxM/Zg8X/tU/tCftJ/HnTP2kf2tP2y/HvgfxP8XPiJ4X%2BHMXwk%2BH/h3wP8I/CMvgr4L/B/%0A4feAIPEPib7H4X%2BGuial4jitvEOpanJ4m8Uza7NqPiybVPEK3%2Bv6v3v7UP7E3hT4j/stftM/BX9m%0Azwr8EP2efif%2B0D8CfiJ8B7f4q6f8J9NtG8NeHfiloz%2BGfE948fgK48F%2BIby6tNHurnUvD8aa9FY2%0Afiyw8P6xqFjqtrpkmm3QB/Jd/wAE6/Hnxw%2BHvww/4NtrP47fsp/sJeOfhD4x%2BIfjLwF%2Byv8AF3S/%0AFnxw1r9pP4VReNPCfivX/HPim78Jz%2BG/Anwr8K%2BLfEmuaP4Y1PUb9Nb%2BK2m3f9hyKmkWOvX9j4r0%0AH7V%2BO3/BxT8UfCXjj9uf4j/DHxX/AME4rH9nv9g79ojUvghd/s5/F/4zaro/7dv7Wun/AA21/wAI%0A%2BGPjF42%2BA2n6d8QNO8MeGfC66prupj4UeJbr4a/FDTPHVxpV9o8kOn3%2BlXjze8aN/wAEXv25fCPw%0As/4JO/D3wp%2B1T%2By2s/8AwSy8TeKvGHhy4174D/FO%2B0v4ta5ql3faZ4dbxJbWHxi07U9HtdG8EX1x%0Apl7Do%2Bq241LxJdXWvYh09bHQLP7A8FfsH/t3/s0%2BLv2uPAf7If7QH7O3hX9mr9rX48/EH9oHRr/4%0Al/D74hat8av2UfGfxzjgufjWfhPp3hnxLpngH4j6bL4q%2B3%2BOvhZo/ii68D6X4I12/bT9Xt/FmixP%0Aa3AB8X/DP4yfC74V/wDBbb/gqJ%2B29421a08F/AnwD/wSi/Za%2BLvxC1nUtDuoPFsXh7UbS58aWeoP%0Aodrp8%2BrahrUPhPwLc6RJ4fhmk17UNXHhvw7pum6pdi0is/iH9vX4/ftlftw%2BD/8Aggt%2B0p8d/gF%2B%0Azn8Kf2SP2k/%2BCu3/AAT98a/CzwHpvijxr41/aO8K6d408Q%2BPbrwJb/EfxFqGkaN8PfEng343fCHU%0AH8aXmmeE9E0G78E33h/w7ZapfeL38UtF4J/XO5/4JDfE74j/ALZv7bnxQ/aH%2BKnwl%2BKX7MP7aX7F%0Aem/sYeJfD1t4D1jTv2j9C8L%2BFNOi0vwj40T4g6zeeIfBEvxBkku9W8V%2BJfEml%2BE9G0a48b2Xg3Xt%0AC8FaLN4VtvO%2BW/ij/wAEi/8Agql4v/Z//wCCen7Mmm/tU/sTav4J/wCCYP7QH7OHxq%2BA3jzWvg38%0AX/DPjH4tn9lnSrjwz8GbP4reG9K8X%2BJNB8HxeEfCN1d6Frel%2BFNX8TS/EOSez1W98S%2BFL/TLmTXA%0AD%2Boe/v7HSrG91PU72007TdOtLi/1DUL%2B4hs7GwsbOF7i7vb27uHjt7W0tbeOSe4uJ5I4YIY3lldU%0AVmH80N5/wWL/AG1/En7KPiT/AIK5fCb4Kfs1eJ/%2BCWvhTx34ogX4Yatq/wASdK/bk8d/AD4ffF/U%0APgl41%2BP3hfxJe6rpnwR0DxKniLR9d8V6R%2Bz94s8M215N4P0OeB/i7N4m1LS9Im/pR8Q%2BH9G8WaBr%0AnhbxHp1vq/h7xLo%2Bp%2BH9e0m7VmtNU0bWbKfTtU065VGRmt72xuZ7aZVZWMcrAMpwR/M6/wDwRd/b%0Ag8Mf8E/vEP8AwSG%2BF37T37PPhv8AYd1/xt44trH4vav8OfGmsftOeF/gN8QvjJr3xk134Vr4Qubu%0A4%2BGfinxCNR1/UdBPxCn8V%2BHb%2B5sb%2Ba%2BstN0K4tLG3jAPjHwP8LPiD%2B3f/wAHI/7Q4/aG%2BEX7IH7S%0A37Kt1%2BwX4AktdD8Xan48%2BIXhKH9lHxR8VvDvxW/Zl8Z%2BBbPxl8KrrTLL41634isNB8VePPhu82gf%0ADea18SfE670Xxh4ju4LZ9Q/sw8QeHtA8WaNqPhzxToej%2BJfD2r27WmraD4g0yy1nRtUtGZXa21HS%0A9RgubG9t2ZEZobmCWMsqkrlQR%2BHn7MP/AATb/ar/AGf/APgq38Tf2ym8V/svJ%2By54t/Zx8Jfsl%2BE%0Afhf4b0z4lt8YvCXwd%2BDljpK/Bu%2BPiDW7ObQZ/FrXXh/TrXx9bW2pW3hrUdMuRLY2D6poWmXU37Lf%0AFtvi2vw38X/8KIj%2BHUnxebSmTwEPi3N4mg%2BG8euSTwolz4vPg6C48TzaVa2zXFy1lowt7zUJ4YbE%0AX%2Bmx3Emo2oB/Av8AsPar8XLL/g1u%2BKWh/AL/AIJX6r8TvFXjT4C/th%2BGPiN%2B1Z4h1T9mbRfCXi/4%0AYax8XviVb%2BNfHfg6zg%2BJOq/tMeP/AIgfBPR7TS5fDHgvV/hP4Y8NT698GTrfhzxrqut%2BHPCXhXxZ%0A%2B0n7Mn7Yjfs8/sVf8ESv2D/%2BCa/xP%2BH/AO0/8R/2yfB/xG8N/DP9pj47%2BGfiVffCTwb8NP2X9Ev/%0AABh%2B0b4y8YfC6z8feEfilHqGg6u%2BofCz4XfCV/HnhSw8OX%2Bly%2BH5PEMFn4LsfDOt637LX/BLX/gr%0AR%2Bzz/wAE09F/4JdeF/2ov2JfhX8OrLRPil4Luf2kPCnwy%2BNHxT%2BNNl4E%2BM3xD8Z%2BOvHWn6B4I8T6%0A/wDDfwDaeKy3xD8SaZo/i651OaOw0W2tNPs9AsfEt1B4%2B0L6U8Lf8EXrj9l74Wf8E4dF/YR%2BOOlf%0ADn4pf8E45fjjZaDr3xx8A6p8SfBHx%2B8L/tTw2k37RejfFDQfBnjf4da1o%2BoeJfFWm6F418E63out%0Aaj/whEvhjSPCttpt7pBjvdPAPmv44/8ABZL9rb4Q/Dj/AIK0fBSLQf2Vbn9vD/gldovwz%2BL1/wCK%0AvE3hX4uaB%2BzB8fv2dviL4e8MfEO28ReG/AmmfE3xH4/8BfFq38H%2BI7Lwwfhprnxl1vR5vHmueF5d%0AK8da5pFx4jtvDPM/tA/8FlP%2BCi37Hn7K%2Bg/Gz9qP9n39irwZ4y/bK8a/s0%2BCf%2BCf0Om/GPxdYfDD%0A4d6n%2B0B4G1fxh4zl/bj8a%2BL7/Q9F0vw1%2BzjFaaV/wnfj3wN4j8IeGvGcmsQpoP8AYPhew8ReP/D/%0AAKn8UP8AgiF8c/i58L/%2BCm3ivxn%2B0f8ABfUv20f%2BCo978PPBfxU%2BJ9v8E/F%2BlfB74Sfs6fCvw94e%0A8JeD/hb8LPAlv8VJ/FN34tTSfDGmXes/EbxV4svrbWbi20W1vvBkt1oV5r/ib6d/aI/4JffGP9qP%0A9hz9kD4N%2BPvj58NfAX7Yv7D3xa%2BFHx7%2BBP7Q/wAL/g3qFx8N9L%2BKH7Plxr%2Bn/CP%2B0vhd8Q/HXi7W%0A77w1qHhWbwxD8SNLu/Gt5p%2BueMNIk8S2WjwaTDpfhSzAPyb/AGrP%2BCk%2Bo/tJ/wDBP/8A4LYfsv8A%0Axf8Aib%2Byt%2B1zefBT9jrQvil4B/aF/YY1a70v4IeMfD3xnuvEPh/T9E1Zj8T/AI/Q6F40%2BCXxL0Kz%0AN54dbxzeT%2BPfCVnbRXb6LfTarqy/1G/sm/8AJrH7NH/Zv/wa/wDVdeHK/LX9qv8AYI/4KLftr/sz%0A/tRfCL4x/tY/s/8Ag7Wf2iPhr4N%2BCekfDj4f/CXx1q/wD%2BHHg/SfEeoeIfiB8T7STXPGek/E/wAR%0A/Gr4jytoenaU%2Bq66ngL4b%2BHNDGhx%2BHvHGpaje%2BKW/XL4FeG/iT4O%2BDvw38J/F/XfA/ib4leGfCWk%0AaD4t1/4beHNe8I%2BBtW1HSbdbFLzw54b8UeJ/Gev6RZS2cFrvtdT8UaxP9pE8guVikjhiAP5w/it/%0AwWj/AGurv9uH4n/sy/BL/hiXwhrfwj/bV%2BGn7NK/shftCaZ8SdC/bE%2BPfwe1/wAW/CTw94h/aP8A%0AgL411340/Bz4M63p/iK0%2BIfiTWvA3gBfD%2Bva7Y%2BFvAmoeOLy88T%2BHp5lsfJvFH/Bx74wl8YftI/G%0ADw74x/4J7eBP2YP2Wf2lrn4CXP7NPxH%2BLeoa7/wUO/an8J%2BEPiTpXgvx78Z/2f8Awx4e%2BI3h7TtB%0AsLzQ9fTxF4A8Eaz8HfHTa7deEfEOlSeNja3x1jw79W/GH/gjr%2B1j8ete%2BJfw3%2BMvx7/Zt%2BLnwG8W%0A/t8Tftn/AAm%2BLfxH%2BFHj3xF%2B1/8AsreCZPj1ZfGv/hmf4IeIr3xjdeE4PBEdnb6x4C0nXL7UbK18%0AO6B498fxWvgq%2B0rUvD2geGPYfh1/wTY/bf8A2WbH9rT4Ffsb/tTfB/wH%2BzF%2B1d8avir8Z/C3iHxz%0A8PfGmp/tCfsaar8cIYZ/iPYfBS40HxLZeBfiZaWGtveax8KR43Hg658A3y2174gvfiPcyXhmAPHf%0Ahh%2B01/wUp%2BJX/Bdf9of4OeFf%2BGXD%2BzH8Nf2ff2atV1Pwlr3xM%2BMdhqafAfxz8SfHPinS/i34Y8IH%0A4eqNZ/ad13w9rOp6D4l0XU4vDnw18Jafp/g/Q5/FurXvnX%2BuR/Cf/gr1%2B0Z8YP29vib%2Bz94S8T/s%0AETeEvhp%2B3P4j/ZK8Tfsm%2BItY%2BJ3w%2B/b30D4a%2BEPHF/4J1L9pjw/d%2BN/GWlfD346eFtTsfD118TE8%0AKfCL4bXy%2BFvh74w0e41/xhNqXg3XYdd%2Bif2pv%2BCZ37Q/xH/bF%2BKnx/8A2ffjt4T%2BHPgT9q34e/sq%0AfDP9pe08S/8ACWW/xF8KaP8AswfG62%2BIkOs/AvUvBtrY2099498Az678PtX0XxrrVtpNpcatfate%0AS6/pl7F4e0vyG5/4JN/tQfE34pfsu6v%2B0R40/Y%2B%2BIVn%2BzP8Att6h%2B2Bon7T9r8N/ifP%2B2fP4csvj%0Ar49%2BN3hf9nWx8W654kubO38AB9b8GfD7U7668YXOiQfDzwlo%2BhN8O9c1Hw3pGv3AB/QldzSW1rc3%0AENpcX80FvNNFYWjWiXd7JFGzpaWz39zZWKXFyyiGFry8tLRZHU3FzBCHlT%2BXL9lD/guH%2B1VrX7WP%0A7I/wM/bQ%2BHv7HPgofts%2BLPjD8PNN/Z/%2BB3xVutc/a6/Yk8d/D3WPEsXw/wBP/ao8Jah498WweNfD%0Afxr0fQ7e58K%2BLfCXg/4Z2ehNdXNx4is7eaCw0fVv6KP2l/hd4m%2BOH7OP7QHwV8FfEDU/hN4y%2BL/w%0AS%2BK3wu8JfFTRV1B9Y%2BGfib4geBNe8J6D8QNJTSdW0DVW1PwZqurWniOwXTNd0XUGutNiFnq2nXJj%0AvIf5Y4f%2BCAf/AAUO8PfBf9gT4O/Cf9qD9jr4Ir/wTs8V/FTWvhV458E/CT4h6n458f8AiP42re2X%0AjD47eL9a16eSLw78T/CdrLFrXgrwPplhr/gjVfFE0UvjnXNc0bw14a0y1APUIf8Agtb/AMFRNU1j%0A/gqd4e0f9hb9lVYf%2BCUH9teO/jZ8Q7/9oT4kr4I8a%2BAfB1rr/jDWfhx8NrD/AIVnbeINZ%2BKHiz4V%0AeFPFfjXw74w1BdJ8M%2BEpdEsNF8beCo9R8YeHreT648e/8FdfjL8b/i1/wTo/ZV/YM%2BF/w80j9oH9%0AvD9kLwL%2B334t8bftNWfirxV8KP2Zv2YPGXhebXdMl13wr8NPE/gLxH8VfiF4h13TtZ8A6VpXh3xz%0A4W07RfENvompa5NJ4e8QS6hofzpon/BH3/goboWsf8Fxtetviz%2BxjLff8Ff/AAnJ4Ms4J9P%2BN7Wn%0Awr0WW3%2BJvw63XJjsoprnVIfg58YvF1x9uibVbSf4l6N4cuv7Kt/C1xqemj2/4U/8EiP2n/hfB/wT%0Aa%2BP/AIT%2BPXwR8Bftu/sH/s3eHv2IPHGo6V4F8beNf2ef2kP2OvB9lNo/hrwF4s0jVtb8LfELw34%2B%0AsRb6f43s/G/h7VLXS9N8c32uo/hLVtBOk2tgAeMf8G69z8Trn9oD/guSfjTD8P4/i1bf8FJPEunf%0AEWb4VW3iWx%2BHGqeL9Js/EGma3r/gvTPGPizxx4o0bw/4jvbOXX9P0LXPFGqX%2BhRakNImNn9iGn2n%0A9NniHxBo3hPQNc8U%2BI9Rt9I8PeGtH1PxBr2rXbMtppejaNZT6jqmo3LIrstvZWNtPczMqswjiYhW%0AOAfx5/4Je/8ABOL9pX9hn46/t3/Ff40/tHfCv46aT%2B258YH/AGgNV0nwV8HNd%2BGOpeDPidqWoa1c%0AaxbaZPffEHxfazeCxpWtJotjpl4l9rBGh6Xqk2sx3Nzq8N99Jf8ABTX9mr9p39sL9ln4ifsyfs5f%0AGL4RfBPT/jl4M8a/C34x%2BLPiZ8OfGvxB1t/hr438PzaBrGnfDs%2BE/iL4EsvD%2Bu6jZXmo6ZqmoeJr%0ADxfptxot/c2sGj292yXqAH813/BIDxx%2B1l4e/wCCdn7VXjf4Vf8ABMn9oD9pn47f8FP/AI1ftJ/t%0AL%2BH/AIo/EzxF%2Bx5pn7L3jfSPjJqF94O8PxfGC%2B8c/tCQ/Ei18FaNZ6V4q8S%2BJ/h/4j%2BFPiK/8a3%2B%0Auazpui6jeeHvH2i6ta/0Lf8ABGz/AIJ96/8A8Ex/%2BCf/AMHv2TvGXj23%2BInjzw3ceKfF/wAQNa0W%0A9vb7wRZeNPHeuXPiDXtB%2BHD6roXhzWR4H0m5uBDps%2BtaRY6tq98%2BqeILyx0uTV/7H076C/4J/fs9%0A/EX9k39jP9nb9mT4oeLvh1468S/AL4X%2BFvhDaeK/hf4I134feGtb8K/D3TLfwr4O1O90DxD4v8aX%0A8vjDUfDOl6ZqHjzWo9UstO13xld63qul6Doen3Vvp0P2LQB/MN8f/jH/AMFStX/4OK/Bf7Pf7Pni%0AH9kqL4WeCf8AgnXqHxp0jwZ8Y9U%2BOsGhx/Av4lftIfDL4c/GDxnqVv8ADqKzXxJ%2B01D4%2B%2BGukp8O%0A9F1mJvhVpnwuil0%2BTV9L8aa1qviGvHPjD4z/AG1fh9/wXO/4Kd6l%2BwZ4K/Z71D4mWv8AwTz/AGcf%0Ail4u8QftE2nxb1zwPead8N9O8UppngTTfAfwav8Aw5rnin4mfEnULvRNL8NeJ9V8T2sfg7w34S8Q%0AQ2mjeJ5dUt9Ki/QT9v8A/wCCVH7Qf7Sn7Z1/%2B15%2Bzr%2B054W%2BDGo/Er/gnv8AEb/gm58WvD/inwd4%0Aq1K%2B0z4UfErxj4x8WX3xS%2BHfiHwf4t8P3x%2BI/hm%2B8avrHhTRtU/svR7LxV4O8Kaxc61cWrahph6F%0Af2Bv28dD/bZ/bv8A2wvA37Rv7MXh3Uf2qvgRovwM%2BGWlTfBX4r3Wr/Ce1%2BGWo62vwc8Za3qkHxnt%0ALbxR4j07RvE2uv42TTbPQNP1bxENC1HQLfQNO0m/0nXgD4A8Vf8AByDL41/Zh/4Ji%2BL/AIReFP2d%0A/gJ8d/8AgpFcfGK6vvEn7ZHxatfD37L37LHgj9nbxz4o%2BHnxQ%2BIHxK8Xw6/8NdU8WW/ijxJ4Q1i2%0A%2BDfhe013wXfeMZ4JdFvtX0vxQ%2Bi6D4h/Tf8A4JCf8FJfEn/BQXwP%2B0RoHxIi%2BB%2BsfF39lL44an8G%0AfHXxS/ZY8Xv45/ZZ%2BNdhd2jeIPBfxN%2BBuuX/AIg8S%2BIIPD%2Bs%2BH5UttS0XW9a1yaw1Gy%2B1Ra7M%2BpX%0AOheHfzp/Z8/4ID/tAfsyfB//AIJxar8NP2oPgw37W3/BMD4j/tBXvwg8d3vwY8UwfDj40fAL9pLW%0A/E/iL4k/Av4v2l54%2B8RaxoGr32o/ET4j6fofxa8G6Tc3PhPw34hitNK8Cz%2BKrf8A4TWv3j/ZZ%2BH3%0A7U/hd/i740/az%2BL3gzx741%2BJ3jjSNY8I/Dn4S6NqmkfBj4EeAvD3gfw34Ys/BXgi68VJJ468W6x4%0Aj8Sad4n%2BIHjfxn4purUalq/ie20XQfDXh3RfDtrHegH8%2Bf8AwdyeAvBPjD9jz9hdfFPh%2B0vW1n/g%0Ap9%2Bzt4C1TWLPQ4dS8VL4J8VfC79oQeJfD%2BlXEOm6jq01pqhs7O6k0O0t7yHU9R0/TXfTb25trZB/%0ASB8dvjl8EP2NfgH4x%2BLvxP1fQfhz8I/g18PfEevf2dYR6TpZPh/4c%2BCdZ8VP4P8Ah/4aE%2Bmw6vry%0A%2BFfCmpR%2BFvBmhoLu%2BXTxZabaiOI%2BX%2BXn/BZv/gmn%2B1J/wU58P/AX4a/Cr4%2B/A34GfDP4H/Gf4f8A%0A7S9nfeMvhN448f8AxJ1L42fDmy%2BIWgeH449U034jeHvB9v8ADdfD/jp5r/QbrwjfeIdQ160huE8S%0A6fpcMumX/AfG/wD4Jsft7/t8/Fb9nux/4KF/HL9lPVP2TPhp4T%2BPUfxN%2BCP7Lfgr47fDnXPiB8QP%0AjN%2Bzx8Xf2dLbxHZ%2BJPHXxI8YrZjwt4d%2BLmraz4dvrgi60CRdW0T7Brlt4kv9QiAPkf8AZR/4OEfi%0AD8ZfjX%2BxJrHxMu/2FrT4Af8ABQf4o6v8IPhv%2Bz/8EfjHf/Er9uT9mPxLfRTaV8K9Z/aSs5vFlh4a%0A1fQPiH4r0q%2BstfHh74YeC3%2BGcev%2BFdLvbzxTrM1xHL4r%2ByN%2B0h%2B2P%2Bwd8HP%2BC0v7Snwf/Zi%2BCHjn%0A9lL4If8ABXP/AIKIfFv4lWfi/wCLniLwV8YPiNpGhfFvSfDfxAT4IeHPC/wgvfBPhjw98OPAfhbT%0ATaXXxA8UeJb7xJ4i0rxPpmnpoukabomlR/sN%2Bxv%2BxZ/wUR/Zp%2BEv7OP7I%2BvftZfCK/8A2cv2X/Fv%0AhO10b4s%2BD/AXig/tJfGf4AfDi8kv/AvwB8aaH48vfFXw4%2BGmmSQQaF4F8X%2BMvBmqeJNVvfhRoZ8I%0A%2BDNN8Ca7qS%2BM9L%2BbviJ/wSs/b11r4Gft6/sb%2BCf2if2VrP8AZy/b9/aT/ad%2BNeveP/FPwp%2BKN38a%0AvgT4U/ai%2BLmoeOvGPgvw14a0nxxaeB/i3faf4funttH1nW9e%2BGr2eu6rqLSrfabYaWigGf8AGz/g%0AsNqXx3/aV%2BB/7MH7Gf7RX7O37JugeNv2Jvh9%2B394v/ai/bF8IjV/D1x4R%2BLZ8N3vwU/Z68L/AA41%0Ar4l/B%2BB/Hfjjw94n07xr478QL401FPCngz7fZ6Db3fi3S9QtLX5J%2BI//AAX%2B%2BN9xov7GXhHxl4t/%0AZ6/4J36p%2B0H8CvjL8QfE/wC1V8efgV8c/jv%2BzH4y%2BPHwX/ag8Ufs%2B6V8AfhHqfhTxd4CHgrwz8Qo%0Afhb4o%2BKviv4nfE7U9Zj8GfA34gfD%2B9s7S18Za/4e1C//AFpuv%2BCYPiX4C/tHfAL9p39gv4h/Dr4Y%0AeJPhf%2Bx/4L/YN%2BJPw2%2BNfw%2B1Pxr4C%2BK37PnwtutG1L4Ta5Fqfw68SfDjxB4Z%2BMvgSXQ4/D3/AAl1%0A4ninRdc8G3NjoX/CPaNbeGrOHUvKvEP/AATl/wCCgmrfF21%2BLniP9r39m/4/Wnjv9j%2B%2B/Z0/aE%2BC%0Af7Sv7NGveNPgT448cap%2B0B8bP2gtJ%2BIXgvwJpHxM03TvDWifCe6%2BI3gz4T%2BAfDPii18b3fiP4TeC%0ABY%2BPtZ1Lxiug%2BL9DAPCPij/wVs/bhtPhN%2BwT4J8BfDL9l34f/tZftMfsJfFP9u74weMvjRP4k8S/%0As4%2BGvB/wY%2BHOmeMda8C/CLwz8LvjHL4h8aa94%2BOsx%2BI9F1ef40XNn4A8CWFrqGuaf4tude1WXwJ8%0Aq6x/wWk/4K26Z/wRbuf%2BCxLfB7/gn7a%2BC9Q%2BIWnaxpnwu1jT/wBpfTvGehfA3Uv2g7j9nDT7C504%0A%2BKpNJ8Z/ELWPGFz4e8Sad4xt/F3gbwg3w/1G48QweEtQ1RLHw3ffG/7c37Bfhv4e/tSfsIf8E7rz%0A9p/4Afsx%2BBP2Qf8AgmB4zfwB%2B01/wUU%2BAn7P3x0%2BBX7SWtH4m2Xif4pw/C/4e/GzQ9X%2BDvw28a/C%0Am48E6L4z8aanq3jQfEyz%2BG2pr4fv/C3iP4TQ%2BJPEHxQ/T/4W/Dj4n/8ABbr/AIJS/t0/sMfF74w/%0AAXxJp/gX4623wI%2BAn7cX7L3w%2Bj0/9nf4s6d8G7v4Q/Gbwjr3w68BXt9abNE8BeJYpPgX431f4fyD%0A4d6xo%2Bm6vbfCvxv4zFnqfiG5APqT40f8FUvjd%2BzX/wAFZ/jV%2Byn8bPB3wYtv2J/hd/wTH8ef8FFb%0Afx94JtfH2u/tDQeE/hfqF5o3jX/hJodR1rSPBTTxa54L%2BI9no/gLwp4V8QX1/oS%2BC/EsvxAtr7Uv%0AEHhDQfgb9kL/AIOHfi38YPiP%2BxHrnxZ1L9gDVfhn%2B3x8e7n4NaF%2BzV%2Bzf8VPEvxE/bJ/Y6s/GGo3%0AWifBbxJ%2B0qU8X61beJIvFHicaV4Z8Tz2fwS%2BEWheGE1/Rdf1bW9JnvX8LWv2t/w64/bH%2BNf/AAUZ%0A179sf9rv4vfst%2BJPg78Xv%2BCePxH/AOCdvxz%2BAvwi%2BH/xO0a71H4N%2BO9X1zxVe6P4Z8XeL/EurT3N%0A5rHjzV38Uan4vvYtGvNM8PzSfD6y8NaibceN7z1n9kD9hD/goR%2By18Pf2af2SbX9sv4Xz/skfsv%2B%0ANdPv7Dx94e%2BEt5F%2B1T8avgv4W8Q614h8D/s4ePm8T33iH4TeBvCljbXug%2BCfGPxD8BaXJ438V/D/%0AAML2%2Bj6DH4C1zV9V8TSgH6ZftU/EP4lfCL9mz46fFb4PeF/CXjX4l/DT4WeNPHvhHwj461zUvDXh%0ATxHqXhHQ7zX30jVtb0jTtUv7BL2zsLmG0eK0WKW/a1t7u90yzmn1K0/mZ8Af8F3f%2BCkA%2BEv/AATs%0A%2BOnxN/Yz/Zau/C//AAU4%2BIH/AAoP9nn4feDPjJ8R9K8c6P8AFLU7e08PeAfib8RvEN34W8aeGNE%2B%0AD/j/AMfX63es%2BF9Os9V8dfCn4c2L6xfar478R3s/h7RP6XP2t/hp8QvjR%2By5%2B0L8HvhTqvgjQ/iJ%0A8V/g58Q/ht4T1n4kaXf634F0vU/HPhfUvDAv/E%2BkaYkt5qOmWsGqTzS2ccF1HO6Rpc2d7ama1m/n%0AZ0b/AIIz/wDBSPwz8Dv%2BCT/ww0v43fsIatrf/BLz4o6/8SfDN94i%2BHfx4n0LxuBD4c0/wRpWs2tv%0A4nF3eajoNrdfEDV9U8RaefCn2/XF8BHSNF0GPS9autSAPrjRP%2BC0WrfBvw7/AMFZbD9tX4W%2BA/DH%0Aj/8A4JN6L8Adf8fal8D/AB5e6x8OPjOP2nvh9ceLfhF4U8B3fxA0zQdc8P8Aiu/8UxWfw8u4Nfin%0AjuNZ1zRri1givJbvQrTjf2IP%2BCv/AMZ/ir8U/hN8KPjtq/7DHxo8V/tO/sxfGX9pr4J%2BGf2CPirq%0AXjfxL8P9V%2BFFj4C8Qj9m34z2fiTxl4viu/iFrHhrxhqw034kaa3g3wvrPijwP4u0DTfB4h0W81SO%0AtrP/AARp%2BPHxg%2BPX/BZfV/jz8W/gnP8As%2B/8FZPC3w%2B8Ivp3g/w54%2B1b4t/CMfs8eD9S8Ffs4eKr%0AK48SajYeEL/UtDt9RtvEXjnQZY7uyvNc8PaRbeFdY0HTHmii%2BjPib%2ByH/wAFMP2of2a/Hv7MPx3/%0AAGjv2evg14W8SfsofE34K3/iz9lrw540tfEnxN%2BLninw1YeG/Bfj3xJB4p0bQ4/hL8L9AjtNQvPF%0Anwo%2BGeo6lfeMrXxFf%2BGovG2g%2BHbSKzuwD5Z/Y/8A%2BC2Xxs%2BIf7Zv7JX7Jv7U/wAM/wBmTwZ4l/ba%0A8DfGjxD4S8AfAz456d4%2B%2BNv7J3jr4M%2BHb7xjJ8Nf2s/CT65qlja23xH8H6HqesfDPx3oE3h%2B88R6%0AvdxeFW%2BHNpN4d8Taza3/ANl3/gtN8Wv2s/2ktO%2BGvw80D9k7TraL9sHxx%2BzV8Qf2VdR%2BJfjaf9uT%0A4U/Dn4W3nxH0nxh%2B0R4m8Mahpnhnw5q3ha4uvBmiapeaPoHhu50r4fWvim38P61458R%2BI4DA/wA2%0AaV/wQ2/4KC%2BHPDH/AATrtfhl%2B0F%2Bxr%2Bz/wCJ/wDgml8L/jr8Hvg7qHwi%2BHXxYhHjK%2B/ai%2BF2rfCz%0A4w/tQ%2BJtSur/AEx9I%2BOtmlv4W%2BI%2Bn%2BDJ9F8aeAPHvxRfxPq3jPV7Pwnqdz4L1P6e%2BHP/AASM/ab8%0AUa3%2Bxdqn7XXiz9ln4l/Fz9j79qbw5%2B0Cf29fAun/ABQ0/wDbH%2BMXgXwFDrtz4R%2BBfj3UfEljc67e%0A%2BG9VutcTwv44vNe%2BOni7wvrvw98P%2BEtHuvhvfa1oA1y%2BAKPjX/grp%2B214a/4Jof8FLv2p5/2bP2c%0AND/ab/4J8/tQePf2etd%2BHq/Fnxt46%2BDtzpPgnw58EvFGp%2BOIdbsfDng3xH411DRrL4wG2ufBy3nw%0A2fWf%2BEfu7hde8O6o6eGXi/bb/wCCnH/BSr4VftY/sB/ss/sofB39jDxz4l/4KCfs9fELxn8PLf4z%0Aa58a9J1zwJ8U/hz8NtG%2BJHia/wDiDH4JkuvD9r8L9N0ptcXRdO8Oa54l8ReLbu2v7TVvEPw40/w3%0AaeIPHF79o/8A4JK/tceOPgb/AMFNf2SPgL8Rv2afh78Hf%2BCkn7WOqftP6/8AF3XdP8aJ8Rvhpa/E%0AbQ/hJY/GT4d23wZ0nwZf%2BCfFuo%2BK9R%2BEVvdD4oTfFXQNUvx488Y6nceFNM8TxaTrsPonjz/gmN%2B2%0A346/bb/4Jm/tf3X7SP7K9lJ/wT5%2BGWo/D6Xwrb/s%2B/Fsp4/f4sfDO3%2BFX7QGoxySftAyNYJP4ck1%0AG6%2BDsLXaHwvrZsr3xq3jewWfSXAPmL4uf8Fmv2s9F%2BLNp%2Byzc/EL/glB%2Bw/%2B0v8ABD9k/wCCvxU/%0AbCtP26/jner4Nh/aa%2BNXhqTxbo/7NfwIttK%2BM/wdPiH/AIRzwhBZeLfH3xA0/wAZfEvwj4Ok8YeF%0A/Cqaz4igudK8beJ/3H/4J8ftbWX7d37Ff7OP7XNl4VfwOfjl8N9N8W6n4OOoPq0XhnxFDdXuh%2BKd%0AEs9WktNPl1XTNO8S6RqttpWqTWFjNqOmx2l5NZ2ss7wR/Nnjj9ib9oX4VftufGT9uD9iT4g/BTSN%0Ab/at%2BH/wj8C/tV/Bn9oDwp4rvfCfjfxB8CLTVfD3wk%2BMngz4heAr5fF3hTxh4T8Ba7qngbU/B1xp%0AGpeD/F%2BktY3t5LpOt6bY6pb/AHr%2Bz74J%2BKnw8%2BD/AIO8J/G/4sRfHH4s2MWs33jz4o2nhGHwFpni%0AbX9f8Rav4huF0LwXBrGvx%2BGPDWhR6rF4b8MaKdb1SWx8P6PpsM95NMsjkA/FjUf%2BCjv/AAUM/ad%2B%0AMP7ca/8ABOb4J/s5a18B/wDgnV8UvEfwJ%2BI//DRuk/G%2B5%2BKH7WPx2%2BFltLrPxo%2BE37PWofDW%2B03w%0A18M9T8MxWY8FaBrni/w/8V/7a8Q%2BKfBPiqTw9beHdWOmH5y1P/guP%2B2p8TPjl/wSq0j9ln9lX4C%2B%0AMfgt/wAFWPg/8Qtc%2BGVv8RfiB8T9D%2BJHwz%2BJXwv0HSLv4xXfj3xHpXg1vB9x8P8A4D3b%2BIde1PTv%0ACnh7xF4n%2BLvg/S4ptP1D4UX91pv9r/W0X/BMr9sL9nv4if8ABQSD9iD9oj4N%2BGPgl/wUc%2BI3jD42%0AeNdE%2BO/hX4oeJvGX7Nfxx%2BLWhalonxn%2BJnwaXwb4z0bw/wCOZ/Gd3eWXiTTdC8Tt4LXQr3QfC%2BiS%0AaxqmieHY4tQ8d8Kf8EWv2gvgl%2B0t/wAEifG/wK%2BMn7Px%2BCP/AASt%2BDPxD%2BGenxfEHwF8Sj8Wfi9r%0AHxz8F3/gf4369rUPhzxnH4G0mz1S21fVvEvw2tNMWzvvBviXUb3/AITHUfijpUlhY6WAfoN/wTc/%0Abs%2BJv7XGr/tifCD48/Cnwf8AC/48/sS/tH6r8A/iDP8ADDxfqnjP4U/ECwuNEsvE/hDx14KvfEmj%0AeHvFeif2lpV5Paar4Y17TruayNjYaums/aNdvfDHhT9P6/HT/gmV%2Bwr%2B1h%2ByV8eP%2BCgHxd/aQ%2BKP%0AwE%2BI8X7Z/wAb9O%2BOVkvwe8L%2BPfDF14e1yHTJ9AfQp9L8W6jqVtpnh7SNBg02y02NdZ8V6xqtx5t7%0AqWtW0lsyan%2BxdABXz78Pv2rP2c/it8aPi9%2Bzv8Nvi/4N8bfGn4BpozfGX4f%2BHNQfU9Y%2BHT%2BIIUm0%0Ai28Sy28DabZ3t0HaJrBb6S9tbq3vLK8t7e8sb2CD6Cr%2BMf4//Fbxn4C/4J7/APBQv4beB/F%2BofD/%0AAFH9rT/g4k%2BMH7Ieu%2BJ7CfxtN4itfAPxr%2BP/AISsviRY%2BH9Tnl8L6xpM3iXwFo/iLw/dyRXEGi23%0AhTWNY0bwJrfkzeEfFMYB/ZHo%2BvaH4itZL7w/rOla7ZQ3dzYS3mj6jZ6naxX1lIYbyykuLKaeJLu0%0AlBiubdnE0EgKSojcVq1/NtpHgH4UfsGf8F6v2S/2e/2Sfhb4P/Z%2B%2BC/7Wn7DXxqb4yfCr4Q6Vp/g%0AL4ZeLPGPwP1yTxB8NviRrfw%2B0WxTwtefEDw/pkXiHwz/AMJpZafpXjHVdL8T3sOveI9Wsof7OuP6%0ASaACvnr4N/tYfs5ftB%2BOvjh8Mfgv8X/B/wAQ/iF%2BzZ4wj8A/HfwdoF5PJr/wv8XT6h4k0q20bxVp%0A91bW01nLe6j4P8U2lhcxiaxv5dB1P7HdTrayMP53vBfwF8PfBf8A4OqPDmmaF8SPjR4ltfFv/BKP%0A4k/He8g%2BL3xY%2BKHxouR4q8fftX%2BMfC9/4P8ACepeN/HF03gP4eeH/DejaDD4V8I2lrq/hTR7PwjY%0A2dt4cTX9QTxdpf7BfDqR7L/grx%2B15Y2dvolvZ63/AME6/wDgnprWsuGgg12/1jSv2j/%2BCmGkaZdx%0AwJNHNf2kWlXE1lql9Ja3TWn2fw5Ztd2kb21vcgH6SVxV78Svh1p3iJPCGoeP/BVh4sku7Cwj8MXv%0AirQrXxFJfaqtu%2BmWSaLPfx6k13qKXlo9hbrbGa8W6t2t0kE8Rbta/kx/4LtfCr9l39lP/gov/wAE%0Agf8AgqR8Tvg98FrXwPF%2B1drXwF/a38feKvhv4f1y38TS/ET4X2GifAj4qfEs3NxYw63q/wCzfo/g%0AXxl4y8GeMdRttZ8S%2BDU8KeG9Q0f%2B0V8B%2BGPDcgB/UpD8Ufhnc6%2B3hW3%2BIvgSfxQuoT6S3huHxd4f%0Al19dVtZZIbrTG0dNQbURqFtNDLDPZG2%2B0wyxSRyRq6MBf0Hx34H8U32oaZ4Y8ZeFPEepaTn%2B1NP0%0AHxFpGr32m7Zmt2/tC00%2B8uLiz23CtAftEceJlaI/OCo/iC8MJ8Av2Tf%2BDoWx/aU1P9mH4VfD/wDZ%0Aa/bB8bfG79jH4MfFm9/4Qu50ax/bZ%2BGWi/B1/jB8afCugBLm3%2BFPjXxR8Y9S1T9nJdQ059E1nxjq%0A3ir4na8ZZNQ1zxrYP%2Bh/7E/iP9kH9kfxDc/tffBD9mPw140/a8/4LT/tS/HW5/Yw%2BGXgfQPhz8Lf%0AGWqfsj/DddN1SS6tfEqC38GfDn4NXngH4a237WfxEvtNtU17xh4l%2BKvgDTvHVh4l8W6RpV94dAP6%0AmaK%2BKP2Qv2ybf9pnW/j58KfGnwy1f4FftJ/sq%2BN/CXgX9oH4K6t4s8MfEK18MXHxG%2BH2g/FX4ZeK%0A/CXxD8GzPofjLwR4%2B8B%2BI7PUNIv7mw8MeJ9N1fTfEWg%2BJPCGi3Ol282o%2BR/tj/8ABSrwr%2Byr8dvh%0Aj%2BzJo3wwufiX8avil8KPiJ8bdHtPEHxb%2BD/wG%2BHNp4C%2BG13aaLqNhcfEn4t%2BK9Hsrn4heJfFOsaF%0Aofg7wfYaRcWd5Bc6tr/iLxJ4X0PQbq8lAPZPFv7HP7Ffi79tr4W/tkeKvh14Ku/23vhV8Otb8JfD%0AP4gnxfr2meNdM%2BHevaf4w8O6qD4H0/xPZeHfE1odM8XeNNGtPEXiDwnrWo6RaatqtjpWq2MamOL7%0APr%2BPn9rv9pvSfiN/wU0/4NrP%2BCgngT9kD4u3/wATfjt8Lf8AgoDocnwd0nw34RsP2lNUjk%2BAel%2BH%0AvCHw58YXfifWPCfh3S/A/wAKPF3xc8b%2BNLrxP4q8X6XoPg3wV4i%2BIvj7UdN0JV1Oyvf2M%2BCf/BWO%0AH47fss/tvfHrwX%2ByT8etX%2BK/7C37Qfxh/Zl%2BIf7KHhKbwp8SvjP4w%2BKXwck8Kxaxp/gCH4caj4os%0APENjqFx4oazs73TVvVubvw14o/sdNbsbCzv9RAP1q1bVtK0DStT13XdT0/RdE0XT73VtZ1nVr220%0A7StJ0rTraS81DU9T1C8khtLDT7C0hmur29upora1topJ55I4o2YeVSftGfs9xeEfCPxAl%2BO/waj8%0AB/EDW5PDXgPxtJ8T/BCeEfG3iOK%2Bv9Ml0Dwj4kbXBo3iTW4tS0vU9Pk0rRr29vkvtOv7RoBPZ3Ec%0Af51fsh/8FDLr9r343fGr9hL9qv8AZg8CfBr41aH8BdI%2BL3iL4XeH/jj4S/a1%2BFvi74J%2BO9d1X4Ve%0AMPCXxA1y3%2BHfw7/4RT4i%2BGPGtprHgf4lfBbxx4FMi6bc2moQajrWlanPFa/zT/tEfDrwJpf/AAbO%0AfsnWdl4a8NWkPwt/4Kf%2BPtP%2BHCXGkx3d54Vs4/8Agp5%2B074SFh4SvpbS8udFlPhiefTLmZLuxF1o%0Acd5p891OZ47O6AP7Opv2N/2Xbn9pu0/bNuPgl4Hn/aksPC7eC7D43TWE8vjey8LvpVzoT6PaXr3L%0AW1tbvo17eaW0kNolw1jd3NsZjFPKr9/45%2BPfwL%2BGOvaL4W%2BJXxo%2BE3w88T%2BJJI4vDvhzxz8RvB/h%0APXtfllnt7aKPRdI1/WdP1DVZJbm7tbeNLG3nZ57m3iUGSaNW9UuJHhgmmjt5bqSKKSRLW3aBZ7l0%0ARmW3ha6mtrZZZmAjja4uIIA7KZpoo9zr/Cz4H8U/EHxd4G/4OJ/j/wDtUf8ABN7x78WfAvjr9pL4%0Aq/Df9qDxX8Ov2lf2f4fiP8J/gh%2BzZ8J/hrr7fCDRPGcurax458Rp4A8B/wDCO%2BJtetfDTP4I0iOK%0A00bw9p8Oq%2BHta0vSgD%2B6uivxe%2BK//BUO4/Z3%2BKP/AAT4/Zi%2BHX7Gnxc%2BLN5%2B3V8FrrW/2aZ9D%2BK3%0Aw6t1F/8ADPwJ4X8Y%2BOvCXjXUPH/iUa7Yab8Nfhv4j8PeK/FPxD168vb3W7GPXv7A0vxZ4g077Hqe%0A5pv/AAWZ%2BA/hT9j39sP9r39pP4dfEv8AZ70j9hr9pHx3%2Byt8bvhzqFrYeN/GN/8AFrwhqHw/sNE0%0A74ZPo8mm2HjbSfiIfil4GufBviGR9D0CXTNXk8R6tqel%2BENPvPEkQB%2BrV9478D6X4s0XwHqXjLwp%0Ap3jnxLaXd/4c8GX3iLSLTxZr9jYQXdzfXui%2BHbi8j1jVbSyttPv7i7uLGznhtoLK7lmdI7aZk6uv%0A5Pfjr4s8e%2BP/APg4G/4IMeOfix%2BxRJ%2Byz8SPEPgD/goPPqnjm88UfCjx7rPxN0A/sd6je6J8N/EX%0AjPwE6eJJ9f8AgLqWo%2BJF1Pw14it7rwn4fm%2BKH274e%2BItfTxD4oey/rCoA881f4vfCfw/qV3o2vfE%0A/wCHmiavYSCK%2B0rV/GvhrTdSs5WRJVju7G81OG6t5GikSQJNEjFHRwNrAnZ8MeO/A/jX7d/whvjL%0Awp4t/sz7N/aX/CMeIdI17%2Bz/ALb9o%2Bx/bv7KvLv7J9r%2ByXX2bz/L8/7NceVv8mTb/Bz%2B1p8RP2Jr%0An/g4G/4KL/tNftA/sNSftPfs6/sF/saaP4Q8Z6J8P/2StX%2BMXw%2B8UftPa34e0vxtD8Q/2mLnQvhf%0A4p%2BHWjR6f4Wu/i/8Ppviz8fJoW0r/hXHh7UPD8utaX8PrKbQ/wB1LP4RfAb/AIJzfsrfFT/gr34H%0A/Zk%2BCX7JvxOu/wDgl7oN78T/AICfAH4ZaT4L8KaL8bLjRtJ%2BJMOkW%2BmeGE0jwvrNufiFqOjeBZ9S%0A1PwtZahZ2Ph%2B31bUNXurbVNVitAD98PD/j7wL4t1nxf4d8K%2BNPCfibxB8PdXtfD/AI%2B0Lw/4j0fW%0AdZ8D69faXaa3ZaJ4v0zTry5vfDWr3mi39jq9rputQWV7caXe2moRQvaXMMz682vaHbW%2Bt3dxrOlW%0A9p4Z83/hJLqbUbOK38P/AGfS7XW5/wC25pJlj0rydFvbLV5ft7W/l6Xd2t%2B%2B20uIpX/Gr9na7%2BKX%0A7E//AAQyvPjpBptp4z/ak0b9iD4nftrfEu88TrpFxcfED9qzx98KvEH7RHj3VviFqmma1b2fiTf8%0ASNYuNJ1nVrHXlW88M6RbWfh17XT7bRtMtvy2/wCCLnjL9jb43fswftJ/tJ%2BFP22v2mv2xf2rNV/Z%0Aa8UfET/gor8O/iH4n/aN%2BG/7M978c/jF8MdNuPFUifDRvA3w/wDhlqXirwT4c%2BFlp8CdF%2BIXwlGt%0Aa3pvw08OWFzbraLqnhF7UA/qz%2BFXxV%2BHHxy%2BHHgv4vfCHxp4f%2BInwy%2BInh%2Bw8U%2BCvGvha/i1PQvE%0AWhanEJbS%2BsbuI/78F1azpDeWF5DcWF/b217bXFvEfFH4q/Dj4J%2BCNV%2BJXxb8aeH/AId/D7QbvQLT%0AX/Gniu/i0jwzoDeJvEWk%2BFNIu9e1i5K2Oi6VJruuaZa32tapNaaPo9vcPqesX1hpdrd3sH8n37KX%0A/BXbwv8A8EvP%2BCOP/BKK88U/s4fGL43eFPFnwS8GXfxM8X/Dx9K03Sfhb4L1z4sJ8P8AT7ixbW4z%0AH8Q/ibr%2Br65d3Pg/4VaRcaTPrOneG9evdZ8U%2BF7WPTZtS/Z34Oft/wDhv9p39rH4x/8ABND9ob9m%0ALWvhH8Sr79ku3/aLufAPinx/8OPizpnij9nrx54lh%2BE/ifwp8WoPAGoahpXw6%2BJUV94m0iDVvhq2%0Aq%2BMLLUfCXiK21%2By8V3Fhc2ougD9X7e4guoIbq1miuba5ijuLe4t5EmgngmRZIZoZo2aOWKWNleOR%0AGZHRlZWKkGsHxd4x8I/D/wAN6r4x8eeKvDfgnwjoUCXWueKvF2uaZ4b8N6NbSTxWsdxquuazdWWm%0AadBJczwW6TXd1DG080UKsZJEVvxr/wCCCmqePfC37IHxZ/ZD%2BIus%2BKPFGs/8E5f2yP2iP2EfD/i/%0Axi5uNc8VfC74Uan4e8afBXUzff2hcJeaPZfB34neBfD3hkQaT4UtdL8M6JpGhWnhu3ttJi1HU/AP%0A%2BDhTxP8AFONP%2BCXfw10D4J3Hxo%2BFPxM/4Kj/ALLmlfEHwLc/Ej4b%2BEfCHxk1TQR488beE/gL450T%0Ax9p1xBqmj%2BKtZ8JWXi23n1K5fwNfS%2BDbjw14qsX1HWfDE6AH9CPg/wAaeDviH4b0rxn4A8WeGvHP%0Ag/XYpp9E8V%2BD9d0vxN4b1iC3uZ7KebStd0W6vdL1CKC8trm0mktLqZI7m3ngdllikRelr8Bf2d/2%0AgLH9n%2Bz%2BJv7EP/BI7/gmr4f8YeHv2NNQ0VP2jvBev/tMeHfgT4b8E/GL4u6FP8UdX%2BFPw%2B8b%2BIPD%0APxr1D45fFXR/tcum%2BK9Z8Uat4I%2BHWiXMnhzw7pvxMm0WzaHw23xf/wAF6/BzR/8ABPjXPgn%2ByB8c%0AvjV4L/4KUW/i3w98BNTi8Z/B7wH4o0P40eA/E%2BueDfHHwe%2BKHhLxD4xuIfCOp%2BBPEGn6fB428Uw%2B%0AItW8O6Tbza%2BNKm1/VPDMuk6gAfv5RXh/7N3xV8a/G74JeAfin8RPgb8QP2bPGfi/T9QvNc%2BCXxSv%0AfDGoeO/A01lrmqaTbW2u3fg/V9b0GT%2B2bDT7XxJpiwX638Oj6zp8Ouadouux6lo1h7hQAUUUUAFF%0AFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUU%0AUAcf42%2BHngD4laVb6F8RvA3g/wAf6JaahFq1ro3jbwzovirSrbVYLa7s4NTt9P12yv7SHUIbS/vr%0AWK9jhW5jtr27gSQRXEyvseH/AA9oHhPRtO8OeFtD0fw14e0i3W00nQfD%2BmWWjaNpdorM622naXp0%0AFtY2VurO7LDbQRRhmYhcsSdiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACv5XPh1%0A%2BxAf%2BCg37GX/AAU8%2BFXg3xb4f0n4zfDr/guf%2B2P8ff2b/iD4us9d/sHwX%2B0F8EfjPo%2Bs%2BFLTWLlk%0A1/xFonh%2B/jXxH8IfEfijw9p%2BsXWg%2BEvEus694T8K6raWum%2BE7v8Aqjr5I%2BAf7C37LX7MHxZ/aC%2BO%0AHwL%2BGdz4E%2BJX7U3jK%2B%2BIXx11eP4g/E7xDpHjTxrqmva54o1XxHbeC/FnjPXfA/g7UtW1/wASa1qe%0AqyeBfDnhlNTuLwC%2BS4itrOO3APz9%2BDX7J/7W/wC0D/wU18C/8FHP2wvhj8L/ANnLw/8As7fswa58%0ABPgJ8BvAnxeu/jb401T4i/FG607Wfiz8UvHHj/TvBvgbwnF4K0nTrzWvh74A8M2ul3eu%2BICr%2BN9f%0AtPBUtrY6Hqv01/wTG%2BN/7UPxr%2BDnxhh/a91z4C%2BLPjJ8GP2pfjV8A9U8Zfs06P420f4ReJrb4a3m%0AiW13/YcXj65n1y41Xwh4n1HxJ8OfFTeXb2%2Bn%2BJ/BeraFJ/aGo6RqGtap%2Bj9Ynh3w14c8IaRb6B4T%0A0DRPC%2BhWkt9cWmieHdKsdF0i1n1O/utV1Ka303TYLazhl1DVL291K%2BkjhV7u/u7q8uGkuLiWRwD%2B%0Ad7Uf2Pf%2BCmHiP/guN8Lf%2BCknir4C/skP8G/BHwIl/Y1j0fRP2pviTN410b4R638ZvjJ4wn%2BPcVpe%0Afs8aPp%2BpfELQfB%2Bs20F98JJ0uPDutXXjXw1pem%2BOtNuT4r8UfDD9GPDD68n/AAU7/bmsvC95aL4t%0A1D/gnL/wTuu/Cc/iPSdQv/CWl6xafHr/AIKr2emJrb6KumTyaV/a9zaX13pj61ba/qlrJrA0a7S3%0AsHOmfpJXzL8Fv2Pf2f8A9nz4rftAfG/4WeFfEul/FH9qHW9C1742eKfE3xX%2BLvxIfxNdeF9c%2BIXi%0APw1YaHpHxL8d%2BL9B%2BHXhrQNX%2BKnj2fRvB/w00vwf4S0yDX20%2Bz0SHTtN0i008A3P2W7X9pqy%2BAXw%0A2s/2x9Q%2BD2q/tL2ukXsHxX1T4At4qb4RajrMWt6oml33g0eNtH0DxNFbXfhsaLPqlvqWkWYtddk1%0AS2shNp0VpczfMP8AwVo/YW0//go3/wAE/wD9or9lYRaYPG3i/wAG3PiD4OanrF/daVpei/GzwXnx%0AJ8L77VdTs7e8uNP0K78UWNnoXie6isNRkj8LaxrXlafeSlIH/RyigD%2BYH9r/AP4IafEL9or/AIIz%0A/sX/ALFlv4o0%2BD9rH9m/4gfA/wCM3if4nQ%2BJJ7rUtZ%2BL3jTxXf337Y/iWy%2BKviGA%2BJppdf1b4s/F%0AX4vQa9d2kmp%2BOvHHhPwjc6ppMWp3dvNpn0V/wVE/4JG%2BH/2hfDf/AATn8SfBb4HfCX42ab/wTd8S%0Av4U8NfsdfGDUrXw18Hfi/wDs4eN/CngT4aeMfAUmsyadfad4d8V%2BA9D%2BH3gfxZ8ONQ1bTdR8MQ3n%0Aha707VtB1dL60sW/fSigD8qf%2BCXn7LOufs36Z8ftVuP2KP2W/wBgrwf8UfHegat4B%2BBvwCvtM8a/%0AECPRdB0m/t77xJ8efitoGk6H4V17X9a1XVbqbwj8OvBdnqvhH4S6ML7S9I8YeKH168uLX4p/4LBf%0AsGfGL9sn9oPwXc%2BJv2CPhP8At/fsmaT%2BzPrXgvSfBOpftBeDv2ZPjd8Lf2mfEnxX07Urb4meDvi8%0A3gSH4gaH8PbPwVpOi2/xE0Gz%2BInizRfEumRS6lpfwh1XxH4OXwr8ZP6LKKAP5PPhZ/wTm/4KofBn%0A43f8EDtS8ReFvhV%2B0l4f/wCCY/wq%2BOfhT40fFnX/ANouXwtr2rXv7XPhe7%2BEfiHwpo2k6j8LNS1a%0A/wDDX7J3w10/wnH4I1KG31TUfjBo%2BhWuh3Z8GarLPqcNLwd/wR//AOCnlx%2Bzf/wV3%2BE174y/ZX%2BD%0Aeq/8FIv2rNb/AGwvA7eAPj/%2B0HqviDw9P8SfjP4Z8a/En9mr4x%2BLbT4H%2BHNM0fwDqHwssNf8CXvx%0AE%2BC3hw%2BNNc17W7tNbl13wLHp2h2X9aFFAH8j3wJ/4JLf8FL/AIB/tyeDf2uP2UPg3/wS4/YA8Cwf%0AssaN%2BzH4m%2BBfwm1vxx8RpNZ8MfDTx14Q%2BK0GpfEHx9f/ALN3hnVPHPj39ovxpoDfD74j/ESKG38b%0AeE/gna6Nrt14v8efGDwloV5f9H4t/wCCVn/BW3xN/wAEovD3/BPa01T9grw5470P9sTVP2gZviJo%0Av7RP7UFn4S1f4eaj%2B0h4s/a6i8KajpNp%2By9aapd%2BJdP%2BMnia10y0N8t94bi8MeHtH1%2BO0j8aWdjf%0AWP8AV9RQBz/hOTxVN4V8MzeOrLw/p3jeXw/o0njHT/Cep6jrXhWw8VSadbN4hsvDOs6xpOgavq3h%0A%2B11c3kGjanqmhaLqN/p0dtdX2k6dcyy2cP8AL94a/YF/4K96R%2ByT/wAFnPhdqHw2/YSj%2BJP/AAU2%0A%2BJvxS8d%2BEbTRf2kPjpc%2BF/h/pP7SHw48PfCr4q6LceHbv4F6J4WvvEfhD4dpNFo3j6zl8K6x45%2BJ%0AunX%2BreObLX/BmleCrG4/qhooA/nC/wCGSP8AgrNq/wC1b/wR1%2BN2rfDH9iPw34I/YN%2BFviP4OfHD%0ASdE/aG%2BLXjXxL4m0P4%2B%2BG/CXwx%2BMPivwamqfs%2BfD/TYLzwR4D%2BEng7xf4V8M3F7dLL468S3fg%2BXx%0AP4o8NWlz4xtfAZ/%2BCRX7f37Wv7PH/BTL9nH9sG1/Zd%2BCE/7U37Xvhf8A4KA/AD4mfCDxj4k%2BO3hr%0Awx8Zjonw70EfB74kfDb4j/Dbwrb%2BNPAWgeFvhF4b8P8AjLxHfxadaavr3iXXNU0Twd4r8OmHR9O/%0Aq7ooA/mU8c/Ab/gsv%2B0d/wAFK/8Agld%2B2T8Wf2Wv2SPhT8Of2I9d%2BOnhHx/4V8K/tV698Qtd1bTf%0A2jPAGhfCz4x/GTSprn4OeEktNHHhGa31X4R/C2OXxH4hTxL4J1nTfHfinS9K8S6PrKf0V/FXWviP%0A4c%2BHHjTXvhD4D8P/ABQ%2BJuj%2BH7/UfBXw68U%2BPZfhfoXjTXbSIzWnhy%2B%2BIMXg/wAe/wDCJ/2pte2t%0AdXn8JaxZwXj2636Wlk9xf2voFFAH8mP7JH7PP/BYL9nfw9/wULi%2BKP8AwS//AGO/2kvF3/BQv9o/%0A4ufHX4gy%2BM/22dD034U6d4V%2BIfhnTfD/AIY%2BCHiT4N6r8B/G%2Bk/ETwJ4I2eIE1K8vPEHh7V/iBof%0AiKXRvEcsdxYW%2BsP9f6N/wTn/AGpv2m/2C/8AgpGv7fifD%2B4/be/4KEfDrxJ4Om8JfCyXTr74YfBf%0Awt8GtL8aw/sdfDXwXe32qaWniG08E%2BONd1X4rX%2Bs%2BJvFUV5qfiDx7qNhrevWMdg91D/QfRQB/Mj8%0AY/CP7Xn7e3/BB/8AYtn/AGb/AAj4P8b/ABg%2BHWofsv8AiD9pv9ljx1dJ4R0P9ox/2N/EsnhP9or9%0AkvU9V1C6s4PBsup/Hj4XfYdZs9T1WygktPBereCbzVlGrPcv9VfBj/gqx4X/AG4fgj%2B3p8INU%2BAP%0A7R/7HX7Wn7MX7PXizV/jH8C/j34UHh3XvDR8XfDnxhceG/FXw/8AGWkXclr4v8H3d1pc0vhzxXNY%0A%2BEdR1yzNh4j0bQpdAv8AT9Vn/XHwR8EfhL8Nrv4o3fgLwD4c8KD41eNNT%2BI/xUsNFsvsmh%2BNvH2u%0A6Rpuh%2BI/GWseHkc6C3iTxVp2kacni/WLbTre%2B8W3VsupeJJtU1J5bt%2BY%2BG/7Kv7L/wAG7Dx3pXwh%0A/Zv%2BAvwq0v4o6fpmk/EzTfhv8H/h74GsPiLpWi6Re6Bo2meO7Pwx4d0u38XafpOhalqOiaZZeII9%0AQtrDSNQvdNtYorO6nhkAP5mP2N/2Cvir%2B3t/wS1/4N79BtfFvgLS/wBl39njxn4R/ad/aa8D%2BJL3%0AX7Lxd8S5/hg3iRvg94d8Hx6P4X1aw1bTP%2BEh1bxHb%2BNtG8QeI/CWnS6TqVnqlrNqmr6Rp1tB7R/w%0ASV8VaL8U/wBtf/guv/wVr%2BNet6J4W%2BHPhz9oDWv2Pfhz4z8T6lq2m6F8Of2eP2G/DUsnxJ8Yrfa9%0AcafZ6J4K8d6bF8PfiH4ot59BsYPDninw34rlbUrm8utfav6OfhV8Kvhx8Dfhx4L%2BEPwh8F%2BH/h38%0AMvh34fsPC3grwV4WsItM0Lw7oWmRCK0sbG0iH%2B/PdXU7zXl/eTXF/f3Fze3NxcS8Vqn7MH7OWsfC%0A/wAQfBG8%2BBvwqj%2BDvi3xW3jjxb8LdN8DeHdG8AeLPFdx44s/iVqmt%2BKfCej2FjofiO78R%2BOrGHxJ%0A4vGtWV9B4yv5L0eKotYt9S1GC6APye/4N9NF8b%2BJP2QPjp%2B19420e38Jj/gpD%2B3P%2B01%2B3p4K8Br4%0AYsfCmp%2BB/hp8WNT8MeCPAGl6/ZaWo0vU9Y8Q%2BG/hbZfEBvEtnNfS%2BJtO8Z6drmsatrXiG91fV770%0A/wD4Kw/spftt/tUa5%2Bwxe/skaf8Asmz2P7KX7W/g79sPxL/w0z8Qvi34Ql1/xp8K/DXi3wv8P/CG%0AgWXwz%2BDnxNhbQr2H4h%2BJ9Y8Ra5f32matp%2BpaP4ZTRLSRX1C4X9gre3gtYIbW1hitra2ijt7e3t40%0AhggghRY4YYYY1WOKKKNVSONFVERVVVCgCpaAPwbb9nH/AIKNfsU/H39rv4tfsUfCH9nr9pHw5%2B3t%0ArfhX4zeK/Cvxp/aI1v4WT/sx/tSR%2BEbzwZ4x1e0v9G%2BA97P8df2eryx07wHfadosN18NfiYX0PXb%0ABTp76tDqy/COi/8ABGL9vv8AZp/4cp/D79mO9/ZU%2BLPw6/4JYj4w/Ebx3r/xr%2BLHxR%2BEGsfGH4vf%0AtL%2BKPHfiP4p%2BG/Dtl4M%2BBnxmbwf4A8HW14lp4M8XXkfiHXdU1Txnpp1XwLHpvhbVZrz%2BtKigAooo%0AoAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKrtNIt3Dbi0uHhlt7mZ79WtBa2%0A0kElokVpMj3KXzXF6tzNNbNb2c9okdhdi8ubSZ7CK9sUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUU%0AUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQ%0AAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFAB%0ARRRQAUUUUAFFFfGH7a37eXwF/YP8EeE/E3xdm8YeKvGnxP8AFdp4A%2BB3wE%2BEPhxfH3x/%2BP3xBvZr%0ASGHwX8H/AIcx6hpk/iTVYjf2TalqF/qWjeGdD%2B3abHruv6bPq%2Bkw3wB9n0Vx/wAPfFlz498AeB/H%0AN54O8YfDu78aeD/DPiy6%2BH/xCstK0zx/4GufEei2WsT%2BDvHGm6FrPiPRNP8AGHhmW8bRfE1lo/iH%0AXdKtNasr2DT9Z1S0jhvp%2BwoAKK4fSfid8Ntf8Xax8P8AQviF4H1rx54dt57vxB4J0nxZoOo%2BLtCt%0ALaeytbm51jw3Z6hNrOmW9vc6jp9vPNe2UEcU9/ZRSMsl1AsncUAFFFFABRRVe7u7Swtbm/v7m3sr%0AGyt5ru8vLuaO2tbS0to2muLm5uJmSGC3ghR5ZppXSOKNGd2VVJABYor598dftY/sw/DT4F2v7Tvj%0Av9oL4O%2BGv2dNR0zQtZ0n44X/AMRPCx%2BFuuaX4plt4PDF94f8b2%2Bpz%2BH/ABBb%2BJJrq2j0B9Fvr/8A%0Atl54hpoud651f2f/ANpP4A/tWfD2L4r/ALNvxh%2BHnxv%2BHEusan4ePjH4beKNL8U6Lb6/oxg/tTQ7%0A650y4mOnaxZRXVldTaZfpbXosNQ07UVgax1GxuLgA9tooooAKKKKACivjb48ft//ALJ/7NPxx%2BB/%0A7Nnxh%2BJGtaH8cP2jplh%2BDnw98N/Cf4y/EzVfFSHxHpXhNr27vPhf8PvGWj%2BEdJi1/W9OsrjWvGup%0AeHdIto3ub64vYtO0/ULu195k%2BN/wXh%2BKFl8EJvi98L4vjRqOn6nq2n/CGTx/4Uj%2BKF/pWi6fpOra%0Axqdl4AbVh4rutP0nStf0LU9TvYNJktrDT9a0m9upIrbUbOSYA9Qoor4f/wCCin7b2h/8E7f2UPiJ%0A%2B1l4p%2BEXxL%2BM3hX4bnSz4i8O/C6XwZb6ro9jrF1/Ztr4l1668aeJ/DVvZ%2BE7XWJ9L0rVbzQovE/i%0AK1uNZsJ7TwtqNlHqN1YAH3BRXm/xR%2BLvw6%2BCvwv8W/Gf4q%2BJrfwP8MfAfhy48XeM/FutWWqR2fhj%0Aw1ZxJPf6zrVpBYz6nYWWm27/AGnVJJ7FTpdtHcXOoLbQWtxJF2Hh7xDoHi7QND8WeE9c0fxP4W8T%0A6PpniHw14l8PanZa1oHiHQNasoNS0bXND1nTZ7nTtW0fVtOuba/0zU7C5uLK/sriC6tZ5YJY5GAN%0AiiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAK%0AKKKACiiigAooooAKKKKACiiigAor8T/jP/wV18R3P7VnjD9lT9gP9k34jf8ABQbxt%2BzhqOmaV%2B2m%0A3w18T/D74aeGvgVrvjlNX0z4b%2BCbb4i/F7xf4N8K%2BIfHVx4h0XUZ/GenaRBrOg%2BFtB0vxCmqeJNP%0A8Q%2BE/F2keH/oX4x/tM/Ffxr%2B3l8Iv2Fv2btRtfDd34H8IeGv2pv2yvinqvhWz8Tad4O%2BB0njZtD%2B%0AH3wR8NQalcxaefiJ%2B0lq3hnxrod1qu3%2B0fhz8PND1rxxpK3OuPoduwB%2BldFV2u7RLuGwe5t0vrm3%0Aubu3s2mjW7ntLKS0hvLmG3LCaW3tJr%2BwiuZkRo4JL20SVka5hD/mB/wUR/bV%2BJX7A3jT9mT45eJd%0AI8O6j%2Bwhrvji6%2BEv7ZXjG90vUl8Q/s9v8RNW8K%2BG/gj8eY9fs7j7BH8PdP8AiBqi%2BBviXo%2Bpafe3%0ArWni3QNa8P7JtJ1OOYA/UeiiigAoor4B/wCCgf7dek/sSeEPgpY6X4Oh%2BJHxy/aq%2BPfgb9lz9mv4%0Af6p4nsPAnhDXvjJ8SDeJ4fuviN491CDUG8H/AA%2B0j7G0uvaloug%2BLPFl9c3Ol6D4V8I67rGr28UQ%0AB9/UV%2BKH/BP3/gpD%2B0J8Tf2y/wBor/gmX%2B3d8Nfgf4L/AGzv2dPhV4F%2BOTeNv2XPFPjfxF%2Bz38Xf%0Ahd4ytvB39o6t4U0v4o6ZpfxJ8F6r4K1n4geEPD%2Bq6X4kfVhq9/f6ldadcW%2Bl6bp99r/7CQeO/A90%0AgktfGXhS5jbxXeeBFeDxFpEyN440%2Be5tb/waGjvGU%2BK7G5sry3vPDoP9r209pcwzWaSQSqgB1dFf%0AxP8A/BSPwR%2ByF4O/4Ke/Bjw/8KP2nPhlp37ZHiD/AIKG/CP9p39pX9rD4m/HPwiv7TH7Jnw81bTP%0Ah18Ivhr%2Bw3%2Bzx8L/AIe6Pp3xa%2BNOhfG60ur3SdP%2BFD6P8SPCfwV8EXR17456Hpfgv4mQfFPQ/wC2%0ACgAooooAKKKKACiiigAor4D/AG4v2k/iB8MtV/Z1/Zs/Z7vfCMH7U37Y/wATL7wJ8NL3xhZS61pX%0Aw0%2BFfgDSU8aftH/tD3fhYXekWvjFfgz8Ozbjw54QvvEGh2Xij4oeMvhv4ev7qbTNUv7Wb78oAKK%2B%0ABP8AgpT8Wf2jf2dv2Xtf/ac/Zvi0XxNqP7MmuWXx0%2BMnwc1jQ4b6f4//ALOXgrR9fPxo%2BFvhjxSZ%0AJLz4a%2BObbwtfv8T/AAN4007RfFE1z4q%2BG%2BleBtQ8PXWg%2BNNXurL6j%2BA/xp8BftIfBL4R/tA/C3UL%0AjVPhx8bPhv4L%2BKfge%2BvbYWWoy%2BF/Hfh7T/EujJqun%2BbM2maxb2Oow22saVLK1xpepw3en3OLi2kU%0AAHrFFFFABX82nxnupvF//B1N%2Bx94S8SsmseHfg5/wSe%2BJ/xe%2BGOmXcUJi8HfEn4k/Gb4pfCzxx4u%0A0qWNI7n%2B0/EXw%2B0jT/Ct4Lma4tU02Ai1t7eee4ml/pLr%2Ba/xpG%2Bqf8HY/wAI20xH1FfDf/BFPUpP%0AETWCm8GgpP8AtUfE6KB9aNuJBpaTS69oUcTX3kCR9a0lEJbUbMTAH9KFeP8A7Qvw28SfGb4BfHH4%0AP%2BDviHqHwj8XfFb4P/Ev4beFfivpOmPreq/DHxJ468F614X0P4h6Zo0er%2BH31fUPBep6pa%2BJLLTE%0A1/Q3v7nTY7VdX00yi9g9grh/ic3xJT4bfEJ/g1D4Hufi%2BngfxY3wpt/idc69ZfDaf4kroN%2BfA0Px%0ACvPCtpf%2BKLTwPL4oGlp4sufDdje69BoLX8uj2lzqCW8LgH8QXxz/AOCYum/sX/Fn/g34/Yt/ZK0v%0A4Za3/wAFNvhj8bbr43fHf9pX4IfCfR/hbqt3%2BzJ4R8VWR%2BIfjT9oseD7WXxl4r%2BG8ltqi/Cvw/4g%0A%2BJPi2OD4lWfhnxz4UOmLqnxCv/Dlh/drX8en7JH/AASj/wCDhL4T/HXVfHHxp/bL/Yd1fR/j/wDH%0Aj4H%2BPf2zPjp4A8QfGm5/a9%2BI3wq%2BE/jTwnq9x8LfAHjq/wD2efBui%2BD/AAu3grw9qXgPw34C0Cfw%0Ad4D0nSPE%2BvolhC2rahNcf2F0AFFFFABXnvxc0XXvEnwo%2BJ3h3wr4e8GeLvFGvfDzxpovhvwn8RtY%0A1nw98PfE%2Bvap4b1Kx0fw9471/wAO6H4m8QaH4M1rUJ7fTfFGsaF4b8QazpmiXN9e6XoerX0EFhce%0AhV5/8VY/ipN8OPGkPwQvfh/p3xdl8P38fw71D4q6Z4j1r4cWHiqSIrpd7400fwjq2geJ9W8P2s5E%0A9/pmh67ouo38SG1t9W055ftkIB/Ff%2B0Z8Q/jj8Xv%2BCVn/BvZq/wW/Zq/Yy%2BFnwL%2BI37eX7DWnaH8%0ACfEXxj%2BMmoaNH8UIfi5460/4O/DeNdc8AeJLzWPgv4z0DSbzX/jPqHiG4%2BKPj7wk2o6pZ6N4W%2BKE%0AGgT%2BPbj%2Bwb9lb4BfD/8AZx%2BCvhTwF4E%2BBPwF/Z0utQtLXxj8Sfhp%2BzT4esdA%2BD1h8W/Eel6dN8Qr%0Anwe9r4Q8BXevaINbt30vw/4h1vwf4c1q%2B8L6XoMF1oWhxWlvoun/AM7/AIf/AOCMn/BSHwD%2BwX/w%0ATx/Yt8CfH39hyGX9gr9tzQf2sNC8YX/w0%2BO2l6Z4r034Va9N8WPhH4f8V%2BHdA8Z26%2BL9T1X40eMP%0Aibd/FtoLnwLDrvhW78Cah4e1HSPGnh7xBq3iz%2BprSf7V/srTP7d/s/8Atv8As%2By/tn%2ByftP9lf2r%0A9mj/ALQ/sz7Z/pf9n/a/O%2Bxfav8ASfs3l%2Bf%2B93UAfztftYf8Fev29/gj8YfhF4M%2BGf8AwT18CeJv%0ACv7Uf7QPxs/Zv/Zb8PfF/wCOHxC%2BD37QvjrVfgSdStPE3xy8XfCO4%2BA2u2/hP4IeJp9E1PxF8Nki%0A8S6l418ZfD278I%2BL7nSfDkPjPRbR/wBjv2J/i58fvjt%2By/8ACr4sftQ/s9XH7Kvx18X2XiS48dfA%0AW78U23jK58CvpnjTxJofhxptftbWySZ/FXhHTNA8anTpraO90AeIx4f1EyahpdzK/wDPB4h/4JY/%0A8FzPGn7VnwX/AGtfir%2B1p%2Bwl8dfHv7MXiP8Aao0b9mKz%2BIfh/wCLGiaP8HfDf7Ss3iCwg%2BNE2j%2BE%0APhta6F4/%2BKfw78O3Hhaw8N/CnXNI0/wWtr4Z0LTJfic974b0vxTN/UD8IPBGt/DX4U/DX4eeJvH/%0AAIl%2BK/iXwR4E8KeFfEnxR8ZtEfFvxI8QaFodlput%2BPPEiWzGzt9b8Xapb3Wv6lZaeI9MsLrUJbLT%0AILfT4LaCMA/mu%2BNH/BQCb9q39rH9sf4DeLP%2BCunwS/4JH/s/fsnfFK1/Z80bQI/Hv7L/AIS/al/a%0AO%2BI/h/R9A8WeP/iVJ4q/aA8Sya78M/hP4V1zULHwZ4ZtPAvw%2B8v4hvB4iF98RHj0/wAQ%2BD7b9lP%2B%0ACbb%2BA/8AhQWtW3w9/wCCit7/AMFN9Hsfib4ht734/wCofEn4I/FO58NatJ4f8JXs3wnPiX4EabYe%0AHIU8PWN3p/itNK8RTap4utR44%2B0T36eHLzw1pmnflJ%2B0Z/wT3/aB/Zn/AGsf2kP2pv2b/wDgnj%2Bx%0An/wVL%2BFn7UvxD8L/ABw1z4AfG64%2BD/wY/aE%2BBXx60XSPB3h7V/FPwl%2BOnxg8FeOvBniHwN4t1Lw3%0AZ/E7VdC19/Deq%2BFfFunTT%2BDETUbq7vdf/Y79g3X/AI1eJPgNFqPx5/Yc8H/8E9vGw8YeI7eD9njw%0AT8Xvhj8a9KttBiXT2sPGdx4w%2BEPhjwr4Ih1DxPdSX7S6LZWd7eWVtY2lzqOotc3z2ViAfCHx21ub%0Awn/wWx8I%2BOLHSdK1jW/Af/BFH9sPxj4dtNYLRWZ1vRP2p/2Z5rVJL2G3ubvToryJ7jTLy9sYXu49%0AOvr2ONJVmeGT%2BbW%2B8FaPo3/BqvpP/BUCbSNKf/goJD%2B0V4c/bO1T9qz%2Bw1s/jR4o%2BOFx/wAFILn4%0AX6Z4o17x3e6Vp3jE2lt8MfEsnhmHTPD%2BoaR4cexQy6C91o2pXF1rH9N/jqz0PxH/AMF6Pht4S1YW%0Aeo2up/8ABGb9pqz13RXuB50uh6/%2B2V%2By/pjC6hhlS6gs9TjttTtILgGITPa3iW8pktpfL/L63/4I%0A5ft%2B6l%2ByVYf8ESvEms/CGf8A4Jl6B8bIvGdp%2B12/xD17UP2kPEX7OGmfHF/2gtI%2BAer/AAbtfDfh%0A/RLf4nf8JhLBocnxHj8SzeBdI0jTbbUNJ8KM9npmjW4B/U18JvHX/C0PhX8NPiZ/Zf8AYf8AwsT4%0Af%2BDfHX9ifbv7T/sf/hLvDmm6/wD2X/aX2PT/AO0P7P8A7Q%2ByfbvsFj9r8nz/ALHbeZ5Kfkj/AMF9%0ALTWfif8A8E5/jb%2BzB4D174C6d8Qf2g9O0TwmZPjp%2B0d4A/Z30Xwf4Ns9dtvEmo/EZdT8c295F4oh%0A0zXfDeh%2BHD4as00uTUJPEbXI17TxpkqT934S%2BJf7U3gj/gsTL%2Byno2s%2BBNa/YOi/4J1%2BHviX4d%2BG%0APhnw1pul%2BJf2eviD4c%2BLNp8OPCGs%2BKPEf9gw6hdaJ8XNJ0zx34f8C%2BGv%2BEn1O0vrb4Y6/faXoPhw%0A%2BBfEWoeLvM/%2BCyn7MV18e7H4F36/CL9jrSPh1o0vxVn/AGmv28f2mvgV%2Byp8d/EP7FnwR8JfDXxF%0A410PxH4N8GftUaRqfhTWPD/iv4k2Oi6b44ENtrX9iaAl9droljNqX/Ca%2BEgD71/Y3/aB8G/ti/sz%0A%2BHfFOo658FfiFd61omteFfiToXw2%2BO/wh/ar8GXMUGr%2BIfCLJ4g%2BIfwl0TQfht4ml8baNo8mreIN%0AK0rwhoGiWF/qWs%2BFY9GSLR546%2BMv%2BCKXiHT9A%2BDn7W/7JGhW9laeDP2Av%2BCg/wC1N%2Byv8KYV1G6v%0ANZ/4UtH4g0b43fDCy1qC9urqe3TwRo3xmn%2BEWhXETraapoPwz0/UFjt72bULO18v/wCCCHjrwhdf%0A8EtfBX7VHxB/ZX/Zt/Ym1bxToXiXxL8V/GPwW%2BC/gv8AZx8A/GjwJ8KTrFp4f/aa1zwv4f8ABvgR%0ANE0LxX4RttQ8U2kuqafHo0enT6h4m8JwaL4O8QaTo9h89/s16l8Yfhd/wSU/4Kbf8FB/gb8Q/ht8%0AG/id%2B2J%2B0D%2B2J/wUK%2BCPxO%2BMWnMPBnhv4Mahr%2Bn6H8DtX8R6Xrfge11G2s/F/wAB/hdovi7wno/i%0APQddvG1n4h2Opavpmm3%2Bs6l4O0QA/pror59/ZK8ZfFv4jfsq/szfEH4/eFn8D/Hbx3%2Bz78GfGXxq%0A8FSaDqPhWTwf8W/E/wAOPDet/Efws/hfWJ7rV/Db%2BH/GN9rOktoOqXNxqOjtaHT76ea5t5ZG%2BgqA%0ACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAK%0AKKKACiiigAooooAK5rxp4q07wL4O8WeN9Xg1C50nwd4a13xVqltpNob/AFW407w9pd1q97Bplirx%0Ate6hLbWcsdlaLIhublo4Q6l9w6WigD%2BUv/giv%2By148%2BP8/jb/gqz8Dv25vEHwK%2BHH7d37WHiP9q/%0A4wfsUfBXwx8Fvib4Fsru28Qa/BdfBr4q/FLx74U8S%2BMIPHWtWmua7c/Fy88GaZ8NL1LzxTY3Gh6V%0AofiLwp4O8Yaf%2Bm/7I%2Bq%2BHv8Ah7r/AMFgdEmso7Xxz/whH/BOTU1v59Z1ea61v4dL8IPiza6DBY6D%0Ae28Oj2Wm%2BHPGknj9rjUdCnvbi7v/ABKYdea1aDR4pfBdP/4IkeDPg9/wUh%2BDP7X37GujfAz9jr4U%0AeFtYGs/GXwv8ArL40/Dnxl8Y9BPhDxtpl18HPE3w70X4mzfsvX3wy1HxpqXhDxkbrS/hD4R8Q6f/%0AAGSdPh%2B0z6Pp2p332x8aP2RfiZZ/t3/Ab9vD9mzWfBWj%2BJ/%2BEUH7Nf7Ynw/8Xx39jZfGr9mfU/EU%0Afibwx4o8Oa3p1tqEemfGD4AeMZL7xF4TW/0pIfHPhHX/ABL4I1HxPocFvosNyAfyzftffDv/AIJt%0Afte/8FA/2D/hv4P/AGirT496h%2B0Z%2B0r/AMFH/jT%2B2/8AtP8Ajz4ga94I13TP2Zv2etE%2BIL3f7M7e%0AMdch8OjSv2cPB%2BkeC/FXw88GeE/BF3afCPVvC3w3134q6jpF74a8bX/i66/Sr9rv4Nfs0fDn/g1a%0A%2BLHw6/Zq8Q/FD4tfsyWf7IFt4/8Agj4r/aM0%2B7ufiVrvhXxN8StL%2BMPgbxDrdj4g8FeBXs/7F1PW%0ANP1L4dvZeDdC07RdA0vwlc%2BEY/7LsdG1GX9ota/4Jdf8E1PEg1ga/wD8E/P2LNZ/4SDxYfHetnUv%0A2YfgteHVPGEjW73XiK9M/gtzPqepG3A1i4ck60txqCasLxNU1Jbvkf2yv2SPH37XPxA/Zf8AhFqe%0AqaJ4W/Yf%2BGvjWw%2BNX7QngvSrq5sfEnxw8U/CW%2B0rXf2dfgzbW%2BlixXw18NPC3xN03QPi34xuo57r%0A/hIn8BeHPB9rZ6VFPcaooB9hfAW28XWXwM%2BC9n8QG1h/Hlp8J/h1beNn8Q3c9/r7%2BLoPB%2BjxeJG1%0Ay/uprm5vdYbWUvTqd3cXE89zemeaWaWR2dv5if22fiD/AGl4k/ae8JyeE/8Ag7OguNJ/ag1GwbxD%0A%2BxN4P/szw3e22oP4lv7S9/Zh122S30zU/wBl%2BwTwq%2BkadcxJJqtumr%2BGm1tJbjxJDqVz/WdRQB/K%0AD/wSTe2s/wBsnwcv2n/g6%2BupdW8AfEBPK/4K2WWqz/sbabcx2FzJ9m8Yz/2etpZ%2BP5LTTftfw/vf%0AtEmlPf39tp39oDW7v%2By4/wB3P%2BCiP7UH7LP7GH7L/jD9p/8Aa1s/Bmr%2BCPgxfaf47%2BH/AIc8UaZ4%0Af1jxB4q%2BN/hyK%2B1f4WeG/hNYeILe5QfF3VNdsmj8G6vp621z4WK6h4t1HVdC8N6Fr2uab9xV89ft%0AFfsmfszftdeHPDnhD9p/4E/C/wCPPhfwh4lh8YeGtA%2BKPhDSPF%2BlaN4khsbzTBqtlZ6tb3EKSy6d%0AqF5ZXUTK1teW0xiu4J1SMIAfypfsZ/tpfsZ%2BKE/4Kbftb/HL9s/w54S/an/au/ZY8TfEv9oz9pz9%0Am3W9a8W/Cf8A4JqfBbR08A/s7/su/s5%2BAfEiWtlr/j39onStb%2BIEfinSdZ%2BHfgS8g%2BK/xU8H6kmj%0ARzXz%2BE7rxv8AA3wC/Yp/4JJ/t3ft/ft3aveaanhT9lHwH8d/%2BCev7G37HHw8/ZhtJPB3xP8AiR8V%0ArKPTrr4m/ETWrDwzaDxDc2%2Bo6h4Q8a618Z/jTrXhiL4maZ8ItT%2BK3i1/iR4C1nwj4h1XUf7k/Df7%0AFv7HPg7Rvil4c8I/sm/sz%2BFfD3xyt9StPjXoPhv4EfC3Q9G%2BMNprLXD6xa/FLS9M8K2tj8QbfVXu%0A7ptSh8WQavHfNc3DXSymaQsfDr9i39jn4P8AjfSfiZ8JP2Tf2Z/hb8R9A8LxeCNC%2BIHw6%2BBHwt8E%0AeN9F8FwWMOlweENJ8V%2BGfCuma9p3heHTLeDTotAs7%2BHSY7GCG0S0FvGkagFC5/YX/YkvfjAv7Qt5%0A%2Bx1%2Byvd/H5PFFp44T443P7Pfwkn%2BMCeNLCSGWw8Xr8TJfCL%2BNF8UWUttbyWmvjWxq1tJBC8N2jRI%0AV6ey/ap/Z31D9o7Vv2Q7P4ueD5v2l9D%2BH8fxT1X4NfbpY/GVr8PprrTLOPxR9jlgS3n09rjWNOXb%0AbXU10I7nz2txBDPLF9AV8Qa1%2Bw54A8Sf8FBfBf8AwUA1q48Pnxn8M/2cPEfwF8C6HovgDRdA8SPq%0AXjvxfFrXjTxv8Q/ipYXg8UfEvT7Lwvo3h7wf8Kfh1rtnB4V%2BFb6v8WvFFidd134mW8ng4A%2B36KKK%0AACiiigAr%2BCnwz%2ByP%2Bz58J/gl/wAHCH7Zvx21fXf2wPhr%2BzT8TfHfwa%2BGfhD41%2BK/Gev%2BCfGP7fvh%0AH9n3w74N/ab/AGlvHHwY8VeMNS%2BDmq6x49/aM%2BLfhfQvhtql7B4w8YfB/RPA15o/hCWTxX4f8P3%2B%0Apf3rV%2BMv/BUX/gmL4j/a/wD2W/Dn7Jv7MUnwb%2BCfw1%2BLH7cvw6/aB/bVsfENp4g0qy%2BKfwnl8e61%0A8WPjjLpdh4P0PUZ/FPxZ8Z/FK38D%2BNIbPXdU8HaT4hvNFvYdT8d%2BG2NrckA%2BIf2cvDPxK8Hf8FGP%0A%2BCGnw%2B%2BKeqa3r%2BufDv8A4IeeO9I1i91fVL7XtOk%2BLGm6R%2BzL4Z%2BJ3iHT9SvLi8tbvxNrR03SY/Eu%0Au21xLqmr2A0Y6pd3NsbIn40/4L/eK/2fvjH8Yvhj8K/hV%2B0fp/jf/goR44/b/wD2MP2QPgDeeAdZ%0A8SeEPHP/AATw0jxPeaT4s8e6/wCAPGeg6z/YGn/FXxj49fw/L8QvGtjc6J44s9C8U/D3wLq1olj8%0AM7K7f%2Bkf9uT9m74i/E/Uf2cv2jf2eLfwi37UP7HHxXufiN8OrDxdJbaTp3xU%2BFfjnw/deA/2i/2b%0A77xnPpOuv4Gs/jL8Pb2C70HxSNI1Cy0D4s%2BAvhT4j1m3Oj6HfSR9b8Tf%2BCc37BHxm8W%2BOfH/AMUv%0A2Of2b/HHj/4ly%2BErrx549134Q%2BCrjxx4pv8AwLdC78J6tqvjBNIi8Rza3ozx21vBra6kmrzabYaX%0ApN3e3Gl6VptnagHxv%2Bw1%2Byn%2Bwr%2Byp%2Bwt%2B2b8PP2ZvG3xd179mDxb8aP21vFHxG1TXdb1nxdqPg3W%0ALI6n8LPi/wCC/g14p8W6JNP4s8JfDSP4cXfhLw54gvtT%2BI91rHjDQvEN14t8Y%2BJ/F8nidIdj/ggP%0AoHiTw3/wRs/4J7ad4qGoLqdx8ANI1%2B1Gp6O%2BhXP/AAjfivXde8UeDTHZOiGfT28IaxobaTrAUp4i%0A0o2XiCN5E1NZG9n/AGuf2IpPiv8AsY6V%2BwJ%2BzbL4Y/Zr%2BAnjuTQPg78VH8A6TbaLdeCP2VVsNW1D%0A4l%2BC/hfoNlHBpP8AbvxVtNOsvg/qtxqC7NP8K/Erxn4u3XevaZaC4%2B6Ph74A8G/CjwB4H%2BFvw58P%0Aaf4R%2BHvw18H%2BGfAHgTwppKSR6V4Y8G%2BDdFsvDvhfw9pkcsksqafouiadY6bZJJLLIttbRq8jsCxA%0APwB/4KBeF/tv7VHxAvf%2BFGf8HF/ib7Vp/gp/%2BEl/4J%2B/tWf8IN%2ByvrnleC9BtvtPw/8AA/8Aw2X8%0AO/8AhFdQsvI/sbxrZf8ACvfBv9qeNNL17xN/Z%2Br/ANu/8JZ4i8n/AGQPBmmT/tM/CLWE%2BAf/AAc7%0A%2BFLlPHGm29x4o/a7/al1HWP2e7VfC9/q0tmPjV4eH7ZHi7UfE/w3g1E37/2XJ4R8TaNrUOp%2BdFpu%0ApaNq80tz/UNRQBieJV8Rt4c19fB8miQ%2BLm0TVV8LTeJYb658OReIzYzjRJNft9LuLXU59ETU/sra%0ArDp11bX0tiJ0tLiG4aORf5fP%2BCGFxDpP7eP/AAUa8Oft0ab4wsP%2BC1Wtahb%2BLfjpf%2BILyw1L4Q%2BI%0A/wBlGHxPBY/CS6/YwuLVTqFl8ANC0/UPh3pWv6d4kkm8Sw6rL4GttbvJNUsL7Q/DH9T1fzX%2BNLeG%0A0/4Ox/hHJaSW%2Bmyap/wRT1K41RYY5YJPEE0f7VHxOtY47trSBkvbiG307TpVk1SSOJLXRLaKOcz2%0AmnWzgH9KFFFeIftHfHrwt%2BzJ8GPGfxr8YaB428X6Z4THh%2Bw03wP8NNBi8U/Ej4g%2BMvGvirQ/AXw8%0A%2BHHw98OXGoaRaa346%2BInj7xP4a8E%2BEdMvtX0jT7vxDr2nQ6hqum2TT3sAB7fRX88fw2/4L9r4w/a%0Ah8Lfsj%2BKP%2BCcn7ZnhX4p2X7Rvhb9mj9oO/8ACqfC340/D39m7xh8S4fDknww1zxp8SPhT4z8ReCN%0AX8L341vUL/4i3lhrtpefDHRPC%2BuXt7Za7qf9naJf/wBDlABRRRQAUUV4V%2B0%2B2tr%2Bzl8c08O/B/Vf%0A2gdYuvhV45sLX4H6F43i%2BG2ufFdNR8O39jd%2BAtG8fTXNmvhDVfEtlcXGl6dr6XllPpt5cQ3FteWt%0AykVxGAeq%2BKfFnhXwNoN94p8beJvD/g/wxpf2X%2B0/EfinWdO8P6Dp3228t9Osvt2r6tc2mn2n2vUL%0Au0sbX7RcR/aLy6t7WLfPNGjHhbxZ4V8c6DY%2BKfBPibw/4w8Map9q/szxH4W1nTvEGg6j9ivLjTr3%0A7Dq%2Bk3N3p939k1C0u7G6%2Bz3En2e8tbi1l2TwyIv8U37ZvxTn%2BK//AASo/wCDf74f/s2fsL%2BNvE/7%0ALHxw/a7/AGLTd/AXWf2gfh9400rxXpfw7k8e6/F%2BxZ491j4zXfh69%2BIOk/FOx8Ka5rGi%2BOfEum%2BH%0AvhhpVv4Dii8ZjwZA3hzwzqX9V/7A/wCyt8Ff2P8A9mPwB8LfgV%2Bz7d/ss%2BGNZtI/ib4k%2BA1/8RNV%0A%2BKt98NfiP8QtP07XPHnhO98c6j40%2BIFjrl34e1159DuLvwr4o1DwXPNpr3PhRzo89tJIAfZVFfz8%0A/tV/8F6rT9mTxl4c0ey/4J%2B/tcfGTwd8UfH3x1%2BEn7Pvin4eP8MZfF3x9%2BJH7OpntviY3gH4H3ni%0A4fGOb4Z6Zqdhq9nYfEifwjs1210tta8P%2BHNW0fWfC95r36%2Bfsk/tAz/tU/s3/CL9oS6%2BEXxV%2BA1z%0A8U/CsfiO4%2BEnxs8OP4U%2BJfgycX17p01lr%2BjSO0i2l9JYNrHhrUnjtH1/wpqOh682n6a2pnT7YA%2Bi%0AqK/nG%2BK37Wf7cv7c/wC1L%2B1H8A/2Nf2x/gL/AME1f2fv2NPiVZ/A3xn%2B0R8SfhX4D%2BPvx2%2BNvx80%0Azw7p/ifx94d8B/Cv4reI/Dfw88NfCrwTb%2BIdC0LUdc1NLvxHq2qSQazoGq3NhqV3o3h79h/2KfCP%0Axm8EfAXRdA%2BPv7Y2mft2/E2DXfEs%2BsftCaN8Kfhf8FtK1mzutTeXR/Dll8PvhFe6p4S0pPC%2BmG20%0A6e6/tXUdU1S8FzqN7NbrcW%2Bn2YB8m/tw/D//AIJH%2BL/2vf2Hr/8Abmn%2BDX/DYOk%2BOtIu/wBiax8c%0A%2BPvFfhnxff8AjSw8baBN4ctNH8O%2BHvEmkaF4qhuviHc6DD4b0X4gabq%2Bia54s8qw0exv9UguLeP6%0Af8b/ALbvwn%2BH/wC2N8F/2IfEPhf4wQ/FX49eFPFvi34eeMD8NNZs/gpqlt4F8N%2BJ/Ffi3Q7f4r6v%0AJpnhzW/GHh7RvDlvda14Q8JP4k8Q6Nb%2BKfCl/q1hY6VqF3f6d%2Bb3i/SdG8Zf8HKXw50Lxd4ZsfEu%0Ak%2BCf%2BCLHjD4h%2BDz4lXwxr%2Bj6B47b9vb4cWaeJfC2g3egvrnhjxnpNtpNh5njBtcuY721uNMsvDln%0AoN5oniG81/h/%2BCgfxz8daN/wV/8A%2BCWOpeE/g38S/Hnwa/ZsH7Sc/wC0B8QPByfs63nhnQtX/ag8%0AB6L8DvhoJ/EHjj4m6F418JWnwx1e217xV8aIGh8GG18Iat4W1PTI/iJLJHoNqAf0MrbwLPJdLDEt%0AzNFDbzXCxoJ5YLZ7iS3hkmC%2BY8UEl1dPDGzFInubhkVWmkLfmh/wVJ/YY%2BOv/BQD4MeBPg38Gv2v%0A4v2R9O0H4q%2BFPiZ49vLv9n7w1%2B0To/xXtPAd9B4g8I%2BAfF/gjxh468HeFdW8CQeMbHSvEnifwp4n%0AsvFnhXxyukWGgeKPD%2BoaA2oaff8A6U2F/Y6rY2Wp6Ze2mo6bqNpb3%2Bn6hYXEN5Y39jeQpcWl7ZXd%0Au8lvdWl1byRz29xBJJDPDIksTsjKx8V/ab8H3Xjr4BfFTw7Y%2BLviB4Ev28KXet6f4n%2BF3j%2BP4W%2BN%0A7HUPCc0HiyxtdK8fz6lo1p4ZtNZu9Eh0TX72/wBZ0fTZfDuo6tZ6lq2m2Fzc3sAB8W%2BCP2d/g3%2By%0AX%2Bwl42%2BF/wDwUv8A2qvA/wC0X4O8f%2BOPG/jL9pr9oL9rS58E/Bv4W/EjxX8YfiTL4gt9EvfCfiTx%0Adc/Dv4b%2BB4by78O%2BDfB3wv0nX/8AhE7O202y0bQNLtLC7g0S3%2B9/ht4w%2BEfxy%2BEfgjxz8MdU8H/E%0AT4L/ABE8H6D4i8DatocNjqfg3xF4N1Oxtr3QbjTrRoPsn9n/AGT7N5VjLawyWEkP2Se1trm2eGL%2B%0Adn/gg74ssPi1/wAG7/h/xp%2B0TrWpftMpe6F%2B2D4m%2BKWkfGC61r4i3Opv4W%2BMHxW1uDw3rF54v1q4%0Au9aeWw0fRvE1lc2%2Bq6YbLUNXimtLyw1izk1MfpF/wQ5/5RCf8E6v%2BzVfhZ/6ZI6AP1UooooAKKKK%0AACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooA%0AKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAo%0AoooAK/m1%2BL32Xwl/wdV/soeIPFsV9pGk/Fz/AIJG/EP4VfCXV5o/K0vxj8UPAPx2%2BKfxO8beELC4%0AeKRbzUvDfwxuZPFOp2cTwS2tpeaXdSSmKYQzf0lV8cftofsFfsz/ALfXgnwR4K/aP8HaxrX/AAq/%0A4haF8V/hZ408F%2BM/Ffw1%2BJfwy%2BIXh5m/s/xP4F%2BIPgfVtE8TaDe%2BW5S4hgv2sp54dO1RrUa3oega%0AnpYB9j18y/tb%2BIf2sfC3wfk139i/4ffCX4pfGex8a/D95vA3xl8Uax4O8Na78On8XaVH8TbfRvEG%0AkGNNO8a/8Ia2qnwle6xOuhadqvl6rqNh4jWwj8Ka99G6ZYjTNN0/TVur29XT7K0sVvdTunvdSuxa%0AQRwC61C9l/e3d7cCPzbu6k%2Be4neSV/mc1doA/mL/AOCT/wCwJ/wVf/ZD8Z/Efw78X/EXwP8AC%2Bk/%0AG/8Abf8AFX7av7TP7QGg/EjW/jj4n%2BMng3xh8KvCemWP7KfhHwd4%2B8E6VqvhybSviK/iO/8AF/x3%0A1bXNG8QQ6foWnad4W0vxNYaqbu0/p0oooAKKKKACuV8c3njjT/CHiO9%2BGvh7wp4t8fWulXU3hLw1%0A458Zav8ADzwhreuIhNjp3iPxxoHgT4n614W0q4kwl1rOmfD3xheWaZkh0G%2BYeUeqooA/lp0z/glr%0A/wAFLtF/YU/4JI/stJ4X/YI8QeN/%2BCan7Zvwn/aa8Ra/qv7Svx9sPBvxS8M/ARPGq%2BBvDWg2w/Yo%0A1rV/C/i3xfH8UfEC6t4ivrfVdP8AhzrHgPw14k0fTPHy%2BMr3w/4D/qB0GbXLjQ9GuPE%2BnaVpHiSf%0AStOm8Q6ToOs3niPQ9L1yWzhfVtO0bxDqGg%2BFr/XdKsr9ri207Wb7wx4cvNUs4ob650HSJp30%2B31a%0AKAP5dfE37Ff/AAXK%2BIH7c37MH7afxpH/AAT0%2BJmvfsZaB%2B1j4H/Z/wDDfhP4sfHTwN4dTxJ%2B1Dou%0AteF9L%2BMnxE0i5%2BClw2q%2BFfAvgDWfDXgjxd4A8O6hpHjDxFJ8ORq2leIr698RW%2Br6X/SL8H/D/wAQ%0AvCXwl%2BF3hX4ufEGL4tfFbwz8OvBPh/4nfFW38L6T4Ig%2BJvxC0bw1pmm%2BNPiDD4L0EDQ/CEXjPxJb%0Aal4jj8L6MBpOgJqS6TpwFnaQ16LRQB/LZ8Vv%2BCcUv7MP7VH7VXxq8T/8Eiv2cP8AgrN8Ff2tP2j7%0Av9oTw1r9j4Y/Zzm/ah/Z08Z%2BO/C/h/wh468C%2BL/D37SWj6fpPxI%2BD/ibxpJb%2BLfCs/hH4lPp/gO%2B%0Au/iD458ZeDdDxba9L%2B4P7BOm%2BFLT9nXwpdeCP2I7j/gnboN/qfim/P7Lz%2BDfgf4DGkXd9qkWPF13%0A4b%2BBeo6x4X0%2B98RW1pHeNHcz6N4h8y5u4vEWiNJBpd6/2lRQB%2BFN2toP%2BDmfQHSbxQ183/BCjxct%0Axb3dtGnguK0T9v8A8EGzm0C8F280/iieZ79PF1s9hbR2mlW3giWK7vWvZobD8gP%2BCxP7Pnws/Z7%2B%0ALf7U/wC15r3/AAR3/Yj8Y/Av4e/Fz9md9Z8R/E%2Bw8eax8e/%2BCkfxH/at8X2C/Fsfs53Xwt8f6Ro/%0Awi%2BIXw28Q65eW3iNPiZ8KPiXc/EXXDrPjLS0lEc2lah/T98Uf2Af2fPiz%2B1p8L/23Nab4seF/wBo%0Aj4U%2BB9G%2BGGleLfhd8a/ih8KrHxb8MNB%2BIb/FWw%2BGXxQ0L4f%2BJ/D%2BmfE34by%2BPX/t7WfAXjODV/CX%0AiV1Sx8SaPq%2BmKLOvh39tX/gmR%2B2v%2B1h%2B1v4Q/aB8I/8ABTmL4E/Cr4VaVrem/BD4HaV%2Bxz4O8d6v%0A8H9Y8f8AgTSPAnxS%2BKvhj4q618Y9KTUfjrq2mR%2BK7T4Z/E7xH8O9Xb4J6J4w1bR/BPh8S6p4v1Tx%0AiAdF/wAFKfjZ8QP%2BCRP/AATw8H%2BO/wBgv4HfAfUfBfwG%2BJHwl8GX/wACPHWo%2BM9GtdQ%2BDPizxHde%0AGrrwJ8C7fwpcXGsX/wAaPE3jjX/B3h/wLZ38Ou6bZxa1rmu3Xh7xNJpFt4f1L9L/ANojXPjT4c%2BC%0AnxE1b9nXwLo/xJ%2BNsOhG2%2BHHhDxB4s03wRol54g1K9tNMTVNV8R6vp%2BqafbWXhi1vLrxRPYT2TnX%0AU0Y6BBLaXGpxXcHYeNfhj8NviVJ4Mm%2BI3w98D%2BP5vhx440b4nfD2Xxr4T0HxVJ4D%2BJPh211Ox8P/%0AABC8GPrthft4X8caFZa1rFno3izQzY69plrq2p29lfwQ390kvcUAfzt/8EkvgF%2B2n/wTt/4JD/Ej%0A9l/9qP4WeCvC3iL9mT4X/H3xT8LfFvhT4p%2BEPiDo/wAQ4fG%2Br/GT4v6hpl3oz%2BAtStPCn/CJa1rd%0Anayy%2BK9M8f2HiFddkkbSJ9N0uXQpfU/%2BCT2tftJ6T/wSx/4I62/wK8DfC3xh4I1v4RfCfSf2iNS8%0Af%2BL9c8NeIvA3wrfwdc3kviz4c6dpekahY%2BK/EUWpWkWlf2Hql5p6tNqmnToWsU1S9039X/2kP2f/%0AAAl%2B1D8HfFvwQ8deJvil4S8JeNU02217Vvg58UPGXwg8cTaXZarZ6hf6FB418Caro%2BvwaF4ns7Wf%0Aw54r0lbs2eveGdU1XSbuMpdCWLS/Z/8AgV8Nv2Yvgl8Lf2e/g9o93oHwv%2BDvgrQvAHgbSL/V9T16%0A%2BsfDvh6yjsrCO91nWLm81LUbtkQy3F1dXDtJNI5RYohHEgB7BRRRQAUUUUAFFFFABRRRQAUUUUAF%0AFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUU%0AUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQBFcNOsEzWscU1ys%0AUjW8NxM9tBLOEYwxzXEdvdSQRPJtWSZLW5eJCzrbzMojaWiigAooooAKKKKACiiigAooooAKKKKA%0ACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAK%0AKKKAP//Z%0A"</p>
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<div class="prompt input_prompt">In [17]:</div>
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<div class="highlight"><pre><span class="n">x</span><span class="p">,</span><span class="n">a</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">symbols</span><span class="p">(</span><span class="s">'x,a'</span><span class="p">)</span>
<span class="n">xi</span> <span class="o">=</span> <span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">**</span><span class="mi">2</span>
<span class="n">half</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span>
<span class="n">eta_0</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Piecewise</span><span class="p">(</span> <span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="mi">0</span> <span class="o"><=</span> <span class="n">x</span> <span class="o"><</span> <span class="n">half</span><span class="p">),</span>
<span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">half</span> <span class="o"><=</span> <span class="n">x</span> <span class="o"><=</span> <span class="mi">1</span><span class="p">),</span>
<span class="p">)</span>
<span class="n">eta_1</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Piecewise</span><span class="p">(</span> <span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">x</span> <span class="o"><</span> <span class="n">half</span><span class="p">),</span>
<span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">half</span> <span class="o"><=</span> <span class="n">x</span> <span class="o"><=</span> <span class="mi">1</span><span class="p">),</span>
<span class="p">)</span>
<span class="n">v</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">var</span><span class="p">(</span><span class="s">'b:3'</span><span class="p">)</span> <span class="c"># coefficients for quadratic function of eta</span>
<span class="n">h</span> <span class="o">=</span> <span class="n">a</span><span class="o">*</span><span class="n">eta_0</span> <span class="o">+</span> <span class="p">(</span><span class="n">eta_1</span><span class="o">**</span><span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">v</span><span class="p">))</span><span class="o">*</span><span class="n">v</span><span class="p">)</span><span class="o">.</span><span class="n">sum</span><span class="p">()</span>
<span class="n">J</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">integrate</span><span class="p">((</span><span class="n">xi</span> <span class="o">-</span> <span class="n">h</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">sol</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">solve</span><span class="p">([</span><span class="n">J</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">v</span><span class="o">+</span><span class="p">(</span><span class="n">a</span><span class="p">,)],</span><span class="n">v</span><span class="o">+</span><span class="p">(</span><span class="n">a</span><span class="p">,))</span>
<span class="n">hsol</span> <span class="o">=</span> <span class="n">h</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sol</span><span class="p">)</span>
<span class="n">f</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">lambdify</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">hsol</span><span class="p">,</span><span class="s">'numpy'</span><span class="p">)</span>
<span class="k">print</span> <span class="n">S</span><span class="o">.</span><span class="n">piecewise_fold</span><span class="p">(</span><span class="n">h</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sol</span><span class="p">))</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">51</span><span class="p">,</span><span class="n">endpoint</span><span class="o">=</span><span class="bp">False</span><span class="p">)</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">t</span><span class="o">**</span><span class="mi">2</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\xi=2 x^2$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,[</span><span class="n">f</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">t</span><span class="p">],</span><span class="s">'-x'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\mathbb{E}(\xi|\eta)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="nb">map</span><span class="p">(</span><span class="n">S</span><span class="o">.</span><span class="n">lambdify</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">eta_0</span><span class="o">+</span><span class="n">eta_1</span><span class="p">),</span><span class="n">t</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\eta(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">(</span><span class="n">ymax</span> <span class="o">=</span> <span class="mf">2.3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">);</span>
<span class="c">#ax.plot(t,map(S.lambdify(x,eta),t))</span>
</pre></div>
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<pre>Piecewise((1/6, x < 1/2), (2*x**2, x <= 1))
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<div class="prompt output_prompt">Out[17]:</div>
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<pre><matplotlib.legend.Legend at 0x4b2b910></pre>
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"></img>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>The figure shows the $\mathbb{E}(\xi|\eta)$ against $\xi$ and $\eta$. Note that $\xi= \mathbb{E}(\xi|\eta)= 2 x^2$ when $x\in[0,\frac{1}{2}]$ . Assembling the solution gives,</p>
<p>$$\mathbb{E}(\xi|\eta) =\begin{cases} \frac{1}{6} & \text{for}\: 0 \le x < \frac{1}{2} \ 2 x^2 & \text{for}\: \frac{1}{2} < x \le 1 \end{cases}$$</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>This example warrants more a more detailed explanation since $\eta$ is more complicated. The first question is why did we choose $h(\eta)$ as a quadratic function? Since $\xi$ is a squared function of $x$ and since $x$ is part of $\eta$, we chose a quadratic function so that $h(\eta)$ would contain a $x^2$ in the domain where $\eta=x$. The motivation is that we are asking for a function $h(x)$ that most closely approximates $2x^2$. Well, obviously, the exact function is $h(x)=2 x^2$! Thus, we want $h(x)=2 x^2$ over the domain where $\eta=x$, which is $x\in[\frac{1}{2},1]$ and that is exactly what we have.</p>
<p>We could have used our inner product by considering two separate functions,</p>
<p>$\eta_1 (x) = 2$ </p>
<p>where $x\in [0,\frac{1}{2}]$ and</p>
<p>$$\eta_2 (x) = x$$ </p>
<p>where $x\in [\frac{1}{2},1]$. Thus, at the point of projection, we have</p>
<p>$$ \mathbb{E}((2 x^2 - 2 c) \cdot 2) = 0$$</p>
<p>which leads to</p>
<p>$$\int_0^{\frac{1}{2}} 2 x^2 \cdot 2 dx = \int_0^{\frac{1}{2}} c 2 \cdot 2 dx $$</p>
<p>and a solution for $c$,</p>
<p>$$ c = \frac{1}{12} $$</p>
<p>Assembling the solution for $x\in[0,\frac{1}{2}]$ gives</p>
<p>$$ \mathbb{E}(\xi|\eta) = \frac{2}{12}$$</p>
<p>We can do the same thing for the other piece, $\eta_2$,</p>
<p>$$ \mathbb{E}((2 x^2 - c x^2) \cdot x) = 0$$</p>
<p>which, by inspection, gives $c=2$. Thus, for $x\in[\frac{1}{2},1]$ , we have</p>
<p>$$ \mathbb{E}(\xi|\eta)= 2 x^2$$<br />
</p>
<p>which is what we had before.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>Example</h2>
<p>This is Exercise 2.6</p>
<p><img src="data:image/jpeg;base64,/9j/4AAQSkZJRgABAQECWAJYAAD/2wBDAAEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEB%0AAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQH/2wBDAQEBAQEBAQEBAQEBAQEBAQEBAQEB%0AAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQH/wAARCABsApwDASIA%0AAhEBAxEB/8QAHwAAAQUBAQEBAQEAAAAAAAAAAAECAwQFBgcICQoL/8QAtRAAAgEDAwIEAwUFBAQA%0AAAF9AQIDAAQRBRIhMUEGE1FhByJxFDKBkaEII0KxwRVS0fAkM2JyggkKFhcYGRolJicoKSo0NTY3%0AODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqDhIWGh4iJipKTlJWWl5iZmqKjpKWm%0Ap6ipqrKztLW2t7i5usLDxMXGx8jJytLT1NXW19jZ2uHi4%2BTl5ufo6erx8vP09fb3%2BPn6/8QAHwEA%0AAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoL/8QAtREAAgECBAQDBAcFBAQAAQJ3AAECAxEEBSEx%0ABhJBUQdhcRMiMoEIFEKRobHBCSMzUvAVYnLRChYkNOEl8RcYGRomJygpKjU2Nzg5OkNERUZHSElK%0AU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6goOEhYaHiImKkpOUlZaXmJmaoqOkpaanqKmqsrO0tba3%0AuLm6wsPExcbHyMnK0tPU1dbX2Nna4uPk5ebn6Onq8vP09fb3%2BPn6/9oADAMBAAIRAxEAPwD%2B/iii%0Avwe8Df8ABRP/AIKAfEL/AIK4/Gv/AIJzad%2Bzf%2Byb4a%2BHfwFsvCXxm8S/FXxB8fviNe%2BLvGv7Lfjv%0AV/Cul%2BHvEHgDwlpHwqcn4x2kWr6o/iXwh4nttE8D6T4ks5PDEPj6%2B0SLw18QPH4B%2B8NFFef/ABZ%2B%0AKHgr4IfCv4l/Gj4lav8A8I/8OfhD8P8Axl8UPH%2BvfZbu%2B/sTwV4A8Oal4r8Vav8AYbCG4vrz%2BzdC%0A0m/vPstnbz3dx5Pk28MszojAHoFFFeH/ALSn7RPwp/ZK%2BAvxV/aS%2BOHiH/hF/hV8HPB%2BpeNPGOrp%0AAbu8FhYBIrXS9IsFeN9U8Qa/qlxY6B4c0iKRJtW17U9O02FllukIAPcKK/BXxX%2B1/wD8FcIf2Ovi%0AL/wUOtvgr%2ByV8K/hh4J%2BEGs/tFeHf2KPiFoHxf8AHX7TvjH4O%2BF9P8S%2BN9UtvF/x38J/FDwl8Ofh%0AB8Vdc%2BF2naLr2h%2BAtO%2BA/wAZf7J8Qahc%2BFvEmp2mtQTadafq1%2ByB%2B054H/bP/Zi%2BCH7VHw20zxBo%0Avgj45/D/AEXx9oOieK7e0tfEmiQ6pG6Xeja1Fp93qGnnUNJ1CC70%2B4n0%2B%2BvLC6e2%2B02VzNbTRSMA%0AfSFFFFABRX5l/wDBVP46/HT4HfBT4Ewfs7%2BPNK%2BF/wAQ/jx%2B3B%2Bx7%2By63xE1XwTo3xD/AOEM8M/t%0AA/GbQfh94k8Q6b4R8RSxaHq%2Bq6dp2pvNZ2%2Bpt5EmJI1ktZ5Ib228v%2BIP7d37SUXxu%2BLH7IX7Hfwk%0A%2BGv7X/xx/Yg%2BBXwh%2BJn7ZHiP4o/EnxB%2BzxZeIvF3xS8L67rXw9%2BDPws0/wAH/CL4rafD8Zfi34a8%0AL6n8UYb7U4bP4WeC9N1Lwt4PbUNVv/Eeuah8NgD9hKK%2BWP2JP2vPhb%2B3p%2Byr8F/2ufgyusQfDv41%0AeF59e0jTfEMFtb6/oGp6PrereE/FvhXXIrG6vbA6x4S8Y%2BH9f8M6nLp97eadcXukzz2F3c2csMz2%0Av2w/j942/Zg/Z%2B8e/G/wH%2Bz549/aY1TwBplz4i1b4bfDrxR8OPCGuR%2BEtEsrzW/F3iqbVfiX4p8M%0AafcWXhvw9pl/eJonhuPxN4z1/VG0zRtB8MXxvbq900A%2BnaK/HXT/ANrD9oL9s3/gjtoX7c37PXiG%0AL9lj43eK/wBnDxV%2B0J4f0S20bwf8atFTW/B3hPxjqEPw71A/EHQ9C0688P8AifWdGshPrMdpoWr6%0AVA0UcOobI7qa7/SP9nD4i6l8X/2efgN8WtZto7PV/ij8GPhd8RdVs4pI5YrXUvG3gfQ/Et9bRyxW%0A1nFJHBdanLEkkdnaRuqhktoFIiQA9noorj/iH4h17wl4A8c%2BK/C3g/UPiH4n8M%2BD/E3iHw54A0m%2B%0As9L1Xxzr2i6Le6lpHg/TNT1H/iX6dqHibULa30Wyvr7/AEO0ub2O4uv3Eb0AdhRXxR4M/bIXTP2X%0A/gB8fP2qPhL4o/ZR8efG7xx8DvhDd/AHxZrejeL/ABn4Q%2BM3x8%2BLvh34K%2BDPA0Op6SulQeKLeTxT%0A4n0/X5tVtNL0zUrP4fRaj4q1zw1ocmj6zpOnfP8A%2B2h%2B3d8bf2U/23f%2BCd3wI034R/Dfxn8Af22f%0AiR4j%2BDXiDx9deOPEOk/FLwJ8Q9K0HU/EFpJp/hWPw1c%2BG73wu2lrpl9Feya1eahqlzba9ot1Z%2BF1%0Aj0bW9UAP1Xoor80P%2BCsX7WX7SX7Dn7H3j/8Aap/Z6%2BG3wP8AiZafBbT7zxr8WtH%2BM3jvx74Quf8A%0AhAbKFbMW/wAOtO8EeB/E0XiPxhea3qGmt5XiXxF4N0Ww0q0vj9q1G7vYFsQD9L6K%2BT/2JPiX%2B0Z8%0AZP2avhn8U/2ovAHwf%2BG3xN%2BIPh/SfGsHhn4J%2BP8Axd8QvCcHg/xdo2meJvCv9pX/AIy8DeCNQ0Xx%0AhZafq39jeKdA06Txl4fh1LSn1XRPGuqWWrpp%2Bk/WFABRX5TfsMft3fHD9of9tP8A4KafscfHX4Q/%0ADT4ca7%2BwZ41/Z9h8MeJPhn478ReNdP8AH/w//aY8IePPiR8P7nW/%2BEi8L%2BGJtO8S6T4F8O%2BGLjxC%0AbK1Sxk17X9V0y1sre30OG%2B1X9WaACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKA%0ACivnr9nX4yePfjRpfxav/H/wC%2BIHwAuPh3%2B0L8afg34UtPH91plz/wALd8BfDHxjd%2BHPCPx98FLZ%0A/Z9TtPh/8VtMgXWvDlr4j0nSdTTyr2XTf%2BEi8JTeGfGvij6FoAKK4T4nah8SdK%2BH/i7Uvg/4W8H%2B%0ANvifZ6Jdz%2BCPCfj/AMaat8PPBuva%2Big2en%2BIfGuh%2BCviJq3h7T5TuL31l4L16UyLHA1vBHM97bfz%0A2fsB/wDBYn/gpD/wUj/Zxv8A9qn9m7/gmF8Bbv4baT4w8beDIfDXjn/goBr3gnx5451rw9a3WoxW%0AXgxrz9ji/wDCdrFow/sfw9qmteItes9J1/xDr9mbCTQbPTPFn/CMgH9J1FfIH7BH7Wth%2B3N%2ByR8H%0Af2orPwNqHwwu/iTp/iiz8UfDXVdUm12/%2BHnxB%2BHfjvxR8LviX4Gn1%2BfQfDDeIP8AhEviF4K8TaBF%0Ar8fh/SLXXIdPj1WzsorS7gz9f0AFFFFABRRRQAUV%2BXHwe/4KP6z8Qf8AgqX%2B0v8A8EyfHP7P9x8L%0AtZ%2BC/wAB/DH7R3wt%2BK03xJXxZH8efhXrWt%2BEPCes%2BI7TwZp3gLT7DwBb%2BHfGPi2Hw21pq3xA1vWt%0ATu7G5uLfR4baK9ex/UegAor5v/a6/al%2BGX7FX7OnxM/ac%2BMKeILjwB8L9P0W41LTPCdnpd/4q8Qa%0Av4p8UaH4H8HeFfDVprms%2BHdDl8QeLfGfibw/4Z0b%2B3PEOg6JHqOrW0ur61pWnJc39v8AIH7Nf/BT%0ADWPib%2B1Gf2K/2m/2RfjL%2BxN%2B0lrnwv1T40fDHwr8SPGPwf8Aip4O%2BLHw58PanZaN4ovPCHxJ%2BCfj%0Abxn4Z/4SXwzf3scuseFdRe1vIdL/ANPhuZ3ju7S1AP1Poor4C/Yh/wCCiHwY/b58V/td6Z8CWTXv%0AA/7Kf7QUn7Pi/E3T9Xh1Xwz8U9b0rwR4Y1/xR4k8ItDZWyf8I5pHi3VvEHg/SdVtLzW9F8X6f4dt%0APG3h3WbvQvEmnpAAfftFFfhz%2B2b/AMFRv2xf2Z/29vgH%2Bw78Mv8Agnf4H%2BOM37WVv8Trn9mj4na1%0A%2B2nH8ILXx3H8DvhFoPxT%2BMS%2BK/Ddx%2BzL49sfAVx4WbUtV0HSLR/GfiG78RR6Xp%2Btyw6LD4i0%2BxjA%0AP3Gor8W/hz/wU6/amtfjB%2B0V%2Bz1%2B09/wTq1b4PfHH4V/sX%2BNP21Pgh4A%2BC/7RmnftSXH7UnhD4ea%0AwvhTxX4E8DX%2Bh/Bz4cJ4Z%2BIEPjjWPA/g7RvDutW15rWv6p4wtbuw0k6RbQ6hqP6yfCbxxf8AxN%2BF%0Anw0%2BJOqeBvGHww1P4g/D/wAG%2BONR%2BGnxCsIdK8f/AA8v/FnhzTdevPA3jjS7e4u7fTfGHhK4v5NA%0A8TWEF1cw2etafe28VxMkayMAegUUV8QfGX9su7%2BDX7Z/7H/7JeqfBHxhq%2Bgftg/8LX0vwt8erPxX%0A4Hg8G%2BFvGvwm%2BFXxB%2BMGr%2BCdW8HS6vJ8Q73ULvwl4BkuI9ag0C28MRSa7pkEWs3l/BqdjZAH2/RR%0AXz18Y/j/AG3w%2B/Z6%2BMPx6%2BHPhDUPjo3wk0T4m38ngXwhr/hvw5qviXWvhDrWveHPH3hmy17xrqGi%0A%2BG9M1Dw7rnhbxJp2oy6nfRRLc6LeQWa31y1rb3IB9C0V8r/sRftT6B%2B25%2Byf8C/2sPC3hXWPA/h7%0A45%2BB7XxvpPhPxBe2Wo6zoVrdXt7ZLY6je6cFsbm4R7J3MlsojKuoAyDX1RQAUUUUAFFFFABRRXwV%0AP%2B3LDpn/AAUb0r/gnnr/AMF/Guj3/iz9mrX/ANpLwD8cG17wpfeBvF%2Bj%2BE/Fnhrwh4n8Lx6BaX7%2B%0AKdG1XRtT8RKk1xqtrCs5ggktbSWw1CDUEAPvWivjX4TftzfBv4lfs0ePv2svFmlfEX9nb4P/AAx1%0Av4waZ431H9o/wpF8N9S0bSfgr4n1zwv4o8apbw6rr9jqfgrVZ9Au9R8Ka1pGp3763ZSwWZs7TxAl%0A5odpu/tZ/tceFP2QvD3wo8ReK/hX8ffixF8XPjl8O/gNpGmfs/fC2/8AijrHhnXPiRd3llpvjLxx%0AZ2l/pkfh/wCHujT2iw65rYuL3VDdX2m6boOg67qt9BYsAfVtFFFABRRRQAUV8IfsP/t%2BeBP25m/a%0AD03wz8HPj18C/GP7Mfxgl%2BB/xZ8AftBaH8NtF8V6X45h8OaR4olj04/C74p/Fnw7qWlLputWQTUU%0A8QxC4lLPawz2Zhu5vu%2BgAr%2BaP4I3lnH/AMHWf7Zmnvd3Cahdf8Eo/hneWtgtvbNa3NnYfFr4MQX9%0A3NdvaPeQXFlPqOnQ21vBfW1tdR6hdy3dpey2llNYf0uV%2Bafjf/gjd/wSy%2BJPxP8AFPxl8f8A7B37%0ANvjH4l%2BOPHp%2BJ/jPxP4i%2BHel6rL4r8dTaheavqmu%2BItMuxJousv4k1a/vdV8Y6fqGm3GleNdSupr%0A/wAW2OtXbmagD8Nf%2BC1H/BRG/wDjR418Afsw/BrxP4Mt/wBmf4Y/8FHP2M/2fv2tvii/7Rfiv4U/%0ACb4tal8UPAnxG%2BKPiD9mv4vfF74XfDDx/r3wV%2BEUD6F8PPDfxG%2BJfhx/GUSa54g8TeG/FOn%2BB3%2BG%0Auty6r4l%2B3R%2ByV49%2BHP7E/wDwXz0D46J%2BzB4K8Af8Kw/ZK/aj/Zu/Yc/Zj%2BN3xO%2BKPhH9jbxJJ4P%2B%0AI3wa8UeNj/wkXwC/Zp0fQtK%2BO%2BsfD3U/E3hzwnonhCbwzb3vh/XzL4c02XStH13xF/VJ8Nf%2BCfn7%0ADvwg%2BCni/wDZv%2BHf7Jn7P/h74CfEPULTVviF8ID8LvCWsfD7x/qtjb6HbWep%2BOvDOvaZqmmeMdQt%0A08NaBJHe%2BJINTuRdaTZXpkN5EJ66/wCEv7H37LnwL%2BF/i34L/Cn4CfC/wj8LviFDq9v8RvBlv4U0%0AzU9L%2BJUGvaMPDWrwfEptbh1O8%2BIcN74VSDwhJH4zutcRPB1lp3hGJY/Dem6fpdsAZ/7L11%2Bzf4b0%0AX4g/AH9miyi0rwz%2BzV410r4feLND0%2Bz8VT6Hofinx58M/h/%2B0HptjpPi/wASC7t/GUV78PPjL4F8%0AQi90XX9dtNIs9dsNCeaw%2BwQ6dbfkt/wcp6n4E8e/8EvP2pf2Wb34qeGvh38WfiH8APGP7TPgDRfF%0A1xHpWlfEPwV%2BxH8Xv2evil8WfCun61e2E1jL4wvbbxJ4P0/wX4UsdQsPE/ifxBqlq%2BmRX2gaN4tW%0A3/ZH9nH9mD9nv9kP4Z2vwc/Zl%2BEPgf4KfDO01jVvEK%2BEfAejQ6Tp9zr%2Buyxy6trmpygy32saxeJB%0AaWb6nqt3e3semafpekQzx6XpWnWdrX%2BPf7KH7L37VFh4a0z9pn9nT4H/ALQen%2BDdQn1TwjafGj4W%0AeCfiZD4Xv7ybTZtRn8Pr4x0TWP7H/tj%2Bx9Mt9di0828Ou2FlDpusR32nhrZgD8cP%2BDgD42eEtI/Z%0Am8D/ALBEPx0039ny%2B/bLk1Hw18QfHt1p0/irXPC/7KPgC78J6V8YbTwb4aDT6p47%2BKHxW1/xv8MP%0AgB8Mvh9aSR6x441v4paneNquh%2BG/CvjPxd4a/bf4E/BH4Z/s1/Bn4X/AD4N%2BG7fwh8LPg94H8O/D%0A3wL4et5JbltP8O%2BGdNg0yxN9f3LSX2r6xeLAb/Xde1Oe51fX9Zur/WtXu7zU7%2B7upc/x1%2Bzf%2Bzv8%0AUPiH4A%2BLvxL%2BAvwX%2BInxX%2BE9xDd/Cz4n%2BOvhb4H8W/EP4aXdvqCatb3PgDxrr%2Bhah4k8G3EGqRRa%0AlDN4d1LTZItQjS8RluUWQe0UAfzJfFP4JeE7r4n/ABHuX07/AIObbl7jx54vna5%2BFn7Vfxq0j4YX%0ADTeIdRkaf4caTB%2B0LpEOl%2BA5Sxk8IadDpOlxWXh5tOto9OskiFtF9d/8E6Phb4e8LfHTXtY0%2Bz/4%0ALZWl1D8MteiR/wDgov8AtAfFD4jfAuUT%2BJPCETxaD4a8TfGTx1o8nxN2Ey6DeXGh77Tw3F4vaC8t%0A5nRZf2yooA/IP/gsrpt/efCH9izUrbQdV17TPDf/AAVi/wCCY2teJI9L0m51dbHRH/a0%2BH2iC41C%0AKCKWOC0vtY1nSNBtnvDFb3esazpekpI15qNrDN8X/wDBOawsv2bv%2BCzf/BwlYfGPxH4d8JReOLf9%0Ai39qbwz4u8W6u/h211j4EQ/Dn4zP4p8XG%2B8Q6nFpMPgL4O6zqn/CF%2BJPEyx2OnaNrEGow6lqB0%2BL%0AS4rH78/4K/8AwB/aE%2BPv7L3gEfsueBrT4nfGn4Hftafsp/tNeFvh3cfErQPhHd%2BM4vgR8ZPDnjbV%0AtC0bx94r0TXfC/h/WrjTLO5lsdR1%2B3jsbDyZNRji1u%2BsrPwrr/0H8UP2Nf2bP2z9G%2BA3j79s/wDZ%0AK%2BDvjb4p/DWx0DxlomgeO9P0P4oP8JPG2pWui614o8HaT4xbS9Nt/F/hzT/Een29lqMc%2BmReFPGZ%0A0Ow1S/8ADhT7PbwAH4cf8G8Hxu%2BFv7FX/BID/gm78Of2rPidonwl8cftj/Fb4p6N%2BzB4J8Wya2Nb%0A8e3fxF%2BMniO78D6NollHplxDYReLk1rRfEWj3ks1v4eu4PH3hV31SPWPE9vYSfvj%2B2x4w8JeCv2S%0Av2jdT8ZeKfDnhLTr74JfFjRbLUPE2t6ZoNleaxqXw88Sx6dpNrd6rdWkFxqd/IjpZWEUj3V06MsE%0AUjKQOQ/a/wD2PtF/at0D9n3wXdTeB/DfhT4OftJfAj47ahLffDyw8ReK4tK%2BAXjjRPihoPhL4U6%2B%0A%2Bo6cPhZe%2BKvFXgvwp4S8XeJNLtLy8uPhXqPjPwvp8VpNrENzF9BfF74IfBf9oPwbP8Ofj38Ifhf8%0Ab/h7c6hYatc%2BBPi94A8KfErwbcarpUjy6Zqc/hfxnpOtaJLqGnSyPJYXsli1zZyO728kbMSQD8IP%0A%2BCV%2BteDfiL/wbY/DrRtC8Vafr9pYfsIfGzwB4tm8FeL5IdV8N%2BIbXwd8QLXXvD1xrvhHVbbW/CHj%0ADR4r6F5UtdQ0nxNoFzNaXkElhdi2nX9if2GJZZ/2Jv2O5ppJJppv2WP2fJZZZXaSWWWT4S%2BEXkkk%0Akcl3kdyWd2JZmJZiSSa%2BY/2qP2a7n9nr9hr9oj4S/wDBMP8AYY/Z9Tx38Z/Dnifwe/wk%2BGF78Of2%0APfCVzqHxG8I3ngDVPinquteE/C2j6VrGu%2BC9Kn02/OnPd%2BG9f8Q6ZpMejaX400C6hsLqL7S/Zk%2BH%0Amv8Awi/Zt/Z8%2BFHit9Pk8UfDD4H/AAn%2BHniSTSbiW80p9f8ABXgPQPDesPpl3Nb2k11p7ajply1l%0AcS2ttLPbGOWS3hZjGoB/K7/wVe8AX/wP/bP%2BNv7M/wANb34/3Pi7/gsT%2BzXo9j%2BxzcJ%2B0n8evD/h%0A74Xf8FAvhz8f8%2BP9Z8Ka3F8StO0D4W/DNfAXxV8D/GT4gaBZWOpWMtj8N9T8C%2BFdDksPH1l8OPE/%0A0F/wSW%2BBelftwaZoXxQ%2BPWgftHaTF%2Byl%2Bx54G/4JmfGTwp44%2BOHx68L2PxM/bP8AhpZeOvAP7aHj%0AnVPC1j480W31%2Bfw1ol54N8L%2BG/ipHaQapeeNvGvxV0uW5l8SfCfwt4hsP6UPE3wo%2BGXjPxr8NfiT%0A4t%2BH/g/xJ8Qvg1qHifVPhP441nw9peoeLPhxf%2BNvDF94L8ZT%2BC/EFzbSar4d/wCEr8Kajd6B4ji0%0Au6toda01orfUY7lbW18nj9X/AGavgDrvwv8Ait8FtW%2BEXgS8%2BFfxz1D4jav8YPBD6DaJonxF1n4u%0Aahf6r8R9b8VRQpHNqmt%2BK9R1O6u9Q1iaY6nHKLU2V1ajT7AWwB/CP4G8GaR8cP8Aghj/AMGzfjb4%0AwX/i/wAW3y/8Fo/hx8NvFvjLUviB4807UNP%2BFXj79rv9q3Q/FkepeJtL8SabeaMI7XwT4H0nQ/Gh%0AvrPxF4Ih0%2B0svB3iDw%2Buo3kd7%2Bt3/BWDwf8AAL4w/Ez/AIIQfB34ZeJPG3gH4H6J/wAFIvix%2BznF%0Ac%2BBvFXjn4WeLNFu/gLP4v%2BCHjX4feH/GWoy6f4406wuPHXw78Q/DWHXdF1KK68T%2BHblrvwP4iks9%0Aa0LXX/cPw7/wS3/4J5%2BFP2dfF/7JOh/si/Bi2/Zu8d%2BOG%2BJnir4QXXhkar4R1P4htpWgaIvjZbfV%0Ari%2Bu9O8UQaT4W0CxtNb0u7stQsoNMhFpPAzzNLieOP8Agkr/AME3PiTH8IY/HX7Hnwd8S/8AChNC%0Aj8M/CGXUNHvjc%2BB9Dj8T3/jUWGmXcGpQ3Vwj%2BL9W1bxNNNqc19c3GuarqmpzzSXeo3ks4B8B/wDB%0AKOVPhx/wVH/4LofsnfD3xH4gX9mz4A%2BMP%2BCffif4OfC%2B/wDGniTxt4c%2BGniz45/s5eLvFfxzfwte%0AeK9a8Q6vpX/CX%2BPNEi1PxBoiar/Zun61YzxWWnac/wBqhb3r/g4R1mz0L/gjD/wUDvb7zfJn%2BCsG%0AjJ5KCR/tniLxz4Q8P6duUsmIv7Q1O2898kxQeZIFcoEb7N/Z7/4J8fsW/spfEn4gfGD9nX9nX4e/%0ACT4m/FS31C0%2BIfjLwnZ39vrHiy21TXk8UahDqsl1qF1DIl34gjTVZRFDETdKGBC/LW/%2B1P8AsRfs%0An/tuaB4V8LftX/AvwP8AHPw94H1i91/wnpPje1vbq10LWdRshp17qNitle2TJcXNiq20hd3UxgAK%0ADzQB5Bq/7WGi/AP9mj9iDV30ifx94v8A2i/Fn7G/wL8DeFbrxE%2Bga/rD/GbVPAOi%2BOfHMk%2BrXHjb%0AVr5fhN8ML/xp8YNe0ya/1y812Lwa3hy98WWE%2BuDxXafyZfs3Wf7e/wC2L/wTL/ah/ag8afEX9mP4%0Ae/tsfBL9vP4m/Gbx5%2B2J%2B0B%2B2T8c/Cvij9hvxH%2Bz94r8Dal46%2BFVn8GPh1%2Bzp8UPD3hr4R3Hw58H%0A6p4Bk%2BCOmfE6bwB4r8H%2BKtD8Zajp2u61ofhnQrL%2BqX4L/wDBG/8A4Jefs8fFDwf8afgv%2BxN8EPAH%0AxS8AahPq3gzxnpOg3txqvhzVZ9PvNMOp6Yuqalf2kOoQ2l/ciyvTbNc6fctFfWMlvfW9tcxe/wDj%0AD9hT9jr4gfEHUvij40/Zu%2BEviTxpr/iDQfF/iy%2B1Twlp8%2Bk%2BPPGXhO1Sz8IeNfiP4U2L4S%2BJHjXw%0AZbwWS%2BDPGHjvQ/EPiXwg%2Bl6LN4b1TS59E0iSyAP56fhH%2BzR%2BzT%2B0T/wWu/4OHb/9qa7uNH0T4deA%0Av%2BCamt/2nZ/Gfxh8KdK8BabdfsZ3eo%2BMPiiniXw14k8Eyadqfw2HhzSdR0D4k6i9nL8PftV5q1pN%0Ao82qyyV%2BYHwn/aT/AG6v2z/BX/BuH4S%2BPst9458NfHfR/wBtyb4hfD/9oz47/En9nn4M/tt%2BOfgl%0Ar%2Bpj4EyfF74keBfA37QfiHX7g%2BG9F8IeLfCOkfEX4W3Nt8TvGdnrlr4e0q00Xx94Y8ST/wBnHxw/%0A4J2/sMftJeKJ/G/xw/ZU%2BCXxD8aX17dX%2Bt%2BL9X8DaTbeKPFUt94a8K%2BDL%2B38ba/pMOnat440y/8A%0AB/gXwV4UvtH8X3et6TeeGvCXhrQrmyl0vRNNtbb034sfsp/s1fHP4R6L8BPi38C/hf46%2BDXhj/hG%0A28I/DfWfB%2BjHwr4Jm8G2J0vwfeeBNNtLW2TwPqHhLS2fTPDGo%2BEjo1/4f02SWw0m5s7SWSFgD%2BTX%0Axt8PP2rPBPijwt%2BzD4a%2BIv7FPxI%2BMHgb/goj%2B0N8V/g1/wAEw/EPxV/aB%2BJ/7KfiL4ZeK/2W/Bfj%0AzWf2NvGfxn8T/BD4GfDqT4j/AAmk%2BKF5%2B07%2Bz/8ADLx/4Mh0cQapq/iDTrx4fAF/4XrwrUPjNoOk%0Afs7fsqfB3xx4X8V/BH4aeCf%2BC7P7Q/wJ/a//AGTda/a48I2Hgq3TQPgl4j%2BLvjD4OaH8efid49%2BA%0Anwr174I%2BGtZ8WaZqPw58I%2BMfilp02t3954U8P6DD4q%2BIpsdFg/rV17/gmT%2BwF4o8K/D/AMG%2BIP2U%0APg/q%2BjfCnW/G3if4cT3vh0y%2BJPBniv4kXeiX/j3xfoPjP7SPGNl4w8XXnhrw7Pr3ixNdPiLUX0HR%0Axcamy6bZrD8Q/tqf8EsNC1bwx%2Bxnof7Kv7L37JnxJ/Z3/Y31f4xeKNT/AOCcnximl%2BFf7OHxw17x%0A34OsNK8KeO5LrR/hp8WPCLfGz4falaeKT4Y8Q/E/4Z%2BKdL8RzfFn4hXHizxJo95rd/r8oB%2BX/h79%0Ag34HWf8AwR1/4KrftH6vBpOt%2BNtQ%2BFP/AAUH8c/AW78CftTeJfipr/wS%2BBPgS08TfG/9kX4O/Evx%0Aj8I/jx8SPhJ4o%2BK3wJh8LfD%2BWS%2B0Hxn8Q7bSfDlh4b8ADxv4r8JWUyaj9lar4h0%2BH/gp9/wb7Xsf%0AjC7/AOE%2B%2BJf7BX7XFn8TNPfxpqU2peL9B8O/Ab4IeJ/h5d%2BKfDcurvDqHleJNZ%2BKuraPrV5pjXmp%0A6hb%2BIpDfXjeHyNN90/4Jr/8ABNfxP8APjf8AtTftC%2BOf2fv2ZP2PPh3%2B1F8Nfh98O7j9gT9mHxX4%0Al%2BKPwVs5vDM/ibUfEPxO%2BJ2oa54J%2BFnwzvPiN4ltvGF/8PZPCHww%2BEGj/DzRvAmi2pg1bXNV8S%2BJ%0AJJfrvwz/AMElv%2BCafg/xl8PfiL4c/Yo%2BAGm%2BPvhN4w0Dx18M/Go8DWN14q8C%2BIPCMemJ4MXwxr16%0A1zqekeH/AAM%2BjaXd%2BBPB1pcx%2BD/BGo2cWqeFNC0fUd90wB/I5%2Bz78HP26Pjz/wAEyfFX7X/iLX/2%0AVvgZ%2B2l8OP2xPij8U/Hv/BRj4%2B/tyftA%2BB/jb8Gfir8KP2gbzwH4n/Z/8dfCjwH%2Bx3420HxL8Grv%0AwnYXvwo8Dfs86d%2B0T42%2BHF%2B3xF07WfDPheHxPrMngBf7svit8NPDPxl%2BG3jj4U%2BM38QReFviB4a1%0AXwrrk/hPxT4h8EeKLXT9XtZLWW88O%2BMPCepaR4l8M63Z71udM1vQ9TsdS068ihubW4SSMGvnLxb/%0AAME7v2HPHvxsh/aL8a/st/B3xT8ZIvFvh74gyeMtd8JWepG/%2BIfhDRp/D/hH4iaxoFz5nhfXPiF4%0AU0ef7F4Y8d6xol94s8PJa6bJo%2BsWc2k6ZLafSvxK%2BG3gj4weAvFfwx%2BJGgW/ifwP410e50PxFotx%0AcX1i1zZXIVlnsdU0q6sNZ0PWNOuY4NS0LxDoWo6br/h3WbSw1zQdT03WNPsr63AP89j4UaP8dLz/%0AAIIzf8EZ/wBpvTP23/2zfB37S/7af/BVHwT%2ByX4y%2BP8Ab/tO/F27134d/s8eNPiV%2B1h8Ir/4feF/%0AD/iT4jwfDeTQtKuPh/YfEW2u/E2lpep4luZra68Q2nh2x0%2B1sP06%2BMXxf%2BPn/BN/4q/8F5vgd%2Byb%0A8S/jd4m8H/Bz/gnd8BP2ofgjpHxk%2BNXxE%2BPeufAb4kapY694R%2BJfjr4c%2BI/jp4k8fX0Om3enahqf%0Axh8X6D4ik8RprPizwxp9ro8mj6DBZ%2BHov251H/ghZ/wSt1X4I6R%2BzjffsspJ8FfDfxC1T4reEfAs%0AXxr/AGibW08DfELXLbwpaax4n%2BH2q2nxcg8QeAL3U4fBehm%2Bh8GatodlczyeILuW1a68X%2BLptc%2Bg%0AvAv/AATN/YU%2BHHxd%2BKfx38K/s5%2BD4/iv8cNE8TeGfi/4r17U/F3jJviR4a8ZNqg8S%2BHPGGjeMvEW%0Av%2BHtb8P6pZ6vd6NJol5pEmmW/huLTPDFna2/h3RdG0ywAPwW/Yx%2BF3xt/Z9%2BKX/BL79riz8RaX8I%0Af2d/Fn7IHxTm/a/1z4kf8FBvGv7UXxD/AG9bzVf2Xbr9oPwD4/8AAHgjxfZQ2njn4g%2BCvEHhHxf8%0AVJ9T8FxWl7p3wp1nxNZeGfD3hr4eeFv7GT89fC/xh1z9ln4yf8Ea/wBrL4Z%2BLv2qvB37PX7Rf7TX%0Axf0f4rftj/tRftJ3njH4lfty/BTx/N4i8QxeJv2gf2Y/Cepah8OdF8NeHfCmpKPhV4y17UbPx74T%0A0HQfCeval8O9P8QeHNATTf64Pg//AMEuP2CvgN44uPH3wt/Z40HQdSOheK/DWjeGdS8WfEbxj8KP%0AA2i%2BPNF03w146tPhZ8EPG3jHxF8GvhE/jXwzpNp4Z8X3nwv8B%2BEb7xN4ba%2B0DW7m%2B0fVdUsrzwC/%0A/wCCCX/BIrV9P1jR9a/Yr8D67oWq67pfiGw8O6/42%2BL2u%2BGvAl5pfiaPxcdP%2BD/hvV/iHe6H8EPC%0A%2BuazBbp408F/B3T/AAN4P8faHa2nhXxvoXiDwva2%2BjxAH4J/CfX/AInaX%2B1d8Dv2nviJ4o%2BInx0/%0AZn%2BP3/BXu90z4Nf8FK/2ev2qPE3irWfFWkeLfiZ8TPgh8Bv2G/2nv2SviL4c%2BG2leE/hTo/iO60z%0Aw74l1z4Y%2BB9P8FfDzXtEl1H4c2ep6b43j8T6v77488G6Zpv/AAS1/wCDiTw5dfGX45eDtO/Ze/bd%0A/ap8S/AHxjL%2B0F8ctT8deBfEfw6/ZO/Zj8ffDTwbpvxE1X4h33jzUPDmv/EfXbrRz4Y1zxJqnh2a%0A48Y3ktxpczeV5f7k/Df/AIJEf8E5vhH8X7D46fD/APZj8OaH8QtK8e6n8WNN87xl8T9d8CWPxc1W%0A/wBa1GT4v2vwi8ReONW%2BEkPxd06XxBqtp4c%2BKSeBx4%2B8J6PNBoHhfxFpGh6fp2nWh%2B0V/wAEjP8A%0Agnp%2B1h49%2BJ3xJ%2BPv7Pz%2BOPFHxp0Lw3oHxbjtvi18cfBnhT4jR%2BDNDHhrwXr/AIx%2BH3gL4l%2BF/AGv%0AePfBegR2ukeDviXf%2BGZviH4VsdN0aDw/4o05dE0j7EAfkV%2B1Z8DdL%2BOn7Wv/AAb9aT4l/aG/aA8A%0A237TX7Pvx8%2BEvx18B/Cr9pr47/CHW/jh4I%2BHv7FmpfGrwxrU3/Cuvid4dS3l8DeP73Uk8beIdP0G%0APVfHOnfEfSvDfj7xTfaVo/g/w/N866/8EvjnqX/BVXx7/wAE/b/wb8WP26/gB%2ByL/wAE8v2dbn4C%0AfDf9o39vH4m/Aa415tf8aafo3jX9pHxp4i8DeFNVX4s/Fi2vrjxn4WsPGy%2BGtJj8IXeheGfCWgWn%0Ahy0bTdc8Kf0Ix/8ABJ7/AIJ5In7P6t%2BzF4PuX/ZW8P2nhb9ne6v9e8ealf8Awk0Ky8Saj4tt7Hwj%0AqOoeLLm/s8a/qt3eTXU9zcXlzCLPTrm4m0zTdOs7TuP2p/8AgnV%2Bxr%2B2l4g8K%2BMf2ivgxb%2BLPHng%0AjR73w54V%2BJPhTxx8S/g98UtG8NahejUbzwvb/E/4L%2BM/h94/k8LzagZr3/hG7nxHPocd1fanNFYJ%0AJqupNdgGJ/wTDHxjtv2D/wBnTQv2g/i74B%2BO3xo8FeFtd%2BG3xF%2BKXw1%2BIOn/ABV8N%2BIvEXww8ceK%0APh5PaXnxE02y06HxZ438LweGYPCPxO1W5sbXVn%2BJeg%2BLYddhXW4dQr%2BUH/gp7r8Pj7wb/wAFhP20%0AP2VZf2qvH/jn9lb9pHRPhvN%2B2D8Rf2ox%2Bz1pH7Evxj/Z3uPCGgfEL4Mfsd%2BA/DNp4j8RfFLw1d6/%0ArOk6P4t0Hxf4Y%2BHPhjxhY/Ea6k8M%2BMdS8VabF4s8Rf2/fDr4eeCfhL4D8H/DD4beG9M8H%2BAfAHhz%0ASfCXg/wvo8TQ6boXh7Q7KHT9L021WR5ZnS3tYI1e4uZp7u6l33N3PPcyyzP8F%2BP/APgkB/wTf%2BKf%0AxM8Z/Ff4ifsu%2BF/GHiH4j%2BKLvx38RPDeueLviZd/B/x98Qb6LUILj4jeNvgA3jYfAnxZ8SDDql5F%0AD8Qtf%2BHOoeM7VGgW11yEWVl9nAP5oP26fEvxH/at%2BIPxo/a0iltf24v2YfhL/wAEyPCPh79qH4ef%0AB39prX/2a/jP/wAE7vFmoab8XtS%2BOv7Rfwb%2BH2sJY/CD4zeP7FPAnxKvtDOq3mqaz4p1n4d%2BJPgJ%0Aq9xpHh3w1bw63/YT%2Byj418IfEr9lv9mz4jfD7xD4j8XeAvH/AMAvg7418EeK/GFpLYeLfE/hDxV8%0AO/Dmu%2BGvEPimxnmuJrLxHrWjX9lqWt2k1xPLb6nc3UMk0roXb4Vh/wCCF3/BKK1j%2BHUNl%2Bx74S06%0AD4XeB7X4ZeHINM8d/GDTINX%2BHFtdzX83gL4lwaf8RLaL4zeDdY1C5utR8UeGfjAnjnRvF2o3V3qH%0Aimz1i8up55P1ftLS0sLW2sLC2t7KxsreG0s7O0hjtrW0tLaNYbe2treFUhgt4IUSKGGJEjijRURV%0AVQAAfxtfEDRdcsP%2BCSP/AAcR6TpPxU%2BMumXH7L3/AAUf/akvfgR4o1X9ov42nxn8PrP4bfCX9mHX%0AvDXh3SfiTqXxBufHWqaVFda1r6ab4P13xHq2ha3rOumC80y7uLxGH2h8KPDl5%2BxL%2B3r/AMEPfhXp%0AnxS%2BPfinxF%2B2j%2ByX%2B3B4W/aQ1f43fHv4pfHDXvjL8Rvhl8Kf2dvj/o/inxZL421zxf4Yj8Y%2BGtfj%0A%2BII0nxJ4YsvBhs/Deq3fhDTdWPhddN8JXv6e6Z/wSI/4Ju6P4T%2BKngXTf2T/AIf2vhH446hY6t8Y%0ANAXUfGclh8StV07U5tYttT8YpN4okfXNQOp3M97dXt7JLc380rm%2BkuQcV1cf/BL/APYGh8Ufs3%2BM%0A4f2ZvAsHiL9kLStJ0T9me9guvFEMHwb03Q/FJ8a6bD4O0uLX00m1lg8SsdRkvLqxu727TFheXFxp%0AoFmAD71r%2BEj/AINqf2Ffi58Z/wBgP9kr9oDwr%2B1h8YtN8J/C3/goVqnxN8S/swazqPg%2BL9nXUfB3%0Aw4tdY0rUr7TdPsvANz8S4vibda14iHiK0vpPiC3gfVLi003S9W8G211Y6f4y0X%2B4rx74E8IfFDwV%0A4q%2BHXj/QbLxR4K8baFqXhrxR4f1ES/ZNW0XV7WSzvrSSS3lgurd3hlYwXlncW19ZXCxXdlc293BD%0APH%2BYFp/wQg/4JC2Gn6rpNj%2BwZ8ErPStdSzj1vTLS38T2%2Bn6zHp10t7p6arZQ%2BI0ttRSxvVW7s1u4%0A5ltbpVuIAkoD0Afk/wD8FLPiTY/tH/sf6z%2B3R%2Bw1%2B1/8Yv2RNb/ZY/a98bfAX4G/DT4R/Hz4keDf%0A2fv%2BCjvi3wZ8Y59M1X4Y6H8N/g5d6VpHjLxx8ffiw3jHSfhp428NWHi%2B18SWVrqep%2BPdVHgTVNX8%0AV/DiP4X%2BP/Cmk/8ABHX9uX/gpl8UP%2BCkn7bngLWv2otB%2BL154rmtfit8Vfi/H/wTg8cfE/4wv4U8%0AMfs0fAj4LeK/FfhXUvDfj79n/wAb%2BK9C%2BDsviKLVvhd4q8ThD4j0LWfhj4Y1DwpL4Z/a67/4JF/8%0AE8LjXP2U9ds/2d7fw0/7EOu2/ij9lrQPA3xO%2BNHw%2B8B/CbxJD49ufidca/p/w18DfEbw98PPEWsa%0A144vLzW/FOo%2BM/DHiO78Xi7utM8Uy6xo9xNp78wf%2BCLv/BN%2BTTP2qNBvPgT4r1Xwx%2B2zqut%2BI/2o%0A/Buu/tI/tS6/4G%2BLfjDxB8QtF%2BKl78QNV8D638a7/wAI6H8SrTx74f0zXvD3xK8JaL4f8e%2BEyt7p%0AnhbxFo%2Bjatq2n3oB%2BUP7JnxJ1v8AZ6/4KOeK/wBkP41eJPip%2ByV8AP2r/wDgnD8LvE3hT4SfHr9r%0AK%2B8U/HSf9o24/aG039k7wx8RPA%2BqeHvG3jvw98B/i98a9B8feFdO1nwB8Ivi3f3h%2BKOh6b4g0S58%0ATaro%2Boa14c%2BBdD8JftD/AA7/AOCUHwb/AGw/C/7en7eOq/tVfEf/AIKUXv7OXw28WeJv2rfjj410%0A/wAP/DDVP29viN%2BzBe/Du4%2BHvxE%2BIvib4D%2BM9T1vw94WvPG1348%2BLPww8WXEWuRaJpsdvbRaHpl8%0Av9HVv/wRE/4J2TeP9M%2BMXi74dfGj4nfHTStH8V6BB8fPih%2B2F%2B1/44%2BNkmjeLvDn/CJ3Gnz/ABK1%0Aj45y%2BJTb%2BGtFWI/DkRXkLfC7xDCvjj4dN4Y8dy3Piaevc/8ABC//AIJiXf7P%2Bnfss3PwI%2BIE/wAA%0ANI%2BMF58fNN%2BGkn7Wf7Y7aPafF6%2B09dNuPG8V6fj/AP23/aHledeQWjao2k2mt32q%2BIrTT4Nf1fVN%0ASuwD5y%2BDHwUsPgl/wVW%2BPv7D3gz9oL9qK%2B%2BA3xr/AOCWehfGTxF4B8eftbfHz4n%2BPvDPxX1v9or4%0Ai/CPxV8bvhp8TfiP8UvFnxZ%2BFnjPxR4e1KM3fiP4e614Sjt/F2m2XiPTrp9Z0vRh4c/Ibwz4B%2BJn%0AwT/4IXa//wAFMtC/bo/bovf2vv2e/jv8avFPhXx38V/2wvjT478F/EnQPg9/wUC%2BIHwP8P8A7Onx%0AF%2BDmt%2BNLP4Q/EDwt8ZNP0K2t9ROs%2BC9R8eat8V/F1vcWOtal4Wh0r4YP/Stcf8E3f2XvhX%2B0J4s/%0A4KDfCn4N/EDx7%2B3Fo/wv8T%2BG/C/iPxd%2B1z%2B0zOvxBs4PBF1oehfCfxHH49%2BK/jP4a2vg/W5rbTLa%0AE%2BIvAOv6F4W8RrYfEe30V/Fuh2OrRfmx/wAEwf8Agi74P8M/BPTbn9v/APZwuNL%2BM/hT9rf4xftE%0A2vw3sP2z/jl8b/2ZvHXi7xb41m8e/DT4267%2BzuviPwz%2Bznp3xI%2BHWheJZvg1YpL8LNTn1aw8AW/x%0AD1m6l1jxZZ6Z4WAPlX/gob%2B0J45/Zk/4KyftsftO/BbwnrEXxm8Kf8GxPiP4veD9O1bS7bUdc8Ie%0AJtM/bJ1XT4fFeueH73w54vsxZfC6AaP4v8ZeG11Kx8LX1t4Q1CPxBZajBPceJ/DeP/wTM%2BF37dtx%0AqP8AwSY/bH8Ma/8AFnw54Z%2BLPw50SH9s7x7%2B1B%2B35afG7wz%2B2j4f%2BP8A8Otb%2BJcOreDvgjd6nqQ8%0AI/Fb4deNIm1D4YaX4dbwl4v8KeCbXSPAvi%2B38aWOgeOYbf8Aob0//gmp%2Bxpp37YWu/t7J8MfFF9%2B%0A1Z4l8O614N1z4k698c/j/wCJdM1HwVr%2BitoGpeB5fhh4i%2BKWqfCNPA/9nMv9n%2BDYfAkXhjRtQgtN%0Aa0bSrDWbO0v4T4Ef8E0v2K/2Z/FfhzxZ8EvhFqfgtvA2u%2BL/ABT8NfBTfFz42eJvg58IfE3j%2Bw8S%0AaT418RfBP4DeL/iNr/wT%2BCmu%2BJ9I8ZeL9G1nV/hR8P8AwdqF7o3ivxLpEtwdN13VLW6APbf2qfAH%0A7OPxM/Z3%2BLnhb9rzw98P/E37NEfg%2B%2B8WfGXT/imlqfh9Z%2BDfh5LB8QbzxN4onvJIbfTtP8GXHhe1%0A8XrrLT28mh3Wh2%2BsQXNvPYxTx/z%2B/s7eOv2W/wBvz/gqv%2By7%2B35q/jnSPDHw58E/CHxv%2Bz5/wSn/%0AAGe5FvB8Q/iNrMngb4ieNfjb%2B1j43%2BGvhK/1T/hRPw4ufhHZaz4M%2BDeh/GbR/Cdz4q0WLwt4r1HT%0APCXjC8%2BE%2Bg6//QV%2B01%2BzN8E/2xfgb48/Zu/aL8H3Hj/4MfE230S08ceD7bxb418Dya5a%2BHfE%2Bi%2BM%0ANKtm8T/DvxH4T8X6fbx%2BIPD2k3V3DpWv2MeqWsE2kaqt7o9/qFhdfEH7Hn/BEf8A4Jf/ALA3xgX4%0A%2Bfsnfsv2/wALPi3H4X1vwbD4vm%2BL3x8%2BIMlt4d8RvYvrVnaaP8UPin418PWlxfDTrWFtVttJi1eG%0A1%2B02dvfQ2l/fQ3IB1/7e1/8AED9oH4B6F8JP2b/BHi74%2B/Dz42fE7Rvhr%2B0n4u/Z%2B%2BK3wW0HW/BP%0A7PWl366n8bvDujeIPG3xr%2BE7N4t%2BKGg6fP8ABGNvB/iFvEngqw8d6/41WbTtV8MaPa6r%2BZX/AAQf%0A8aaPH%2B1p/wAFpfhD4W/Z38afAvw74Q/bbtNXtfDN9pPwd8N%2BD/hppkHw08IeAvC/w0ttA%2BHHxF8R%0Aw21/d2vgzWtf8O/8IPo%2Bs/DeHwfZWkn/AAl9lqV/pmiXX7m/Bb9kb9nb9nX4QeKvgN8EvhvafDj4%0AUeNPEHxM8U%2BIvCvh/wAQeLUN34g%2BL%2Braprfj/UrTXbrX7rxLo8urahrF6dOj0XWdPg8LWa2GmeEo%0AtD0zSdKs7L5d/Zt/4I%2Bf8E9v2SPG3xJ%2BIfwF%2BC3i3wt4r%2BMXgzX/AIf/ABVvPEP7Rv7T/wAUNO%2BI%0AXhXxOtimtWXi3w58VvjN438Na3e3MWnWsMGvX2kTeINOgWaDTdUs4bq7ScA/Q/wj4w8JfEDw1o3j%0ATwH4p8OeNvB3iOyj1Pw94s8I63pniTw1r2mzFhFqGja7o11e6XqllKVYR3djdTwOVYLISDj%2Bbz/g%0Ar14H%2BKnjD/gs9/wQD0z4K/F64%2BDXxI8R2/8AwU80nRfHt54X0X4g6Z4I07R/2bfAupeK9Y0XwN4k%0Ahk8P6j4o1zwff69odrPrS3emLqMfhq8vbKa30eSG5/oB/Z7/AGffhB%2Byr8F/h9%2Bz18AvB0Xw/wDg%0A98LNEbw94E8HQ6z4i8QpoekPfXmpyW7a54t1fXvEuqyzahf3l3Pfa1rOo3889xI811ISMfFv7S3/%0AAAR2/wCCev7Xvx4g/ab/AGgfg3448Y/HOx0e08P6R490X9pr9qv4bXXh3RrTR20Ead4U0b4XfG3w%0AX4b8J295pclxBq6%2BGtG0ptdlvdQvNaa/vdRvrm4APlXQv2OPi38E/wDgiz8e/hF%2B1Z8UPFHxB/ab%0A%2BHXwn/4KLeI779p/4f8Ajrxd8OPiv4iTxv8AH74x/tGeFvGeh/EPwH4h0jxx4NsviLb6V8JvFfj7%0A4ZWfiebwrqkWlxfDTx3pvizwnp15pF58FfG7wD4w%2BP37Cf8AwbSeNdY/aE/ay%2BHnjz45%2BOP2B/gp%0A8ZvFvw2/ae%2BNvw/1v4n/AAz%2BJ37FHxF%2BJ3xYh8WHwT8RtNs9V8W/E6/%2BHdpa33xQu1b4lWui%2BI9a%0At5dbikvbnT4/6fPjv8Dfhn%2B0t8IfHnwI%2BMmjar4j%2BF3xN0RvDnjfQNF8ZeNvh/qGtaFJdW13caWv%0Ai34c%2BIvCfjLS7S%2Be1jttSj0bxBp/9p6bJd6RqBudKv76zuPzif8A4ISf8Eun8H/A7wAP2f8AxxF4%0AO/Zr8UP42%2BBmhW37Vv7Ydpa/DnxcddHiS28R6K9t8fYbmbWNK1kPdaBfapPfz%2BHY57qy0F9OsLu6%0AtZgD8t9f/aU%2BJv8AwTI%2BHv8AwcU6Z8BY/if4r8EfsPzfsneJ/wBlfRviv8Vfib%2B0bZeAPEn7QP7L%0AnwvuvGUcOs/Gbx1428dzaB4T%2BIetT/GDxX4Z1rxlPYx6ZrTz6fpNnpl9DHqHR2X7LurfBT9vr/gg%0At8YNX/av/a9/aY8TfGvxb%2B0vr3jix%2BOnxtl%2BJXw5PjPxF/wTo%2BM3iHxF8Qvhj4N1m2Sw%2BE9lqV3q%0Al9NF4J%2BHmsab8P8ASPDXk6R4b8ImSy07d%2Byvwx/4JYfsK/CH4zftHftA%2BCvg1rH/AAtb9rrwv4x8%0AF/tI694w%2BNPx7%2BJekfFrwv491K21TxLo/iTwR8Svih4t8BpbzSWq6fpL6V4Z0648M%2BHbjUPCvhib%0AR/DOqalpF38/6d/wRE/ZLsPHv7NXxAl%2BKH7aWq337HeseHb39mTQdZ/a/wDjJqHhX4OeHvDpmtf%2B%0AEC8J6RLrgaPwj4g8Nmx8CeLDqF1feJ/Efw90XRvBGpeJpfDll/Z8gB%2Bs/izQP%2BEs8K%2BJvC39t%2BIP%0ADX/CS%2BH9Z0D/AISPwnqX9jeKtA/tnTrnTv7b8M6v5Nz/AGV4g0r7T9u0bUvs8/2HUYLa68mXytjf%0Ayvf8ErNE%2BGXwQ/4Jg/FHTvE37TPxt8YfGL9rbRv%2BCufibwf8KPjP8Xdd%2BIWl6tH8AP2gPjb4H%2BI/%0AjH4e6VrVjNd6frElhD4M8WfFHV77Xbq78TeOfHXiLxbdCXU/EuoMn9YNflVdf8Ecf2OZ9e/aA1y3%0Am%2BOulxftC%2BFP2hfB%2BqeG9I%2BO/wAQLDwf8K9K/ax8W%2BDvHf7ST/ATwxFqX9lfCG7%2BM3ivwH4d1Txu%0A3hiCK21KOPVNJS1g0TX9d0zUQD%2BY39hPxN8bP2J/2Jf%2BDdr46W/7Uv7ROteM/wBsX9tb4VfsueMf%0Ag3J8RL7WP2X7H9lb4s%2BIfiR4P8OeB/D/AMDxZeFfh7onirw5aQfDjxRqfxKv/DfiT4vSeONb8f3E%0AHxN1bwvY%2BGPDsH2JpniP9uj9uzx1/wAFlPHnhGT/AIKC6J8Zf2dv2v8A4sfss/sNa58G/wBqzwZ8%0AA/2NPgfdfsm2eh674Ck%2BJPwc1H9qb4N6h8Vbv47ar4o0zxX%2B0L4k%2BK/wF%2BNPhK4%2BE/iLw/p3ww1q%0Aa706/wDDmn/qVrn/AAQH/Yj134J/sy/AObx5%2B2Lp/gT9j/4i%2BI/it%2Bz7LpH7V3xP0/xH8O/HOtye%0AEZtH1rw5rkd60%2BgS/Dm48HWt38KF8OR6KPhzf674x1LwybHUfF2vXV79I/FP/glL%2Byp8U/it8avi%0A29z8bfhxqf7T%2BmeHdE/ao8IfB345/ET4afDn9pXRvDWlJ4btdP8Aix4N8O6xBp9y2o%2BD/tngzX9W%0A8JN4S17XvDWr61Z6pq1zcate3coB%2Bhvhq91vUvDmgaj4l0SLwz4jv9E0q91/w3Bq0WvQeH9burGC%0AfVdEh1yC1soNai0m%2BknsI9WhsrSLUUt1vI7W3SZYk/mM/Yn8BftH/wDBRz4TfED9vfVv29f2hfgN%0A8YtF/wCCinjO807wrp3xc%2BIdt%2BzV8I/2Yv2XPjVdeGNW/Z2u/wBnTwh8UfBvwJ8T/wDCdfDnQLif%0Axz8Sfir4Y8c6zql/e2%2BqXUhgmu9Uvf6hrS0tLC1trCwtreysbK3htLOztIY7a1tLS2jWG3tra3hV%0AIYLeCFEihhiRI4o0VEVVUAfnLff8ErP2Vr34nfEbx8jfF3S/CHxh%2BNFn%2B0V8Wv2dtB%2BLfizQf2aP%0AiT8cbT%2Bx72T4leLvhDpNzaaTqOu6z4p8OeG/H3iy3S6t9E8bePNA03xJ4x0rXbpblLkA/BP4SfEj%0A9oz9uf8A4Jzf8FPv%2BCp2v/tb/tQfCP8AaJ%2BCviv9ru%2B/Zh%2BGXw1%2BN/i74QfBP9mfwt%2Bx74Y17xb4%0AC%2BGXj34C6N4yh%2BCnxX8S%2BOLmOeX40eM/2gvDfja8v9G1PQbPSk0DS/DcN7rB8M9a/at/4KK/8FB/%0A2DdQm/a6/bQ/Zu%2BEf7ZH/BFzwj%2B1z%2B1F8FPhR8b7/wCHHh%2By1TS/Gnw58MW9/wDAKz02HW2%2BEcvj%0Ar4ha34G8TS/EXwXeaL8Wta%2BGF7rfh0%2BKfDGjeMvFfhef9tPi5/wRn/Yq%2BMvjD40654itfjVoPgX9%0Apvx14d%2BJ37T/AMAvAvx2%2BI/hD4AftDfEHw1Is1v4t%2BI/w40vWUt4NW1m4svDl34wPgXUPBlr42vf%0ACPh298WW2sXUWpTan2vi3/gld%2BzL4q/at%2BHX7YFprnx7%2BH3xH%2BFPwq034GeBPCvwi%2BOfjr4TfCfw%0A38HNMsWsY/hlongDwDfaDYaB4KudtnqF14f8PXOkac2s6VouswQW%2Bp6Rp91bgH4d%2BCf21f21bH/g%0AlJ4h%2BHll8aPGvhr42Xf/AAV5vP8AglB4I/av%2BJ2o%2BFfG3j7wL8Htd/ag0z4RWPxe1XxJe6rqV342%0A8YeB/B9/qPgaTx744tofHOoeJdJn8cakmqyQ2PjnWff/AIZ/sp%2BH/wBkv/gvD8HvBOhftFftg/Fy%0AP44/8Er/ANpvw/NqX7Qvxy8dfH3xn4Ml0j9oL4Jawl14Y%2BJnxBt/FviPw5ot9HYvd22gnUPDvh/R%0AfFOiafqunahcap4q1TQPEP3h8Ff%2BCIf7DvwT%2BAn7Uf7MtnF8dfiR8GP2wfEug%2BPPjN4S%2BLvx5%2BIP%0Ajdbn4l%2BH9WHiGD4r%2BE9Sk1HT9S8IfFXU/EVj4W8QeIfiJo11D4r8Q6x8P/h5eavql23g7Rlt%2Bj8N%0Af8Egf2d9A/aT%2BHP7XOo/Gb9tPxr%2B0D8M/C9z8PtI%2BIHjT9rH4papf6z8L7jyZx8LPFFtZ6hpllq3%0AgJNYW88T3GjrBaz6x4r1O68ReIb7WNUttJuNNAP53v2w/CHjrxV/wbq/8FVPB/jr9pT9qX4j3X7K%0Av/BQr9qb4e%2BDfHvxH%2BM3jn4j/Ejx38PPgj%2B17ZfCPwd4A%2BMHifVrm71Lx34Kn8L3YudX0S9Wy8N2%0AXiKx0rxYmn2EWgxQp9z/APBWfT7n9lv9ir4Q6H%2BzP%2B21%2B2lfeOP2af8Ago5%2Bzj8IfiN4i1n9rX4w%0A%2BIviNfXXx41X4b/ELX/hf8bfG8%2BtWfib4kaEPhh8S/BniHwv4a8Uatq%2Bj6J4f8Y2NtbxG3W3srH9%0ALfhb/wAEX/2MvhZ%2By/8AtS/sgwXPx78f/Bb9sLVdU8SfGfS/it8efH3j7WLvxjrT/bNX8c%2BG9W1e%0A/LeHfGusa5Fp/ibxF4mtbd9U8Wa9o2i3Xi2fXYNKs7WLnPjD/wAEP/2OPjt8EoPgP8SfF37VGt%2BF%0A7z40ar%2B0L4/8QH9pX4iL43%2BMfxjvvD3gzwhpXjr4v6/LdzJ421jwX4R%2BHfgfw54Fku7C3j8LWPhy%0A0utORNYu9U1O/APlj4dJ8Qf%2BCj//AAVI/wCCpvwS%2BNvxv/aq%2BC/wm/YF0/8AZr%2BE3wF%2BFf7Mn7SP%0AxY/ZeguZv2hPhX4o8f8Air45fEzVPg1408GeOPiZ41vtT0TRdQ%2BFqeMo/wDhVnhnwpDFp%2BmeH/HF%0AzqnjDUrj82tE/aB/a1/bl/Zn/wCCVnxLvP2w/wBqX4G/E26/4KqfED/gmR8VPib8DvHE3w/8C/tQ%0A/AW38IfH6/8AEfxpuPh9p1gngPVPi34h8JfCvQ/Cvhr4n2kev6D8MPiknjnx54C0DTNZttM0Hw7/%0AAER/Ef8A4JTfs2%2BO/Fmj/Ezw/wCNP2lfgx8aIPg/4S%2BAvjf45/Bn9ob4heFPi78b/hZ4L0y10fRd%0AG/aA8Xapf%2BIn%2BM/iWPTbU2z/ABS8eWesfFyFpjdWHjyyvbTTLmxofE7/AIJD/sf/ABE8GfsmfDbQ%0Af%2BF4fAz4a/sT%2BIJvFfwC%2BH37PHx8%2BJ/wj8K%2BHvE7yTm28UapZeH9f%2B0a14w0631Xxbp%2Bn%2BOLm7Xx%0ArBp3xA8f2beIZIfFmqLKAfGvwu%2BL9j%2ByJq//AAWy%2BCnxL/az/aF0X4EfsleCfgT8U/h78aPjR4g8%0Af/tI%2BPv2fvD/AMfv2Zr9taufDfxB%2BI%2BrePPiZ8XNb0b4peB9b8e6P4G1rU9StdHufEXh7wzoGk21%0AjqrC7%2BUf2Jv2r/j98Kv2uv8Agl94D8XfGz9tPxv8Bf20v2PPjTL8TPFX7b2mfDDRPB/xI%2BKv7P3w%0A5%2BFPjrRv2g/gZYSa34g%2BO/wh0jxtJ4/1KDWPC/xXufCXhfVtD8X%2BALHT7fxV410nUJNB/XWD/gkL%0A%2ByFefED9tLx948uPj78W0/b90LWfC/7SXgT4nftF/FzX/ht4j8Oanq8GpaVpfh/whZeJ9ItfCb/D%0A/T7W18IfCTWdHmh8R/CPwJC/g74caz4Z0K%2B1KzvfHfEf/BD34C/Ejxh8FPHfxz/ag/bl%2BPniT9nz%0A%2Bw9C%2BFz/ABF%2BMvgTSNC0H4VaZb6Nba38GW8HfCj4QfDTwbceFPiIvhrws3xP8eNoCfH/AMd/8It4%0Abi134zXFto1jBEAcp/wSKWD/AIat/wCC4jLJKbk/8FL71ZYmhRYEgX4CfCY28kdwJ2klllka6WaF%0ArWFIEhgdJ7lrmSO1/c%2Bvzg/YC/4Jp%2BAf%2BCfOu/tKeJ/B/wC0R%2B1V%2B0F4h/ar%2BIGh/FP4oaz%2B0/47%0A%2BHvjnUE8f6RZ6zpt/wCJfDlx4F%2BFHwxewu/E%2Bm6npmm69b6idYsI9N8IeEdP8P2mg2emXEN9%2Bj9A%0ABRRX8SfivT/%2BCmX7XnxP/wCCjn7RvwZ8M6v4g%2BLX7MH/AAUN%2BO/wk/ZX/aK13/gpz4x%2BA3we/Y18%0AGfsv%2BLtC0a/8Na7%2BxRZ%2BHJvg94p0j4hfDzwroes/GofFu41TTvjR4E%2BJtzqerjQtLtrS78TAH9tl%0AFfx4/wDBWfUvjX/wT2%2BJ%2BifGz4eftqftX%2BJfh7%2B3b8H/AIufDD9qnw7P%2B0P%2B0N8Tvhz%2BwV4R%2BMPx%0Ae%2BDEOrf8FHP2WPB2j%2BNLFPhr4f8A2e/EXxD07wN8OPDH9veHNEh07xN4O8HeG9b06y1LxU0vqH/B%0AQz4A%2BM/2evip/wAELfgT8O/%2BChX/AAUKXwr8V/2irz9nn4ufEvUP22fi9N4s%2BPvgTV4tc%2BKEuoeP%0ANa0/xPp%2Bi%2BIPGvinW/El34R8NeMbSwj1nw34Qu/C/hbwVNYWfhTwfFpYB/V9RX8Wn7bX7f8A%2B1p/%0AwSu%2BMH/BVv8AZu/ZM%2BKvxd/ab8CfB/8AYj%2BDf7YvgbVvj5rPxC/aN8d/sV/GX4o/tG/CL4H%2BIfhb%0AP8V/i9q/jnxZ8RvhvL8G/G8/7RWhaN8Q9X8Vnw7HpkltA8WjaV8Sddue3/Yw%2BGf/AAUo%2BD3xA/ZS%0A/bBsvDX/AAq74R6v%2By/8a/FP7QFt8af%2BCp/xa/bE1f8A4KRfEbxT%2Bztc/Fb4N%2BKPCXwa1H4c6H4U%0A8IfGrw/qnhLXPiM2m/D%2Bw06zb4P2fjDwzpGseHdG8LQW%2BuAH9ilea/Dz4xfDL4saj8UNJ%2BHPjHSv%0AFuofBf4l3/wd%2BKFvpRuX/wCEO%2BJuleFfCPjXUvB2pSzW8MEuq2Hhnx34T1O8FjJd21v/AGxHZS3C%0AahbX1pbfh1/wTd/ZM8CfHf8AZr/YC/b%2B8d/tS/tC%2BI/jl8aPhJrvxG/aC8YS/tA/FLTLX43eLP2q%0AfhDrPhzxL8FbeCXxzFH8J/B/wK8Y%2BKGh%2BCnhP4CJ8OofA3i34b%2BH9d8KW2nawL29ufjv/gg5pn7N%0Av7FDf8FNNR%2BI3xt%2BJa%2BINH/4Kk/8FPfhzoFx4v8AjN%2B0P488L3/wo/Zl0f4Q%2BLviD42%2BJHw4fxJr%0A3wff41af4f8ACzeLtd%2BNHijwrP8AG/xR4W1ODQLbx3rOlajN4dAB/WnXC6P8TfAfiH4geOPhZoni%0AbT9U8f8Aw10TwT4g8eeHLL7RNc%2BFtN%2BIzeKB4L/ta6WE6dBqGuQ%2BDtevo9GF42s2elR6Zq%2Bo2Fnp%0AfiHw9eap/EX8Evil8Vf2RvjN/wAEa/2sfEc%2Bt/s7/su/tF6f8f8AxN8Q/ib8Wf21fiJ8aPjl%2B1h%2B%0Aza/7P/jb4%2Bad8ef2v/gdo/gvSvgB4D1DSPBWoeFviTZWfgTxZ8Q/Evw78T3k2ja1qt7JFpn2b7w/%0AYm8C/sVfBL/grz/wVE/aY13xl8So5R%2B0f/wTh%2BFf7OPiXWf2iP2jPF%2Bl%2BKvGP7dP7KXge2tdL13R%0AfEHxL1XTfixF458UfEqO78NQfF%2By8Z6V8L9K07Qz8NbTwTa%2BFNFgswD%2Bkb4EfGPxP8Y4/i/J4m%2BB%0A3xS%2BB/8AwrD48fEn4OeH1%2BKNno1k3xc8MeArrT7bSfjj8PF0jVNSluPhb8Qhezy%2BD7zWY9K1m6TT%0AL6afS4rRrK6vPeK/iub4v/HH9kn9gz/gqh4E%2BEXxX%2BJmj/DLTf8AgvLefsZt8U/Gvjvxv8S/HX7L%0Av7MHxeg/Zq8N/ETxNY/Gj4uePfEer%2BGbGxTx%2Bvh7wn40119U1jwh408aXni%2B41i/8X%2BIbfxT4d/S%0Ajx9%2Byv8ADn9lv9vz9i79ij9lrRvGHhT9mT9uP9lD9uf4VftofCfTPjj8YtRu9P8Ahr8JPhR8L9E%2B%0AEPx80HUvE3i7X9e8LfErRde8Wx/BS5%2BKOi%2BJ9I%2BI/iHT/Hfh55rzxDF8PLfUfCQB/RTRX8BX7N3w%0Ah1b4P/8ABJD9nX/goVa/HH42z/tU/D//AIKeeAvAPhjxiv7Y3xm8WaNcfBfS/wBvbVfgVrnwE%2BHv%0Ah34gfE7wh8K/EXgLxXo2t%2BONd8RWGq%2BB54fHuia34r8RePraDQLrWH8Nf360AfjR/wAFQv8Agpx%2B%0A0J%2BwF8Uv2S/hr8H/ANiLR/2t7n9sD4j6Z8FfhzHb/tKzfBvxRD8WtSurt30q%2B0C4%2BAfxL0O28D6T%0AoY0vW9c%2BIOseNNDs9Ht72/m1DSLfRtB1PXI%2Bo%2BFX7ef7aq/tE/s/fBT9q/8A4J16J8CfBv7Sfiv4%0AmfDrwN8bPhF%2B2F4V/ab8KeG/iV8MvhT8Q/i/e%2BEviVo1v8HPhHq3hf8A4SHw18LPHGn%2BH9S02TxI%0Ap8R6DeaRq9ppTLHNL%2BZH7Z/ws%2BJX/BQH/g4a%2BAvwY%2BHPxs8R/AbSv%2BCa/wCw74m/aJtPip4W8EfC%0AT4nat4U%2BOn7Q/ji08C/2LZ%2BEPjB4X%2BJXw8XU9Z%2BG8Pg7xFpF94o8FNrGkS%2BGLrxBoJ07VbDwlr5/%0AQX9h628N/spfE/4R/wDBN74k67pv7T37Xep6b%2B2J/wAFDfjl8dlTw4dR8Eaj8Q/j9e6D4R%2BIPirT%0A30Dw7FoPjT4r%2BGvjvrXwv8Nw%2BBNF8PRW3hn4e%2BPNOsdCs/A9wDIAfan7bH7Yel/si%2BAvB0mheBb3%0A42/tB/G3x7o3wg/Zh/Zw0DxBZ%2BFvEvxw%2BK%2BuCS7OlHxTqGn6tpfgHwF4L8N2msePfir8UfEGn3Hh%0A3wB4F0DVNSng1XXbrw74a1/W/a3/AG1vgf8AsY%2BHvAuqfFfU9c1XxV8UvG%2Bi%2BA/hX8Ivhz4c1n4i%0AfGv4q6xd6tpNv4kX4X/CbwhZav46%2BIEngXw9qUvi/wAU2nhbRNSu7XSLNLWKKXWNX0PTtS/Pr9qe%0A9vdV/wCC63/BJ/w7qllb3fh7wt%2BzF/wUM8c%2BGHurBJl0/wAb6to3wb8H6trGn3cyOkWsWvhG4m0S%0AK6tfLvrDRvFGvWAmWx8SX0Fz81fGz4efGH9tn/gtzo3xa/Y68QfC34d%2BI/8AglJ8J9S%2BCXxS%2BLP7%0AQvw3tvjr8L/EPxM/al8LaP4qm8D/AA5%2BG/gH4x/CH4naf4o8AfB7xlql9rfj3/hPvDWnX/iTU734%0AXXNvp9jYeJbjxEAfrf8A8Nnadr/xi/Y38B%2BCfBHjKLwh%2B1Xpv7SOqS6r8Wvht8W/gP8AEbwknwA0%0ArR3Fte/Bz4z%2BBfAPxH0Z/Emr6pMbS48XeG9CF54ftLHxDoMGsaH4i0nV2%2B4q/L39oSe7sP8Agon/%0AAMEqrPxLq2j3XiG68G/tw6fdXen2UmgadrmvwfCH4X3mpvoOh3%2Bsa7fWVvOtjqWp2ujSa5rt7p2m%0AW8q3GqaiLK4v5PwB%2BG2iftOa/o3/AAXV%2BM37Wn/BTD9s2/8A2ff2H/GXxB0fwdbfCj44%2BJP2cPD%2B%0AoftN%2BFf2WrXxH8Xv%2BEW8W%2BE9Ki%2BLHgzwF4J%2BJPiTwIfgT%2Bz94G8WJ8MTrl7p1v4r8FfGJtdu7fxM%0AAf1I/ET9py2%2BEf7T3wd%2BCPxK0rQ/C/w8/aI8Navonwa%2BK994jeBde/aL8MXN9repfAXUtFl0tbOw%0A1rxd8Mo7jxx8NNTbXFHiW48D/EHw19iTWYfDUGtfV9fy9/8ABQP4ofFey/4I5f8ABJD9o74zXEXi%0AP9p7wl%2B0L/wSK%2BNut6v4j8NWUc118cr6%2B8DT%2BO5tet9G1PRm8Py65beI/G%2Bg%2BLp/CMtiNUsdZ13w%0A1Y2Oj6D4iuk07%2Bn%2B4hS5gmt5GlWOeKSF2t7ie1nVJUZGaG6tZIbm2lCsTHcW80U8L7ZIZEkVWAB8%0AD6n/AMFEfhj4O/Zc/ag/a5%2BL/wAHf2nfgH8Mv2U/FHxU8O%2BLtI%2BNfwbvPBPj/wCIGn/C86fFB49%2B%0AEPhZ9Zv/APhMfhv8S7vVLKy%2BGHjK7v8AQtM1yZ7ibWm8OWNhqN3aaH7B/wC1Z8c/2rfBnjzxJ8dv%0A2NfiF%2BxrrHhvxRo9t4O8O%2BOvGWk/ECD4ieAPFPhLRfGXhnxvo/ibQNA0PQkvUstZXQ/GPhnTZ9fh%0A8I%2BLtN1XQv8AhItZW2jv5/5Ovir4O%2BKHxY/4IY/8HAXw5%2BNn7S/7WvxMb9kH/gqp%2B1j4S%2BGHin4o%0A/GvxT4u8ea/8L/gRH8A4vhr8OPHmseJ4r2PXfhVqUesP4wvvBOk6doPha78Q3sPijw9Y6Ncm3mX%2B%0Aur/gn98EG/Z5/Y8%2BAfwxl8a/Fjx1c6b8N/CWqX%2BqfGb4k6z8WPGOnanrvh/TNU1Tw3B4x166v9Rm%0A8L%2BHtRuLrTPCmji%2Bu9O8PaFDZ6JoskejWNhbwAH2RRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRR%0AQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFA%0ABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABXwb4v8A%2BCZf7FHjf4n/ABO%2BLms/%0ACPWLLxP8cXs3%2BPOheEfjH8c/h/8ACf8AaBFjo9z4fji/aB%2BA/gL4leG/gf8AHW3n0S%2Bv9L1K2%2BLX%0Aw88Y2%2Br2eoajDqsV4uoXnn/eVFAHxJqH/BOf9jXV/E/x68Z6t8HTqvin9pv4Xan8EPjfrep/EP4q%0A6he%2BNPgxq9w0158JY3vPHEyeFvhskLPpOn%2BB/BqeHvDWkeHHk8LaTptl4bkfSm/Hb/go5/wSu%2BJv%0Ajjxn/wAExfg9%2ByT%2Byh4K8afsZfsWfGWP4xeJ/DfiT9rvx58P/EGm20iHw7B8Ovh%2Bmv6f4n8WeE9J%0A8HWNtpPjfwdrPhX4gW1vYarZt4c0rw34YWGHXrn%2BmSigD5n%2BFv7HX7NXwd0b4taL4M%2BFOj3cfx%2Bd%0AD8eNa%2BIeqeJPjD4z%2BN8UXhgeC4LL4w%2BPvi7rXjjxx8TtMtfCRm8OWmleN/EOuabaaPe6lY21pFBq%0AmopdcP8ACD/gnp%2Bx/wDAjX/hj4k%2BGfwiOmah8D7XxRZfAu08Q/ED4o/EDw78CbTxto0vhzxhbfA3%0Awh8Q/G3irwn8GofE/h6e50LXY/hnovhZdT0a6u9MvBLZXdzBL9oUUAfGnw1/4J9fshfB/wAXaB4y%0A%2BHHwgj8MXXg/xV4p8d%2BB/CsHjj4lX/wo%2BH3jnxu%2BvP4u8bfDn4J6t4yv/g78PfGPiI%2BKfEY1XxT4%0AL8C6Frt1HrWoxSX5jupEL9B/4J7fsT%2BGvih8QvjJpH7NPwti8f8AxWf4rz/EHULvQf7V0PxLqHx5%0At9As/jprV14I1ae98DW3iL412HhXw5pvxd8S6f4btNf%2BJel6Lp%2Bl%2BNdS1vT7ZLYfZFFAH5tX/wDw%0ASA/4Jpa14d0Dwh4l/ZB%2BF/jPwn4PuxceBPDPj1vE/j3QPhxavfaZf3/h34YaP4y8Qa5p/wAMvBWu%0AnRtL03xT4C8BW/h3wV4t8PWNv4X8TaBq3htP7KP0Jrn7Fn7LHiP41%2BHP2idV%2BCnhB/jJ4Uh8HRaP%0A4ysk1LSHL/Dq28Q2Xw7vtX0LSNQsfDHiTVvh9p/izxPpvgPWvEei6rq/g3TNf1fTPDd7plhf3NvJ%0A9Q0UAfFHw9/4Jy/sQ/C64/aAn8G/s4%2BAbeL9qm0nsf2itM11NZ8aeH/jBa3iXkeoxeNPDXjTVvEH%0Ah7U/7XXULv8AttxpkUmuNLv1d7144mTR%2BGn/AAT%2B/Y8%2BEWr%2BP/EfgX4H6BY%2BKvih8Prz4S%2BOPGWu%0Aa14v8aeN9Y%2BFN8kST/DCHxr428ReIvFWjfDtGggmtPBGg6vpnhmwu4Yryx0u2u40mH2NRQB%2Bamu/%0A8Eef%2BCaPiL4Y%2BHPgtf8A7I/w7tvhH4S8e618T/Dfw10HUPGfhXwVpHjvxFYeEtO1rxDaeHfDHijS%0ANMFxfweBvC80tq9u%2BnpqOnSavDaR6tqWq3t9%2Bj2madZaPpun6RplulppulWVpp2n2sZdktrKxgjt%0AbS3RpGeQpDBFHGpd2chQWZmyTdooA/ODxF/wSM/4J2%2BKPGOtfEPU/wBm7Sbbx/4lu9Wu/Evjfw94%0A/wDi14R8XeJW1nVH1ie08R%2BJPCnj7RdY17SrC%2Bdh4d0XVry80fwpZbdK8L2Oj6WiWa%2B0/sufsI/s%0Aj/sWXnxS1L9l/wCB/hb4Sat8bde03xP8Wtb0e78Qavr3j/XNHl12fSr/AMSa74o1nXdXv20658Ue%0AJbuzga9W1t73xBrV5HAt1qd7LN9bUUAflR/wU9/ZZ%2BLXxLk/Zh/bI/ZZ8P6P4t/bB/4J8/FLXvit%0A8IvAWuXVppVl8aPhn8S/DDfDf9pL4AweJL/UNNsPCWu/E74YXLXHgzxHqM50iHx54U8LaVrr6doe%0Ar6lrukfJH/BTD/gn18UvEln4j/bu/wCCa3hz9qr4L/8ABSf4ww/D6z8Y33wY/aD%2BHXwz0PxnF4W%2B%0AHFxo/wAOF/a2%2BFHxw8Y6p8APiX4L%2BFU2keH/AAfreleCRD48s7XV77WdCu/iDZ6LL4a1v%2Bg6igD8%0As/iN8JP2jNQ/az/4JIeOfFOgR/Eu6%2BD/AIN/aZ0D9p/4q%2BEdUuPCfg/w3418Z/s4%2BG9Ih8Wx%2BGbe%0AC2fXNB8dfEbw/f6RoHha%2BsLTS9IOr2Wvytaat4b0Wyuvnb/gqh/wTz8d/EH9iLxb%2Byf%2BwR8LbXQb%0Ab9r79ujwT8Uv2wX0rxnpGhPqHgP4p/F6P4oftQfFPxFqXxA8WaZquvW%2Bs6lo%2BkR694H8D6/Y65q3%0AhmZvBfhTRn8KR3Hhif8AdWigD82P2t/2e/GP7Wf7U37HPw%2B8R%2BENYX9lr9m3xnL%2B2N8VfEWoX2hR%0AeEvib8afBttq/g/9mj4OW2iC8ufEPiVPCXibV/FPx18dHU9LsfCOh3Xgf4TW4k8Q634ogbwf%2Bkcs%0AazRSQuZAksbxsYpZYJQrqVYxzwPHNDIASUlhkSWNsPG6uAwkooA/LK0/4Iuf8E6bT4b/ALRfwh/4%0AVF8Tr74bftbeNB8R/wBpDwnrX7XP7ZHiDTvi78QX1p/EV7438Uya78ftRvH8Wa1rhg1TxJ4hsLqy%0A1XxRd6dosniK61P%2BwtGFj9%2B/Bf4PeCPgB8L/AAf8HfhsPGEfgTwHp8%2Bk%2BF7fx58TPiX8YPE1jpUu%0AoXmow6ZdfEH4v%2BLvHXxC1nT9La9fTvD9lrfinUbbw14etdK8LeH49M8NaLo%2Bk2PqFFABRRRQAUUU%0AUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUU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SPHHhVx4a8X%2BKtb0OOOwT7noAx7TxDoF/rOseHLHXNHvfEPh630q7%0A1/QbTU7K41nQ7TXVvX0O51jS4Z3vtMt9ZXTdRbSpr2CCPUVsL02jTC1nMexXwF8Hv2Sl8Kft%2B/ta%0A/tpax8Pvhv4J1f4vfDf4JfBfw7rPgjUtQvfFvxQ0X4cW2sarrvxJ%2BLnnaRpNgniOSPUfA3wz8K6f%0AF/aV9onhj4VlJtY1DTdV0hbX79oAKKKKACsTxLZa3qXhzX9O8Na3F4Z8R3%2BiarZaB4kuNJi16Dw/%0Ard1YzwaVrc2hz3VlBrUWk30kF/JpM17aRailu1nJdW6TNKm3RQB/n2eLtc/ah0P/AIJKf8FKf2v/%0AABT%2B23%2B2X8ePiF8I/wDgon4i/Z3/AOCcFpY/tBfHrSb/AEvWPD3x%2B%2BHXwYj%2BMTweBfHt34k8Z%2BNv%0AG9hrPjHU9C%2BEXxF1rxh8HvBs%2BiNp3w48I6UnjyXT2/pc%2BHI1hP8Agv18VNJ8b3uma3r3hb/gjx%2Bz%0A7ZeDNdnt9MtNZ1fRdY/at%2BL/APwsnxHZaXa3M66EnifxnoGhxeKbPSLfTLC4Xw14GivrW4j0rQLu%0ATvP2i/2Arrxf8Uf%2BCZfwJ%2BDfwp%2BHvgv/AIJ/fs4ftD/Ev9rP49eEvD%2Bq/wDCJXMXxX%2BHWm6r43/Z%0Al0vQ9C0%2B4ivfEul%2BJP2jviJ4l%2BLvxKguFudO1vUfB8X/AAlT3cevS2uo9N%2B0j%2BzB8bbH/gpX%2Bxf%2B%0A358DtGh8b6L4a%2BHPxD/Y4/aj%2BH9pq3hfw14oPwJ%2BLfibQPGnhf4qaFqnijUdC07xDpPwb%2BJ%2BgWHi%0A3x14IfV5vEuqeE2vbz4a6Jr3jC1HhnxGAfl1/wAFS/Cd/wCKv2qf2L/Dnws/bv8Aj5q3xu/ai/4K%0AbeGv2f8AV9J%2BB37TXj34T%2BHv2av2Sfh18GdV1/8Aap%2BC3hTwN8I/ifo/gXTPjTNpep%2BG/EmvfFL4%0Aj%2BFPEHxqtrzx5okWiXnh3T9D8BabZ/Tnxj%2BA/wAOPh1/wb8ft6/C3QP2oPjB%2B3R4E8M/sgft4Xuh%0AfHH9oHx5F8T/ABlqevfDTwn8Vfs2k2njK307TG1bw/8ADH4heB303wjIZdVs4U0CBdH1O68NLpFt%0AB96eMf8AglF/wTt8e%2BNfEvxK8Ufso/DW9%2BIvi34lap8YNa8fWY8Q6H41b4k%2BILSaw8VeKdL8VaDr%0Amma74eu/G1jLDZ%2BPrTw9f6Xpnj210zQbfxlZa5F4c0FNN5L9uD9izXP2lvg38G/2DfhhoHgL4O/s%0AUeI9T0Ww/aft/Cmk23hc2X7OPwl1DwnrGh/syfBPwv4M1PwufBN78Z9SttP8JXnirSLOLQfh38Jv%0AD3j5LBbbxZqXgbT9RAPcv%2BCd2t%2BLPE3/AAT%2B/YY8R%2BPtMl0Tx14g/Y6/Zk1vxpo0%2BmXWiT6R4s1X%0A4KeCb/xHpk2jX3%2BnaTLYaxPeWsmmXn%2BlWDxNa3H76J6/Kj9oX9vD9nHwZ8cviz4Q8R/8FZP%2BCpnw%0AX17wv498S%2BH9X%2BFfw6/4Jw%2BFfGnw/wDA2oaRqc9hdaD4I8Z%2BIf8Agjl8XdV8Y%2BEraWBm8O%2BLJfil%0A8QrXxJpMlprWm%2BMNf0y%2Bs9SuP6EbCwsdKsbLTNMsrTTtN060t7DT9PsLeGzsbCxs4Ut7SysrS3SO%0A3tbS1t444Le3gjjhghjSKJFRVUW6APwr/YJ/bR%2BAnxm/aOb4feAf%2BCkP/BRr9pzxNPaeIRF8Lv2j%0A/wBiTwh8HPhTqS6f4Q0nXLrxJN4%2B8Lf8Ey/2Zr3wnLolk4h0rR9a%2BKvhGLWteWfb4W8Rf23o9zrH%0A7qUV5j8Yvgz8Mf2gPh5rvwo%2BMPhKx8cfD7xLJpUuteG9QuNRs7e9l0PWLDXtKk%2B2aReafqVtJaat%0AplldK9pewNIIWt5jJazTwSgH85//AAaG/wDKG3wX/wBnAfHb/wBPmlV%2Bv/7b37Qf7eXwK/4QeX9j%0AD/gnfp/7eFvrf9ox%2BOYZP2wPhZ%2By5qvgaaPa2kS2Vt8UfCGu6Z4u0%2B9SK6TU57XW9Jv9KuZdJjtN%0AL1q3u9Uu9D9I/ZV/YT/ZS/Yit/Gunfsq/Ce3%2BDeg/EB/Ds3iTwj4e8X%2BP9Q8CC68LnXjpt/4c8Be%0AI/FeteDPBOp3beJdVfxHqfgrQvD%2BoeMZP7Nl8XXOtyaFobad9b0AfjB8Gf21P%2BCxfjr4xeDfA/xZ%0A/wCCI3h/4GfCrVfGGj6V44%2BN99/wU7/Z9%2BIlh4Q8Gza5Z2PiDxfo3w%2B8F/Cufxn4w1DTdClvdd0X%0Awq9v4ak12azi0i/1nw3Ld/a7f0v/AILjf8ohP%2BCiv/ZqvxT/APTJJX6qV8Af8FS/hN4z%2BPX7A/7R%0AXwT8B%2BF9V8Z658WvD/hX4d33hrQhH/bGpeDfF/xE8H6F8Rv7OkkkiW2u7P4e3vii/ivQ/mWLWv2y%0AJJJYEjYA83/4KBfEj4G/sm6l8N/2qrf4HW/7RH/BQXU/C/ib9lf9iP4eadLdJ8UPjB4u%2BJM%2Bk%2BKd%0Ab8CaKbC1uPD3hDwvAPCEfjP4pfFrU9BsNJ%2BGXw%2B03xW0%2BvaNpXiS80XxB8Ifsf8A7MkPwR/Yx8a/%0ACC5/bOtvB/xs8U/t0fFD4uf8FePiH8L/ABJ4j8N%2BOvG/7RHxS%2BBP/C8fjL8BP2a/HF4nhzW/Aniy%0A80nWfgFb6f42%2BGUlh4tg%2BGugeN9e8Jan4W%2BKeqXGreHv0i/bY/4JPfsCf8FFp/hfdftmfAq7%2BNVz%0A8GdK8QaL8OLi8%2BL/AMdPBc%2Bgaf4qfQZPEKzzfDf4meDpPEF3qsnhnQ3udS8StrGpF7ENHdo090Z/%0ALdQ/4IX/APBKHVfgf8HP2cdR/Y68G3fwc%2BAXxJ1z4ufCvwtN42%2BLZvdB8feJ105PE%2Bsap4vHxBHj%0AbxnZeJo9G0KHxH4c8ceI/EnhjXoPD/h2DVtGvIfD%2BjJYgH5Wf8EDNJ%2BPXxT8BfsV/tpftKf8FIP2%0Ar9e8dftCfC/9qbSvhj%2Bw98bPia3jXwx40%2BA/wv8Aizc%2BGPDnjW9stb0bQNX8beOvAFvd%2BDPFerft%0ABahpWpeN/F2j/EDwnp1/4ug8Ia5pvhmv6Av2uviv%2B1z8JvC/hDVf2Rf2RPC/7X/iXVvFFtpHi/wp%0A4l/aX8O/s0L4P8O3lxY2g8XWuveJfhz8Q7HxNb6a11cahrWjW0OnavFo2mXk2hw%2BIdal0/QL7J/Z%0Ag/4J8/sh/sbapqOs/s6/CV/BGpXvhyXwVptxrHxB%2BKPxKHgb4f3PinUfHF38NPhNB8U/G3jW2%2BDP%0AwtvvGWqXXirUPhh8JofBfgDUPEEen6ve%2BHJ77SNKnsvs2gD8gP8Ahqv/AILJf9IhfgB/4tP0L/6D%0ASv13tGu3tbZ7%2BG3tr57eFry3tLmS9tYLto1NxDbXk1pYTXdvFMXSG5lsLKSeNVle0tmcwpYooA/m%0A%2B/4JWxXVh/wXC/4ONLCaJ7pLjxn/AME59SbUra90e8srUXPwN%2BLN5ZadcPZXpmF7La6kyrbx2k76%0Ac%2Bl39h4gm0/WIo7a7%2Bwv%2BCUH/JRf%2BCwv/aX/AONf/rLP7HFen/AL/gkB/wAE/v2Yvjxqn7TXwV%2BE%0Anj/wx8cNemW48SeONY/ag/av%2BID%2BKpobG40yyk8XeHviN8b/ABb4T8WHSbC6ntdCHiPQtTXQIX26%0AKLDZGVs/8E2fgd8VPg5Zft0eIvjD4SuPCPin9oH/AIKT/tcfHHQ4p9S0W%2Bj8QfCu71/w78Lfgl4t%0AtLbRNS1NNHt/EPwd%2BFfga9bTdTay1qS6Nzqup6bYz6n5IAP0PsPEOgarqeu6JpmuaPqOs%2BF7ixtP%0AEukWGp2V5qfh271PTrfV9NttdsLeeS60m41DSbu11Sxhv4reS7065t723WS2mjlbYrwT4V/svfAH%0A4I/Ez4%2B/GT4V/DDw94N%2BKH7Ufizw142%2BPnjXTm1GfWviR4j8HeG4/Cfha51WbUb68isLDQ9HF42n%0A6FocWlaBDrOt%2BKPE39mHxL4t8T6tq3vdABRRRQAUUUUAFFFFABRRRQBXubS0vY1hvLa3u4UuLS7S%0AK5hjnjS7sLuG/sLlUlV1W4sr62t7y0mAEltdwQ3ELJNEjrYoooAKKKKACiiigAooooAKKKKACiii%0AgAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKA%0ACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAK%0AKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAoo%0AooA//9k%3D%0A"</p>
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<div class="prompt input_prompt">In [18]:</div>
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<div class="highlight"><pre><span class="n">x</span><span class="p">,</span><span class="n">a</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">symbols</span><span class="p">(</span><span class="s">'x,a'</span><span class="p">)</span>
<span class="n">xi</span> <span class="o">=</span> <span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">**</span><span class="mi">2</span>
<span class="n">eta</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">-</span> <span class="nb">abs</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span>
<span class="n">half</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span>
<span class="n">eta</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Piecewise</span><span class="p">(</span> <span class="p">(</span><span class="mi">1</span><span class="o">+</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">-</span><span class="mi">1</span><span class="p">),</span> <span class="mi">0</span> <span class="o"><=</span> <span class="n">x</span> <span class="o"><</span> <span class="n">half</span><span class="p">),</span>
<span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">-</span><span class="mi">1</span><span class="p">),</span> <span class="n">half</span> <span class="o"><=</span> <span class="n">x</span> <span class="o"><</span> <span class="mi">1</span><span class="p">),</span>
<span class="p">)</span>
<span class="n">v</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">var</span><span class="p">(</span><span class="s">'b:3'</span><span class="p">)</span> <span class="c"># assume h is quadratic in eta</span>
<span class="n">h</span> <span class="o">=</span> <span class="p">(</span><span class="n">eta</span><span class="o">**</span><span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">v</span><span class="p">))</span><span class="o">*</span><span class="n">v</span><span class="p">)</span><span class="o">.</span><span class="n">sum</span><span class="p">()</span>
<span class="n">J</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">integrate</span><span class="p">((</span><span class="n">xi</span> <span class="o">-</span> <span class="n">h</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">sol</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">solve</span><span class="p">([</span><span class="n">J</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">v</span><span class="p">],</span><span class="n">v</span><span class="p">)</span>
<span class="n">hsol</span> <span class="o">=</span> <span class="n">h</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sol</span><span class="p">)</span>
<span class="k">print</span> <span class="n">S</span><span class="o">.</span><span class="n">piecewise_fold</span><span class="p">(</span><span class="n">h</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sol</span><span class="p">))</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">51</span><span class="p">,</span><span class="n">endpoint</span><span class="o">=</span><span class="bp">False</span><span class="p">)</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">5</span><span class="p">,</span><span class="mi">5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">t</span><span class="o">**</span><span class="mi">2</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\xi=2 x^2$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,[</span><span class="n">hsol</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">t</span><span class="p">],</span><span class="s">'-x'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\mathbb{E}(\xi|\eta)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="nb">map</span><span class="p">(</span><span class="n">S</span><span class="o">.</span><span class="n">lambdify</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">eta</span><span class="p">),</span><span class="n">t</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\eta(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
</pre></div>
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<pre>Piecewise((2*x**2 - 2*x + 1, x < 1/2), (2*x + (-2*x + 2)**2/2 - 1, x < 1))
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"></img>
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>The figure shows that the $\mathbb{E}(\xi|\eta)$ is continuous over the entire domain. The code above solves for the conditional expectation using optimization assuming that $h$ is a quadratic function of $\eta$, but we can also do it by using the inner product. Thus,</p>
<p>$$ \mathbb{E}\left((2 x^2 - h(\eta_1(x)) )\eta_1(x)\right)= \int_0^{\frac{1}{2}} (2 x^2 - h(\eta_1(x)) )\eta_1(x) dx = 0$$</p>
<p>where $\eta_1 = 2x $ for $x\in [0,\frac{1}{2}]$. We can re-write this in terms of $\eta_1$ as</p>
<p>$$ \int_0^1 \left(\frac{\eta_1^2}{2}-h(\eta_1)\right)\eta_1 d\eta_1$$</p>
<p>and the solution jumps right out as $h(\eta_1)=\frac{\eta_1^2}{2}$. Note that $\eta_1\in[0,1]$. Doing the same thing for the other piece,</p>
<p>$$ \eta_2 = 2 - 2 x, \hspace{1em} \forall x\in[\frac{1}{2},1]$$ </p>
<p>gives,</p>
<p>$$ \int_0^1 \left(\frac{(2-\eta_2)^2}{2}-h(\eta_2)\right)\eta_2 d\eta_2$$</p>
<p>and again, the optimal $h(\eta_2)$ jumps right out as</p>
<p>$$ h(\eta_2) = \frac{(2-\eta_2)^2}{2} , \hspace{1em} \forall \eta_2\in[0,1]$$ </p>
<p>and since $\eta_2$ and $\eta_2$ represent the same variable over the same domain we can just add these up to get the full solution:</p>
<p>$$ h(\eta) = \frac{1}{2} \left( 2 - 2 \eta + \eta^2\right) $$</p>
<p>and then back-substituting each piece for $x$ produces the same solution as <code>sympy</code>.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>Example</h2>
<p>This is Exercise 2.14</p>
<p><img 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u7fX9HmiTUbYA89b/AIKOfsGWugah4q1z9rr4A%2BDPDel6p4O0W91v4i/Enw58NdNi1T4hafd6%0At4Hso7vx/e%2BG4rmXxXpGn3%2Br6F9laZNR0qyu9Tt2extpp07fRv20/wBknxH8MPiT8aPDv7R3wc8Q%0A/C74Oz3Vp8VPGehePfD%2BsaV8P7%2B3ht7mLTPFQ029ubvR9V1a0vdMvfDWlXdtHqPiux1rQb7wza6t%0Aaa/o019/NN%2B1j8G/i3%2By9%2BwJ%2B1rf%2BOfDHx6m%2BH%2Bv/FD/AII7aL8Bfh3%2B2r8av2bPiT%2B0h4i8SfCv%0A9uj4Ya18Q/hjF8Zvhv448YaHrPwW0rS9X8G6f8P0%2BLHjLVfEnhjxM/xs8R6pd2vw81PQJY/rT9qf%0A9mr9vH4zeKv27vj58If2VdP8CJ8ftB/YJ%2BA1j8DviNr/AOzH4k%2BK3xG8J/s%2BfFjx18Q/i5%2B1NY2M%0A/wASfH/7NenfGjwt4c%2BIei%2BCv2cT8TfHt3c2lx8LdO8ReKdG0X%2Bx/COl6iAftHoH7aH7LfiT4PeM%0A/j5YfGrwfafCf4b%2BJpvBPxG8U%2BIpNQ8IyfD3xxBeaDYN4G8d%2BHvFdhovinwd43kvPFXha3tfB/iP%0ARNN8SXknifw4LTS5xrulG7w/EH7e/wCxv4V8GaX498RftE/DbSNC1nx2vws0%2BzvNXkj8YSfFL/hH%0ALHxhcfDC4%2BHYtm%2BIFh8SdN8I6ppvivWfAV/4YtvFuh%2BGr%2B017WNHsdKuIrtv5xLz/gm//wAFKfE/%0AjX4p%2BNJ/g7eN8NfDf/BRn9jv/gpH4E%2BFfxt/a0tfif8AGf4/L%2Bzn4O%2BFnwdHwJ8e/EW18V6p4Ss/%0AEsHgfwH4j%2BJMy%2BNda1f4a2PxB074BeE9L8X%2BKfBvhfxbLp/3p%2B2N%2Bxp%2B0Z8T/wBtf9iT9uj4e/Aj%0AxDH4Z%2BBusfHfw/40%2BDfwq%2BJfwb%2BHf7SWi6f%2B0t8KfhF4b1n48ab4g8S%2BJ7H4J3HxY8O%2BIPDniX4e%0A/E/SNL%2BKt5deLvgvoegHw7471/ULhfBsAB%2B6fgHx/wCCPip4L8MfEb4beLPD/jrwF400ay8Q%2BE/G%0AHhXVbPW/DviLRNRiE9lqek6rYSz2l5aTxtlZIZW2sHjcLIjovyH8Jf8Ago7%2ByH8Zf2hv2nP2ZfCX%0Axt%2BGLfFP9lLUpNP%2BJHh29%2BIXhK01vydD0LT9V%2BIuuWHh%2B41SHWJPDHwp1XUofB/xE8SfZX0fwt4u%0AhvND1i7sb228p73/AATs/Z7s/wBl79krwB8ILD4Wal8ErPTvE/xh8W2fws1r4pXPxi1/whYfEv4z%0AeP8A4jaZaeKfHBvdT0W58c6jpnimz1v4iaP4L1bXvAHh34gan4o0TwP4p8ZeGrHTvF%2Bu/n78SPg9%0A%2B3v4I%2BMf/BWuX4Cfs4%2BBvHth%2B19faX8Q/hb40%2BLuufDDWfgz4xsfBv8AwTV8HfATTPhPrnhGT4ta%0AD8QrLxt4o/aQ8A%2BGtGvofGHw%2BPwlvPhPdeKL7V/Gun393p0MQB9l/FP9uz4XfEH9mX9rfxz%2BxV%2B0%0AJ8D/AB98WPgB8CPEvxasLowS/FHwraWUXgrX/GfgjxZN4d0Dxh4HufGnw68f2fhnWrHwX4/8L%2BLH%0A8E65qFhqc%2Bka14iPhzWdFflv2U/%2BCmP7P/xB/Zt/Zp8V/tB/Hv4HfDn9oDx7%2BwX8Cf20fjX4KufE%0Atn4N0zwR4R%2BIPwi8O%2BPfGHjyS08R6tdnwx8ONP1O91v7Bea5rt19j0yxZLnUrx7We5b8L/hr%2BxR/%0AwUx8K%2BNfGnjPwd%2Byp451nwV8Zf8Agk7a/sGeJbX45fGv9m7wb408B%2BNfC3hr4v6xpOgfDj4HfB7x%0AL4f%2BBXwf%2BB2k%2BNPiB4R%2BGfwy0XwRPqtp4V8B%2BAfEPiq98Na74g%2BJWo%2BKrm18O/8Aglr/AMFCfCP/%0AAAT5%2BK//AAT7tvhxrE1p8c/2af2O/iJe/HHxh8c/hTH4x%2BHH7W37KujfCzQfFf7MU%2BveFb7Vdf8A%0AEX7Lfj3wp%2BzR8NPCnwR8UxaLqWq/B2w%2BInibwv4j0fxJ4TgS68NAH9UHwe/aB%2BDHx/sPEOo/B34i%0AeHfHkPhHV7bQfFVvpE80eqeGtWvtIsPEGmWev6JqEFlrOkNrHh/VdM8QaHLqFhbwa3oOo2GtaTLe%0AaZd291J4B%2B37%2B2Fd/sa/Bnwt4l8IfD5vi98cfjb8Yvh3%2BzV%2BzR8Im1yHwpYfEj49/Fm61CLwloni%0ADxfd29zY%2BEvCej6Ponibxt4x8Q3kTJYeFvCmsfZw9/JZxvwP7D/wc13wr8Uf2jfjZ4u%2BAXxy%2BDnj%0AD4y6B8A/D3iPxD%2B0T8fvh58YviN8QLv4VaV8RFht7Tw78H/GvxE%2BGvg3wP4Hg8dro3hzxBD4nsvH%0AHxA1K98RnxX4O8P6N4T8F6j4j8z/AOCg/wAL/wBp341ftH/sBeEvAP7OGifE79n/AOGX7TXwo/aU%0A8afHCL4jeC/DHiL4BePPhJ4rl026vNT8HeK/Feiaz418O%2BL/AIR%2BNfH2haVb%2BANA8Va3ba/LM%2Brx%0A6VYixe/AMnxR%2B0d%2B3d%2Bzj%2B0t/wAE/fhN8efEn7MPxm0X9tD4hfET4Z%2BPfCvwQ%2BBvxY%2BHnin4Ra/4%0AP%2BBvif4rHxX4H8Y%2BKf2g/iXF45%2BGfhq98ISaP4uuvFHw08MaxdQa0vixL/wpbS2ngrT/ANVvGvjv%0AwP8ADXw5f%2BMfiN4y8KeAPCOleR/anirxr4i0jwr4c037VPHa232/XNdvLDTLP7RcyxW8H2i6j82e%0AWOGPdI6qfyh%2BDvwk/ad%2BIv8AwVc%2BM/7SX7QPwM8SeHPhR8HPAnjL4Kfsk%2BOvF/xD%2BG2veCtO%2BHvi%0ABPhLf6vrfwb%2BGngT4ha/4n0H4kfG3xVZeP8AVvjh8S/ir4b0DU4vhx4B%2BAHwu8J6Sk9v4%2BmsPvb9%0Arr4F6v8AtK/s/ePPgno198FrGfxuugQ3D/tCfs%2BaR%2B1L8JZ7PRvEmkeIpLbxR8D9f8aeANE8ZLLL%0ApEJ03%2B0vEdvFouqJZ65bwTX2nWmwAdpH7ZX7IPiC%2BtNM0H9qv9m7W9Sv0Mtjp%2BkfHL4YalfXkazP%0AbtJaWln4omuLhFuI3gLwxuomR4id6lR9I1/Orof/AAQx1rw/4r0fxBa23/BFy%2B0/Sb3T74aPqX/B%0ABH4R29xPNaRwtM6a/wCGP2ufDOqWEwvUkvdMudOe2uNNcWiSyX/2eV7v%2BiqgD%2BeD9kP/AIKG/tZf%0AGr4a/wDBP74FeDfHHwY%2BL37av7WP7GS/8FC/jF41%2BLXhh/B/w3%2BBv7P19qHgHwlaaLp/w2%2BFN14d%0A8ReLdW1r4o%2BO28FeAbmTVtMiudP8G%2BM9Z1/WL57C3WX9Rf8AgnN%2B1zdft1fsZfBb9qDU/B9n4B8Q%0AfEC28caJ4v8AB%2Bmarc65pGheO/hV8S/Gfwg8e2miave2GmXmoaC/jPwFrtxoVxdWUVw%2BkTWQme4k%0ADXM34j/sG/s6fEH4Kx/8EjP%2BCk3wr%2BCnj743%2BHvEX/BBT4K/sP8AxL%2BF/wAMLz4YaL4g8H67LH8F%0AP2k/hX8TLCw8Y%2BK/B9hrul/EjxVc/Ejwr4%2B1Aa2tz4V1DUfBfizV7P8Asu58Z6jo/wBvfsG%2BFvjf%0A/wAE5fhN/wAExP2C/GHw88MeN7341Sftga/8evGvhz4kaZPefAv4haxqPjv9rSK10XwjPpVjrPxE%0A%2BFq%2BKPHPiX4Qav8AEbT4bHT/AAn4lm%2BFVpqwubj4l6QzgH0P%2B1Z%2B2B%2B0b8B/2zP2E/gD4I%2BAnw88%0Ab/Bb9rP4l33w28Q/FHUPilrVn8S/DuuaD8NvjF8VPGsOg/Cq38CroH/CP%2BCPAnwxsPFV74y1b4mT%0AXeuya3d%2BF9L8BQy6XJ4ik9Wt/jt478Gft6z/ALM3j97PUfh78c/gLq3xy/Zy1%2BHTrTTLvRPEHwU1%0A/wAFeA/2gfhNq09vPLJr7CD4kfCv4qeEdXvYLPUGh8S/EHQz9s0zwrpr23y7/wAFBNA/aV1H9rz/%0AAIJtfEv4Mfsg/F/9ojwB%2BzD8YPir8W/ix4h%2BHPxN/Zm8CSWumfEf9m343/AHSvDnhzTfjT%2B0B8I9%0Ac8Q%2BJNG1rx/pPi3xDZS2eneFLzwoV0208R6xqt1quiaZ6R8StSn%2BJP8AwVW/ZZ8IeDNTsb2D9mP9%0AmP8AaY%2BJ/wAd0tL5HuvDL/H7W/g38PPgF4Z1eC3ivfIn%2BIH/AAgvxn8V2FhqA0ieez%2BFzapZ300E%0AUlhqAB9q/Gj9oD4D/s3%2BE7Tx5%2B0R8bPhH8BfA1/rlp4YsfGfxo%2BJHg34W%2BE73xLqFlqOpWHh608R%0AeONZ0LR7jXL3TtH1a/tNJhvHv7my0vUbqG3eCxuZIvn2w/4KU/sB6wZzoP7XvwF8SW1t8XvC/wAA%0ApdS8L/ELQ/E%2BjD41eNVum8J/DBNc0C41HR5vGevmxvYtN0OG%2BkvLi4tLm1VBcwSxL0n7e/7LHhn9%0Atn9jT9o79lrxTbLLbfGD4X6/oXh698yaKbw38RdMSLxN8KfG9k0OoaVu1PwD8TtE8I%2BNtJiuL%2B3s%0AJtT0C0h1IyafJdQyfkZ%2BxH/wTM/at8MftS%2BDv2kv2n/Eum2Xhv4veH/B37cX7TfwY0PVtFu9K0X/%0AAIKhQab8Qfhvb6P4eu/C99FY3fwZ%2BG/wV%2BImnWMGgOfGWmat8WfhZ4J%2BIK%2BOfEF9pVvqF8AexJ/w%0AWd%2BFHwW/Zqf49ftQfFj9lfxLJ4y/bo1P9lv4O2v7M3xC8ReIdA8SeCpfjF4P8HnV9a1HxLpt3NZ/%0AEX4VfCzxNf8AxS%2BLWkP/AGd4Zhi0S20WHVfD2o%2BIrG1tPUviD/wU58H/AAT/AG9dJ%2BEHxo%2BNf7MP%0AgX9kL4h/sL6X%2B1H8MfiD4o1HUvBXxI1Dxnd/FjSfBUWkafrmseOdQ8LfEXwRqngu51Xx1dS6H4I8%0AK634J07T4b/VbjXdAOravo35YaP%2Bwh%2B31e/s9eN9T1L9lrWvD3xH8Kf8HF%2Bk/wDBVfR/g5L8U/2c%0AvE%2Bv/Fn9nC/%2BOuj/ABUv9D8BeNJ/iqngXwP4/wBB06/voL2y8U%2BI/AZ1O78L6hYWV5eab4vnef6a%0A%2BLvwZ/bw8f8A7TPjX41eOf2LdX8e%2BIPHX/BE34ofss6l4t%2BGnj39nTwr4Nsf2lPiv4ym%2BKOqfBTT%0APCXxH/acu/FK%2BGrJrLw74J1H4lahdaj4b1TxVpX9q2b2HhG%2Bmv7MA/oc0bWdI8RaRpXiDw/qum67%0AoOu6bY6zomt6NfWup6RrOkanaxXum6rpWpWUs9lqGm6hZTw3djfWk01rd2s0VxbyyRSI5/O61/4K%0AT/s6fGfx9%2B2b%2BzZ%2BzP8AHH4VeJP2kf2W/hbrOuXYk1jQPHHhWz8cp4a1%2B4mgm0Xw/wCM9EvfFOm/%0ACzxLbaBo/wAVtKt/EnhifStdvr7wVd63oWu6VrE2l%2Bpf8E5PDHxY8B/sDfsb/Dn47/D7xP8AC74z%0AfC39m74QfCb4n%2BC/F/iXwP4y1uy8bfCrwTo/w78Qav8A8JV8OvHvxM8La/o3i3UPDU3i7wxqkXjP%0AUdcuvDOu6RJ4vsfDvi8674b0j81dX%2BDf7fPw7/b7/wCCiOr/AAq/Z48K%2BK/2f/2wb7w38SPEnxQ1%0AHxJ4G0/XPE/hb4ff8E49O/Z88CfCPwNc3vjPRNW0X4lXP7UfhbRdQ1L/AITzRZPAGj/Ca51/XLHX%0ArLXPEC20gB9Af8Eqf%2BCq3wX/AG6/2aP2Y7/xh8Zfgwn7W3xF%2BBUXj74l/Cbwn4jtLa7j1zwgljpn%0AxI1PQ9FurqV4rTStRu7PU9d0C0v9Uu/BQ1Q6Vq0xbTLq4X7o%2BDH7Y37K/wC0V4m1vwd8Cfj/APCz%0A4s%2BJNA0Q%2BJ7zTPAvi7SvED3nhRfEGoeEpvF/h65sZ5bPxZ4PtfFulal4UvvFnha41nw7YeJ7G58P%0AXmpwaxE1kP5m/Af/AATY/bfk8Nf8EfPgTefsy%2BOfhToP7Nf/AAT1/bz/AGVv2jPj3a/Gj9mzxbcf%0AC34m/tbfCfSvhf4e8UeGtJ0b4jaJ4t8YeH/CfinwvP8AESztdA0LU7rQvDPjPT9FS88Q%2BKNA1qXV%0Aftb/AIJffsGfGv4BeOP2NB8cf2dfH2geNP2Pv2O/HP7NGs/HLxR%2B0v8ADbxj8ILya51L4aaHaab%2B%0AzJ8KPh3FP4w1Lwn8Sovh/N8SNe1z44%2BHPhH4m8Ci8tPDseneOtZ1C/vtEAP3v%2BIPxF%2BH3wl8GeIf%0AiP8AFXx14N%2BGXw88JWJ1TxX48%2BIPifRPBngzwzpgmitzqPiHxR4jvtN0PRbEXE8EBu9Svra386aK%0ALzN8iKfJ/gd%2B2B%2ByV%2B05feINM/Zs/ai/Z1/aE1LwlaWF/wCKtP8Agd8bfhp8Wb7wzY6pNcW%2BmXvi%0AC08BeJtfuNGtNRuLO7gsLjUY7aG8mtbiK3eR4JVX4o/4KXfAT4s/Gnx/%2BxPr%2BnfAm9/ay/Zr%2BCvx%0AX%2BJnxP8A2gf2WNE1H4P6XrfxT8X6f8IvEFh%2BzZrtyfjz8SPhn8MfEXhP4dfFG8l1vWfB2t69DNee%0AIb3wb4ujkuLDwRqnh7xHwf7EP/BPD4rfDv8AZD%2BIXhfxj481T9iH9oP9pn9pP4vftT/Fo/sTQ/BB%0A5vhefij4q12%2B8PfAfw94p%2BJvwS%2BJHgzxFoPgfwXd6FpWp%2BJrDwDYajJ4utdX1XwrrdtBcy6pq4B4%0Ab/wUw/4K067%2Bz3%2B2J4P/AGJPhV8a/wBnf9nvWx%2By98ZP2i/iZ8a/2hPDuueLPD2ja/4esbS1%2BDHw%0An0%2B103xN4Z0nwm3i3WLiXxN498a%2BKE16x0v4fWsseh6fBrsq3dv9weAf%2BCmP7OHg39nX9mX4hftd%0AftMfsr%2BAvif8bvgv4c%2BKGpQ/DP4hXmvfDTVLN7G0j8YfELwFc6rHL4ni%2BB%2Bi6xLNHcfEfxRDD4U8%0APWyldc8WSR276lP%2Bf/x0/ZV/bA0n/gotqvxh8FfAD4w/tC/CbwD/AMEkPjx%2ByP4Q%2BOPiL4r/ALLr%0A/En4jfHn4neN7b4maJJDp/ir4jfCzU/D0tvDpMvgS%2B8Xz%2BHfCGmxX2uW8EDDwi%2BtatB8I/s0f8Et%0Af2wvhLb/ALMHxG%2BM37MPx18aWEn/AASEh/4J%2BfHP4HfBb4//ALKGh/FXwb4r%2BGvxI%2BIWsHwLf%2BKf%0AGvxN0P4ca9%2Bz5%2B0V4b8Rafr95F4F%2BKWoeItA8c2Og3XjCwj059S07QAD9y/hR%2B3R4117/goV%2B2N%2B%0Az74/8UfAGL9ln4Mfstfs3/tQ/Cj4r6DBq3h/VYPCfxhHjez8S3fxJ%2BIutfErXvh14j8NWk3gS98S%0AeHvFHhvw54J02Dw1q9kl/JqT2cuq3fnv7MX/AAU0b9p3/gqF8a/2V/hN8VP2XPjp%2Byp4a/Y5%2BHv7%0ASPw9%2BIXwRudX8QfEHQPGXiH4lTfDXX/AHxM8YWvxI8ReCNU86LSJviR4XTRvAvhO6uvh948%2BHerx%0AXWtaXfW2v%2BIfxp%2BOf/BK/wDbq%2BMPgT9u/wDZ8%2BGv7Mmt/s7eF/iF/wAEn/8Agnz%2Byb8BvFzfHv4O%0AeOvh3qfjz9jjxNofjDxp8FrTxKnxKg%2BNFx4W8YaJqepfCXw18RfiL8KPClv4o0vRPEniLxxbeE01%0AXRdP1v8AUz9nPwd%2B0t4n/wCCx3iD9p/xd%2Bwx8Vf2bfgRrX/BM74d/sxafr/ibxd%2BzZq%2Bl6Z8S/h1%0A8d/FvxI1fSr3S/hX8Y/FeqQ6NBYeK4vBfg6%2B0vRdRj1m08Lxa5Hb6V4TvdMu0AP2M%2BMfxs%2BE37Pv%0Aga8%2BJXxp8feHPhv4HstS0XRX1/xNfLaW91r3iXVLbRPDfh3SbZFlv9b8R%2BI9ZvbTSdA8PaNaX%2Bta%0A1qdzBY6ZY3V1KkR/Kr9v/wD4KzfDT4Ofsx/CH4jfsz/GT4N3Xi79oj9rT4P/ALI3hjxr8RrLVb/w%0Az8GtY8bePNC0L4t%2BM/il8PJ9b8CeMtG1X4J%2BCtSufEeteD/E76BqWkave%2BFn8T6S2i6hJBde9/8A%0ABRf4GfGj4lan%2BxV8Y/gz4X1P4rXP7If7XXhn9oDxl8A9G8ReDvCuu/FzwgPhx8SPhjqUfg7WfiN4%0Ai8JfDsfELwL/AMLDHjXwdpfjnxV4T8N63c6XeaTceL/C15d2Os2v5VfGT9kr9u%2Bb4heMv2u/Bf7M%0AHxE8T6/8df8Agqr%2BxZ%2B18/7JVh8Qf2WLPxN8Dfgn%2Byp8C/DfwR8Y%2BKPFviLXf2gvCPws8RfHP4u2%0A2mS%2BIT4I8B/Frxd4Njfw74X06/8AiPpM%2Boa1qtsAf0AfAQftI3n7OHgEftAeIvgZcftOXfgEf8Jp%0A4r%2BC%2BmeMfEHwCl8c3Fvdf2br/hLR/E2q%2BF/GWu%2BD2R9L1C7sLnWfDV7q3%2BnQafqGhwXFpPafy63n%0A/Bcj/got8PP2PPit%2B2v4/k/4J0eNNJ/Z9/a78Ufst%2BMf2W/CHg341%2BAfj78c4/C/x80P4NnxZ8Ct%0AX1T9pL4nRaVresQ6/Lq0HgbX/hp4qnsNL8IeJPFC69qtlLBolp/Xlf6pfWfh691q38OazqupWujX%0AGqQeErCfw9H4h1G%2BgsXu4vDllc6nruneFI9ZvJ1XTLefUPE1h4eW%2BkSS712100SahH/EfY/8ErP2%0AhPFv7EX7UPwg8R/8EV2079vf48ftRfG34ufBv9s3xT8Sv%2BCffhyf4Bad8UPjKPHHw88Zaz%2B0J8Mf%0A2j/Gn7TsWp/B7R7mbVm8G%2BB/h34i03WtWsIvDdpc/wBlajNqlsAf10fFP9tL9l34E/8ACKRfHT40%0A%2BCPgxqPi3QdE8URaN8S9Xt/C%2BqeGfDviC%2BTSNN8Q/EGG8Yw/Dnw0/iF/%2BEVk8V%2BN7jQvCyeLFbw0%0ANabWSlm3Iaj/AMFF/wBg/R/jR/wznq37XPwB0z48Hx14T%2BGS/CK/%2BJXhq08fyePfHiWLeC/C8Xhq%0Ae%2Bj1KXU/FUmqaXZ6BHHA8Wqalqul6XZyy6jqVjaz/i5rv7JP7f3wQ8c/8FWfDGnfAjWv22/GH/BS%0Ar4Dfs4/DjwT%2B0QPHXwP%2BFnws8IXvgz9nDV/2cvidbfH6Dxz8QIfih4f8N%2BDdY8Rat8TfB/hv4Y/D%0AP4uS%2BNtO8UahpVomla1L4w1rT/Wv2E/2IPjn8Mf25v2oU%2BPP7Pfju7%2BEn/DIP7CX7PXwG/bA8TeN%0A/wBnHxVc%2BMPGX7Gng3xZ4N8Q/FS38H6N471/4m%2BAfiD451nxlo3jz4ea1f8AwwsodEu/BviCfXZ/%0ADuoXXh3S9UAP008Tftu/Bnx/4V%2BPfhP9lz4%2BfBPxh%2B0F8Nvhf8X/ABP4a0rVf7S8b%2BEU8S/CyTU/%0ADevSajpnhvxH4OuPHug%2BA/iDb6d4Y%2BKOl%2BBfHFpqvhS91PTNL1vU/Duoa5orXfo37EXxr8VftKfs%0AX/siftGeOtP8P6T42%2BP37MHwC%2BNfjHS/CdrqNj4V03xV8VPhT4T8deIdP8M2Osarr2r2fh%2Bz1fXb%0Ay30a11TXNZ1G306O2hvtV1G5SW8m/BX9k39gj9o/9m34Afs9/DLxJ%2ByB8aviP8Xf2N/2ef2vvhN4%0AY8b2/wC0x%2Bz/AKb8G/FL%2BOPDGoeC9Ai/Z38An4oadq0upftHXE3g7xf4gg/aA074UaH8Nx4e8VXm%0Ar6zqviq08KDxH%2B3H/BOH4d/EX4Pf8E%2B/2I/g58XfBWpfDr4pfBv9lH4A/B/4h%2BDNV1jwf4gutE8Y%0A/Cv4X%2BGPAPiGOLXPAXiXxf4T1XTbzU/DtzqOjX2k%2BIb0XGkXdjJexafqJvNMswDtdI/bV/ZA1/x9%0A4r%2BFmj/tPfAa/wDiJ4Hg8YXPivwfD8VPBf8AbejQfDqA3PxJee0fWEM3/Csodj/E1bVp3%2BHX2izX%0AxquhPe2iz/LHxl/4LE/sB/CT9mP44/tQWnx88FeP/DnwP8I6d4lv/BPhvWINO%2BIfi%2B%2B8X6f4ivPh%0ANoXhLwr4nXRdX1KP4zN4X1qX4Z%2BKUsm8H%2BKfD%2Blaz410rXLvwboer65afhJ4p/4Jbft0fEi7%2BMfw%0AX%2BEHw9%2BKvwF8EfF/4Xf8FRtE%2BImi/tJeO/gp8Sf2avhj4z/alPiIfC3xD%2Bxd45%2BHHxE8V/tHeALH%0A9pHxHHZ6t8fPC3jew%2BIujeDfAniHx3puseFbHxNqHheLxH9K/ED/AIJ/ftIfHr9kz9px4f2TvjV8%0ANf2mfGf7E3wH%2BAtrbftH/tZfCX4q33j3xH8HvjSfja3wi%2BGmh%2BAfib48%2BD1v8I9L1ubxZD4G%2BJvx%0Ae8S%2BAPG/neNV0K98CeGvCh1jyQD9yNa/4KCfsQeHPCXw88e65%2B1X8DNM8EfFXwtqvjnwH4uuviH4%0AeTw3rfgfw/qNpo/iXxrLrIvDp%2BjeD/Cur3sGk%2BLfFGvXGl6F4V1Xz9O8Q3%2Bm3trdQQ/YNfytftv/%0AALOX/BRL9q3VPHHxa1H9lv8AaD%2BD/wAZNc/Zu/aN%2BD3wF1H9nj41/suXGlappPiX49/EE/BX9nD/%0AAIKL/Df4jfGPxz8KfiJ8JvGHw%2Bsvhp48%2BJN58I9N8SaN4TsfEnxM8L3/AI48VapqGjaTon9O8R8b%0A23gOMzJ4f1L4kW/hFDLHFJeW3ha/8bxaMDIkcrxRX8Hh%2B615SqSNDHeRadIGaJJlKAA8I%2BG/7cP7%0AG/xgu/H9j8Lf2pfgD4%2BufhZomveKPiIvhb4r%2BCtWj8I%2BE/Ct7PpnifxjrNxa6y9vb%2BDPDmq2t3pO%0AveMBK/hrSNXsr3StQ1W31CzuraLN%2BFX7fH7EXx08faZ8Lvgt%2B1r%2Bzv8AFnx9rumalrPhvw18Ofi7%0A4H8Z3vivTNFs4NR1268IS%2BHta1Cz8Wr4e0%2B5t77xFH4budUm8P2k0dxrMdjC6uf5iv2TP2bf%2BCiX%0A7Mvxv/4JxfGOy/Yb%2BO2sXHwR8DftPfBz9pH4W%2BH/APhiD4H/ALPHwdh%2BPXxO%2BFvxH8c6T%2By54H%2BE%0AXjSw1/xL8MfCuk6D8VvHPhnVfHuq%2BL7j4w/GCfwBo2na74XfxLr954U6L/gnb%2Bw3%2B2t8Kvgz/wAE%0AKPhtf/sX/Fj4Up%2ByV8Sf2/db/am8c33xA/Zu0TVvhrL%2B0P8AD/49fCz4f%2BM7HT4vifb%2BLfiHqMcP%0Axc0TxDNqOh%2BD7vUrLSPCSWX9n3dw1jbKAfsD%2B0f/AMFOfDHhf9sP9iX9nH9mv9oP9kT4pXvxb/ag%0A1T9nz9pn4NR6t/wnfx68H6U3wt%2BJfj2Hxr4IvPCPxg8PaT4UT4f658LZvB3xR03xT4E8c3Glal8Q%0A/A0d2nhzUbjTdL8Vfph8dvjh8L/2a/g98RPjx8aPF2jeBfhf8LfDGoeK/F/ibXdS07SrGzsLJVS3%0As4LnVbyws59Z1vUZrLQvDmlfao7rXPEOpaXounrNqGoWsEn8037On7KP/BQD4B/BT/gjz%2ByBrX7D%0A/i/xXp/7B37bnjD4y/H79oPwX8bv2ZD4H8TeE9X0X9qrwToPxF8DeGvFvxd8I/EPxVceMZ/2jp/i%0AR8UbLXdA8P8AjrwrPoOoWul6B8T9Z8TW9zD/AEDft3/CHxZ%2B0H%2Bw9%2B2V8BPAMVtP46%2BN/wCyn%2B0P%0A8IfBcF5c21lZzeLPiV8IvF/gzw5Fd3l5Pa2dpbSaxrVmk9zdXNvbQRF5Z54okaRQD81vjH/wWO8N%0A6f8AA7/gl9%2B2R8E/FH7OFz%2Byd%2B2d%2B0F8Kvgv%2B0HqvxW%2BI9hY%2BJvgOnxb%2BGWrfEee2uvGnhLxVffD%0Abwh47%2BB8PhDxno/xm8L%2BNri9Gma7ZWujvcaEtrquq2/6j3/7YX7KGnfBW3/aQl/aS%2BB118AL2%2Bsd%0AK0z4zaR8UfBmu/DTWtY1TVY9B0rRNA8Y6JrGoaHr%2Buarrs0Oi6Xomi3t9qupavLHpljaT3siQH8F%0A/Fnw2/av%2BL/wd/4JJfCv4if8E4PiL4a8C/srft%2BfsleLofBmg3/7Pl3a/DL4Lfs4/sb6X8OL3xV8%0ARNOPxR1PSrE6F%2B1V8QfG9t8PrTwHpkWmar8EfhVp3iua58L32s6Jpvifz5v2JP2uvjBZaj8SrP4H%0AftHfByb4O/8ABan4yft0WfwYsvE/7PHgn4l/F/4TfGfwHqfgGw8a/C7xTqPxU8Q/CfTPir8N18Sa%0Ax4sjsvHfiLQNK1Sz1bxHolr4ktvEl6slmAf07/C/4rfDb41%2BC9L%2BIvwl8b%2BG/iF4H1mfVbPTvE/h%0AXU7fVtKmv9B1a90HXtMknt3Y2uq6Drum6jomuaVdrBqOj6xp97pmpW1tfWk8EZXyx%2BwD8Fpfgr8K%0APiHa3Hw0%2BKXwuuviT8fPin8Xr/SvjZ8UfBfxS%2BK/ifVvG97psmt/EXxzc/DZbr4e%2BDNZ%2BIOtafqH%0AikeBPCPiLxXZaTZajaanq2q6Z4o1rXvCPhkoA%2B56%2BL/jN/wUO/Yx/Z5%2BOHgv9m34z/Hjw14A%2BOPx%0AH0/StR%2BH/wAO9X0nxbNrPjWPXNZOgaTZ%2BF59N8PX2l65rV7qwW0j8PadqFxryiWC4k01LW4gnk%2B0%0AK/Ab9sOe5vv%2BDgj/AII3adLdLPZaD%2Bzn/wAFD9ds9PvbJbmDTr7WPh/4R0a%2B1LRZzdodP1fUra0s%0ArLULsWsm/SbD7ABIL9pLEA/fmvDZv2l/gJb/ALQdn%2BylN8VPCMX7Rt/8PLj4s2fwefUCPGs/w2tt%0ATGjS%2BM00zysf2H/anmWEV20oFxc219HbrKbC9%2Bz/AI1/tz/8FQPira/trfs7fsC/sm%2BC/jFr%2Br/E%0AHx58ftD%2BLvxJ%2BAN1%2ByfqPxgaX4AfAzwB8U9Y8BfAW2/a28YaL%2Bzk2sWWp/F7wnH8Tdd%2BJ2oXGu%2BF%0ALPwP498J6F8N9S12/wBI8Saf8d%2BHbX9vf42/8FA/%2BCUkPx5%2BI2l/sw/thL%2BwZ/wUV0b4xeKPCXgv%0A4L/FG9Y%2BCfjz8PfB3hm6h0HQvGvxM%2BCnhnxT4utNI%2BH/AI0%2BI2haJr3xA0rwtcT%2BNvhlod3o2tPY%0A%2BLfCoB/WVVe7tLS/tbmwv7a3vbG9t5rS8s7uGO5tbu0uY2huLa5t5leGe3nhd4poZUeOWN2R1ZWI%0AP8l3wd/4K/ft1/tg3v7FX7PXw%2B8H694V%2BJfxW%2BEP7Z/xF%2BL3xn/Zu8MfAfWdY%2BI837Lvxp8Zfs0e%0AE9U%2BBWj/ALV/jXT/AIMeGfCfinxloUPxA8fHxLf%2BNdYs2ksfCPhKKDTU1DXry3a/t4/8FStD8f8A%0Awz1n9oXx5Z%2BFfhj4Iuf2Cvhv%2B0H44/Y5X9lD47/BH4RfG7xB8Y7fSv2jdI/a68H%2BJPDfiP8AaFh8%0AGfGvwL4g8K6P4V8b/s1eLvDng34S65danaf8JNca54d8Xf8ACOAH9RPwx%2BD/AMJfgnoF54U%2BDXwu%0A%2BHXwk8L6hrd94lv/AA38MfBPhrwFoF94j1SK1h1PX7zR/CumaVp1zreow2NlFfarNbPfXcVnax3E%0A8i28QT0Wv50fhL8ZP%2BCkP7WH7Xs9v8Nvj58WPhr8CvCf7X/7d/wr%2BNGnaP8As8fBeb4G%2BBPhD%2Byb%0A8Vrv4UfAgeAPin8TPgrqXif4i/Eb42%2BK/DOt6D8bfDeifFbxNqvh%2BSbxLrPhe3%2BDll4IsIPGVnwf%0A%2B2p%2B17Y/D39nXWvEHxy0bxFrV1/wXU%2BPP/BPv4iTat8L/AGmn4p/Afwz8f8A4%2B/DDwzpdpa6HY6R%0AF4R8V%2BHNB%2BHWh6na674ejnv763sNRXXhq01xPqsQB/RLRX84X7JX7bH/AAUi/an%2BJEXxt0HwX4pt%0AP2fl/wCClHxd/Zs8WeEr2z/ZE8OfAXwZ%2ByR8MtW8W/CHT/FeneKdZ%2BLk/wC1b4m/aQn%2BL%2BkeH7jx%0AHcQ6RJ4G8Tzavd%2BDvAHwo0vT7bT/ABxq33b/AMFWfif%2B0/8ADTwR%2ByLYfso/GWy%2BCnj74y/t5fs5%0A/s/%2BJPEWsfDfwX8UNDuPhx8VLzxHY%2BMjfeHfF1ukv2vRdP00eIdJbQda0DUdRvNMHh99U0%2BDWW1f%0ASgD9UKK/ks%2BM/wC3h/wUS%2BCenf8ABQ/x/pn7T0nj7wX/AME%2B/wBvj9kD9nLwN4K8R/Cb4E2Xjb49%0AeDPjz4p%2BANv8StM%2BL3iLw18LtF0%2B51Kx0z4u2%2BmfC%2Bb4J%2BEPhZrOiFPEWr%2BMn8erFpEWkehaB/wU%0Ab/4KXfG79pv9pzX/AIPfDPxLpvwS/ZF/4Ke2f7Fnifw/r95%2BxZ4C/Zkvv2ffBvi7wB4c%2BL3xD%2BMP%0Aj340fGjwv%2B2BZ/HfxHo/jE%2BNPgdqfwa8Gz/B59Nv/Cvhy48HfFfxNqerW%2BhgH9SNFfy/fAD/AIKt%0A/G23/at/Z01P4p/F688afsbftYfE3/gonpkPxr%2BI3w5%2BDP7O37Mvg/4Z/syaB4l%2BIXwq1z9nDU9T%0A1ew/aY8ZTaJ4Z8D3Wj/FTxl8cdNl%2BHXi5PFGpat8NNbkHgC8uNY8Xsf%2BCgX/AAUg8T/sx/tc6t48%0A%2BLfxc/Zf/aV%2BE3/BMD4xftt%2BGovEHwK/Zh8Z/CDx14s%2BE13o3jJvHn7IPjzw34T%2BIfgzx9%2BzXq%2Bg%0AzxfDy/0v4xeJ9b%2BLOhweOfA3jaxuPEwudTn0wA/q08JfDX4c%2BANQ8a6t4E8AeCvBWqfEnxXP47%2BI%0AupeEvCuheG9Q8feOLrTdN0a58ZeNbzRrCyufFXiu40jR9I0qfxDrsl/q82m6VptjJeNa2NrFF2tf%0AgT8Wv2pP2qtD%2BMn7RvhLRP2jL/RNCk/4Isah%2B238PtJsPhl8HdSv/g78ctCvX8OJ4n0h9Z8H3l34%0Av8Pa9e%2BHdR1efwv49n1bR3utR1vTdLl0%2B1ttLOjfGfxB/bc/4KifCz4E/wDBLy5v/jXffHPxf/wV%0AT8T/AAE1f%2B1PgX%2Bzr%2Bzz8Pvir8A/Cmqfs26f8Wfir8Lfg7rP7QnxC0D9nDxV8R/HniXUBN8KvGHx%0A90TTtK0Hw5o/iXTtR8DeLdbt7DzAD%2Bl342fs5/s9/tLeHNL8H/tG/Aj4NfH/AMJaJrcfiXRfC3xs%0A%2BGHgj4q%2BHNI8Rw2N9pcWv6XonjvQ9e0zT9bi0zU9S06PVbS1hvksdQvrRZxb3dxHJ7Bb28FrBDa2%0AsMVtbW0Udvb29vGkMEEEKLHDDDDGqxxRRRqqRxoqoiKqqoUAV8K/8E6vGX7WXi39n7UbP9tXw/L4%0Ae%2BOvgn4t/FLwdPFrWufA3VPiDq3w0g8SSeIPgp4h%2BLuj/s4eJfFnwb8FfFfXPhF4i8F3/jPwp4I1%0AqbQ0vZY9f0q2sdK1%2Bws4fx5%2BFn/BTD9rrV/gF%2ByP/wAFGfEnxA0LVfhR%2B2L/AMFHW/ZAtf2Jh8Mv%0AC9mfh38H/Gn7S3xA/Zr%2BH0vg74i2gsviX4q/aZ8E6x4Dn8Y/EXUtZ17WvhT4%2B0uXU/Dfg74Z%2BAW0%0AmHxtqYB/TrXlXxy%2BM/gT9nb4QfEX44/E671Wy8A/C7wpqvjDxTPoOgax4q1w6XpUBlkt9G8N%2BH7P%0AUNa1vVbyQxWenabp1nNcXd5PDEoVWaRf5bv2W/24v%2BCkWjaL%2ByR8Y/jP%2B11c/Grwv8av%2BC13xS/4%0AJsal8NNb%2BCX7PvhDR9X%2BBOkp8c9H0T4gapr3w3%2BGfhLxVYfFrQfF3wfvXgn0PVtK8Hal4dudOstT%0A8KXOpQXutap%2B33/BYDVZNG/4Ju/tS3cV9FpzTeE/C2lG4mEZSSPXfiT4L0SaxXzLa7XzdUh1CTTY%0ACIkkE93GYrmyl2XkAB9I6x%2B2J%2BzP4R8R/AvwH8S/jL4B%2BDvxU/aR0vRdR%2BDvwa%2BMXijRPhp8YPF9%0Azr0VkLPw9Z/DjxXf6d4nHilNS1C38OT%2BHvsDaivist4Zihm1pfsR%2Bl6/lj/a6tPDl54F/wCDqbxV%0A8SbW20zxp4c%2BEPwu0vwPrsIvdT1fTPBvgv8A4J%2B%2BCviD%2BzwdA1C50OC40/T4P2h9Y8b%2BKLO80q0W%0APwv8Rb7xVJa%2BJGufDVv4itf6N/gD4j1vW/hB8KoPHOoSyfFSD4SfC7VPibo%2BrrFYeLdJ8U674PsL%0AnU38U%2BHzHa32g6hfaxb6xvtL3T7Ei6tL6CKBDaTRxAHRWfxc%2BF2o/FDWvglp/wAQvBt/8YPDfhHS%0A/H3iP4ZWXiLS7vxz4f8ABOuajPpOieKdd8NW9zJq2kaHrWpWt3ZaPqOo2ttb6pPZ3iWL3H2S4MUH%0Awn%2BLngT42eEj40%2BH2qXeoaVb%2BIPE/hHV7PVdG1jw14i8NeL/AAVr1/4Y8XeE/FHhjxFY6Zr3h/xB%0A4f17TL2wvtP1TT7Z5Fjg1Cya70q%2B0%2B/uvyD%2BGPwv%2BGnw%2B/4OEPj7f%2BAvA3hjwLe%2BMP8Agk78HvHX%0Ai5fBfhLSPDVj4y8b%2BNP22/2iZvGHjzxlPoelWkWuePNfl0LRRf8AiLWrqXX9chtZXuptRWw8yy98%0A/Z48UeEfhn%2B3R/wVhgv/ABtpvhv4b6bd/sifHXx2/iPUtM8P%2BDPAvxC8YfAC%2B8EeOtdutV1S5trb%0ASU1j4e/Bb4Ta14hvLie00TzgupZ/te78Q3U4B%2BkvjTxj4W%2BHXg7xZ8QfHGuaf4Y8FeBfDWu%2BMfGH%0AiXVphbaV4e8LeGNLutb8Qa5qdwQRBp%2Bk6TY3d/ezEERW1vI%2BDtxX5t2H/Bbn/gkVqN9Zafb/APBR%0Af9kqO4v7u3soJL/4w%2BFtKsY5rqZIInvdT1O8s9N060V5Fa4v9Qu7Wxs4Q9xd3EFvHJKv6k1/IH/w%0ARR%2BCH7Tn7VP7DGgfC/4u2n7O3iT/AIJ1/tAfFb/go9YftB%2BDfB%2Bs6xqfxc8UTeO/id8V/CNv4T8R%0APrGnWb/D/SLDxzJeeN/BHiz4aeKYPiNpkWmeA9VuY4LTW59T0gA/rM8J/ELwB49ufGNn4G8ceD/G%0Al38O/GF78PfiBa%2BE/E2i%2BI7nwN4/0zStG13UvA/jGDR728l8M%2BMNP0TxH4e1i98M60tlrVppWu6N%0AqE9lHaapYzT9hX8jH7Qn7VP7Z3wa%2BCv/AAWsn/Zq%2BLvgH4M/GP8AZg/4Kgfstaf8LtU0b9nv4OaN%0AY%2BPfDn7V%2Bt/s5eBpdH/aFuNT%2BHXi1/iNFeaN8XNMgu/i3bafD8dNStPhl4Xvbzxzf6bePo1z7H48%0A/wCCg37b/wCz5rX/AAUi/ZuuPjFp/wATvGnwK/a6/wCCX37N3wd/a6%2BOnwr%2BHfh7wJ8L4v8AgoP4%0AT%2BFMvxH8V/EHwz8JtD%2BFvhPW/DPwM1/xbqmu%2BHtK1Rre80zw54x%2BG%2BleN/iH471ODWL3UQD%2BgjwJ%0A%2B1F8A/iZ8bfjH%2Bzj4G%2BJWjeIPjd%2Bz9aeEb/4wfDyC01m11nwVY%2BPLGXUvCd5dzahplppmpWmr2kM%0AjpcaHf6pDayr9lv3tborCdP4%2BftGfAn9ln4d3fxb/aM%2BK/gj4MfDSx1TS9Eu/G3xA1y08P8Ah6HV%0A9bnNtpOmtf3jrGbzUJldLa3XdJJskYLsjdl/EH/gn/4b%2BKHhf/gub/wVW8OfEz4wa98fbzwj%2By1%2B%0AwdpGl/EXxb4W8E%2BEPFVlo2t2XxM8UWfhjWtN%2BGnh3wT4Du5Y9W1DxJe2epaJ4O0otpH9l2d%2B1xrF%0Arq2p6v6n/wAHCcni34j/ALCvhz9iX4X%2BILHw78X/APgox%2B0r8Av2PfAmq3t1cRQ6HpvifxvafEP4%0Al%2BKdXtrGOfUJ/B2ifDP4deKbfxm9tbTO2j6s2m2iT6zquj2N6Aep/sL/ALbX/BGfwFo/wq/Yq/Ye%0A/a0/Z3uLRrm58PfCT4PaF8dNV8fa9d3LwXWpjw/4ev8Ax94o8SeJdQZLa0mj0fRX1m4WGCC30nRb%0AVEW0sq/Sbx98Y/hn8OPG/wAJfBHirUpR8QvjNr2teEfhf4c0nw9rXiPxBrTaRpS%2BJfGOoyLoem6g%0AfDngzwzo%2Bn2mpeLvGHiKfR/CWl3D%2BHdJvdV/4SHX/Cuk6t%2BYn7M3hL4kL8Yvhh8Jv2/JPhxH8fvh%0A98WPiF8fP2FfhJ%2BzbLqdh8H/AINfs%2BfAf9mzwP8Asn%2BIPG9peSPB8RJ9F8WX37TPjXQ9T0H4t65r%0AOnz6l4y8L2WgeFdJHhue6k9V0rVND1T/AIK5/H3XPGeqW1nbfs5/8E2/2dLnwHDrs0NuNC039oX9%0AoT9qrVPj14u8Nwyu0kmkatafsx/AnSPFusQpEttd%2BENI0%2B4JUxGQA%2B/L/wCL3wt0v4naH8FdU%2BIX%0Ag/TPi54n8NXvjLwz8N9R1/TbHxl4l8LaZcXNrqmueHtAuriLUdc0/SprS4GqzaXBd/2ZHGJr8W8M%0Akcj8N%2Bz%2Bn7PXjDQNX/aI/Z%2B8P%2BE1sP2k7nSviH4o%2BI%2BieDp/CviL4rXumaPa%2BEtD8QeMptY0fRvF%0Amq3mkaBotn4f0dfEtutxpOj2Nvp9jBbWSpGfxG/4JIfsz/F7xt8dPjd/wUR%2BMnw4/Zc%2BNPwg/bN%2B%0ALPxH/a1/Y3/aO8ea74i8Wft0/BX4F/EtraH9nL4KWug6z8NvEPgf4Y/DXRPhDrfiXUINB8C/G6z8%0AQ%2BGZ/HuoaD4m8OQ6pe%2BIdA8P/NvwQ%2BNP/BQPTvhb/wAEKP2fv2QPjL8Ffhh4L/ap%2BBXjuHx3N8Qv%0Ahzq/xT8X2Phr4Y%2BDNd8a/FH4qjTdZu7HUIP%2BEB0/UPhf4d%2BGcNx4uurfxL8VPiba2PxWtm8NWkNx%0AqAB/UXr3xR8A%2BF/H3w/%2BF/iDxLZ6T46%2BKln4zvvh5oN5FeRP4rT4fWek6n4utNL1D7MdKfV9J0rW%0ArXWBoU19DrWo6La65rGlWF9pnhrxHd6Ve%2BIPxA8E/CjwL4x%2BJ3xK8U6J4I%2BHvw%2B8Na14x8beMfEl%0A/DpmgeGPC/h3T7jVdc1zWNQuGWG00/TdOtbi6uZnOFiibAZiFP8ANH8Vf2ifip8cP%2BCY/wACP2xP%0AiD4lkf4lfBT/AIK9fDX/AIZU%2BKl7Y6f4G8Y/E34T6f8A8FFLv9kbQPEHinwt4LttD0Gy1T4m/s0e%0ANviN4V8feHNHstM8L%2BLvCN1q2u2limna3ZoP3U/b8XWv%2BGHf2vZ/D2p6VpWr6f8As3fGbWLafX/B%0AXg74j%2BHL2PRfh/r2rXmheJvAXxB0bxB4L8X%2BGPEtlZXPh3xJ4f8AEejahpup6Hql/aywhpEkQAx/%0Aj7/wUT/Yi/Zc0j4R6/8AHr9pP4a/DrRfjxZ2%2BqfCLVNR1G81Sy8c6FdP4ajh8T6VceHrHV44vCBl%0A8ZeE4ZPF%2BoGz8MQz%2BJdDhm1aOXU7NJftCv5i/gX8M/iP8ff%2BCpn7FHxO8V/FbTf%2BEIuv%2BCHXwl8Z%0A%2BMfg9P8ACH4Yah4H1yz8ZfGH4Qa1r/g3RdIk0CHSPCWh6p478EeGfHSajpOlDXPDt54e0nw/4Wu9%0AJ8Ow29lYf06UAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUU%0AAFFFFABRRRQAUUUUAFfIHxq/YB/Yw/aM%2BI2n/F744/s4fDH4m/E/SdDTwzpXjvxRohvPE2meH0ad%0AhpGn6qk8NzZ2Lm6n86C3eNZ/NYT%2BYMAfX9FAHxvD/wAE9v2J7T4Waf8ABXTP2a/hfofw20f4i3/x%0Af0TQfDuiv4butB%2BLGqrNFqnxM0DxJoVzp3inQvH2o21xcWN94u0jW7PX7vTbifTZ7%2BSwmltnb4y/%0A4J6fsU%2BPdK8AaT4i/Zw%2BHAX4U6br%2BkfDPV/D%2Bn3ng3xb4C07xhrE3iDxxb%2BEvG/g6%2B0HxloH/Cwt%0AZur7UfiPNpuu28/xEuNU1dvG0uvLrGprd/ZVFAHyd4v/AGFP2PvHHgX4VfDbW/2efhpaeDPgXY6l%0ApPwY0nwnoY8Ay/CrQ9e02PRfE/hz4fav4Em8Oa34S8MeMtEiTRfHXhnRNRstB8c6Mv8AZPi7Tta0%0A8m2MNh%2BwR%2BxVpGo%2BFNV0f9ln4GaLe%2BCYvANv4abRfhz4b0e3sbf4UW2i2vwqtrix02wtbHVbb4Xp%0A4d0WX4b2%2BsW1/D4DvbCPUfCkekag81zJ9b0UAfzn/Cf/AIIpeMvhz8VPhr4qLfswQeNfBP7ZV7%2B0%0A/wCM/wBvjw34X8caX%2B3d8bfAY%2BLg%2BMQ%2BCPxEuSg0l5/H05uPhZ8ZfHzfFPV/D3jn4V319p8Xwh0v%0AUr7UptR/VbxX/wAE4P2GvHPxCuPil4u/Zo%2BG2veNbj4hW3xbXUL%2By1B9MsPivbzTXM3xQ0Xwsmop%0A4U0H4i6xd3E174n8b6Jomn%2BJ/Ft8wvvE%2Bqavdoky/bVFAHzLp37Gn7LukfF65%2BO2kfBTwZpPxPvf%0AEup%2BOb/XNLtrzT9M1D4hazawWOqfE3UfB1neQ%2BCtQ%2BKd7YW0djP8T7zw7P4/ayae0HiMW91dRzbn%0Ax7/ZX/Zx/aksvB2m/tFfBf4e/GbT/h/4lh8Y%2BCrP4geHrPxDbeGvE9v5Xk61pcN4jpBeoYIDvwyM%0A0ELOjNFGV9%2BooA%2BDte/4Jff8E9PFOn/EfSvEv7IHwN17T/i98QND%2BKvxMttW8GWV%2BnjT4keGh4iX%0AQfG2uNcmSS58R6VH4t8RxWeph0uIodWuYN5hESR%2Blaz%2BxD%2ByR4h%2BLNh8c9Z/Z%2B%2BGl/8AFaxu/DOp%0Az%2BL5dAiW58Q654Hgt7fwB4m8b6dE8ei/EDxd8OYrWD/hW3i/xzpviHxP8O5I/O8FatoUzPI31PRQ%0AB%2BeGrf8ABJn/AIJxa9q2n6vrP7Ifwl1OXRfFl7418Oabe6fqs/hnwjrWrnWH8RQeCvB76qfCfgrw%0A14wm1/Vrvxz4J8KaLo/gzx3fXf2/xhoGt3tvazw%2BsfC39hD9kf4NWviDTvh/8D/C1ho3ib4bTfBn%0AUfDmv3fiDx14Vtfg3cqVvPg/4d8KePNZ8S%2BHPCPwo1BNkepfDfwppei%2BC9SjgtY7/Q7hLO1WH63o%0AoA%2BDNH/4Jd/8E8vDz63Non7H3wM0y68S/Dqf4ReI9QtPBdlFqmvfDC4gsbQ%2BAtZ1UH%2B0dT8LQWGm%0Aafpljo97dTWmm6TZ2%2BlafHa6dEtqPWZv2M/2Wbj4CeGv2XpPgZ4BHwC8FS2N14I%2BGUWlNb6D4Gv9%0AK1a613RdW8ESwTR6p4O1vQNXvbq/8Oa34a1DS9X8Ozyk6Je2CKir9NUUAeefC74S/Db4K%2BE4fA3w%0Aq8GaH4G8LRalq2tyaVodr5A1DXtfv5tU8QeItZvJWm1DXfEmv6ncT6lrviHWbu/1rWdQmlvdTvrq%0A5keU/Ptv/wAE%2B/2LbX4pf8Lkh/Zx%2BG3/AAnCeMNf%2BI9t52l3Fz4N034oeLFkXxX8WdG%2BGFxeS/DH%0ARPi94o%2B0XzeJPivpHg%2By%2BImvSaprMuq%2BJruTWtWa8%2BxaKAPgzS/%2BCXX/AATs0XTPDmj6V%2Bxj%2Bz1Y%0A6Z4P%2BKF38avCdpD8ONBEfhn4s303hue7%2BIOgs1s0mk%2BKZpPCHhwNqlg8E4j01YVKw3F5HcdL/wAF%0ACvgT8RP2lv2OfjZ8FvhO3hR/iH4u0rwxc%2BGNO8da74h8MeDfEN54S8d%2BFvG0/hDxR4i8J2l94j0T%0AQ/Gdh4cu/CepappFnc3Vla6zJcCF0RxX2dRQB8WeLf2H/wBnD9oPX/C/xp/aP/Zz%2BHV98b9S8EfD%0AnSfiXp%2Bm%2BKPE3ifwZrU3grWLPx7ongPxo8dl4A0D9oHwR8N/iKs%2BreArn4sfDVraK7tbbxDZeEvD%0AN7cvp9r2/gn9mPwp4N/au%2BPP7WFqNKt/F/xx%2BGPwQ%2BFWqWWjaINIebSvgvefEjUrXxJ4s1SG9dvG%0APivWH%2BIsXh22vdSs408M%2BEPA/hnSdD8hr/xHLqX05RQB8nWH7Cv7Iem/tDXH7WVv8Afh/J%2B0pcz3%0A08nxqv8AT7nVfH6jULWSzuLO31zVLu8ntNKjglYWOjWoh0nTJRHcabZWlxDFKnL/AAi/Y18Mad4I%0A/aY0r9ou28IfHfxH%2B2T8SvGHjz9oTTta8Pzaj8N9a8OajoekfDf4efCXSfCviq710t4J%2BHHwW8H%2B%0ABfA8hvJI4/FnijTfFPxFGjeGbnxjJ4c0f7ZooAxvEfh/R/Fvh7XfCviKxj1Pw/4m0bVPD%2Bu6bLJN%0AFFqGj6zYz6bqdjJLbSQ3Ecd3ZXM9u8kE0UyLIWikjcKw%2Bf8A9nX9jb9l39kiHxLa/s1fBTwV8GLL%0Axjdx3/iXTvA1nc6Vpeq30bySC8m0v7XLp8V3JNNPPcXFtbQzXVxPPcXLyzzyyP8ATNFAHwL4i/4J%0Abf8ABPvxlqnxj1fxz%2Byv8M/iBP8AtAfEzQ/jJ8YbL4h2%2Bs%2BP9A8cfE/w2muw6H4z1Dwv4x1fW/Dd%0Alq%2BnWniXWtNgbR9J063Oj3g0aWCTSrWzs7frtN/4J2fsKaVL8ZpbT9kr4Cv/AMNE6Jpfhv45wX/w%0A48OatZfFXQNDuLe70TR/G1hqtleWOuafo93aWV3pVtdW7R6fd2NjdWiwz2VrJD9mUUAfKPwY/YZ/%0AZF/Z3%2BJviv40fBP4AfDv4c/Frx34f/4RXxp8R/D%2BlSR%2BMfFfh0XWjXqaT4h166ubnUNXtIrnw9ok%0AkC308zW40y0jgeOKPYc/9pT9gn9kT9sDU9F1r9pP4KaD8VdY8M2dlY%2BF9U1nWPFunah4Ujsb3Ur%2B%0AO58KXnh7xBo9x4Y1S5l1fUbbVNZ0GTT9X1nS7j%2BxdXvr7R4YLGL6%2BooA%2BP8A9nf9gj9kj9lLxz45%0A%2BJ3wH%2BD1j4N%2BI/xK0PQvDHjjx5qHirx5468X%2BIPDvhqe4utE0O58RfEPxT4r1a30qxuLlphY2N3a%0A21w8NkbqOf8As%2Bw%2BzW/Gv7OOtX37YvwV/az8B%2BKNN0C%2B0D4Y/EL4A/HbwxqlnqFzF8Sfg74hcePv%0Ah9No01ndxWth45%2BGHxk0XTrrQr3VbO7sj4E%2BInxa06J7XU9S04y/WtFAH5F/sbf8E1/Ff7GH7Y3x%0Aj%2BI3wl%2BIvhT4f/sQ%2BMPh/r%2BhfD39jb4eyfHGPwh4Z8feJPFPw/8AFf8AwsGTwz8RPjR8QvhT4E1L%0AwzNovxL0ewsPgP4D%2BFXhrxFpPxMUax4XtLnwpYSXvcfsy/sEa18BP2Iv2dfhbd6l8PNT/bO/Z1/Y%0At8Tfs5%2BAP2hLTS9QuNJ8DeNvHnhTw8/i/UPBst/YQ6rD4RvPiB4S8JX91fjQNK1zXdF8Lac9zo%2Bl%0A/aZ9Ci/TyigD8mfDP/BNJdN%2BFf8AwTR/Zm1bXPDEn7Nn7CFj8PPiZ408P6e3iCPxR8af2lvgjomg%0AQfBzW72RVsrK2%2BH%2BmfES98cfHPxnHqV5fal4s%2BImm/D%2ByuNHi0SDxC13%2Bk3xU%2BFXw5%2BOHw88WfCb%0A4ueDtD%2BIPw18d6VJoXjLwV4ms11Dw/4l0aaWKabS9XsXIjvLGeSCIz20mYplXy5VeNmU%2BgUUAfHH%0Agv8A4J7/ALEvw68f/Dn4qeBP2ZfhN4S%2BI3wi8KweBvhp4y0Lw3Dp%2BveDPB1tLfTw%2BGtEvYJFe30Z%0AZ9RvZzYN5kDzzedIjSxxOn2PRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQA%0AUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUVFJG7vbstxLCsMrSSRxrAUukME0It7gzQyyLEskqXS%0AtavbT%2BfbQq0zWzXFvOAS0UUUAFFFFABRRRQAUUUUAFFFFABRRWfp2raVq6XUmk6np%2BqR2OoXuk3z%0A6de216lnqumzta6jpl01tJKtvqGn3KPb3tlMUubWdWinjjkUqADQorG1DxH4e0nU9B0XVdd0bTNZ%0A8VXd7YeF9J1DVLGy1PxJfabpl3reo2Wg2FzPHdaxd6fo1hfave2%2BnxXE1rplld386R2ltNKmzQAU%0AUUUAFFFc74v8V%2BH/AAH4T8UeOfFmorpHhXwZ4d1vxX4m1Z7e7u10zw/4d0251jWdRa00%2B3u7%2B6Wy%0A06zubk29la3N3OIjFbW80zJGwB0VFfFH7Jv7dfwu/a61nxz4V8LeDfiZ8MvGvgXwf8J/ifc%2BC/ix%0AZ%2BArXXfEHwe%2BPGl%2BINa%2BDHxc8NzfDr4g/EjQrnwf8QtN8La9NZWGp61pPjnw1c6dLpPj3wd4S1lo%0AtOk%2B16ACiiigAooooAKKKytK17Q9d/tL%2BxNZ0rWf7G1W80HV/wCytRs9Q/srXNP8v7fo2pfZJpvs%0AOq2PnRfbNOuvKvLbzY/OhTeuQDVorKvde0PTdR0bR9R1nStP1bxHPeWvh7S73UbO11HXrnTrCfVd%0AQt9Gsp5o7nVJ7DTLW51G8hsYp5LWwt57ydUt4pJF1aACiiigD4h/aJ/bb0H4JfHv9nv9ljwX8Kfi%0AL8ff2hv2hZdV8Sad8P8A4cv4Y02z%2BGnwP8GaxoemfE39oD4s%2BL/GeuaB4d8IfD7wcdestP0m1%2B1X%0A3in4heMbrTvBHgnQ9U1q9b7L9vV%2BPd7Y3Mn/AAX78M6kmnafLZ2n/BHvxzYz6tJFqJ1WyudQ/bS%2B%0AHlxaadZzRONIj0/VItMvbnU4r6NtRmudH0l9LdLaDWFk/YSgD5h8T/tq/si%2BC/j1oH7Lni39pP4L%0A%2BHP2iPFDaVDofwc1j4heG7Dx7e33iCFLrwzox0K4v0ubXxF4ts2e%2B8HeG70W%2Bv8Ai/T7a%2Bv/AA1p%0A2q2en309v9PV/JV/wWo%2BFf7Jer2v7Y37Gfw8%2BLHhb4bfGf8Abv1D4Ffte/t7/F74qfFa/wBX8Gfs%0Aefs//sl618M5z8Y9D8CzT6jrT/FPx94Z8GWPg/4Z/CLwkmnar8RbeHU2l1nwv4c0fwzFd9L/AMFC%0A/jh%2B17c/tj/Dj4R/s5f8FUNJ%2BB/hP4kWn7Z8fxGb4W6f8Cvjj8OP2Ov2bv2Of2a/AvjfWvFnxm0b%0Axb4O1f40eJv2mPEnxW8ZaD4k1mbTfiD4butA8CaxZ%2BDdJ0C%2BttL8S/EfXAD%2BrCiuI%2BGcOt23w3%2BH%0A1v4m8ar8SvEdv4I8KQ%2BIPiKvh6y8JL4%2B1uLQbCPVfGq%2BFNNZ9O8ML4qv1uNdHh6wd7LRRfjTbVmg%0Atoye3oAKKKKACiiigAr46/aQ/wCCgH7Hf7I3xA%2BDfwq/aI%2BOnhr4c/Ef4/eItN8MfCnwZcaZ4p8R%0A6/4h1DV9d07wtpl9qFj4P0DxDJ4Q8LXvifVtM8NweNvGv/CO%2BDX1%2B%2Bt9G/t4alILavsWv5Gf%2BCoO%0Ar/GD4Q/8FufDH7QOk/FPWvE3hL4Gf8Egv2s/2nNH%2BAkvgf4T63pXivS/gZfr4n8VfAqS71Xwdrfi%0Aa38NfGvxH4U8MapqXjGSG68eaDqVrc2HgDxJptnCLDTQD%2Buaiv5C/CP7TH7Z/gr4d/8ABLLxvp/7%0Af/xW%2BKHjj/grL%2Byb%2B0J8QvjL4bvfDvwC1%2BX4NfEmx/ZTl/aT8M/GX9nTw1eeB7nT/h74Z/Z18RwQ%0AfBrxx4DutM8RfD3xLoeoR6t4i0LTvir9m1PUP1S/4IW67%2B198V/2HfhD%2B09%2B1f8AtiX37UV5%2B0h8%0AIPhb408MeF7z4WfDDwTH8GtT0xvGWmeKLG08Y/D3SdC1Hx5J4r0l/A0XiKDxjp0%2Bp%2BHvG/hLxXqF%0Anq91b%2BLZdN0YA/Rnx3%2B11%2Byz8L/ij4b%2BCXxH/aK%2BC3gT4u%2BLm0yPw58OPFnxJ8J6B4v1ObXr610v%0Aw3ax6Jqeq217DeeK9UvINM8IWV1FBd%2BLtR8%2Bx8Nw6pdWt1FD9E1/Hr%2B1Je%2BEY/jF/wAHhcvxQuVn%0AeD9hD9kXSvAVn4nsprm3tNGuf2Dvik/gybwcs1i8x0%2Bf496uz3N1ps02maf4zt7K71E2F3bm4b%2Bi%0Af9mST9ocf8E3/wBm6XSW8I6r%2B1N/wxj8EWEvxg1DxhD4Lv8A41n4MeFf7Qn%2BImpaVpDeOv7Gm8Xm%0A6m8QXFrosXiSWI3B%2BxW1%2B7JEAfXHjXx34H%2BGvhy/8Y/Ebxl4U8AeEdK8j%2B1PFXjXxFpHhXw5pv2q%0AeO1tvt%2Bua7eWGmWf2i5lit4PtF1H5s8scMe6R1U9XX8q/wALdZ%2BFfgT48eA9V/4L9/Cj4jP%2B1b4z%0A8baja/Bv44ftAWvh3xx/wSs8K%2BJbx5JtA8A/svWPh74h%2BOvhB8F9fsbPfH4Y8a/tNeA/Afx48UPI%0AIl8c3%2BrXMuh2v9Ff7TPxGk%2BGvwB%2BIfjvSvi/8EvgZe2ui2FloPxn/aD1W0sPgt4B1fxXrGl%2BGNA8%0AUeMJLjxF4TtNbtbfU9bsf7D8Lt4u8Lp438QzaN4STxNoR1xdWtAD32iv50v%2BCXXxL/bS%2BL3x71/x%0Ab8Sv%2BClt98aP2ftC/aD/AG9Pgt8OvgV44/Z%2B/Z607xr8fvAH7OPjjwn4Dtfj3b/FD4Q%2BB/AupeGZ%0A/CfxO8V6tpeqaXbaTpHgy%2B8HXHw00zTPC9leX9xrGq/bH/BU/wDbc/aR/Yl8CfDHxV8D/wBnOP4k%0AeDfGPifWNC%2BMf7RXiHTvH/jz4afsj%2BHbDSV1ew%2BJXxP%2BDHwU0fW/jX8QfB17Baa5Fqep%2BHJPB3hP%0AwlJptmfF/jzQzrukpcAH6q18Vf8ABQ/9rrUv2D/2PPjV%2B1pYfCqT4zW/wU8OL4s1rwLF43tPh9Lf%0AaHHeW9pe3MfiS88P%2BKVgktWuoCltDoOoXFy0gVI0RZJU8P8A%2BCeXwD0RrOT9sHV/2/viZ/wUT8c/%0AGLw7c2ui/GJPG9j4e/Z10LwXq93pGqXnhj4Lfs8fC3Wn%2BEPgrT21DRNJlv77XovGnxJsr2wmsLrx%0Afa/adYsbvE/4LN61ra/sP/FD4evrh8CfDP40eGfGPwh%2BNHxOb4s/svfCCHwr8OPiB4P1rwrreit4%0Aj/ay8ReF/hwk3jq11i48O291Z6xYeItKLvfaTf6fdrDeQgH6q2F2L%2Bxsr4QyW4vbS3uxBLLaTywC%0A4hSYQyTafc3thNJGH2PLZXl3aSMpe2uZ4Skr26%2BDP2PfipqnhD9lXwJe/tBQw/Cfw14E0HwL4I8N%0A/FH4o/FX9nLUdH%2BJuhz6fpGi%2BE/Fdt4h%2BCGvyfCvS4tdu7rT/D%2Bi6ZYSae2p3n2UadaXJvYJLj7z%0AoAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAr8Lf%2BC6HxZ%2BN%2Bkf%0ADb9lf9m79n%2B2%2BOHiTW/2nf2j9O0r4/8Aw7/Zc0iHUv2l/Gv7EXw58Laz4h/acsvhLqmp33hfw/4B%0AvL3%2B1Phz4N1D4jar8RvAlzoNx440fR9HvNUv/EwsK/dKvyQ/4Kc/se/tGfFXUfhd%2B1Z%2Bw38YfiF8%0AGf2y/gL4d8XfDnwvJ4N0D4Q%2BMPDvxU%2BE/wAVPFvw28R%2BNfh18RvCfxs13wl4P1Pw7aa98NPC3iiC%0Ae38Z%2BGdXtksNVj0%2BbUNUuNJjtADQ/aG%2BJ6/8E8/%2BCQXx0%2BNHw90P48%2BG9d%2BD/wCy54%2B8feAvCv7R%0AfxP1j4%2B/GnwL8R/E%2Bg3134J8M/Evx54v%2BLHxXbXI/AfjrxHo2l63ZaP8UvGPh7QPDejXGkeCb3V9%0AK03SLS55b9pz4seK/wDgkT/wT50V/hL4J0j4heEv2Xf2cPHPj34h/FD4x%2BJ/ESaZ4o1r4aaf4YiX%0AQtc1bw5p2teL/Ef7Qv7W3xg8eJc6Z4i1TT7Pwna30/xQ%2BI3jjxAl7o2i%2BEvGv2L49%2BBXij9r/wDY%0AM1/9nj9rTSfBvhv4iftC/suN8N/j9Y%2BAbV/EPgrwR8UvH/wzj0nx3qHw2TxDeX91eab4E8eX9/q/%0Aw71DUdTn1SB9F0LUjqo1GFb9fiH4k/8ABNTUP%2BCj/wAB/gJa/wDBQTxt%2B1D8KfF/hf4J678Kfjh%2B%0Azx8Hv2kbO0%2BCHxL%2BJOla7o03hv47%2BIbbwrpt/pnjjV/DfxC%2BH/hv9oD4GXmtJpM2nzS%2BE9I%2BL/w/%0AuRYeLPhswBwP7Uer/tS/tIf8Esf20Pit%2B1b%2Bzz4f/ZM%2BLXwK%2BH/iz9rL9kSf4efGvVPF/jLwd4y%2B%0AB3wjg%2BPHwi8aa54hk8E%2BA/EHgHxl4X8f6df/AA5%2BImjHS9KuvEHhuPxjoU9hpmg%2BIXW%2B/ZzRvGur%0AeIvhPpXxF0HwrJqmua78PLHxro3gmLWrK1l1HVtT8NRa7p3hWPxDqUNlp1tJeXk8OkJrV/BaWULS%0AC%2BuobeBZET81f2k/2F/HEn7Gni79iT9muTVdcvf2wvFd74Z/bC/ak%2BMvi7SL/wCIGqfD/wAe6Jp2%0AgftF/F7x9faVY6TqXj74z/Ej4T%2BHofgp8LtB8GeDtI8A%2BA473wVo2n6X8O/g98NdC8NWX6lap4O8%0AI654R1H4f614V8N6v4C1fw3d%2BDtW8Eapoemah4R1TwjqGmSaJf8AhXUfDV3azaNe%2BG73RppdIu9D%0AubKXTLnTJZLCa1e1doiAfnL4R/bA/by8QafLe6h/wS/1dC8unS6c3hH9tb9l/wAYafeaVPey22oX%0A8uoT6v4bMEtslte/2dbwWt7bane2ktlPqGlhJrmH7L%2BIvh34kfGb9mrx34S0LWtV/Zz%2BL3xW%2BBvi%0Afw7o3iHZoPjfXPgR8SPHXgG%2B03Tta2aXqMvhjxdqvww8T6pDfbNO1eTQdfvNDxaai%2Bn3cdyfnL/h%0A07/wSy/6Rp/sAf8AiG/7Ov8A87mvrPUvhR4XX4N3/wADfAQl%2BDHgtfhndfCjwYPg/p/hzwfP8KPC%0A48LSeEPDo%2BF2lnQdR8I%2BFpfAuk/Yx4J08%2BGL/wAOaI%2Bl6ZbnQbrTLX%2Bz3AP52f2Tvgpb/s6/tyeP%0A/FX7Dfhnwz42%2BHX7LX7AF5%2Bz1%2B1NafCLS/h58IfCn/BRX/goT4NOg%2BM7a28OeHdL8UWnw5t/j34H%0AtrbUJ/jX8X9VPiPVvAWt/Fj/AIVH4k1fUr9PEcvh/wC8/hL%2B2v8At3/GH9p34p/Avwf%2Byd%2Bz54i%2B%0AFX7P/jyP4CfHT9prQP2oNUuNC8K/HO7/AGavhD8fbpvDfwp1D4Taf4h8XeFfB3iD4sWPwl8ReHrb%0AxdZ%2BLZdYsp/EOq3Hgr7Pc%2BG089%2BEX/BA79k34MeAPE3w28NfHr9vPV/DmqfAXxt%2Bz34Jj8V/tY%2BM%0ANYb4E%2BGPiRrngzxP498U/s%2B2MWm2OhfCPx14w1/wLol/4l1Xwxo8Gj65az614d1Tw/d%2BFNc1PQrn%0A6i/ZR/4Jd/s9/se/EnU/id8OPGXx88W6tqU/j3xCNI%2BLnxZ1D4iaJa/E/wCMv/CG3Xx9%2BNW/U9Oh%0A8Qav8VfjvrngXRPEfj/XPEuv67pOm6nNrVp8NtB8AeHtd1XQ7oA85/4Jo/skftmfs7%2BMP2mPiX%2B2%0An8Sf2e/i18Tv2g9R%2BHuu6x4/%2BEWj/EZfFuvaz4Sfx3aNY6/qvjuWzsfDvw18JeHNe8N%2BHPhN8KPB%0A2hQeHfBq23i/Xftk%2Bt%2BMtcuL/wDWWvFv2kfA/iH4mfs8/Hb4deENR8U6R4t8d/B34leEPCuq%2BB/H%0Ad38L/Gem%2BJfEXg3WdJ0G/wDCnxJsbPUbvwD4itNWu7SfR/GMOnak3hy/jg1ZtOv0tGs5k/Zr8BfE%0AH4V/s6fAL4YfFv4gXfxZ%2BK3w4%2BCvwr8BfE34qX91qd7ffEv4g%2BD/AALoXh7xn8QL281qa41m7u/G%0AXiPTtS8R3F1q082pzzak8t/NJdPK7AHtVFFFABX5Vftfft6/tE/AH9rL9mP9mj4R/shaV8XdP/aS%0A8VH4feH/AIh%2BO/jgvwg0W78a/wDCtvi78V9fk8PW%2BkfDP4satfeDvg94J%2BEUmtfGPX9R0fTtSt4v%0AiB4QsvhxoHxA1ex8S6bpn6q18LftOfsyXniP4kWH7aXw7Pjjx3%2B01%2BzZ%2BzP%2B0j4J/Ze%2BDVz4t8Ka%0AJ8INU%2BJvxV8PaddW%2BuajZ%2BIdHgWx8d%2BIL7wloHgC38Var4x07wppXhXWNVTWdJlBTU7IA%2BO/GH/B%0ASr4hax/wSO/bb/bW0vwN4b8L/Fv9nHTf21vhvZWfhfxVqfiPwBq3xB/Zr%2BIHxD%2BD2l/EbwZ4i1rw%0At4e1vUPB2u6v4WsvG%2Bnabq/hvTtTt4pJ/DN3cPLbHWpcv4lfHDx7/wAEqv8Agl98FfFfwc%2BDOg/F%0AvwH8BP2TNW%2BLHxm8e/ED4hP4FsL7UvCPgbw54iv4rqaKz8Y%2BLvE/xw/ak%2BMHja81KDVZ7K%2B0a11C%0A4%2BInjLxj4hvNci8PeG/GvuP7OH/BPDR9G/4JCeBf%2BCcnxpR7a78Y/sl6r8KfjreWF3b6xc2vxU%2BM%0AXhrVtZ%2BNPijSdRhuZbTUNRt/iv4z8WeJtI1KC7aCe/S1vIZhGUYfN/xG/wCCRNr/AMFC/wBlb4F/%0ACr9vv4hftM/DDxl8Jvghrf7M3xD8J/AH466Ho3wn%2BNNh4Z8Q%2BGYtB%2BN1/wCF77SfiTp0954s1f4W%0A%2BAPjP4R0/wAQ/ZvEnhm9Nj4K%2BI1j4msdGu9PvQDmP26vi3%2B1J8QP%2BCQX7RH7Wfxl%2BDfgj9l34/fs%0A2ah4c/a8/Zd/4QP4tXfxrisbT4VW3gP4qeC/EGp%2BJtO8P/DC%2B0nVPFuk6x8RPgF8T/DFvCLLXfBN%0A94sdprnw146bw9b/ALa33xKk/wCFLz/F7QPDN3r0j/DU/ETR/CM2u%2BG/DV3qfneGR4jsdDufEvif%0AVNK8J6C84eOzu9b1vVrPRNMBlvry7S0hd6/OP9qL9gfX/Hf7HDfsDfDvVfiR418NftO/EPSrb9rf%0A9p74neP9Cn%2BJZ%2BHEviDQ/Fnxp8Z%2BJLzR18KXXib4n/Fnwt4Wh%2BDPw/0DwH4EsPhv4Mg1zTW1Dw/4%0Ac%2BG3gu08Kan%2Bnni/4dfD74g%2BBta%2BGHj3wL4N8b/DXxJocvhjxF8PPF/hjRPEvgbX/DU8C2s3h7Wv%0ACWs2N7oGq6HNbKttLpN9p89hJAqwvbmMBaAPhKw/bH/ar1lVbw//AME6fiB4iY%2BS5i0P9qr9jm/k%0AW1kvVspr5tnxgWP7JZyGV7l1dpCLeaG2iubzyrWX788HavqniDwj4V17W9LsdD1rW/Deh6vq%2BiaX%0Artv4p0zR9U1LTLW9v9L07xNaWtjaeI7HT7uaW0tNdtbKzt9Xt4Y9QhtbeO4WFPiP/h07/wAEsv8A%0ApGn%2BwB/4hv8As6//ADua%2B7dG0bSPDukaV4f8P6VpuhaDoWm2OjaJomjWNrpmkaNpGmWsVlpulaVp%0AtlFBZafpun2UENpY2NpDDa2lrDFb28UcUaIAD8N/%2BCePxcvP%2BCgX7fn7UX/BQzwdosXgr9nf4W/C%0AmH/gnd8JtD8S3OlQ/GLxz4x%2BGvxk8TfEX40fEH4k%2BAre5k8TfBfTrDxFeaF4X8BfD74h2%2BneOdW0%0AVLzxrqnh7wzaazp1vqP7tV%2BJHw20vTPh7/wcIftLeG/A%2BnWPhXQvjf8A8Eq/2ePjj8W9L0S0t7C0%0A%2BIHxf8I/tS/HH4UeHfiL4ljhjX7f4r0z4c29v4Pj1Vis82i21rbXJnNtA8f7b0AfBPxT/wCCWf8A%0AwTi%2BOHxS1r42fF79iP8AZp%2BI/wAWPEt5pmpeI/Hvi74TeEtY8ReIdS0hbGPT9R128utOf%2B19Qig0%0Ayws5L7UEubq60%2B2j067lnsd1u3ay/wDBPX9gm58eax8VL39in9lDU/idr/xCuPizrPxG1b9nv4T6%0Av471D4n3V5JqM/xAn8W6n4Tu9f8A%2BEwk1Ge41E%2BIE1BNTXULq7vluVu7u5ml/Gj/AIKZftnf8FTv%0A2OrP45ftdWg8AfD/AOC3wR/aZ%2BAfwW/Zx/ZT1Lw/8MfHQ/4KD/Df4ua74K8N%2BJtbtvHWjeKNa%2BMX%0Aw7%2BOOma/q3iqfwd4ctLHwHoGmeDPDl7faz4N%2BIN5E2qxfb/7V/8AwWw/Y9/Y9%2BJeu/Cz4h%2BGv2j/%0AABrrmiy%2BJNFttW%2BC/wADPEfxS8LeJfiD4G8J6H8RPiN8K/DOt%2BH7kx6j47%2BGPwx1%2B0%2BI/wASY5ob%0APwx4I8L22oxeIfE9j4isZdAoA/Xmiuf8J%2BJtM8aeFfDPjHRU1KPRvFnh/RvE2kprOkan4f1dNM17%0ATrbVLBNV0HW7Sw1nRNSW1uohfaRq9jZ6npt0JbO/tLe6hlhToKACivkj9u74h/Gj4Pfsh/tC/GT4%0AB%2BIPhn4e%2BIvwc%2BEPxK%2BLFhN8WvAXiv4ieD9Ts/h34E8R%2BLLrRLjRfBfj34fa7ZXupSaVbx2utw6h%0ArkWnhZRJ4V1tp0jh6L9jDx18QPij%2Bx5%2Byh8TPizd2OofFT4i/s1fArx18TL/AEu3tLTTL34geLvh%0Ad4W8QeM7vTrXT7TT7C2sbnxHqGpTWlvZWFlaQ27xx21pbQqkKAH0rRRRQAV83eJP2Pf2XPGH7QPh%0Ar9qzxV8B/hp4i/aO8HeHz4U8LfGXWfDVlqHjvQvDZs9dsBomm6zcpJLb6atr4m19EtVXy431a8mQ%0ALPL5g%2Bka/mq/4KHftz/Hxf8AgrH8CP2HPhl%2B1B8Sf2KPhP8ADv8AZo8Sftb/ABh%2BKtr%2ByjafF74e%0A%2BOz4Q%2BIXgvWdVsPi34x8a2%2Blab4S/Zs0L4N%2BHviXo3iT4teDfGXhbQvB3xL19NC8canres2Xh/Rd%0AIAP230f9iT9jHw8/jqTQP2Rf2YdDk%2BKPhPXvAXxMk0f4B/CrTH%2BIvgXxTealqPifwX46ay8JwN4u%0A8J%2BI9Q1nV77XvDniA6ho%2Br3mq6ldahZ3E99dPL6b8H/gd8FP2evB0fw7%2BAXwf%2BFvwO%2BH8OpX2sxe%0ABfg/8P8Awn8NPB0Wr6oYm1PVY/DHgvSNF0RNS1FoYTfXy2IurswxG4lkMaY/OHQf%2BCzH7NOo6LFr%0Anib4XftPfDWDxj%2Bzh8SP2tf2eLDx58K9Ht9e/a0%2BAvwn0K38UeNfFX7P2geGvHHibWL7xDp3hq%2B0%0AzxND8LfifZ/C/wCLd34S1TTvGkPgU%2BDbka%2Bv0H%2Bwt/wUQ%2BAf/BQ7wh4h8e/s/aN8ZbPwhoVn4M1S%0Ay1/4rfCDxj8LtL8ZaL4507VLrStc8A33iiyt7TxfpFpqWgeIvDurXmkzTJaatokl1ELnw7rPhbXt%0AeAPQPjj%2Bx/8Asc/HDxv4P%2BJP7QHwJ%2BDXj/xzos2leHPDfiXx94b0O81DVo7bUZta8O%2BD9WN8qQ%2BO%0AdI07XfP8QeG/B3ieHXtI0rxF5uv6NpNrrAa9Hv3j2y8eaj4O8RWPwv8AEnhHwf8AEG502WLwn4n8%0Ae%2BCdZ%2BJHg7RdXYr5F94i8C%2BHviB8Ktb8UabGocS6VpvxF8IXUxZWTWYAhV/yH/4LE3Gn6Vqn/BOf%0AxlD%2Bz/8AG741eI/hT%2B338I/i9da78C/2Xfi3%2B0X4m%2BGHwr8D2%2BqL8VNdur74S%2BAPG2reDLTU4NZ8%0AMQLpDS6fqnj0WM0Giabr3/CN6kunftNQB%2BVfjT9mT/gqJ8R75bfxV/wUB/ZGsvA1xeabZ%2BJ/hro3%0A/BM/UNc8EeOPC8D211qljqVj8TP21/H2u6RqOsO11plw58TeIdKitrWy1Sz021uriext/wBC/iX8%0AH/hb8bPhrrnwg%2BNPw48A/Ff4Y%2BKtP0/TvFXw88eeENF8U%2BA/EFvpd7Y6rp8d/wCE/EFvqukzRaZr%0AGmadq%2BjrcRXEukanp2nahZTx39hbXUf576z/AMFH9a%2BNfxP%2BJ/wC/wCCdHwV/wCGr/iD8H9dXwh8%0AVvjr4s8Yp8Kv2IPhL43Wxlu9S8BeIvjvp%2Bi%2BO/FnxK%2BJWgGTTIfEXgH4DfC34mSeFrjUHsPHviLw%0ATqlnLYv%2BgPwYsPjJpvwz8L2n7QXib4deL/jCsWpT%2BNdd%2BEvhDxD4E%2BHM13ea1qV5pun%2BFPDPizxh%0A488SWmn6FodxpehSX2seKb%2B91280251%2BS30gaoui6cAcV8F/2Sf2Wf2cNU8Ra3%2Bz7%2Bzl8D/gjq3i%0AuKC18QX/AMKPhb4L%2BH9zqVhapaiDSZH8LaLpfk6JHPaR6iNEthDpB1ma/wBcayOsanqV9d/Qtfip%0A49/4L/f8E5vhl8UH%2BEvjbxN8c9H8SXF5ps3h3Uk/Zs%2BM%2Bp%2BEvGXgbVfixf8AwIsvjD4V8UaT4Svt%0AK1D4PXvxn024%2BGmjePrmax0rxV4km00%2BDx4g0nWdI1O%2B/Z3VNU0zQ9M1HWta1Gx0jR9IsbvVNW1b%0AVLu30/TNL0zT7eS7v9R1G/u5IbWysbK1hlubu7uZYre2t4pJppEjRmAB8%2B/D/wDZC/Zr%2BE3xl8Y/%0AH/4V/CHwt8M/ip8RtEvtC%2BIuufD4aj4L0b4hx3%2BpaFqr654%2B8C%2BG7/TfAPjXxvZ3Ph2xh0b4ieJ/%0ADGqePdA0278Q6NofiTTtH8WeKbHWfmz/AIKkTeItD/Z38NfET4ffsX%2BK/wBuH4xfCv4ueGvGnwT%2B%0AHnhOXQnn%2BG3xIu/C/jbwLbfHLV9F17xb4Si8X6H4C8L%2BNPFNnd%2BCbObUbjxbc%2BIrPQ7xPDWiXmr/%0AABA8G8/4R/4Ka%2BHf2kPizF8Lf2B/hNrn7XGh%2BF/GWmaV8Zf2kU166%2BF37IfgLwrFrkGieLp/Anx8%0AvfCHjDTf2hfiVo7y3c%2BifD74L6B4n8N6odA8SW3in4oeAZ7LTBrX2Z%2B01f8Axg0f4E/EvXvgT4y%2B%0AHvgP4meHPC2qeJND8R/FD4Z6/wDFzwfbw%2BH7SXV9Us73wR4Z%2BKnwZ1W9vdS02zurDStQXx1a2mi6%0AlcW2rX%2Bj%2BI7O0m0K/APym/Z2/wCCcOj/ABS/4IpfDH/gn/8AEXwRdfDvTfEXhTXNI8W6X%2B0z8KfC%0AHxA8f6PY%2BJvjJ4p8X%2BLvGy/Drw38QpPCPwX%2BMni7Q9e1vxb8LrXwX8QfF2m/sl%2BNPE/hq28Kal8Q%0Abb4XWKa9%2B6FfnZ/wSW%2BP3xd/an/4JwfsiftE/HnVNP1r4t/GD4UWnjfxnqulWmhadp1/e6rrOsNY%0A3FppvhrT9M0fSon0dNOxpUFmLjSyDp%2BpXF9qdteX1z%2BidABRRRQAUUUUAFFFFABRRRQAUUUUAFFF%0AFABRRRQAUUUUARxmUqTMkaP5koCxyNKpiEriByzRQkSSQiOSWMIVhlZ4UlnRFnkkoooAKKKKACii%0AigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKAPw0%2BJXi7wx8AP8A%0Agvv8OviR8ZNe0v4efD79qj/gmbo37MHwS8b%2BKbuLRvCfi39oX4XftW%2BK/iTqnwdTxBftb6Va%2BPvE%0Avgv4o6NrPgTw/LdfbPFx0nX9N01G1eDS7DVv3LrlfGHgTwP8Q9Lt9E8f%2BDfCnjnRbTVdN1600jxh%0A4e0jxNpdrrmjXAu9H1m3sNas720h1XSrtVutN1GOJbyxuFE1rNFIA1dVQB%2BDP7Vv7G3/AAVa%2BNP7%0Aa1z8efh58VP2GdO%2BEXw18KXnhb9kbR/iboHxf8V%2BNf2cfEvi%2Bz07R/iP%2B0bpvhNfD9z8LfF/7St3%0AoX9uaP8ADbxX41tfE3g34aaFejw3a%2BC9b03xB8SpfH9Lxf8AsD/8FKPip%2B2b4H/ai8bfFz9hrwRY%0AfANf2wfh9%2Bytp3wn%2BFfxPbWPg34a/amufDkFx%2B0/rul/EHUPFPhT4h/tQ23huy8Q6P4k8IpbeDvh%0Apruoa1d6tf8AiXU9O1vxD4av/wB86KAPM/jJ8GfhR%2B0N8MfGPwX%2BOPw98J/FX4U%2BP9LXR/GPgLxv%0Ao1nr3hvXrKK7ttRs/ten3sckaXul6rY2GtaJqlsYNT0LXdO03XNHvLHVtOsryD4B8Pf8ESv%2BCTXh%0ALxz4P%2BJPhf8AYH/Z18O%2BNPAOq%2BDNe8JazovgwaYuka58PxpDeFtZXTLO7h0e91W1uND07UdS1HUd%0APu7zxHrC3eteJJtX1bU9Tvbz9SqKAPkX9vT4U/Gn49fse/tD/AX4A3Hw60v4kfHT4UeOPgzY%2BJfi%0Ajr3iPQvC3gvRvih4a1TwV4k8aoPC3g/xrqeua94W0TWr3WPDfhqTTtO03XdbtbKy1XX9I097i4pv%0A7B/w7%2BPvwd/ZN%2BB/wb/aSsPhHafEr4PfDjwP8KpLv4KeNvHnj3wVr2hfD7wd4f8ACuk%2BIf7Z%2BJHg%0AbwD4pg1nVhpdxPqenXmlahHA4inTW76S6mitfryigAooooAK/C39uD/gmd%2B0L%2B15%2B2T8WPiVbeLP%0Ag74S/Z4%2BMX/BML45/wDBOjXbtvFPxCX406A3xr8W6N8RG%2BKOleGtP8Dx%2BDdUh8MeK/DmkabJ4Pv/%0AAB5aWvinwy%2BqwahqGnyX4tYP3SooA/nW%2BJf/AASc/bC%2BPI/Z%2B1741fGD9m3UvGX7EH7H/wC1Z%2BzF%0A%2BzE/gHw78TfCWg/EXxX%2B1N8GPDX7Oev/ABg%2BN8er3Piy88H2mjfCPR9Ra2%2BGnggeMYX8dalBryeN%0A49Gsv%2BEcn/Ub/gmz%2Bzj8Sv2QP2Hv2cf2XfixqvgbX/F3wG%2BHGi/DKTxD8PL7X77w54g0zwxALHTN%0AbQ%2BJND8PanaX2p2yC6v9Nawkt9OnkNpb3%2BoxxC8m%2B46KAP5pP%2BC/X7Mng3xPaeHP2k9S/ZK/ZS%2BO%0Avinwl8MbzwV4Bl%2BMHgX4l/Gb4u/Hj47T/ELwza/Av9jT4efCX4d%2BK/AGq2mm/FG08ZfFXxRe/E3w%0A7rPivxH4fu/CNuNV8OaP4T0291W%2B/e79mzwJ4a%2BGfwA%2BDngfwh8NdE%2BDfh/QPh14UhsvhV4cv11b%0AQ/h9Nd6RbajqPhPTdWVnXV7bRtUvL2zXVw7f2o0bX%2B4/aM18yftt/wDBMP8AZS/4KDa18Nte/aR0%0A/wCLWp33wq0/xho/hWH4f/Hr4wfCrSho3xB05dG8b6bqOi/D/wAY6DpF4vivQfO8O69qQsodc1Pw%0A3dXHh%2B71SXR2Syj%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<div class="prompt input_prompt">In [19]:</div>
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<div class="highlight"><pre><span class="n">x</span><span class="p">,</span><span class="n">a</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">symbols</span><span class="p">(</span><span class="s">'x,a'</span><span class="p">)</span>
<span class="n">half</span> <span class="o">=</span> <span class="n">S</span><span class="o">.</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span>
<span class="n">xi</span> <span class="o">=</span> <span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">**</span><span class="mi">2</span>
<span class="n">eta</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">Piecewise</span><span class="p">(</span> <span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="p">,</span> <span class="mi">0</span> <span class="o"><=</span> <span class="n">x</span> <span class="o"><</span> <span class="n">half</span><span class="p">),</span>
<span class="p">((</span><span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">-</span><span class="mi">1</span><span class="p">),</span> <span class="n">half</span> <span class="o"><=</span> <span class="n">x</span> <span class="p">),</span>
<span class="p">)</span>
<span class="n">v</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">var</span><span class="p">(</span><span class="s">'b:3'</span><span class="p">)</span>
<span class="n">h</span> <span class="o">=</span> <span class="p">(</span><span class="n">eta</span><span class="o">**</span><span class="n">np</span><span class="o">.</span><span class="n">arange</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">v</span><span class="p">))</span><span class="o">*</span><span class="n">v</span><span class="p">)</span><span class="o">.</span><span class="n">sum</span><span class="p">()</span>
<span class="n">J</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">integrate</span><span class="p">((</span><span class="n">xi</span> <span class="o">-</span> <span class="n">h</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span>
<span class="n">sol</span><span class="o">=</span><span class="n">S</span><span class="o">.</span><span class="n">solve</span><span class="p">([</span><span class="n">J</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">v</span><span class="p">],</span><span class="n">v</span><span class="p">)</span>
<span class="n">hsol</span> <span class="o">=</span> <span class="n">h</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sol</span><span class="p">)</span>
<span class="k">print</span> <span class="n">S</span><span class="o">.</span><span class="n">piecewise_fold</span><span class="p">(</span><span class="n">h</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">sol</span><span class="p">))</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">51</span><span class="p">,</span><span class="n">endpoint</span><span class="o">=</span><span class="bp">False</span><span class="p">)</span>
<span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">(</span><span class="mi">5</span><span class="p">,</span><span class="mi">5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">t</span><span class="o">**</span><span class="mi">2</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\xi=2 x^2$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,[</span><span class="n">hsol</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">t</span><span class="p">],</span><span class="s">'-x'</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\mathbb{E}(\xi|\eta)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="nb">map</span><span class="p">(</span><span class="n">S</span><span class="o">.</span><span class="n">lambdify</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">eta</span><span class="p">),</span><span class="n">t</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">r'$\eta(x)$'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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<pre>Piecewise((2*x**2 + x + 1/4, x < 1/2), (x + (2*x - 1)**2/2 - 1/4, 1/2 <= x))
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"></img>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>As before, using the inner product for this problem, gives:</p>
<p>$$ \int_0^1 \left(\frac{\eta_1^2}{2}-h(\eta_1)\right)\eta_1 d\eta_1=0$$</p>
<p>and the solution jumps right out as
$$h(\eta_1)=\frac{\eta_1^2}{2} , \hspace{1em} \forall \eta_1\in[0,1$$</p>
<p>where $\eta_1(x)=2x$. Doing the same thing for $\eta_2=2x-1$ gives,</p>
<p>$$ \int_0^1 \left(\frac{(1+\eta_2)^2}{2}-h(\eta_2)\right)\eta_1 d\eta_2=0$$</p>
<p>with<br />
</p>
<p>$$h(\eta_2)=\frac{(1+\eta_2)^2}{2} , \hspace{1em} \forall \eta_2\in[0,1]$$ </p>
<p>and then adding these up as before gives the full solution:</p>
<p>$$ h(\eta)= \frac{1}{2} +\eta + \eta^2$$ </p>
<p>Back-substituting each piece for $x$ produces the same solution as <code>sympy</code>.</p>
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<div class="highlight"><pre><span class="n">xs</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">rand</span><span class="p">(</span><span class="mi">100</span><span class="p">)</span>
<span class="k">print</span> <span class="n">np</span><span class="o">.</span><span class="n">mean</span><span class="p">([(</span><span class="mi">2</span><span class="o">*</span><span class="n">i</span><span class="o">**</span><span class="mi">2</span><span class="o">-</span><span class="n">hsol</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">))</span><span class="o">**</span><span class="mi">2</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">xs</span><span class="p">])</span>
<span class="k">print</span> <span class="n">S</span><span class="o">.</span><span class="n">integrate</span><span class="p">((</span><span class="mi">2</span><span class="o">*</span><span class="n">x</span><span class="o">**</span><span class="mi">2</span><span class="o">-</span><span class="n">hsol</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span><span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span><span class="o">.</span><span class="n">evalf</span><span class="p">()</span>
</pre></div>
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<pre>0.266422848567040
0.270833333333333
</pre>
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<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Summary
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>We worked out some of the great examples in Brzezniak's book using our methods as a way to show multiple ways to solve the same problem. In particular, comparing Brzezniak's more measure-theoretic methods to our less abstract techniques is a great way to get a handle on those concepts which you will need for more advanced study in stochastic process. </p>
<p>As usual, the corresponding <a href="www.ipython.org">IPython Notebook</a> notebook for this post is available for download <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Conditional_expectation_MSE_Ex.ipynb">here</a><br />
</p>
<p>Comments and corrections welcome!</p>
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<div class="text_cell_render border-box-sizing rendered_html">
<h2>References</h2>
<ul>
<li>Brzezniak, Zdzislaw, and Tomasz Zastawniak. Basic stochastic processes: a course through exercises. Springer, 2000.</li>
</ul>
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</body>
Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-3158267096357571992012-12-11T18:00:00.000-08:002012-12-11T16:05:33.181-08:00Conditional Expectation as Projection<div class="text_cell_render border-box-sizing rendered_html">
<h2>Introduction</h2>
<p>In these pages, I have tried to distill and illustrate the keys concepts needed in statistical signal processing, and in this section, we will cover the most fundamental statistical result that underpins statistical signal processing. The last sections on <a href="http://python-for-signal-processing.blogspot.com/2012/11/conditional-expectation-and-mean.html">conditional expectation</a> and <a href="http://python-for-signal-processing.blogspot.com/2012/11/the-projection-concept.html">projection</a> are prerequisites for what follows. Please review those before continuing.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Inner Product for Random Variables
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>From our previous work on projection for vectors in $\mathbb{R}^n$, we have a good geometric grasp on how projection is related to minimum mean squared error (MMSE). It turns out by one abstract step, we can carry all of our geometric interpretations to the space of random variables.</p>
<p>For example, we previously noted that at the point of projection, we had</p>
<p>$$ ( \mathbf{y} - \mathbf{v}_{opt} )^T \mathbf{v} = 0$$</p>
<p>which by noting the inner product slightly more abstractly as $\langle\mathbf{x},\mathbf{y} \rangle = \mathbf{x}^T \mathbf{y}$, we can express as</p>
<p>$$\langle \mathbf{y} - \mathbf{v}_{opt},\mathbf{v} \rangle = 0 $$ </p>
<p>and, in fact, by defining the inner product for the random variables $X$ and $Y$ as </p>
<p>$$ \langle X,Y \rangle = \mathbb{E}(X Y)$$ </p>
<p>we remarkably have the same relationship:</p>
<p>$$\langle X-h_{opt}(Y),Y \rangle = 0 $$ </p>
<p>which holds not for vectors in $\mathbb{R}^n$, but for random variables $X$ and $Y$ and functions of those random variables. Exactly why this is true is technical, but it turns out that one can build up the <strong>entire theory of probability</strong> this way (see Nelson,1987), by using the expectation as an inner product.</p>
<p>Furthermore, by abstracting out the inner product concept, we have drawn a clean line between MMSE optimization problems, geometry, and random variables. That's a lot of mileage to get a out of an abstraction and it is key to everything we pursue in statistical signal processing because now we can shift between these interpretations to address real problems. Soon, we'll see how to do this with some examples, but first we will collect one staggering result that flows naturally from this abstraction.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Conditional Expectation as Projection
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p><a href="http://python-for-signal-processing.blogspot.com/2012/11/conditional-expectation-and-mean.html">Previously</a>, we noted that the conditional expectation is the minimum mean squared error (MMSE) solution to the following problem:</p>
<p>$$ \min_h \int_{\mathbb{R}} (x - h(y) )^2 dx $$ </p>
<p>with the minimizing $h_{opt}(Y) $ as </p>
<p>$$ h_{opt}(Y) = \mathbb{E}(X|Y) $$</p>
<p>which is another way of saying that among all possible functions $h(Y)$, the one that minimizes the MSE is $ \mathbb{E}(X|Y)$ (see appendix for a quick proof). From our discussion on <a href="http://python-for-signal-processing.blogspot.com/2012/11/the-projection-concept.html">projection</a>, we noted that these MMSE solutions can be thought of as projections onto a subspace that characterizes $Y$. For example, we previously noted that at the point of projection, we have</p>
<p>$$\langle X-h_{opt}(Y),Y \rangle = 0 $$</p>
<p>but since we know that the MMSE solution</p>
<p>$$ h_{opt}(Y) = \mathbb{E}(X|Y) $$</p>
<p>we have by direct substitution,</p>
<p>$$ \mathbb{E}( X-\mathbb{E}(X|Y), Y) = 0$$ </p>
<p>That last step seems pretty innocuous, but it is the step that ties MMSE to conditional expectation to the inner project abstraction, and in so doing, reveals the conditional expectation to be a projection operator for random variables. Before we develop this further, let's grab some quick dividends:</p>
<p>From the previous equation, by linearity of the expectation, we may obtain,</p>
<p>$$ \mathbb{E}(X Y) = \mathbb{E}(Y \mathbb{E}(X|Y))$$<br />
</p>
<p>which we could have found by using the formal definition of conditional expectation,</p>
<p>\begin{equation}
\mathbb{E}(X|Y) = \int_{\mathbb{R}^2} x \frac{f_{X,Y}(x,y)}{f_Y(y)} dx dy
\end{equation}</p>
<p>and direct integration,</p>
<p>$$ \mathbb{E}(Y \mathbb{E}(X|Y))= \int_{\mathbb{R}} y \int_{\mathbb{R}} x \frac{f_{X,Y}(x,y)}{f_Y(y)} f_Y(y) dx dy =\int_{\mathbb{R}^2} x y f_{X,Y}(x,y) dx dy =\mathbb{E}( X Y) $$</p>
<p>which is good to know, but not very geometrically intuitive. And this lack of geometric intuition makes it hard to apply these concepts and keep track of these relationships. </p>
<p>We can keep pursuing this analogy and obtain the length of the error term as</p>
<p>$$ \langle X-h_{opt}(Y),X-h_{opt}(Y) \rangle = \langle X,X \rangle - \langle h_{opt}(Y),h_{opt}(Y) \rangle $$</p>
<p>and then by substituting all the notation we obtain</p>
<p>\begin{equation}
\mathbb{E}(X- \mathbb{E}(X|Y))^2 = \mathbb{E}(X)^2 - \mathbb{E}(\mathbb{E}(X|Y) )^2 <br />
\end{equation}</p>
<p>which would be tough to compute by direct integration.</p>
<p>We recognize that $\mathbb{E}(X|Y)$ <em>is</em> in fact <strong>a projection operator</strong>. Recall previously that we noted that the projection operator is idempotent, which means that once we project something onto a subspace, further projections essentially do nothing. Well, in the space of random variables, $\mathbb{E}(X|\cdot$) is the idempotent projection as we can show by noting that</p>
<p>$$ h_{opt} = \mathbb{E}(X|Y)$$</p>
<p>is purely a function of $Y$, so that</p>
<p>$$ \mathbb{E}(h_{opt}(Y)|Y) = h_{opt}(Y) $$</p>
<p>since $Y$ is fixed and this is the statement of idempotency. Thus, conditional expectation is the corresponding projection operator in this space of random variables. With this happy fact, we can continue to carry over our geometric interpretations of projections for vectors ($\mathbf{v}$) into random variables ( $X$ ). </p>
<p>Now that we have just stuffed our toolbox, let's consider some example conditional expectations obtained by using brute force to find the optimal MMSE function $h_{opt}$ as well as by using the definition of the conditional expectation.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>Example</h2>
<p>Suppose we have a random variable, $X$, then what constant is closest to $X$ in the mean-squared-sense (MSE)? In other words, which $c$ minimizes the following:</p>
<p>$$ J = \mathbb{E}( X - c )^2 $$ </p>
<p>we can work this out as</p>
<p>$$ \mathbb{E}( X - c )^2 = \mathbb{E}(c^2 - 2 c X + X^2) = c^2-2 c \mathbb{E}(X) + \mathbb{E}(X^2) $$ </p>
<p>and then take the first derivative with respect to $c$ and solve:</p>
<p>$$ c_{opt}=\mathbb{E}(X) $$ </p>
<p>Remember that $X$ can take on all kinds of values, but this says that the closest number to $X$ in the MSE sense is $\mathbb{E}(X)$.</p>
<p>Coming at this same problem using our inner product, we know that at the point of projection</p>
<p>$$ \mathbb{E}((X-c_{opt}) 1) = 0$$</p>
<p>where the $1$ represents the space of constants (i.e. $c \cdot 1 $) we are projecting on. This, by linearity of the expectation, gives</p>
<p>$$ c_{opt}=\mathbb{E}(X) $$ </p>
<p>Because $\mathbb{E}(X|Y)$ is the projection operator, with $Y=\Omega$ (the entire underlying probability space), we have, using the definintion of conditional expectation:</p>
<p>$$ \mathbb{E}(X|Y=\Omega) = \mathbb{E}(X) $$</p>
<p>Thus, we just worked the same problem three ways (optimization, inner product, projection).</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>Example</h2>
<p>Let's consider the following example with probability density $f_{X,Y}= x + y $ where $(x,y) \in [0,1]^2$ and compute the conditional expectation straight from the definition:</p>
<p>$$ \mathbb{ E}(X|Y) = \int_0^1 x \frac{f_{X,Y}(x,y)}{f_Y(y)} dx= \int_0^1 x \frac{x+y}{y+1/2} dx =\frac{3 y + 2}{6 y + 3} $$</p>
<p>That was pretty easy because the density function was so simple. Now, let's do it the hard way by going directly for the MMSE solution $h(Y)$. Then,</p>
<p>$$ \min_h \int_0^1\int_0^1 (x - h(y) )^2 f_{X,Y}(x,y)dx dy = \min_h \int_0^1 y h^2 {\left (y \right )} - y h{\left (y \right )} + \frac{1}{3} y + \frac{1}{2} h^{2}{\left (y \right )} - \frac{2}{3} h{\left (y \right )} + \frac{1}{4} dy $$ </p>
<p>Now we have to find a function $h$ that is going to minimize this. Solving for a function, as opposed to solving for a number, is generally very, very hard, but because we are integrating over a finite interval, we can use the Euler-Lagrange method from variational calculus to take the derivative of the integrand with respect to the function $h(y)$ and set it to zero. Euler-Lagrange methods will be the topic of a later section, but for now we just want the result, namely,</p>
<p>$$ 2 y h{\left (y \right )} - y + h{\left (y \right )} - \frac{2}{3} =0 $$</p>
<p>Solving this gives</p>
<p>$$ h_{opt}(y)= \frac{3 y + 2}{6 y + 3} $$</p>
<p>Finally, we can try solving this using our inner product as</p>
<p>$$ \mathbb{E}( (X - h(Y)) Y ) = 0$$</p>
<p>Writing this out gives,</p>
<p>$$ \int_0^1 \int_0^1 (x-h(y))(x+y) dx dy = \int_0^1 \frac{y\,\left(2 + 3\,y - 3\,\left( 1 + 2\,y \right) \,h(y)\right) }{6} dy = 0$$</p>
<p>and if this is zero everywhere, then the integrand must be zero,</p>
<p>$$ 2 y + 3 y^2 - 3 y h(y) - 6 y^2 h(y)=0 $$</p>
<p>and solving this for $h(y)$ gives the same solution:</p>
<p>$$ h_{opt}(y)= \frac{3 y + 2}{6 y + 3} $$</p>
<p>Thus, doing it by the definition, optimization, or inner product gives us the same answer; but, in general, no method is necessarily easier because they both involve potentially difficult or impossible integration, optimization, or functional equation solving. The point is that now that we have a deep toolbox, we can pick and choose which tools we want to apply for different problems.</p>
<p>Before we leave this example, let's use <code>sympy</code> to verify the length of the error function we found earlier:</p>
<p>\begin{equation}
\mathbb{E}(X- \mathbb{E}(X|Y))^2 = \mathbb{E}(X)^2 - \mathbb{E}(\mathbb{E}(X|Y) )^2 <br />
\end{equation}</p>
<p>that is based on the Pythagorean theorem.</p>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">sympy.abc</span> <span class="kn">import</span> <span class="n">y</span><span class="p">,</span><span class="n">x</span>
<span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">integrate</span><span class="p">,</span> <span class="n">simplify</span>
<span class="n">fxy</span> <span class="o">=</span> <span class="n">x</span> <span class="o">+</span> <span class="n">y</span> <span class="c"># joint density</span>
<span class="n">fy</span> <span class="o">=</span> <span class="n">integrate</span><span class="p">(</span><span class="n">fxy</span><span class="p">,(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span> <span class="c"># marginal density</span>
<span class="n">fx</span> <span class="o">=</span> <span class="n">integrate</span><span class="p">(</span><span class="n">fxy</span><span class="p">,(</span><span class="n">y</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span> <span class="c"># marginal density</span>
<span class="n">h</span> <span class="o">=</span> <span class="p">(</span><span class="mi">3</span><span class="o">*</span><span class="n">y</span><span class="o">+</span><span class="mi">2</span><span class="p">)</span><span class="o">/</span><span class="p">(</span><span class="mi">6</span><span class="o">*</span><span class="n">y</span><span class="o">+</span><span class="mi">3</span><span class="p">)</span> <span class="c"># conditional expectation</span>
<span class="n">LHS</span><span class="o">=</span><span class="n">integrate</span><span class="p">((</span><span class="n">x</span> <span class="o">-</span> <span class="n">h</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="o">*</span><span class="n">fxy</span><span class="p">,</span> <span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),(</span><span class="n">y</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span> <span class="c"># from the definition</span>
<span class="n">RHS</span><span class="o">=</span><span class="n">integrate</span><span class="p">(</span> <span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="o">*</span><span class="n">fx</span><span class="p">,</span> <span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span> <span class="o">-</span> <span class="n">integrate</span><span class="p">(</span> <span class="p">(</span><span class="n">h</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span> <span class="o">*</span><span class="n">fy</span><span class="p">,</span> <span class="p">(</span><span class="n">y</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">))</span> <span class="c"># using Pythagorean theorem</span>
<span class="k">print</span> <span class="n">simplify</span><span class="p">(</span><span class="n">LHS</span><span class="o">-</span><span class="n">RHS</span><span class="p">)</span><span class="o">==</span><span class="mi">0</span>
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<pre>True
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<h2>Summary</h2>
<p>In this section, we have pulled together all the projection and least-squares optimization ideas from the previous posts to draw a clean line between our geometric notions of projection from vectors in $\mathbb{R}^n$ to general random variables. This resulted in the remarkable realization that the conditional expectation is in fact a projection operator for random variables. The key idea is that because we have these relationships, we can approach difficult problems in multiple ways, depending on which way is more intuitive or tractable in a particular situation. In these pages, we will again and again come back to these intuitions because they form the backbone of statistical signal processing. </p>
<p>In the next section, we will have a lot of fun with these ideas working out some examples that are usually solved using more general tools from measure theory. </p>
<p>Note that the book by Mikosch (1998) has some excellent sections covering much of this material in more detail. Mikosch has a very geometric view of the material as well.</p>
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<h2>Appendix</h2>
<h4>Proof of MMSE of Conditional expectation by Optimization</h4>
<p>$$ \min_h \int_\mathbb{R^2} | X - h(Y) |^2 f_{x,y}(X,Y) dx dy $$</p>
<p>$$ \min_h \int_\mathbb{R^2} | X |^2 f_{x,y}(X,Y) dx dy + \int_\mathbb{R^2} | h(Y) |^2 f_{x,y}(X,Y) dx dy - \int_\mathbb{R^2} 2 X h(Y) f_{x,y}(X,Y) dx dy $$</p>
<p>Now, we want to maximize the following:</p>
<p>$$ \max_h \int_\mathbb{R^2} X h(Y) f_{x,y}(X,Y) dx dy $$ </p>
<p>Breaking up the integral using the definition of conditional expectation</p>
<p>$$ \max_h \int_\mathbb{R} \left(\int_\mathbb{R} X f_{x|y}(X|Y) dx \right)h(Y) f_Y(Y) dy $$ </p>
<p>$$ \max_h \int_\mathbb{R} \mathbb{E}(X|Y) h(Y)f_Y(Y) dy $$ </p>
<p>From properties of the Cauchy-Schwarz inequality, we know that the maximum happens when $h_{opt}(Y) = \mathbb{E}(X|Y)$, so we have found the optimal $h(Y)$ function as :</p>
<p>$$ h_{opt}(Y) = \mathbb{E}(X|Y)$$ </p>
<p>which shows that the optimal function is the conditional expectation.</p>
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<h3>References</h3>
<p>This post was created using the <a href="https://github.com/ipython/nbconvert">nbconvert</a> utility from the source <a href="www.ipython.org">IPython Notebook</a> which is available for <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Conditional_Expectation_Projection.ipynb">download</a> from the main github <a href="https://github.com/unpingco/Python-for-Signal-Processing">site</a> for this blog. </p>
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<h3>Bibliography</h3>
<ul>
<li>
<p>Nelson, Edward. Radically Elementary Probability Theory.(AM-117). Vol. 117. Princeton University Press, 1987.</p>
</li>
<li>
<p>Mikosch, Thomas. Elementary stochastic calculus with finance in view. Vol. 6. World Scientific Publishing Company Incorporated, 1998.</p>
</li>
</ul>
</div>
Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-17975961463846614352012-11-30T17:01:00.003-08:002012-11-30T17:02:22.836-08:00<body>
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<h2>Projection in Multiple Dimensions</h2>
<p>In this section, we extend from the <a href="http://python-for-signal-processing.blogspot.com/2012/11/the-projection-concept.html">one-dimensional subspace</a> to the more general two-dimensional subspace. This means that there are two vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$ that are not colinear and that span the subspace. In the previous case, we had only one vector ( $\mathbf{v}$), so we had a one-dimensional subspace, but now that we have two vectors, we have a two-dimensional subspace (i.e. a plane). The extension from the two-dimensional subspace to the <em>n</em>-dimensional subspace follows the same argument but introduces more notation than we need so we'll stick with the two-dimensional case for awhile. For the two-dimensional case, the optimal MMSE solution has the form</p>
<p>$$ \hat{\mathbf{y}} = \alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 \in \mathbb{R}^m $$</p>
<p>where $\mathbf{y}$ exists in the m-dimensional space of real numbers. We want to project this vector onto the two m-dimensional $\mathbf{v}_i$ vectors. Here, the orthogonality requirement extends as</p>
<p>$$ \langle \mathbf{y} - \alpha_1 \mathbf{v}_1 -\alpha_2 \mathbf{v}_2 , \mathbf{v}_1\rangle= 0 $$ </p>
<p>and </p>
<p>$$ \langle \mathbf{y} - \alpha_1 \mathbf{v}_1 -\alpha_2 \mathbf{v}_2 , \mathbf{v}_2\rangle= 0 $$ </p>
<p>Recall that for vectors, we have</p>
<p>$$ \langle \mathbf{x} , \mathbf{y}\rangle = \mathbf{x}^T \mathbf{y} \in \mathbb{R}$$</p>
<p>This leads to the linear system of equations:</p>
<p>$$ \begin{eqnarray}
\langle \mathbf{y}, \mathbf{v}_1\rangle = & \alpha_1 \langle \mathbf{v}_1, \mathbf{v}_1\rangle & +\alpha_2 \langle \mathbf{v}_1, \mathbf{v}_2\rangle \\
\langle \mathbf{y}, \mathbf{v}_2\rangle = & \alpha_1 \langle \mathbf{v}_1, \mathbf{v}_2\rangle & +\alpha_2 \langle \mathbf{v}_2, \mathbf{v}_2\rangle
\end{eqnarray}
$$</p>
<p>which can be written in matrix form as</p>
<p>$$ \left[ \begin{array}{c}
\langle \mathbf{y}, \mathbf{v}_1\rangle \\
\langle \mathbf{y}, \mathbf{v}_2\rangle \\
\end{array} \right] =
\left[ \begin{array}{cc}
\langle \mathbf{v}_1, \mathbf{v}_1\rangle & \langle \mathbf{v}_1, \mathbf{v}_2\rangle \\
\langle \mathbf{v}_1, \mathbf{v}_2\rangle & \langle \mathbf{v}_2, \mathbf{v}_2\rangle \\
\end{array} \right] \left[
\begin{array}{c}
\alpha_1 \\
\alpha_2 \\
\end{array} \right]$$ </p>
<p>which can be further reduced by stacking the columns into </p>
<p>$$ \mathbf{V} = \left[ \mathbf{v}_1, \mathbf{v}_2 \right] \in \mathbb{R}^{m \times 2} $$</p>
<p>and </p>
<p>$$ \boldsymbol{\alpha}= \left[ \alpha_1, \alpha_2\right]^T \in \mathbb{R}^{2}$$</p>
<p>which gives</p>
<p>$$ \mathbf{V}^T \mathbf{y} = (\mathbf{V}^T \mathbf{V}) \boldsymbol{\alpha} $$</p>
<p>Note that by writing this using vector notation, we have implicitly generalized beyond two dimensions since there is nothing to stop from stacking $\mathbf{V}$ with more column vectors to create a larger subspace. By solving we obtain,</p>
<p>$$ \boldsymbol{\alpha} = (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T \mathbf{y} $$ </p>
<p>and so the optimal solution is then,</p>
<p>$$ \hat{\mathbf{y}} = \mathbf{V} \boldsymbol{\alpha} \in \mathbb{R}^m $$ </p>
<p>Note that the existence of the inverse is guaranteed by the non-co-linearity of the $\mathbf{v}_i$ vectors. Whether or not that inverse is numerically stable is another issue.</p>
<p>Then, we can combine these to obtain</p>
<p>$$ \hat{\mathbf{y}} = \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T \mathbf{y} $$ </p>
<p>when then makes the projection operator for this case:</p>
<p>$$ \mathbf{P}_{V}= \mathbf{V} (\mathbf{V}^T \mathbf{V})^{-1} \mathbf{V}^T \in \mathbb{R}^{m \times m} $$ </p>
<p>As a quick check, we can see this reduce to the 1-dimensional case by setting</p>
<p>$$ \mathbf{V}= \mathbf{v} \in \mathbb{R}^m$$</p>
<p>so then,</p>
<p>$$ \mathbf{P}_{v}= \mathbf{v} \frac{1}{\mathbf{v}^T \mathbf{v}} \mathbf{v}^T $$ </p>
<p>which matches our <a href="http://python-for-signal-processing.blogspot.com/2012/11/the-projection-concept.html">previous result</a>. The point of all these manipulations is that we can construct another projection operator with all the MMSE properties we had before, but now in a bigger subspace. We can further verify the idempotent property of projection matrices by checking that</p>
<p>$$ \mathbf{P}_V \mathbf{P}_V = \mathbf{P}_V$$</p>
<p>The following graphic shows that when we project the three dimensional $\mathbf{y}$ vector onto the plane, which is spanned by the two $\mathbf{v}_i$ vectors, we obtain the MMSE solution where the sphere is tangent to the plane. The point of tangency is the point $\hat{\mathbf{y}} $ which is the MMSE solution.</p>
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<div class="highlight"><pre><span class="c">#http://stackoverflow.com/questions/10374930/matplotlib-annotating-a-3d-scatter-plot</span>
<span class="kn">from</span> <span class="nn">mpl_toolkits.mplot3d</span> <span class="kn">import</span> <span class="n">proj3d</span>
<span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">as</span> <span class="nn">plt</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">([</span><span class="mi">8</span><span class="p">,</span><span class="mi">8</span><span class="p">])</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">,</span> <span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'x-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'y-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlabel</span><span class="p">(</span><span class="s">'z-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">y</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span><span class="o">.</span><span class="n">T</span>
<span class="n">V</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.25</span><span class="p">],</span> <span class="c"># columns are v_1, v_2</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.50</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.00</span><span class="p">]])</span>
<span class="n">alpha</span><span class="o">=</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">y</span> <span class="c"># optimal coefficients</span>
<span class="n">P</span> <span class="o">=</span> <span class="n">V</span><span class="o">*</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span>
<span class="n">yhat</span> <span class="o">=</span> <span class="n">P</span><span class="o">*</span><span class="n">y</span> <span class="c"># approximant</span>
<span class="n">u</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">xx</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">yy</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">zz</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">size</span><span class="p">(</span><span class="n">u</span><span class="p">)),</span> <span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">sphere</span><span class="o">=</span><span class="n">ax</span><span class="o">.</span><span class="n">plot_surface</span><span class="p">(</span><span class="n">xx</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">yy</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">zz</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="n">rstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">cstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s">'gray'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">0.25</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'r-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'ro'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'g--'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'go'</span><span class="p">)</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{y}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">20</span><span class="p">,</span> <span class="mi">20</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\hat{\mathbf{y}}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">40</span><span class="p">,</span> <span class="mi">10</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_1$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">120</span><span class="p">,</span> <span class="mi">10</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_2$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">30</span><span class="p">,</span> <span class="mi">30</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
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</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h3>
Weighted Distances
</h3>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>As before, we can easily extend this projection operator to cases where the measure of distance between $\mathbf{y}$ and the subspace $\mathbf{v}$ is weighted (i.e. non-uniform). We can accomodate these weighted distances by re-writing the projection operator as</p>
<p>$$ \mathbf{P}_{V} = \mathbf{V} ( \mathbf{V}^T \mathbf{Q V})^{-1} \mathbf{V}^T \mathbf{Q} $$</p>
<p>where $\mathbf{Q}$ is positive definite matrix. Earlier, we started with a point $\mathbf{y}$ and inflated a sphere centered at $\mathbf{y}$ until it just touched the plane defined by $\mathbf{v}_i$ and this point was closest point on the subspace to $\mathbf{y}$. In the general case with a weighted distance except now we inflate an ellipsoid, not a sphere, until the ellipsoid touches the line.</p>
<p>The code and figure below illustrate what happens using the weighted $ \mathbf{P}_v $. It is basically the same code we used above. You can download the IPython notebook corresponding to this post and try different values on the diagonal of $\mathbf{Q}$.</p>
</div>
<div class="cell border-box-sizing code_cell vbox">
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<div class="prompt input_prompt">In [9]:</div>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">mpl_toolkits.mplot3d</span> <span class="kn">import</span> <span class="n">proj3d</span>
<span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">as</span> <span class="nn">plt</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">([</span><span class="mi">8</span><span class="p">,</span><span class="mi">8</span><span class="p">])</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">,</span> <span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'x-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'y-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlabel</span><span class="p">(</span><span class="s">'z-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">y</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span><span class="o">.</span><span class="n">T</span>
<span class="n">V</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.25</span><span class="p">],</span> <span class="c"># columns are v_1, v_2</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.50</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.00</span><span class="p">]])</span>
<span class="n">Q</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">3</span><span class="p">]])</span>
<span class="n">P</span> <span class="o">=</span> <span class="n">V</span><span class="o">*</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Q</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Q</span>
<span class="n">yhat</span> <span class="o">=</span> <span class="n">P</span><span class="o">*</span><span class="n">y</span> <span class="c"># approximant</span>
<span class="n">u</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">xx</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">yy</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">zz</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">size</span><span class="p">(</span><span class="n">u</span><span class="p">)),</span> <span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">xx</span><span class="p">,</span><span class="n">yy</span><span class="p">,</span><span class="n">yz</span><span class="o">=</span><span class="nb">map</span><span class="p">(</span><span class="n">squeeze</span><span class="p">,</span><span class="n">split</span><span class="p">(</span><span class="n">tensordot</span><span class="p">(</span><span class="n">dstack</span><span class="p">([</span><span class="n">xx</span><span class="p">,</span><span class="n">yy</span><span class="p">,</span><span class="n">zz</span><span class="p">]),</span><span class="n">Q</span><span class="p">,</span><span class="n">axes</span><span class="o">=</span><span class="mi">1</span><span class="p">),</span><span class="mi">3</span><span class="p">,</span><span class="n">axis</span><span class="o">=</span><span class="mi">2</span><span class="p">))</span>
<span class="n">ellipsoid</span><span class="o">=</span><span class="n">ax</span><span class="o">.</span><span class="n">plot_surface</span><span class="p">(</span><span class="n">xx</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">yy</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">zz</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="n">rstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">cstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s">'gray'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">0.25</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'r-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'ro'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'g--'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'go'</span><span class="p">)</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{y}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">20</span><span class="p">,</span> <span class="mi">20</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\hat{\mathbf{y}}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">40</span><span class="p">,</span> <span class="mi">30</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_1$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">120</span><span class="p">,</span> <span class="mi">10</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_2$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">30</span><span class="p">,</span> <span class="mi">30</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
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</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Summary
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>In this section, we extended the concept of a projection operator beyond one dimension and showed the corresponding geometric concepts that tie the projection operator to MMSE problems in more than one dimension. </p>
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<div class="text_cell_render border-box-sizing rendered_html">
<h3>
References
</h3>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>This post was created using the <a href="https://github.com/ipython/nbconvert">nbconvert</a> utility from the source <a href="www.ipython.org">IPython Notebook</a> which is available for <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Projection_mdim.ipynb">download</a> from the main github <a href="https://github.com/unpingco/Python-for-Signal-Processing">site</a> for this blog. The projection concept is masterfully discussed in the classic Strang, G. (2003). <em>Introduction to linear algebra</em>. Wellesley Cambridge Pr. Also, some of Dr. Strang's excellent lectures are available on <a href="http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/">MIT Courseware</a>. I highly recommend these as well as the book.</p>
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<div class="text_cell_render border-box-sizing rendered_html">
<h3>
Appendix
</h3>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>Below is some extra code to handle the more general case where there is a rotation as well as a weighting of the axes. Note the projection operator is constructed exactly the same way.</p>
</div>
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<div class="prompt input_prompt">In [15]:</div>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">mpl_toolkits.mplot3d</span> <span class="kn">import</span> <span class="n">proj3d</span>
<span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">as</span> <span class="nn">plt</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_size_inches</span><span class="p">([</span><span class="mi">8</span><span class="p">,</span><span class="mi">8</span><span class="p">])</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">,</span> <span class="n">projection</span><span class="o">=</span><span class="s">'3d'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlim</span><span class="p">([</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'x-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'y-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_zlabel</span><span class="p">(</span><span class="s">'z-axis'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">y</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">])</span><span class="o">.</span><span class="n">T</span>
<span class="n">V</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.25</span><span class="p">],</span> <span class="c"># columns are v_1, v_2</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.50</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mf">0.00</span><span class="p">]])</span>
<span class="k">def</span> <span class="nf">rotation_matrix</span><span class="p">(</span><span class="n">angle</span><span class="p">,</span><span class="n">axis</span><span class="o">=</span><span class="s">'z'</span><span class="p">):</span>
<span class="n">angle</span> <span class="o">=</span> <span class="n">angle</span><span class="o">/</span><span class="mf">180.</span><span class="o">*</span><span class="n">pi</span>
<span class="k">if</span> <span class="n">axis</span><span class="o">==</span><span class="s">'z'</span><span class="p">:</span>
<span class="k">return</span> <span class="n">matrix</span><span class="p">([[</span><span class="n">cos</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="n">sin</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="n">sin</span><span class="p">(</span><span class="o">-</span><span class="n">angle</span><span class="p">),</span><span class="n">cos</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">]])</span>
<span class="k">elif</span> <span class="n">axis</span><span class="o">==</span><span class="s">'y'</span><span class="p">:</span>
<span class="k">return</span> <span class="n">matrix</span><span class="p">([[</span><span class="n">cos</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="n">sin</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="n">sin</span><span class="p">(</span><span class="o">-</span><span class="n">angle</span><span class="p">),</span><span class="n">cos</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="mi">0</span><span class="p">]])</span>
<span class="k">elif</span> <span class="n">axis</span><span class="o">==</span><span class="s">'x'</span><span class="p">:</span>
<span class="k">return</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="n">cos</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="n">sin</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="n">sin</span><span class="p">(</span><span class="o">-</span><span class="n">angle</span><span class="p">),</span><span class="n">cos</span><span class="p">(</span><span class="n">angle</span><span class="p">),</span><span class="mi">0</span><span class="p">]])</span>
<span class="n">S</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">3</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">]])</span>
<span class="n">R</span> <span class="o">=</span> <span class="n">rotation_matrix</span><span class="p">(</span><span class="mi">30</span><span class="p">)</span><span class="o">*</span><span class="n">rotation_matrix</span><span class="p">(</span><span class="mi">30</span><span class="p">,</span><span class="s">'x'</span><span class="p">)</span><span class="o">*</span><span class="n">rotation_matrix</span><span class="p">(</span><span class="mi">40</span><span class="p">,</span><span class="s">'y'</span><span class="p">)</span>
<span class="n">Q</span> <span class="o">=</span> <span class="n">R</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">S</span><span class="o">*</span><span class="n">R</span> <span class="c"># apply 3-D rotations</span>
<span class="n">P</span> <span class="o">=</span> <span class="n">V</span><span class="o">*</span><span class="n">inv</span><span class="p">(</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Q</span><span class="o">*</span><span class="n">V</span><span class="p">)</span><span class="o">*</span><span class="n">V</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Q</span> <span class="c"># build projection matrix</span>
<span class="n">yhat</span> <span class="o">=</span> <span class="n">P</span><span class="o">*</span><span class="n">y</span> <span class="c"># approximant</span>
<span class="n">u</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">2</span><span class="o">*</span><span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">np</span><span class="o">.</span><span class="n">pi</span><span class="p">,</span> <span class="mi">100</span><span class="p">)</span>
<span class="n">xx</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">yy</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">u</span><span class="p">),</span> <span class="n">np</span><span class="o">.</span><span class="n">sin</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">zz</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">outer</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">ones</span><span class="p">(</span><span class="n">np</span><span class="o">.</span><span class="n">size</span><span class="p">(</span><span class="n">u</span><span class="p">)),</span> <span class="n">np</span><span class="o">.</span><span class="n">cos</span><span class="p">(</span><span class="n">v</span><span class="p">))</span>
<span class="n">xx</span><span class="p">,</span><span class="n">yy</span><span class="p">,</span><span class="n">yz</span><span class="o">=</span><span class="nb">map</span><span class="p">(</span><span class="n">squeeze</span><span class="p">,</span><span class="n">split</span><span class="p">(</span><span class="n">tensordot</span><span class="p">(</span><span class="n">dstack</span><span class="p">([</span><span class="n">xx</span><span class="p">,</span><span class="n">yy</span><span class="p">,</span><span class="n">zz</span><span class="p">]),</span><span class="n">Q</span><span class="p">,</span><span class="n">axes</span><span class="o">=</span><span class="mi">1</span><span class="p">),</span><span class="mi">3</span><span class="p">,</span><span class="n">axis</span><span class="o">=</span><span class="mi">2</span><span class="p">))</span>
<span class="n">ellipsoid</span><span class="o">=</span><span class="n">ax</span><span class="o">.</span><span class="n">plot_surface</span><span class="p">(</span><span class="n">xx</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">yy</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">zz</span><span class="o">+</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span>
<span class="n">rstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">cstride</span><span class="o">=</span><span class="mi">4</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s">'gray'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.3</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">0.25</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'r-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'ro'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'b-'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">]],[</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">]],</span><span class="s">'bo'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">],</span><span class="s">'g--'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mi">3</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot3D</span><span class="p">([</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]],[</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">]],</span><span class="s">'go'</span><span class="p">)</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">y</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{y}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">20</span><span class="p">,</span> <span class="mi">20</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">yhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">yhat</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\hat{\mathbf{y}}$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">40</span><span class="p">,</span> <span class="mi">30</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_1$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="mi">120</span><span class="p">,</span> <span class="mi">10</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">,</span> <span class="n">_</span> <span class="o">=</span> <span class="n">proj3d</span><span class="o">.</span><span class="n">proj_transform</span><span class="p">(</span><span class="n">V</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span><span class="n">V</span><span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span> <span class="n">ax</span><span class="o">.</span><span class="n">get_proj</span><span class="p">())</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span>
<span class="s">"$\mathbf{v}_2$"</span><span class="p">,</span>
<span class="n">xy</span> <span class="o">=</span> <span class="p">(</span><span class="n">x2</span><span class="p">,</span> <span class="n">y2</span><span class="p">),</span> <span class="n">xytext</span> <span class="o">=</span> <span class="p">(</span><span class="o">-</span><span class="mi">30</span><span class="p">,</span> <span class="mi">30</span><span class="p">),</span> <span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span>
<span class="n">textcoords</span> <span class="o">=</span> <span class="s">'offset points'</span><span class="p">,</span> <span class="n">ha</span> <span class="o">=</span> <span class="s">'right'</span><span class="p">,</span> <span class="n">va</span> <span class="o">=</span> <span class="s">'bottom'</span><span class="p">,</span>
<span class="n">bbox</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">boxstyle</span> <span class="o">=</span> <span class="s">'round,pad=0.5'</span><span class="p">,</span> <span class="n">fc</span> <span class="o">=</span> <span class="s">'yellow'</span><span class="p">,</span> <span class="n">alpha</span> <span class="o">=</span> <span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span> <span class="o">=</span> <span class="nb">dict</span><span class="p">(</span><span class="n">arrowstyle</span> <span class="o">=</span> <span class="s">'->'</span><span class="p">,</span> <span class="n">connectionstyle</span> <span class="o">=</span> <span class="s">'arc3,rad=0'</span><span class="p">))</span>
<span class="n">plt</span><span class="o">.</span><span class="n">show</span><span class="p">()</span>
</pre></div>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-20626852802162941142012-11-21T17:00:00.000-08:002012-11-21T17:00:04.250-08:00The Projection Concept<div class="text_cell_render border-box-sizing rendered_html">
<p>On the road to a geometric understanding of conditional expectation, we need to grasp the concept of projection. In the figure below, we want to find a point along the blue line that is closest to the red square. In other words, we want to inflate the pink circle until it just touches the blue line. Then, that point will be the closest point on the blue line to the red square.</p>
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<div class="prompt input_prompt">In [1]:</div>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span>
<span class="kn">from</span> <span class="nn">matplotlib</span> <span class="kn">import</span> <span class="n">patches</span>
<span class="n">fig</span><span class="p">,</span> <span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_figheight</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="mi">6</span><span class="p">)</span>
<span class="n">y</span><span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">2</span><span class="p">],</span>
<span class="p">[</span><span class="mi">3</span><span class="p">]])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">x</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="o">*</span><span class="n">y</span><span class="p">,</span><span class="n">marker</span><span class="o">=</span><span class="s">'s'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">patches</span><span class="o">.</span><span class="n">Circle</span><span class="p">(</span><span class="n">y</span><span class="p">,</span><span class="n">radius</span><span class="o">=.</span><span class="mi">5</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.75</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'pink'</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span><span class="s">'Find point along</span><span class="se">\n</span><span class="s"> line closest</span><span class="se">\n</span><span class="s"> to red square'</span><span class="p">,</span>
<span class="n">fontsize</span><span class="o">=</span><span class="mi">12</span><span class="p">,</span><span class="n">xy</span><span class="o">=</span><span class="p">(</span><span class="mf">2.5</span><span class="p">,</span><span class="mf">2.5</span><span class="p">),</span>
<span class="n">xytext</span><span class="o">=</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span><span class="mf">1.5</span><span class="p">),</span>
<span class="n">arrowprops</span><span class="o">=</span><span class="p">{</span><span class="s">'facecolor'</span><span class="p">:</span><span class="s">'blue'</span><span class="p">})</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">xmax</span><span class="o">=</span><span class="mi">5</span><span class="p">,</span><span class="n">ymax</span><span class="o">=</span><span class="mi">5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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<p>It may be geometrically obvious, but the closest point on the line occurs where the line segment from the red square to the blue line is perpedicular to the line. At this point, the pink circle just touches the blue line. This is illustrated below.</p>
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<div class="prompt input_prompt">In [2]:</div>
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<div class="highlight"><pre><span class="n">fig</span><span class="p">,</span> <span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_figheight</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="mi">6</span><span class="p">)</span>
<span class="n">y</span><span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">2</span><span class="p">],</span>
<span class="p">[</span><span class="mi">3</span><span class="p">]])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">x</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="o">*</span><span class="n">y</span><span class="p">,</span><span class="n">marker</span><span class="o">=</span><span class="s">'o'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span><span class="n">patches</span><span class="o">.</span><span class="n">Circle</span><span class="p">(</span><span class="n">y</span><span class="p">,</span><span class="n">radius</span><span class="o">=</span><span class="mi">1</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="mf">2.</span><span class="p">),</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.75</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'pink'</span><span class="p">))</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">],[</span><span class="mi">1</span><span class="p">]])</span>
<span class="n">Pv</span> <span class="o">=</span> <span class="n">v</span><span class="o">*</span><span class="n">v</span><span class="o">.</span><span class="n">T</span><span class="o">/</span> <span class="p">(</span><span class="n">v</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">v</span><span class="p">)</span> <span class="c"># projection operator</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="o">*</span><span class="p">(</span><span class="n">Pv</span><span class="o">*</span><span class="n">y</span><span class="p">),</span><span class="n">marker</span><span class="o">=</span><span class="s">'s'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'g'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_line</span><span class="p">(</span> <span class="n">Line2D</span><span class="p">(</span> <span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="mf">2.5</span><span class="p">),</span> <span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mf">2.5</span><span class="p">)</span> <span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'g'</span><span class="p">,</span><span class="n">linestyle</span><span class="o">=</span><span class="s">'--'</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_line</span><span class="p">(</span> <span class="n">Line2D</span><span class="p">(</span> <span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="mi">0</span><span class="p">),</span> <span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">)</span> <span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">,</span><span class="n">linestyle</span><span class="o">=</span><span class="s">'--'</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span> <span class="s">'Closest point is</span><span class="se">\n</span><span class="s">perpedicular</span><span class="se">\n</span><span class="s">to line and tangent</span><span class="se">\n</span><span class="s">to circle'</span><span class="p">,</span>
<span class="n">fontsize</span><span class="o">=</span><span class="mi">12</span><span class="p">,</span><span class="n">xy</span><span class="o">=</span><span class="p">(</span><span class="mf">2.6</span><span class="p">,</span><span class="mf">2.5</span><span class="p">),</span>
<span class="n">xytext</span><span class="o">=</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span><span class="mf">1.5</span><span class="p">),</span>
<span class="n">arrowprops</span><span class="o">=</span><span class="p">{</span><span class="s">'facecolor'</span><span class="p">:</span><span class="s">'blue'</span><span class="p">})</span>
<span class="n">ax</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="o">.</span><span class="mi">7</span><span class="p">,</span><span class="mf">1.5</span><span class="p">,</span><span class="s">r'$\mathbf{y}$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>Now that we can see what's going on, we can construct the the solution analytically. We can represent an arbitrary point along the blue line as:</p>
<p>$$ \mathbf{x} = \alpha \mathbf{v} $$ </p>
<p>where $\alpha$ slides the point up and down the line with</p>
<p>$$ \mathbf{v} = \left[ \begin{array}{c}
1 \
1 \
\end{array} \right] $$ </p>
<p>Formally, $ \mathbf{v}$ is the <em>subspace</em> on which we want to <em>project</em> the $\mathbf{y}$. At the closest point, the vector between $\mathbf{y}$ and $\mathbf{x}$ (the dotted green <em>error</em> vector above) is perpedicular to the line. This means that</p>
<p>$$ ( \mathbf{y}-\mathbf{x} )^T \mathbf{v} = 0$$ </p>
<p>and by substituting and working out the terms, we obtain </p>
<p>$$ \alpha = \frac{\mathbf{y}^T\mathbf{v}}{|\mathbf{v}|^2}$$</p>
<p>The <em>error</em> is the distance between $\alpha\mathbf{v}$ and $ \mathbf{y}$. Because we have a right triangle, using the Pythagorean theorem, we compute the squared length of this error as</p>
<p>$$ \epsilon^2 = |( \mathbf{y}-\mathbf{x} )|^2 = |\mathbf{y}|^2 - \alpha^2 |\mathbf{v}|^2 = |\mathbf{y}|^2 - \frac{|\mathbf{y}^T\mathbf{v}|^2}{|\mathbf{v}|^2} $$</p>
<p>where $ |\mathbf{v}|^2 = \mathbf{v}^T \mathbf{v} $. Note that since $\epsilon^2 \ge 0 $, this also shows that</p>
<p>$$ |\mathbf{y}^T\mathbf{v}| \le |\mathbf{y}| |\mathbf{v}| $$ </p>
<p>which is the famous and useful Cauchy-Schwarz inequality which we will exploit later. Finally, we can assemble all of this into the <em>projection</em> operator</p>
<p>$$ \mathbf{P}_v = \frac{1}{|\mathbf{v}|^2 } \mathbf{v v}^T $$</p>
<p>With this operator, we can take any $\mathbf{y}$ and find the closest point on $\mathbf{v}$ by doing</p>
<p>$$ \mathbf{P}_v \mathbf{y} = \mathbf{v} \left( \frac{ \mathbf{v}^T \mathbf{y} }{|\mathbf{v}|^2} \right)$$ </p>
<p>where we recognize the term in parenthesis as the $\alpha$ we computed earlier. It's called an <em>operator</em> because it takes a vector ($\mathbf{y}$) and produces another vector ($\alpha\mathbf{v}$).</p>
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<div class="text_cell_render border-box-sizing rendered_html">
<h3>
Weighted distances
</h3>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>We can easily extend this projection operator to cases where the measure of distance between $\mathbf{y}$ and the subspace $\mathbf{v}$ is weighted (i.e. non-uniform). We can accomodate these weighted distances by re-writing the projection operator as</p>
<p>$$ \mathbf{P}_v = \mathbf{v}\frac{\mathbf{v}^T \mathbf{Q}^T}{ \mathbf{v}^T \mathbf{Q v} } $$</p>
<p>where $\mathbf{Q}$ is positive definite matrix. Earlier, we started with a point $\mathbf{y}$ and inflated a circle centered at $\mathbf{y}$ until it just touched the line defined by $\mathbf{v}$ and this point was closest point on the line to $\mathbf{y}$. The same thing happens in the general case with a weighted distance except now we inflate an ellipsoid, not a circle, until the ellipsoid touches the line.</p>
<p>The code and figure below illustrate what happens using the weighted $ \mathbf{P}_v $. It is basically the same code we used earlier. You can download the IPython notebook corresponding to this post and try different values on the diagonal of $\mathbf{S}$ and $\theta$ below.</p>
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<div class="prompt input_prompt">In [3]:</div>
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<div class="highlight"><pre><span class="n">theta</span> <span class="o">=</span> <span class="mi">120</span><span class="o">/</span><span class="mf">180.</span><span class="o">*</span><span class="n">pi</span> <span class="c"># rotation angle for ellipse</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">1</span><span class="p">],[</span><span class="mi">1</span><span class="p">]])</span>
<span class="c"># rotation matrix</span>
<span class="n">U</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span> <span class="n">cos</span><span class="p">(</span><span class="n">theta</span><span class="p">),</span> <span class="n">sin</span><span class="p">(</span><span class="n">theta</span><span class="p">)],</span>
<span class="p">[</span><span class="o">-</span><span class="n">sin</span><span class="p">(</span><span class="n">theta</span><span class="p">),</span> <span class="n">cos</span><span class="p">(</span><span class="n">theta</span><span class="p">)]])</span>
<span class="c"># diagonal weight matrix</span>
<span class="n">S</span> <span class="o">=</span> <span class="n">matrix</span><span class="p">([[</span><span class="mi">5</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="c"># change diagonals to define axes of ellipse</span>
<span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">]])</span>
<span class="n">Q</span> <span class="o">=</span> <span class="n">U</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">S</span><span class="o">*</span><span class="n">U</span>
<span class="n">Pv</span> <span class="o">=</span> <span class="p">(</span><span class="n">v</span><span class="p">)</span><span class="o">*</span><span class="n">v</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="p">(</span><span class="n">Q</span><span class="p">)</span><span class="o">/</span><span class="p">(</span><span class="n">v</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="p">(</span><span class="n">Q</span><span class="p">)</span><span class="o">*</span><span class="n">v</span><span class="p">)</span> <span class="c"># projection operator</span>
<span class="n">err</span> <span class="o">=</span> <span class="n">sqrt</span><span class="p">((</span><span class="n">y</span><span class="o">-</span><span class="n">Pv</span><span class="o">*</span><span class="n">y</span><span class="p">)</span><span class="o">.</span><span class="n">T</span><span class="o">*</span><span class="n">Q</span><span class="o">*</span><span class="p">(</span><span class="n">y</span><span class="o">-</span><span class="n">Pv</span><span class="o">*</span><span class="n">y</span><span class="p">))</span> <span class="c"># error length</span>
<span class="n">xhat</span> <span class="o">=</span> <span class="n">Pv</span><span class="o">*</span><span class="n">y</span> <span class="c"># closest point on line</span>
<span class="n">fig</span><span class="p">,</span> <span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_figheight</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="o">*</span><span class="n">y</span><span class="p">,</span><span class="n">marker</span><span class="o">=</span><span class="s">'o'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">x</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="o">*</span><span class="p">(</span><span class="n">xhat</span><span class="p">),</span><span class="n">marker</span><span class="o">=</span><span class="s">'s'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_patch</span><span class="p">(</span> <span class="n">patches</span><span class="o">.</span><span class="n">Ellipse</span><span class="p">(</span><span class="n">y</span><span class="p">,</span><span class="n">err</span><span class="o">*</span><span class="mi">2</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">S</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]),</span><span class="n">err</span><span class="o">*</span><span class="mi">2</span><span class="o">/</span><span class="n">sqrt</span><span class="p">(</span><span class="n">S</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">]),</span>
<span class="n">angle</span><span class="o">=</span><span class="n">theta</span><span class="o">/</span><span class="n">pi</span><span class="o">*</span><span class="mi">180</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'pink'</span><span class="p">,</span>
<span class="n">alpha</span><span class="o">=</span><span class="mf">0.5</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_line</span><span class="p">(</span> <span class="n">Line2D</span><span class="p">(</span> <span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span> <span class="mi">0</span><span class="p">),</span> <span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="mi">0</span><span class="p">)</span> <span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">,</span><span class="n">linestyle</span><span class="o">=</span><span class="s">'--'</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">add_line</span><span class="p">(</span> <span class="n">Line2D</span><span class="p">(</span> <span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">xhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">]),</span>
<span class="p">(</span><span class="n">y</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">xhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">])</span> <span class="p">,</span>
<span class="n">color</span><span class="o">=</span><span class="s">'g'</span><span class="p">,</span><span class="n">linestyle</span><span class="o">=</span><span class="s">'--'</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span> <span class="s">'''Closest point is</span>
<span class="s">tangent to the </span>
<span class="s">ellipse and </span>
<span class="s">"perpendicular" </span>
<span class="s">to the line </span>
<span class="s">in the sense of the </span>
<span class="s">weighted/rotated</span>
<span class="s">distance</span>
<span class="s">'''</span><span class="p">,</span>
<span class="n">fontsize</span><span class="o">=</span><span class="mi">12</span><span class="p">,</span><span class="n">xy</span><span class="o">=</span><span class="p">(</span><span class="n">xhat</span><span class="p">[</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">],</span><span class="n">xhat</span><span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">]),</span>
<span class="n">xytext</span><span class="o">=</span><span class="p">(</span><span class="mf">3.5</span><span class="p">,</span><span class="mf">1.5</span><span class="p">),</span>
<span class="n">arrowprops</span><span class="o">=</span><span class="p">{</span><span class="s">'facecolor'</span><span class="p">:</span><span class="s">'blue'</span><span class="p">})</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">xmax</span><span class="o">=</span><span class="mi">6</span><span class="p">,</span><span class="n">ymax</span><span class="o">=</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_aspect</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
</pre></div>
</div>
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"></img>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>Note that the error vector ( $\mathbf{y}-\alpha\mathbf{v}$ ) is still perpendicular to the line (subspace $\mathbf{v}$), but it doesn't look it because we are using a weighted distance. The difference between the first projection ( with the uniform circular distance) and the general case ( with the ellipsoidal weighted distance ) is the inner product between the two cases. For example, in the first case we have $\mathbf{y}^T \mathbf{v}$ and in the weighted case we have $\mathbf{y}^T \mathbf{Q}^T \mathbf{v}$. To move from the uniform circular case to the weighted ellipsoidal case, all we had to do was change all of the vector inner products. This is a conceptual move we'll soon use again.</p>
<p>Before we finished, we will need a formal property of projections:</p>
<p>$$ \mathbf{P}_v \mathbf{P}_v = \mathbf{P}_v$$</p>
<p>known as the <em>idempotent</em> property which basically says that once we have projected onto a subspace, further subsequent projections leave us in the same subspace.</p>
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<h2>
Summary
</h2>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>In this section, we developed the concept of a projection operator, which ties a minimization problem (closest point to a line) to an algebraic concept (inner product). It turns out that these same geometric ideas from linear algebra can be translated to the conditional expectation. How this works is the subject of our next post.</p>
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<h3>
References
</h3>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>This post was created using the <a href="https://github.com/ipython/nbconvert">nbconvert</a> utility from the source <a href="www.ipython.org">IPython Notebook</a> which is available for <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Projection.ipynb">download</a> from the main github <a href="https://github.com/unpingco/Python-for-Signal-Processing">site</a> for this blog. The projection concept is masterfully discussed in the classic Strang, G. (2003). <em>Introduction to linear algebra</em>. Wellesley Cambridge Pr. Also, some of Dr. Strang's excellent lectures are available on <a href="http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/">MIT Courseware</a>. I highly recommend these as well as the book.</p>
</div>Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com2tag:blogger.com,1999:blog-2783842479260003750.post-40204043743501253182012-11-14T16:00:00.000-08:002012-11-14T16:00:03.890-08:00Conditional Expectation and Mean Squared Errors<div class="text_cell_render border-box-sizing rendered_html">
<p>There are lots of statistics in statistical signal processing, but to use statistics effectively with signals, it helps to have a certain unifying perspective on both. To introduce these ideas, let's start with the powerful and intimate connection between least mean-squared-error (MSE) problems and conditional expectation that is sadly not emphasized in most courses. </p>
<p>Let's start with an example: suppose we have two fair six-sided die ($X$ and $Y$) and I want to measure the sum of the two variables as $Z=X+Y$. Further, let's suppose that given $Z$, I want the best estimate of $X$ in the mean-squared-sense. Thus, I want to minimize the following:</p>
<p>$$ J(\alpha) = \sum ( x - \alpha z )^2 \mathbb{P}(x,z) $$</p>
<p>Here $\mathbb{P}$ encapsulates the density (i.e. mass) function for this problem. The idea is that when we have solved this problem, we will have a function of $Z$ that is going to be the minimum MSE estimate of $X$.</p>
<p>We can substitute in for $Z$ in $J$ and get:</p>
<p>$$ J(\alpha) = \sum ( x - \alpha (x+y) )^2 \mathbb{P}(x,y) $$</p>
<p>Let's work out the steps in <code>sympy</code> in the following:</p>
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<div class="input hbox">
<div class="prompt input_prompt">In [14]:</div>
<div class="input_area box-flex1">
<div class="highlight"><pre><span class="kn">import</span> <span class="nn">sympy</span>
<span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">stats</span><span class="p">,</span> <span class="n">simplify</span><span class="p">,</span> <span class="n">Rational</span><span class="p">,</span> <span class="n">Integer</span><span class="p">,</span><span class="n">Eq</span>
<span class="kn">from</span> <span class="nn">sympy.stats</span> <span class="kn">import</span> <span class="n">density</span><span class="p">,</span> <span class="n">E</span>
<span class="kn">from</span> <span class="nn">sympy.abc</span> <span class="kn">import</span> <span class="n">a</span>
<span class="n">x</span><span class="o">=</span><span class="n">stats</span><span class="o">.</span><span class="n">Die</span><span class="p">(</span><span class="s">'D1'</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span> <span class="c"># 1st six sided die</span>
<span class="n">y</span><span class="o">=</span><span class="n">stats</span><span class="o">.</span><span class="n">Die</span><span class="p">(</span><span class="s">'D2'</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span> <span class="c"># 2nd six sides die</span>
<span class="n">z</span> <span class="o">=</span> <span class="n">x</span><span class="o">+</span><span class="n">y</span> <span class="c"># sum of 1st and 2nd die</span>
<span class="n">J</span> <span class="o">=</span> <span class="n">stats</span><span class="o">.</span><span class="n">E</span><span class="p">((</span><span class="n">x</span> <span class="o">-</span> <span class="n">a</span><span class="o">*</span><span class="p">(</span><span class="n">x</span><span class="o">+</span><span class="n">y</span><span class="p">))</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span> <span class="c"># expectation</span>
<span class="n">sol</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">solve</span><span class="p">(</span><span class="n">sympy</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">J</span><span class="p">,</span><span class="n">a</span><span class="p">),</span><span class="n">a</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span> <span class="c"># using calculus to minimize</span>
<span class="k">print</span> <span class="n">sol</span> <span class="c"># solution is 1/2</span>
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<pre>1/2
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<p>This says that $z/2$ is the MSE estimate of $X$ given $Z$ which means geometrically ( interpreting the MSE as a squared distance weighted by the probability mass function) that $z/2$ is as <em>close</em> to $x$ as we are going to get for a given $z$.</p>
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<p>Let's look at the same problem using the conditional expectation operator $ \mathbb{E}(\cdot|z) $ and apply it to our definition of $Z$, then</p>
<p>$$ \mathbb{E}(z|z) = \mathbb{E}(x+y|z) = \mathbb{E}(x|z) + \mathbb{E}(y|z) =z $$</p>
<p>where we've used the linearity of the expectation. Now, since by the symmetry of the problem, we have </p>
<p>$$ \mathbb{E}(x|z) = \mathbb{E}(y|z) $$</p>
<p>we can plug this in and solve</p>
<p>$$ 2 \mathbb{E}(x|z) =z $$ </p>
<p>which gives</p>
<p>$$ \mathbb{E}(x|z) =\frac{z}{2} $$ </p>
<p>which is suspiciously equal to the MSE estimate we just found. This is not an accident! The proof of this is not hard, but let's look at some pictures first</p>
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<div class="highlight"><pre><span class="n">fig</span><span class="p">,</span> <span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">v</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)</span> <span class="o">+</span> <span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)[:,</span><span class="bp">None</span><span class="p">]</span>
<span class="n">foo</span><span class="o">=</span><span class="k">lambda</span> <span class="n">i</span><span class="p">:</span> <span class="n">density</span><span class="p">(</span><span class="n">z</span><span class="p">)[</span><span class="n">Integer</span><span class="p">(</span><span class="n">i</span><span class="p">)]</span><span class="o">.</span><span class="n">evalf</span><span class="p">()</span> <span class="c"># some tweaks to get a float out</span>
<span class="n">Zmass</span><span class="o">=</span><span class="n">array</span><span class="p">(</span><span class="nb">map</span><span class="p">(</span><span class="n">foo</span><span class="p">,</span><span class="n">v</span><span class="o">.</span><span class="n">flat</span><span class="p">),</span><span class="n">dtype</span><span class="o">=</span><span class="n">float32</span><span class="p">)</span><span class="o">.</span><span class="n">reshape</span><span class="p">(</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">pcolor</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">8</span><span class="p">),</span><span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">8</span><span class="p">),</span><span class="n">Zmass</span><span class="p">,</span><span class="n">cmap</span><span class="o">=</span><span class="n">cm</span><span class="o">.</span><span class="n">gray</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">([(</span><span class="n">i</span><span class="o">+</span><span class="mf">0.5</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticklabels</span><span class="p">([</span><span class="nb">str</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_yticks</span><span class="p">([(</span><span class="n">i</span><span class="o">+</span><span class="mf">0.5</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_yticklabels</span><span class="p">([</span><span class="nb">str</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">):</span>
<span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">):</span>
<span class="n">ax</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="n">i</span><span class="o">+.</span><span class="mi">5</span><span class="p">,</span><span class="n">j</span><span class="o">+.</span><span class="mi">5</span><span class="p">,</span><span class="nb">str</span><span class="p">(</span><span class="n">i</span><span class="o">+</span><span class="n">j</span><span class="p">),</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'y'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'Probability Mass for $Z$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'$X$ values'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$Y$ values'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">);</span>
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<p>The figure shows the values of $Z$ in yellow with the corresponding values for $X$ and $Y$ on the axes. Suppose $z=2$, then the closest $X$ to this is $X=1$, which is what $\mathbb{E}(x|z)=z/2=1$ gives. What's more interesting is what happens when $Z=7$? In this case, this value is spread out along the $X$ axis so if $X=1$, then $Z$ is 6 units away, if $X=2$, then $Z$ is 5 units away and so on.</p>
<p>Now, back to the original question, if we had $Z=7$ and I wanted to get as close as I could to this using $X$, then why not choose $X=6$ which is only one unit away from $Z$? The problem with doing that is $X=6$ only occurs 1/6 of the time, so I'm not likely to get it right the other 5/6 of the time. So, 1/6 of the time I'm one unit away but 5/6 of the time I'm much more than one unit away. This means that the MSE score is going to be worse. Since each value of $X$ from 1 to 6 is equally likely, to play it safe, I'm going to choose $7/2$ as my estimate, which is what the conditional expectation suggests.</p>
<p>We can check this claim with samples using <code>sympy</code> below:</p>
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<div class="highlight"><pre><span class="c">#generate samples conditioned on z=7</span>
<span class="n">samples_z7</span> <span class="o">=</span> <span class="k">lambda</span> <span class="p">:</span> <span class="n">stats</span><span class="o">.</span><span class="n">sample</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">sympy</span><span class="o">.</span><span class="n">Eq</span><span class="p">(</span><span class="n">z</span><span class="p">,</span><span class="mi">7</span><span class="p">))</span> <span class="c"># Eq constrains Z</span>
<span class="n">mn</span><span class="o">=</span> <span class="n">mean</span><span class="p">([(</span><span class="mi">6</span><span class="o">-</span><span class="n">samples_z7</span><span class="p">())</span><span class="o">**</span><span class="mi">2</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">100</span><span class="p">)])</span> <span class="c">#using 6 as an estimate</span>
<span class="n">mn0</span><span class="o">=</span> <span class="n">mean</span><span class="p">([(</span><span class="mi">7</span><span class="o">/</span><span class="mf">2.</span><span class="o">-</span><span class="n">samples_z7</span><span class="p">())</span><span class="o">**</span><span class="mi">2</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">100</span><span class="p">)])</span> <span class="c">#7/2 is the MSE estimate</span>
<span class="k">print</span> <span class="s">'MSE=</span><span class="si">%3.2f</span><span class="s"> using 6 vs MSE=</span><span class="si">%3.2f</span><span class="s"> using 7/2 '</span> <span class="o">%</span> <span class="p">(</span><span class="n">mn</span><span class="p">,</span><span class="n">mn0</span><span class="p">)</span>
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<pre>MSE=9.99 using 6 vs MSE=2.97 using 7/2
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<p>Please run the above code repeatedly until you have convinced yourself that the $\mathbb{E}(x|z)$ gives the lower MSE every time.</p>
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<p>To push this reasoning, let's consider the case where the die is so biased so that the outcome of <em>6</em> is ten times more probable than any of the other outcomes as in the following:</p>
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<div class="highlight"><pre><span class="c"># here 6 is ten times more probable than any other outcome</span>
<span class="n">x</span><span class="o">=</span><span class="n">stats</span><span class="o">.</span><span class="n">FiniteRV</span><span class="p">(</span><span class="s">'D3'</span><span class="p">,{</span><span class="mi">1</span><span class="p">:</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">15</span><span class="p">),</span> <span class="mi">2</span><span class="p">:</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">15</span><span class="p">),</span> <span class="mi">3</span><span class="p">:</span> <span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">15</span><span class="p">),</span>
<span class="mi">4</span><span class="p">:</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">15</span><span class="p">),</span> <span class="mi">5</span><span class="p">:</span><span class="n">Rational</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">15</span><span class="p">),</span> <span class="mi">6</span><span class="p">:</span> <span class="n">Rational</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">)})</span>
<span class="n">z</span> <span class="o">=</span> <span class="n">x</span> <span class="o">+</span> <span class="n">y</span>
<span class="c"># now re-create the plot</span>
<span class="n">fig</span><span class="p">,</span> <span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">foo</span><span class="o">=</span><span class="k">lambda</span> <span class="n">i</span><span class="p">:</span> <span class="n">density</span><span class="p">(</span><span class="n">z</span><span class="p">)[</span><span class="n">Integer</span><span class="p">(</span><span class="n">i</span><span class="p">)]</span><span class="o">.</span><span class="n">evalf</span><span class="p">()</span> <span class="c"># some tweaks to get a float out</span>
<span class="n">Zmass</span><span class="o">=</span><span class="n">array</span><span class="p">(</span><span class="nb">map</span><span class="p">(</span><span class="n">foo</span><span class="p">,</span><span class="n">v</span><span class="o">.</span><span class="n">flat</span><span class="p">),</span><span class="n">dtype</span><span class="o">=</span><span class="n">float32</span><span class="p">)</span><span class="o">.</span><span class="n">reshape</span><span class="p">(</span><span class="mi">6</span><span class="p">,</span><span class="mi">6</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">pcolor</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">8</span><span class="p">),</span><span class="n">arange</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">8</span><span class="p">),</span><span class="n">Zmass</span><span class="p">,</span><span class="n">cmap</span><span class="o">=</span><span class="n">cm</span><span class="o">.</span><span class="n">gray</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticks</span><span class="p">([(</span><span class="n">i</span><span class="o">+</span><span class="mf">0.5</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xticklabels</span><span class="p">([</span><span class="nb">str</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_yticks</span><span class="p">([(</span><span class="n">i</span><span class="o">+</span><span class="mf">0.5</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_yticklabels</span><span class="p">([</span><span class="nb">str</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">)])</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">):</span>
<span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">7</span><span class="p">):</span>
<span class="n">ax</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="n">i</span><span class="o">+.</span><span class="mi">5</span><span class="p">,</span><span class="n">j</span><span class="o">+.</span><span class="mi">5</span><span class="p">,</span><span class="nb">str</span><span class="p">(</span><span class="n">i</span><span class="o">+</span><span class="n">j</span><span class="p">),</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'y'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">r'Probability Mass for $Z$; Nonuniform case'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'$X$ values'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$Y$ values'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">);</span>
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<p>As compared with the first figure, the probability mass has been shifted away from the smaller numbers. Let's see what the conditional expectation says about how we can estimate $X$ from $Z$.</p>
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<div class="highlight"><pre><span class="n">E</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">Eq</span><span class="p">(</span><span class="n">z</span><span class="p">,</span><span class="mi">7</span><span class="p">))</span> <span class="c"># conditional expectation E(x|z=7)</span>
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<pre>5</pre>
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<p>Now that we have $\mathbb{E}(x|z=7) = 5$, we can generate samples as before and see if this gives the minimum MSE.</p>
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<div class="highlight"><pre><span class="c">#generate samples conditioned on z=7</span>
<span class="n">samples_z7</span> <span class="o">=</span> <span class="k">lambda</span> <span class="p">:</span> <span class="n">stats</span><span class="o">.</span><span class="n">sample</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">Eq</span><span class="p">(</span><span class="n">z</span><span class="p">,</span><span class="mi">7</span><span class="p">))</span> <span class="c"># Eq constrains Z</span>
<span class="n">mn</span><span class="o">=</span> <span class="n">mean</span><span class="p">([(</span><span class="mi">6</span><span class="o">-</span><span class="n">samples_z7</span><span class="p">())</span><span class="o">**</span><span class="mi">2</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">100</span><span class="p">)])</span> <span class="c">#using 6 as an estimate</span>
<span class="n">mn0</span><span class="o">=</span> <span class="n">mean</span><span class="p">([(</span><span class="mi">5</span><span class="o">-</span><span class="n">samples_z7</span><span class="p">())</span><span class="o">**</span><span class="mi">2</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">100</span><span class="p">)])</span> <span class="c">#7/2 is the MSE estimate</span>
<span class="k">print</span> <span class="s">'MSE=</span><span class="si">%3.2f</span><span class="s"> using 6 vs MSE=</span><span class="si">%3.2f</span><span class="s"> using 5 '</span> <span class="o">%</span> <span class="p">(</span><span class="n">mn</span><span class="p">,</span><span class="n">mn0</span><span class="p">)</span>
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<pre>MSE=3.27 using 6 vs MSE=2.92 using 5
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<h3>
Summary
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<p>Using a simple example, we have emphasized the connection between minimum mean squared error problems and conditional expectation. Next, we'll continue revealing the true power of the conditional expectation as we continue to develop a corresponding geometric intuition.</p>
<p>As usual, the corresponding ipython notebook for this post is available for download <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Conditional_expectation_MSE.ipynb">here</a>. </p>
<p>Comments and corrections welcome!</p>
</div>Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-30477469729127913632012-11-01T16:00:00.000-07:002012-11-07T12:31:23.336-08:00<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Expectation Maximization
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<p>Expectation Maximization (EM) is a powerful technique for creating maximum likelihood estimators when the variables are difficult to separate. in the following, we set up a Gaussian mixture experiment and derive the corresponding estimators using this technique.</p>
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<h3>
Experiment: Measuring from Unseen Groups
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<p>Let's investigate the following experiment: You have two distinct groups and you can randomly pick an individual from each group (you don't know from which) and then measure that individual's height. Group <strong>a</strong> is normally distributed as</p>
<p>$$ \mathcal{N}_a(x) =\mathcal{N}(x; \mu_a,\sigma) $$</p>
<p>and likewise for group <strong>b</strong></p>
<p>$$ \mathcal{N}_b(x) =\mathcal{N}(x; \mu_b,\sigma) $$</p>
<p>Note that the standard deviation, $\sigma$ is the same for both groups, but the means ($\mu_a,\mu_b$) are different. The problem is to estimate the means given that you can't directly know which group you are picking from.</p>
<p>Then we can write the joint density for this experiment as the following:</p>
<p>$$ f_{\mu_a,\mu_b}(x,z)= \frac{1}{2} \mathcal{N}_a(x) ^z \mathcal{N}_b(x) ^{1-z} $$</p>
<p>where $z=1$ if we pick from group <strong>a</strong> and $z=0$ for group <strong>b</strong>. Note that the $1/2$ comes from the 50/50 chance of picking either group. Unfortunately, since we do not measure the $z$ variable, we have to integrate it out of our density function to account for this handicap. Thus,</p>
<p>$$ f_{\mu_a,\mu_b}(x)= \frac{1}{2} \mathcal{N}_a(x)+\frac{1}{2} \mathcal{N}_b(x)$$</p>
<p>Now, since $n$ trials are independent, we can write out the likelihood:</p>
<p>$$ \mathcal{L}(\mu_a,\mu_b|\mathbf{x})= \prod_{i=1}^n f_{\mu_a,\mu_b}(x_i)$$</p>
<p>This is basically notation. We have just substituted everything into $ f_{\mu_a,\mu_b}(x)$ under the independent-trials assumption. Recal that the independent trials assumptions means that the joint probability is just the product of the individual probabilities. The idea of <em>maximum likelihood</em> is to maximize this as the function of $\mu_a$ and $\mu_b$ after plugging in all of the $x_i$ data. The problem is we don't know which group we are measuring at each trial so this is trickier than just estimating the parameters for each group separately.</p>
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Simulating the Experiment
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<p>We need the following code to setup the experiment of randomly a group and then picking an individual from that group.</p>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="kn">from</span> <span class="nn">scipy.stats</span> <span class="kn">import</span> <span class="n">bernoulli</span><span class="p">,</span> <span class="n">norm</span>
<span class="c">#np.random.seed(101) # set random seed for reproducibility</span>
<span class="n">mua_true</span><span class="o">=</span><span class="mi">4</span> <span class="c"># we are trying to estimate this from the data</span>
<span class="n">mub_true</span><span class="o">=</span><span class="mi">7</span> <span class="c"># we are trying to estimate this from the data</span>
<span class="n">fa</span><span class="o">=</span><span class="n">norm</span><span class="p">(</span><span class="n">mua_true</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span> <span class="c"># distribution for group A</span>
<span class="n">fb</span><span class="o">=</span><span class="n">norm</span><span class="p">(</span><span class="n">mub_true</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span> <span class="c"># distribution for group B</span>
<span class="n">fz</span><span class="o">=</span><span class="n">bernoulli</span><span class="p">(</span><span class="mf">0.5</span><span class="p">)</span> <span class="c"># each group equally likely </span>
<span class="k">def</span> <span class="nf">sample</span><span class="p">(</span><span class="n">n</span><span class="o">=</span><span class="mi">10</span><span class="p">):</span>
<span class="s">'simulate picking from each group n times'</span>
<span class="n">tmp</span><span class="o">=</span><span class="n">fz</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="c"># choose n of the coins, A or B</span>
<span class="k">return</span> <span class="n">tmp</span><span class="o">*</span><span class="p">(</span><span class="n">fb</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="n">n</span><span class="p">))</span><span class="o">+</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">tmp</span><span class="p">)</span><span class="o">*</span><span class="n">fa</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="c"># flip it n times</span>
<span class="n">xs</span> <span class="o">=</span> <span class="n">sample</span><span class="p">(</span><span class="mi">1000</span><span class="p">)</span> <span class="c"># generate some samples</span>
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<p>Here's a quick look at the density functions of each group and a histogram of the samples</p>
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<div class="highlight"><pre><span class="n">f</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="n">mua_true</span><span class="o">-</span><span class="mi">2</span><span class="p">,</span><span class="n">mub_true</span><span class="o">+</span><span class="mi">2</span><span class="p">,</span><span class="mi">100</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">fa</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">x</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">'group A'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">fb</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">x</span><span class="p">),</span><span class="n">label</span><span class="o">=</span><span class="s">'group B'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hist</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span><span class="n">bins</span><span class="o">=</span><span class="mi">50</span><span class="p">,</span><span class="n">normed</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span><span class="n">label</span><span class="o">=</span><span class="s">'Samples'</span><span class="p">);</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">(</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">);</span>
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<p>Just from looking at this plot, we suspect that we will have to reconcile the samples in the overlap region since these could have come from either group. This is where the <em>Expectation Maximization</em> algorithm enters.</p>
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<h2>
Expectation maximization
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<p>The key idea of expectation maximization is that we can somehow pretend we know the unobservable $z$ value and the proceed with the usual maximum likelihood estimation process.</p>
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<p>The idea behind expectation-maximization is that we want to use a maximum likelihood estimate (this is the <em>maximization</em> part of the algorithm) after computing the expectation over the missing variable (in this case, $z$). </p>
<p>The following code uses <code>sympy</code> to setup the functions symbolically and convert them to <code>numpy</code> functions that we can quickly evaluate. Because it's easier and more stable to evaluate, we will work with the <code>log</code> of the likelihood function. It is useful to keep track of the <em>incomplete log-likelihood</em> ($\log\mathcal{L}$) since it can be proved that it is monotone increasing and good way to identify coding errors. Recall that this was the likelihood in the case where we integrated out the $z$ variable to reconcile as its absence. </p>
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<div class="highlight"><pre><span class="kn">import</span> <span class="nn">sympy</span>
<span class="kn">from</span> <span class="nn">sympy.abc</span> <span class="kn">import</span> <span class="n">x</span><span class="p">,</span> <span class="n">z</span>
<span class="kn">from</span> <span class="nn">sympy</span> <span class="kn">import</span> <span class="n">stats</span>
<span class="n">mu_a</span><span class="p">,</span><span class="n">mu_b</span> <span class="o">=</span> <span class="n">sympy</span><span class="o">.</span><span class="n">symbols</span><span class="p">(</span><span class="s">'mu_a,mu_b'</span><span class="p">)</span>
<span class="n">na</span><span class="o">=</span><span class="n">stats</span><span class="o">.</span><span class="n">Normal</span><span class="p">(</span> <span class="s">'x'</span><span class="p">,</span> <span class="n">mu_a</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span>
<span class="n">nb</span><span class="o">=</span><span class="n">stats</span><span class="o">.</span><span class="n">Normal</span><span class="p">(</span> <span class="s">'x'</span><span class="p">,</span> <span class="n">mu_b</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span>
<span class="n">L</span><span class="o">=</span><span class="p">(</span><span class="n">stats</span><span class="o">.</span><span class="n">density</span><span class="p">(</span><span class="n">na</span><span class="p">)(</span><span class="n">x</span><span class="p">)</span><span class="o">+</span><span class="n">stats</span><span class="o">.</span><span class="n">density</span><span class="p">(</span><span class="n">nb</span><span class="p">)(</span><span class="n">x</span><span class="p">))</span><span class="o">/</span><span class="mi">2</span> <span class="c"># incomplete likelihood function </span>
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<p>Next, we need to compute the expectation step. To avoid notational overload, we will just use $\Theta$ to denote the $\mu_b$ and $\mu_a$ parameters and the data $x_i$. This means that the density function of $z$ and $\Theta$ can be written as the following:</p>
<p>$$ \mathbb{P}(z,\Theta) = \frac{1}{2} \mathcal{N}_a(\Theta) ^ z \mathcal{N}_b(\Theta) ^ {(1-z)} $$</p>
<p>For the expectation part we have to compute $\mathbb{E}(z|\Theta)$ but since $z\in \lbrace 0,1 \rbrace$, this simplifies easily</p>
<p>$$ \mathbb{E}(z|\Theta) = 1 \cdot \mathbb{P}(z=1|\Theta) + 0 \cdot \mathbb{P}(z=0|\Theta) = \mathbb{P}(z=1|\Theta) $$</p>
<p>Now, the only thing left is to find $ \mathbb{P}(z=1|\Theta) $ which we can do using Bayes rule:</p>
<p>$$ \mathbb{P}(z=1|\Theta) = \frac{ \mathbb{P}(\Theta|z=1)\mathbb{P}(z=1)}{\mathbb{P}(\Theta)} $$</p>
<p>The term in the denominator comes from summing (integrating) out the $z$ items in the full joint density $ \mathbb{P}(z,\Theta) $</p>
<p>$$ \mathbb{P}(\Theta) = (\mathcal{N}_a(\Theta) + \mathcal{N}_b(\Theta))\frac{1}{2} $$</p>
<p>and since $\mathbb{P}(z=1)=1/2$, we finally obtain</p>
<p>$$ \mathbb{E}(z|\Theta) =\mathbb{P}(z=1|\Theta) = \frac{\mathcal{N}_a(\Theta)}{\mathcal{N}_a(\Theta) + \mathcal{N}_b(\Theta)} $$</p>
<p>and which is coded below.</p>
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<div class="highlight"><pre><span class="k">def</span> <span class="nf">ez</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">mu_a</span><span class="p">,</span><span class="n">mu_b</span><span class="p">):</span> <span class="c"># expected value of hidden variable</span>
<span class="k">return</span> <span class="n">norm</span><span class="p">(</span><span class="n">mu_a</span><span class="p">)</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">/</span><span class="p">(</span><span class="n">norm</span><span class="p">(</span><span class="n">mu_a</span><span class="p">)</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">x</span><span class="p">)</span><span class="o">+</span> <span class="n">norm</span><span class="p">(</span><span class="n">mu_b</span><span class="p">)</span><span class="o">.</span><span class="n">pdf</span><span class="p">(</span><span class="n">x</span><span class="p">))</span>
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<p>Now, given we have this estimate for $z_i$, $\hat{z}_i=\mathbb{E(z|\Theta_i)}$, we can go back and compute the log likelihood estimate of</p>
<p>$$ J= \log\prod_{i=1}^n \mathbb{P}(\hat{z}_i,\Theta_i) = \sum_{i=1}^n \hat{z}_i\log \mathcal{N}_a(\Theta_i) +(1-\hat{z}_i)\log \mathcal{N}_b(\Theta_i) +\log(1/2) $$</p>
<p>by maximizing it using basic calculus. The trick is to remember that $\hat{z}_i$ is <em>fixed</em>, so we only have to maximize the $\log$ parts. This leads to</p>
<p>$$ \hat{\mu}_a = \frac{\sum_{i=1}^n \hat{z}_i x_i}{\sum_{i=1}^n \hat{z}_i } $$</p>
<p>and for $\mu_b$ </p>
<p>$$ \hat{\mu}_b = \frac{\sum_{i=1}^n (1-\hat{z}_i) x_i}{\sum_{i=1}^n 1-\hat{z}_i } $$</p>
<p>Now, we finally have the <em>maximization</em> step ( above ) and the <em>expectation</em> step ($\hat{z}_i$) from earlier. We're ready to simulate the algorithm and plot its performance!</p>
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<div class="highlight"><pre><span class="n">out</span><span class="o">=</span><span class="p">[];</span><span class="n">lout</span><span class="o">=</span><span class="p">[]</span> <span class="c"># containers for outputs</span>
<span class="n">Lf</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">lambdify</span><span class="p">((</span><span class="n">x</span><span class="p">,</span><span class="n">mu_a</span><span class="p">,</span><span class="n">mu_b</span><span class="p">),</span> <span class="n">sympy</span><span class="o">.</span><span class="n">log</span><span class="p">(</span><span class="nb">abs</span><span class="p">(</span><span class="n">L</span><span class="p">)),</span><span class="s">'numpy'</span><span class="p">)</span> <span class="c"># convert to numpy function from sympy</span>
<span class="n">mu_a_n</span><span class="o">=</span><span class="mi">2</span> <span class="c"># initial point</span>
<span class="n">mu_b_n</span><span class="o">=</span><span class="mi">1</span> <span class="c"># initial point</span>
<span class="n">niter</span><span class="o">=</span><span class="mi">10</span> <span class="c">#</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">niter</span><span class="p">):</span>
<span class="n">tau</span><span class="o">=</span><span class="n">ez</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span><span class="n">mu_a_n</span><span class="p">,</span><span class="n">mu_b_n</span><span class="p">)</span> <span class="c"># expected value of z-variable</span>
<span class="n">lout</span><span class="o">.</span><span class="n">append</span><span class="p">(</span> <span class="nb">sum</span><span class="p">(</span><span class="n">Lf</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span><span class="n">mu_a_n</span><span class="p">,</span><span class="n">mu_b_n</span><span class="p">)))</span> <span class="c"># track incomplete likelihood value (should be monotone)</span>
<span class="n">out</span><span class="o">.</span><span class="n">append</span><span class="p">((</span><span class="n">mu_a_n</span><span class="p">,</span><span class="n">mu_b_n</span><span class="p">))</span> <span class="c"># keep track of (pa,pb) steps</span>
<span class="n">mu_a_n</span><span class="o">=</span><span class="p">(</span><span class="nb">sum</span><span class="p">(</span><span class="n">tau</span><span class="o">*</span><span class="n">xs</span><span class="p">)</span><span class="o">/</span><span class="nb">sum</span><span class="p">(</span><span class="n">tau</span><span class="p">))</span> <span class="c"># new maximum likelihood estimate of pa</span>
<span class="n">mu_b_n</span><span class="o">=</span><span class="p">(</span><span class="nb">sum</span><span class="p">((</span><span class="mi">1</span><span class="o">-</span><span class="n">tau</span><span class="p">)</span><span class="o">*</span><span class="n">xs</span><span class="p">)</span><span class="o">/</span><span class="nb">sum</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">tau</span><span class="p">))</span>
<span class="n">fig</span><span class="o">=</span><span class="n">figure</span><span class="p">()</span>
<span class="n">fig</span><span class="o">.</span><span class="n">set_figwidth</span><span class="p">(</span><span class="mi">12</span><span class="p">)</span>
<span class="n">ax</span><span class="o">=</span><span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">121</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">array</span><span class="p">(</span><span class="n">out</span><span class="p">),</span><span class="s">'o-'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">legend</span><span class="p">((</span><span class="s">'mu_a'</span><span class="p">,</span><span class="s">'mu_b'</span><span class="p">),</span><span class="n">loc</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hlines</span><span class="p">([</span><span class="n">mua_true</span><span class="p">,</span><span class="n">mub_true</span><span class="p">],</span><span class="mi">0</span><span class="p">,</span><span class="nb">len</span><span class="p">(</span><span class="n">out</span><span class="p">),[</span><span class="s">'b'</span><span class="p">,</span><span class="s">'g'</span><span class="p">])</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'iteration'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$\mu_a,\mu_b$ values'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">)</span>
<span class="n">ax</span><span class="o">=</span><span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">122</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">array</span><span class="p">(</span><span class="n">lout</span><span class="p">),</span><span class="s">'o-'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'iteration'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Incomplete likelihood'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
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<p>The figure on the left shows the estimates for both $\mu_a$ and $\mu_b$ for each iteration and the figure on the right shows the corresponding incomplete likelihood function. The horizontal lines on the left-figure show the true values we are trying to estimate. Notice the EM algorithm converges very quickly, but because each group is equally likely to be chosen, the algorithm cannot distinguish one from the other. The code below constructs a error surface to see this effect. The incomplete likelihood function is monotone which tells us that we have not made a coding error. We're omitting the proof of this monotonicity.</p>
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<div class="prompt input_prompt">In [6]:</div>
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<div class="highlight"><pre><span class="n">mua_step</span><span class="o">=</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">10</span><span class="p">,</span><span class="mi">30</span><span class="p">)</span>
<span class="n">mub_step</span><span class="o">=</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">10</span><span class="p">,</span><span class="mi">20</span><span class="p">)</span>
<span class="n">z</span><span class="o">=</span><span class="n">Lf</span><span class="p">(</span><span class="n">xs</span><span class="p">,</span><span class="n">mua_step</span><span class="p">[:,</span><span class="bp">None</span><span class="p">],</span><span class="n">mub_step</span><span class="p">[:,</span><span class="bp">None</span><span class="p">,</span><span class="bp">None</span><span class="p">])</span><span class="o">.</span><span class="n">sum</span><span class="p">(</span><span class="n">axis</span><span class="o">=</span><span class="mi">2</span><span class="p">)</span> <span class="c"># numpy broadcasting</span>
<span class="n">fig</span><span class="o">=</span><span class="n">figure</span><span class="p">()</span>
<span class="n">ax</span><span class="o">=</span><span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span>
<span class="n">p</span><span class="o">=</span><span class="n">ax</span><span class="o">.</span><span class="n">contourf</span><span class="p">(</span><span class="n">mua_step</span><span class="p">,</span><span class="n">mub_step</span><span class="p">,</span><span class="n">z</span><span class="p">,</span><span class="mi">30</span><span class="p">,</span><span class="n">cmap</span><span class="o">=</span><span class="n">cm</span><span class="o">.</span><span class="n">gray</span><span class="p">)</span>
<span class="n">xa</span><span class="p">,</span><span class="n">xb</span><span class="o">=</span><span class="nb">zip</span><span class="p">(</span><span class="o">*</span><span class="n">out</span><span class="p">)</span> <span class="c"># unpack the container from the previous block</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">xa</span><span class="p">,</span><span class="n">xb</span><span class="p">,</span><span class="s">'ro'</span><span class="p">,</span><span class="n">mua_true</span><span class="p">,</span><span class="n">mub_true</span><span class="p">,</span><span class="s">'bs'</span><span class="p">)</span> <span class="c"># true values in blue</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">xa</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="n">xb</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="s">'gx'</span><span class="p">,</span><span class="n">ms</span><span class="o">=</span><span class="mf">15.</span><span class="p">,</span><span class="n">mew</span><span class="o">=</span><span class="mf">2.</span><span class="p">)</span> <span class="c"># starting point in green</span>
<span class="n">ax</span><span class="o">.</span><span class="n">text</span><span class="p">(</span><span class="n">xa</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="n">xb</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span><span class="s">'start'</span><span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'g'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mf">11.</span><span class="p">)</span> <span class="c"># points per iteration in red</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'$\mu_a$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'$\mu_b$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">24</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Incomplete Likelihood'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">fig</span><span class="o">.</span><span class="n">colorbar</span><span class="p">(</span><span class="n">p</span><span class="p">);</span>
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<p>The figure shows the incomplete likelihood function that the algorithm is exploring. Note that the algorithm can get to the maximizer but since the surface has symmetric maxima, it has no way to pick between them and ultimately just picks the one that is closest to the starting point. This is because each group is equally likely to be chosen. I urge you to download this notebook and try different initial points and see where the maximizer winds up.</p>
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<h2>
Summary
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<p>Expectation maximization is a powerful algorithm that is especially useful when it is difficult to de-couple the variables involved in a standard maximum likelihood estimation. Note that convergence to the "correct" maxima is not guaranteed, as we observed here. This is even more pronounced when there are more parameters to estimate. There is a nice <a href="http://www.cs.cmu.edu/~alad/em/">applet</a> you can use to investigate this effect and a much more detailed mathematical derivation <a href="http://crow.ee.washington.edu/people/bulyko/papers/em.pdf">here</a>.</p>
<p>As usual, the IPython notebook corresponding to this post can be found <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Expectation_Maximization.ipynb">here</a>. I urge you to try these calculations on your own. Try changing the sample size and making the choice between the two groups no longer equal to 1/2 (equally likely).<br />
</p>
<p>Note you will need at least <code>sympy</code> version 0.7.2 to run this notebook.</p>
<p>Comments appreciated!</p>
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Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-90226073864762761382012-10-24T16:00:00.000-07:002015-02-21T06:17:42.899-08:00<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Maximum Likelihood Estimation
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Maximum likelihood estimation is one of the key techniques employed in statistical signal processing for a wide variety of applications from signal detection to parameter estimation. In the following, we consider a simple experiment and work through the details of maximum likelihood estimation to ensure that we understand the concept in one of its simplest applications.<br />
<br />
<a href="http://nbviewer.ipython.org/github/unpingco/Python-for-Signal-Processing/blob/master/Maximum_likelihood.ipynb">Click here to see the math. Blogger broke it somehow.</a></div>
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<h3>
Setting up the Coin Flipping Experiment
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Suppose we have coin and want to estimate the probability of heads ($p$) for it. The coin is Bernoulli distributed:<br />
$$ \phi(x)= p^x (1-p)^{(1-x)} $$<br />
where $x$ is the outcome, <em>1</em> for heads and <em>0</em> for tails. The $n$ independent flips, we have the likelihood:<br />
$$ \mathcal{L}(p|\mathbf{x})= \prod_{i=1}^n p^{ x_i }(1-p)^{1-x_i} $$<br />
This is basically notation. We have just substituted everything into $ \phi(x)$ under the independent-trials assumption. <br />
The idea of <em>maximum likelihood</em> is to maximize this as the function of $p$ after plugging in all of the $x_i$ data. This means that our estimator, $\hat{p}$ , is a function of the observed $x_i$ data, and as such, is a random variable with its own distribution.</div>
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<h3>
Simulating the Experiment
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We need the following code to simulate coin flipping.</div>
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<pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span>
<span class="kn">from</span> <span class="nn">scipy.stats</span> <span class="kn">import</span> <span class="n">bernoulli</span>
<span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="n">p_true</span><span class="o">=</span><span class="mi">1</span><span class="o">/</span><span class="mi">2</span> <span class="c"># this is the value we will try to estimate from the observed data</span>
<span class="n">fp</span><span class="o">=</span><span class="n">bernoulli</span><span class="p">(</span><span class="n">p_true</span><span class="p">)</span>
<span class="k">def</span> <span class="nf">sample</span><span class="p">(</span><span class="n">n</span><span class="o">=</span><span class="mi">10</span><span class="p">):</span>
<span class="s">'simulate coin flipping'</span>
<span class="k">return</span> <span class="n">fp</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="n">n</span><span class="p">)</span><span class="c"># flip it n times</span>
<span class="n">xs</span> <span class="o">=</span> <span class="n">sample</span><span class="p">(</span><span class="mi">100</span><span class="p">)</span> <span class="c"># generate some samples</span>
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Now, we can write out the likelihood function using <code>sympy</code></div>
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<pre><span class="kn">import</span> <span class="nn">sympy</span>
<span class="kn">from</span> <span class="nn">sympy.abc</span> <span class="kn">import</span> <span class="n">x</span><span class="p">,</span> <span class="n">z</span>
<span class="n">p</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">symbols</span><span class="p">(</span><span class="s">'p'</span><span class="p">,</span><span class="n">positive</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">L</span><span class="o">=</span><span class="n">p</span><span class="o">**</span><span class="n">x</span><span class="o">*</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">p</span><span class="p">)</span><span class="o">**</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">x</span><span class="p">)</span>
<span class="n">J</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">prod</span><span class="p">([</span><span class="n">L</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">xs</span><span class="p">])</span> <span class="c"># objective function to maximize</span>
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Below, we find the maximum using basic calculus. Note that taking the <code>log</code> of $J$ makes the maximization problem tractable but doesn't change the extrema.</div>
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<pre><span class="n">logJ</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">expand_log</span><span class="p">(</span><span class="n">sympy</span><span class="o">.</span><span class="n">log</span><span class="p">(</span><span class="n">J</span><span class="p">))</span>
<span class="n">sol</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">solve</span><span class="p">(</span><span class="n">sympy</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">logJ</span><span class="p">,</span><span class="n">p</span><span class="p">),</span><span class="n">p</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span>
<span class="n">x</span><span class="o">=</span><span class="n">linspace</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">100</span><span class="p">)</span>
<span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="nb">map</span><span class="p">(</span><span class="n">sympy</span><span class="o">.</span><span class="n">lambdify</span><span class="p">(</span><span class="n">p</span><span class="p">,</span><span class="n">logJ</span><span class="p">,</span><span class="s">'numpy'</span><span class="p">),</span><span class="n">x</span><span class="p">),</span><span class="n">sol</span><span class="p">,</span><span class="n">logJ</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">p</span><span class="p">,</span><span class="n">sol</span><span class="p">),</span><span class="s">'o'</span><span class="p">,</span>
<span class="n">p_true</span><span class="p">,</span><span class="n">logJ</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">p</span><span class="p">,</span><span class="n">p_true</span><span class="p">),</span><span class="s">'s'</span><span class="p">,)</span>
<span class="n">xlabel</span><span class="p">(</span><span class="s">'$p$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ylabel</span><span class="p">(</span><span class="s">'Likelihood'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'Estimate not equal to true value'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
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Note that our estimator $\hat{p}$ (red circle) is not equal to the true value of $p$ (green square), but it is at the maximum of the likelihood function. This may sound disturbing, but keep in mind this estimate is a function of the random data; and since that data can change, the ultimate estimate can likewise change. I invite you to run this notebook a few times to observe this. Remember that the estimator is a <em>function</em> of the data and is thus also a <em>random variable</em>, just like the data is. <br />
Let's write some code to empirically examine the behavior of the maximum likelihood estimator using a simulation of multiple trials. All we're doing here is combining the last few blocks of code.</div>
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<pre><span class="k">def</span> <span class="nf">estimator_gen</span><span class="p">(</span><span class="n">niter</span><span class="o">=</span><span class="mi">10</span><span class="p">,</span><span class="n">ns</span><span class="o">=</span><span class="mi">100</span><span class="p">):</span>
<span class="s">'generate data to estimate distribution of maximum likelihood estimator'</span>
<span class="n">out</span><span class="o">=</span><span class="p">[]</span>
<span class="n">x</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">symbols</span><span class="p">(</span><span class="s">'x'</span><span class="p">,</span><span class="n">real</span><span class="o">=</span><span class="bp">True</span><span class="p">)</span>
<span class="n">L</span><span class="o">=</span> <span class="n">p</span><span class="o">**</span><span class="n">x</span><span class="o">*</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">p</span><span class="p">)</span><span class="o">**</span><span class="p">(</span><span class="mi">1</span><span class="o">-</span><span class="n">x</span><span class="p">)</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">niter</span><span class="p">):</span>
<span class="n">xs</span> <span class="o">=</span> <span class="n">sample</span><span class="p">(</span><span class="n">ns</span><span class="p">)</span> <span class="c"># generate some samples from the experiment</span>
<span class="n">J</span><span class="o">=</span><span class="n">np</span><span class="o">.</span><span class="n">prod</span><span class="p">([</span><span class="n">L</span><span class="o">.</span><span class="n">subs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">xs</span><span class="p">])</span> <span class="c"># objective function to maximize</span>
<span class="n">logJ</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">expand_log</span><span class="p">(</span><span class="n">sympy</span><span class="o">.</span><span class="n">log</span><span class="p">(</span><span class="n">J</span><span class="p">))</span>
<span class="n">sol</span><span class="o">=</span><span class="n">sympy</span><span class="o">.</span><span class="n">solve</span><span class="p">(</span><span class="n">sympy</span><span class="o">.</span><span class="n">diff</span><span class="p">(</span><span class="n">logJ</span><span class="p">,</span><span class="n">p</span><span class="p">),</span><span class="n">p</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span>
<span class="n">out</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="nb">float</span><span class="p">(</span><span class="n">sol</span><span class="o">.</span><span class="n">evalf</span><span class="p">()))</span>
<span class="k">return</span> <span class="n">out</span> <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">out</span><span class="p">)</span><span class="o">></span><span class="mi">1</span> <span class="k">else</span> <span class="n">out</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="c"># return scalar if list contains only 1 term</span>
<span class="n">etries</span> <span class="o">=</span> <span class="n">estimator_gen</span><span class="p">(</span><span class="mi">100</span><span class="p">)</span> <span class="c"># this may take awhile, depending on how much data you want to generate</span>
<span class="n">hist</span><span class="p">(</span><span class="n">etries</span><span class="p">)</span> <span class="c"># histogram of maximum likelihood estimator</span>
<span class="n">title</span><span class="p">(</span><span class="s">'$\mu=</span><span class="si">%3.3f</span><span class="s">,\sigma=</span><span class="si">%3.3f</span><span class="s">$'</span><span class="o">%</span><span class="p">(</span><span class="n">mean</span><span class="p">(</span><span class="n">etries</span><span class="p">),</span><span class="n">std</span><span class="p">(</span><span class="n">etries</span><span class="p">)),</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
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" />
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
Note that the mean of the estimator ($\mu$) is pretty close to the true value, but looks can be deceiving. The only way to know for sure is to check if the estimator is unbiased, namely, if<br />
$$ \mathbb{E}(\hat{p}) = p $$</div>
<div class="text_cell_render border-box-sizing rendered_html">
Because this problem is simple, we can solve for this in general noting that since $x=0$ or $x=1$, the terms in the product of $\mathcal{L}$ above are either $p$, if $x_i=1$ or $1-p$ if $x_i=0$. This means that we can write<br />
$$ \mathcal{L}(p|\mathbf{x})= p^{\sum_{i=1}^n x_i}(1-p)^{n-\sum_{i=1}^n x_i} $$<br />
with corresponding log as<br />
$$ J=\log(\mathcal{L}(p|\mathbf{x})) = \log(p) \sum_{i=1}^n x_i + \log(1-p) \left(n-\sum_{i=1}^n x_i\right)$$ <br />
Taking the derivative of this gives:<br />
$$ \frac{dJ}{dp} = \frac{1}{p}\sum_{i=1}^n x_i + \frac{(n-\sum_{i=1}^n x_i)}{p-1} $$<br />
and solving this leads to<br />
$$ \hat{p} = \frac{1}{ n} \sum_{i=1}^n x_i $$<br />
This is our <em>estimator</em> for $p$. Up til now, we have been using <code>sympy</code> to solve for this based on the data $x_i$ but now we have it generally and don't have to solve for it again. To check if this estimator is biased, we compute its expectation:<br />
$$ \mathbb{E}\left(\hat{p}\right) =\frac{1}{n}\sum_i^n \mathbb{E}(x_i) = \frac{1}{n} n \mathbb{E}(x_i) $$<br />
by linearity of the expectation and where<br />
$$\mathbb{E}(x_i) = p$$<br />
Therefore,<br />
$$ \mathbb{E}\left(\hat{p}\right) =p $$<br />
This means that the esimator is unbiased. This is good news. We almost always want our estimators to be unbiased. Similarly, <br />
$$ \mathbb{E}\left(\hat{p}^2\right) = \frac{1}{n^2} \mathbb{E}\left[\left( \sum_{i=1}^n x_i \right)^2 \right]$$<br />
and where<br />
$$ \mathbb{E}\left(x_i^2\right) =p$$<br />
and by the independence assumption,<br />
$$ \mathbb{E}\left(x_i x_j\right) =\mathbb{E}(x_i)\mathbb{E}( x_j) =p^2$$<br />
Thus,<br />
$$ \mathbb{E}\left(\hat{p}^2\right) =\left(\frac{1}{n^2}\right) n
\left[
p+(n-1)p^2
\right]
$$<br />
So, the variance of the estimator, $\hat{p}$ is the following:<br />
$$ \sigma_\hat{p}^2 = \mathbb{E}\left(\hat{p}^2\right)- \mathbb{E}\left(\hat{p}\right)^2 = \frac{p(1-p)}{n} $$<br />
Note that the $n$ in the denominator means that the variance asymptotically goes to zero as $n$ increases (i.e. we consider more and more samples). This is good news also because it means that more and more coin flips leads to a better estimate of the underlying $p$.<br />
Unfortunately, this formula for the variance is practically useless because we have to know $p$ to compute it and $p$ is the parameter we are trying to estimate in the first place! But, looking at $ \sigma_\hat{p}^2 $, we can immediately notice that if $p=0$, then there is no estimator variance because the outcomes are guaranteed to be tails. Also, the maximum of this variance, for whatever $n$, happens at $p=1/2$. This is our worst case scenario and the only way to compensate is with more samples (i.e. larger $n$). </div>
<div class="text_cell_render border-box-sizing rendered_html">
All we have computed is the mean and variance of the estimator. In general, this is insufficient to characterize the underlying probability density of $\hat{p}$, except if we somehow knew that $\hat{p}$ were normally distributed. This is where the powerful <a href="http://mathworld.wolfram.com/CentralLimitTheorem.html"><em>central limit theorem</em></a> comes in. The form of the estimator, which is just a mean estimator, implies that we can apply this theorem and conclude that $\hat{p}$ is normally distributed. However, there's a wrinkle here: the theorem tells us that $\hat{p}$ is asymptotically normal, it doesn't quantify how many samples $n$ we need to approach this asymptotic paradise. In our simulation this is no problem since we can generate as much data as we like, but in the real world, with a costly experiment, each sample may be precious. In the following, we won't apply this theorem and instead proceed analytically.</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h3>
Probability Density for the Estimator
</h3>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
To write out the full density for $\hat{p}$, we first have to ask what is the probability that the estimator will equal a specific value and the tally up all the ways that could happen with their corresponding probabilities. For example, what is the probability that<br />
$$ \hat{p} = \frac{1}{n}\sum_{i=1}^n x_i = 0 $$<br />
This can only happen one way: when $x_i=0 \hspace{0.5em} \forall i$. The probability of this happening can be computed from the density<br />
$$ f(\mathbf{x},p)= \prod_{i=1}^n \left(p^{x_i} (1-p)^{1-x_i} \right) $$<br />
$$ f\left(\sum_{i=1}^n x_i = 0,p\right)= \left(1-p\right)^n $$<br />
Likewise, if $\lbrace x_i \rbrace$ has one $i^{th}$ value equal to one, then<br />
$$ f\left(\sum_{i=1}^n x_i = 1,p\right)= n p \prod_{i=1}^{n-1} \left(1-p\right)$$<br />
where the $n$ comes from the $n$ ways to pick one value equal to one from the $n$ elements $x_i$. Continuing this way, we can construct the entire density as<br />
$$ f\left(\sum_{i=1}^n x_i = k,p\right)= \binom{n}{k} p^k (1-p)^{n-k} $$<br />
where the term on the left is the binomial coefficient of $n$ things taken $k$ at a time. This is the binomial distribution and it's not the density for $\hat{p}$, but rather for $n\hat{p}$. We'll leave this as-is because it's easier to work with below. We just have to remember to keep track of the $n$ factor.</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h4>
Confidence Intervals
</h4>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
Now that we have the full density for $\hat{p}$, we are ready to ask some meaningful questions. For example,<br />
$$ \mathbb{P}\left( | \hat{p}-p | \le \epsilon p \right) $$<br />
Or, in words, what is the probability we can get within $\epsilon$ percent of the true value of $p$. Rewriting,<br />
$$ \mathbb{P}\left( p - \epsilon p \lt \hat{p} \lt p + \epsilon p \right) = \mathbb{P}\left( n p - n \epsilon p \lt \sum_{i=1}^n x_i \lt n p + n \epsilon p \right)$$<br />
Let's plug in some live numbers here for our worst case scenario where $p=1/2$. Then, if $\epsilon = 1/100$, we have<br />
$$ \mathbb{P}\left( \frac{99 n}{100} \lt \sum_{i=1}^n x_i \lt \frac{101 n}{100} \right)$$<br />
Since the sum in integer-valued, we need $n> 100$ to even compute this. Thus, if $n=101$ we have<br />
$$ \mathbb{P}\left( \frac{9999}{200} \lt \sum_{i=1}^{101} x_i \lt \frac{10201}{200} \right) = f\left(\sum_{i=1}^{101} x_i = 50,p\right)= \binom{101}{50} (1/2)^{50} (1-1/2)^{101-50} = 0.079$$<br />
This means that in the worst-case scenario for $p=1/2$, given $n=101$ trials, we will only get within 1% of the actual $p=1/2$ about 8% of the time. If you feel disappointed, that only means you've been paying attention. What if the coin was really heavy and it was costly to repeat this 101 times? Then, we would be within 1% of the actual value only 8% of the time. Those odds are terrible.<br />
Let's come at this another way: given I could only flip the coin 100 times, how close could I come to the true underlying value with high probability (say, 95%)? In this case we are seeking to solve for $\epsilon$. Plugging in gives,<br />
$$ \mathbb{P}\left( 50 - 50 \epsilon \lt \sum_{i=1}^{100} x_i \lt 50 + 50 \epsilon \right) = 0.95$$<br />
which we have to solve for $\epsilon$. Fortunately, all the tools we need to solve for this are already in <code>scipy</code>.</div>
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<pre><span class="kn">import</span> <span class="nn">scipy.stats</span>
<span class="n">b</span><span class="o">=</span><span class="n">scipy</span><span class="o">.</span><span class="n">stats</span><span class="o">.</span><span class="n">binom</span><span class="p">(</span><span class="mi">100</span><span class="p">,</span><span class="o">.</span><span class="mi">5</span><span class="p">)</span> <span class="c"># n=100, p = 0.5, distribution of the estimator \hat{p}</span>
<span class="n">f</span><span class="p">,</span><span class="n">ax</span><span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">stem</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">101</span><span class="p">),</span><span class="n">b</span><span class="o">.</span><span class="n">pmf</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">101</span><span class="p">)))</span> <span class="c"># heres the density of the sum of x_i</span>
<span class="n">g</span> <span class="o">=</span> <span class="k">lambda</span> <span class="n">i</span><span class="p">:</span><span class="n">b</span><span class="o">.</span><span class="n">pmf</span><span class="p">(</span><span class="n">arange</span><span class="p">(</span><span class="o">-</span><span class="n">i</span><span class="p">,</span><span class="n">i</span><span class="p">)</span><span class="o">+</span><span class="mi">50</span><span class="p">)</span><span class="o">.</span><span class="n">sum</span><span class="p">()</span> <span class="c"># symmetric sum the probability around the mean</span>
<span class="k">print</span> <span class="s">'this is pretty close to 0.95:</span><span class="si">%r</span><span class="s">'</span><span class="o">%</span><span class="k">g</span><span class="p">(</span><span class="mi">10</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">vlines</span><span class="p">(</span> <span class="p">[</span><span class="mi">50</span><span class="o">+</span><span class="mi">10</span><span class="p">,</span><span class="mi">50</span><span class="o">-</span><span class="mi">10</span><span class="p">],</span><span class="mi">0</span> <span class="p">,</span><span class="n">ax</span><span class="o">.</span><span class="n">get_ylim</span><span class="p">()[</span><span class="mi">1</span><span class="p">]</span> <span class="p">,</span><span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">,</span><span class="n">lw</span><span class="o">=</span><span class="mf">3.</span><span class="p">)</span>
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<pre>this is pretty close to 0.95:0.95395593307064808
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<pre><matplotlib.collections.LineCollection at 0x93d9570></pre>
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The two vertical lines in the plot show how far out from the mean we have to go to accumulate 95% of the probability. Now, we can solve this as<br />
$$ 50 + 50 \epsilon = 60 $$<br />
which makes $\epsilon=1/5$ or 20%. So, flipping 100 times means I can only get within 20% of the real $p$ 95% of the time in the worst case scenario (i.e. $p=1/2$).</div>
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<pre><span class="n">b</span><span class="o">=</span><span class="n">scipy</span><span class="o">.</span><span class="n">stats</span><span class="o">.</span><span class="n">bernoulli</span><span class="p">(</span><span class="o">.</span><span class="mi">5</span><span class="p">)</span> <span class="c"># coin distribution</span>
<span class="n">xs</span> <span class="o">=</span> <span class="n">b</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="mi">100</span><span class="p">)</span> <span class="c"># flip it 100 times</span>
<span class="n">phat</span> <span class="o">=</span> <span class="n">mean</span><span class="p">(</span><span class="n">xs</span><span class="p">)</span> <span class="c"># estimated p</span>
<span class="k">print</span> <span class="nb">abs</span><span class="p">(</span><span class="n">phat</span><span class="o">-</span><span class="mf">0.5</span><span class="p">)</span> <span class="o"><</span> <span class="mf">0.5</span><span class="o">*</span><span class="mf">0.20</span> <span class="c"># did I make it w/in interval 95% of the time?</span>
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<pre>True
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Let's keep doing this and see if we can get within this interval 95% of the time.</div>
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<pre><span class="n">out</span><span class="o">=</span><span class="p">[]</span>
<span class="n">b</span><span class="o">=</span><span class="n">scipy</span><span class="o">.</span><span class="n">stats</span><span class="o">.</span><span class="n">bernoulli</span><span class="p">(</span><span class="o">.</span><span class="mi">5</span><span class="p">)</span> <span class="c"># coin distribution</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">500</span><span class="p">):</span> <span class="c"># number of tries</span>
<span class="n">xs</span> <span class="o">=</span> <span class="n">b</span><span class="o">.</span><span class="n">rvs</span><span class="p">(</span><span class="mi">100</span><span class="p">)</span> <span class="c"># flip it 100 times</span>
<span class="n">phat</span> <span class="o">=</span> <span class="n">mean</span><span class="p">(</span><span class="n">xs</span><span class="p">)</span> <span class="c"># estimated p</span>
<span class="n">out</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="nb">abs</span><span class="p">(</span><span class="n">phat</span><span class="o">-</span><span class="mf">0.5</span><span class="p">)</span> <span class="o"><</span> <span class="mf">0.5</span><span class="o">*</span><span class="mf">0.20</span> <span class="p">)</span> <span class="c"># within 20% </span>
<span class="k">print</span> <span class="s">'Percentage of tries within 20 interval = </span><span class="si">%3.2f</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="mi">100</span><span class="o">*</span><span class="nb">sum</span><span class="p">(</span><span class="n">out</span><span class="p">)</span><span class="o">/</span><span class="nb">float</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">out</span><span class="p">)</span> <span class="p">))</span>
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<pre>Percentage of tries within 20 interval = 96.20
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Well, that seems to work. Now we have a way to get at the quality of the estimator, $\hat{p}$. </div>
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Summary
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In this section, we explored the concept of maximum likelihood estimation using a coin flipping experiment both analytically and numerically with the scientific Python tool chain. There are two key points to remember. First, maximum likelihood estimation produces a function of the data that is itself a random variable, with its own statistics and distribution. Second, it's worth considering how to analytically derive the density function of the estimator rather than relying on canned packages to compute confidence intervals wherever possible. This is especially true when data is hard to come by and the approximations made in the central limit theorem are therefore harder to justify.</div>
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References
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This <a href="https://www.blogger.com/www.ipython.org">IPython notebook</a> is available for <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Maximum_likelihood.ipynb">download</a>. I urge you to experiment with the calculations for different parameters. As always, corrections and comments are welcome!</div>
Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-19442460458841326242012-09-20T19:30:00.000-07:002012-09-20T19:30:00.396-07:00Investigating the Sampling Theorem Part Two<div class="text_cell_render border-box-sizing rendered_html">
<p>We left off with the disturbing realization that even though we are satisfied the requirements of the sampling theorem, we still have errors in our approximating formula. We can resolve this by examining the Whittaker interpolating functions which are used to reconstruct the signal from its samples.</p>
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<div class="highlight"><pre><span class="kn">from</span> <span class="nn">__future__</span> <span class="kn">import</span> <span class="n">division</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">5</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">300</span><span class="p">)</span> <span class="c"># redefine this here for convenience</span>
<span class="n">fig</span> <span class="o">=</span> <span class="n">figure</span><span class="p">()</span>
<span class="n">fs</span><span class="o">=</span><span class="mf">5.0</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span> <span class="c"># create axis handle</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span><span class="n">fs</span> <span class="o">*</span> <span class="n">t</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span><span class="s">'This keeps going...'</span><span class="p">,</span>
<span class="n">xy</span><span class="o">=</span><span class="p">(</span><span class="o">-</span><span class="mi">4</span><span class="p">,</span><span class="mi">0</span><span class="p">),</span>
<span class="n">xytext</span><span class="o">=</span><span class="p">(</span><span class="o">-</span><span class="mi">5</span><span class="o">+.</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span><span class="o">=</span><span class="p">{</span><span class="s">'facecolor'</span><span class="p">:</span><span class="s">'green'</span><span class="p">,</span><span class="s">'shrink'</span><span class="p">:</span><span class="mf">0.05</span><span class="p">},</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span><span class="s">'... and going...'</span><span class="p">,</span>
<span class="n">xy</span><span class="o">=</span><span class="p">(</span><span class="mi">4</span><span class="p">,</span><span class="mi">0</span><span class="p">),</span>
<span class="n">xytext</span><span class="o">=</span><span class="p">(</span><span class="mi">3</span><span class="o">+.</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span><span class="o">=</span><span class="p">{</span><span class="s">'facecolor'</span><span class="p">:</span><span class="s">'green'</span><span class="p">,</span><span class="s">'shrink'</span><span class="p">:</span><span class="mf">0.05</span><span class="p">},</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
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<pre><matplotlib.text.Annotation at 0x2dda950></pre>
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<p>Notice in the above plot that the function extends to infinity in either direction. This basically means that the signals we can represent must also extend to infinity in either direction which then means that we have to sample forever to exactly reconstruct the signal! So, on the one hand the sampling theorem says we only need a sparse density of samples, this result says we need to sample forever. No free lunch here!</p>
<p>This is a deep consequence of dealing with band-limited functions which, as we have just demonstrated, are <strong>not</strong> time-limited. Now, the new question is how to get these signals into a computer with finite memory. How can we use what we have learned about the sampling theorem with these finite-duration signals?</p>
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<h2>
Approximately Time-Limited Functions
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<p>Let's back off a bit and settle for functions that are <em>approximately</em> time-limited in the sense that almost all of their energy is concentrated in a finite time-window:</p>
<p>$ \int_{-\tau}^\tau |f(t)|^2 dt = E-\epsilon$</p>
<p>where $E$ is the total energy of the signal:</p>
<p>$ \int_{-\infty}^\infty |f(t)|^2 dt = E$</p>
<p>Now, with this new definition, we can seek out functions that are band-limited but come very, very (i.e. within $\epsilon$) close to being time-limited as well. In other words, we want functions $\phi(t)$ so that they are band-limited:</p>
<p>$ \phi(t) = \int_{-W}^W \Phi(\nu) e^{2 \pi j \nu t} dt $ </p>
<p>and coincidentally maximize the following:</p>
<p>$ \int_{-\tau}^\tau |\phi(t) |^2 dt$</p>
<p>After a lot of complicated math that I'm skipping, this eventually boils down to solving the following normalized eigenvalue equation:</p>
<p>$ \int_{-1}^1 \psi(x)\frac{sin(2\pi\beta(t-x)}{\pi(t-x)} dx = \lambda \psi(t)$</p>
<p>with $\psi(t) = \phi(\tau t )$ and $\beta=\tau W$. Note that $\beta$ is proportional to the time-bandwith product. Fortunately, this is a classic problem and the $\psi$ functions turn out to be the angular prolate spheroidal wave functions. Also, it turns out that the $\phi$ functions are orthonormal on the real-line:</p>
<p>$ \int_{-\infty}^\infty \phi_k(t) \phi_n(t)^* dt = \delta_{k,n} $</p>
<p>and <em>orthogonal</em> on the finite interval $[-\tau,\tau]$</p>
<p>$ \int_{-\tau}^\tau \phi_k(t) \phi_n(t)^* dt = \lambda_k \delta_{k,n} $</p>
<p>This is a lot to digest at one sitting, but all we have to do here is watch the largest eigenvalue because it represents the fraction of energy contained in the interval $[-\tau,\tau]$. In particular, we want to track the largest eigenvalue as a function of the time-bandwidth product because this will tell us for a fixed $2W$ bandwidth, how large a time-extent we must sample through in order to approximately represent signal in terms of these angular prolate spheroidal functions.</p>
<p>Since we have the Python-based tools at our disposal to motivate this investigation, let's look into the eigenvalues and eigenvectors of this form of the equation:</p>
<p>$ \int_{-1}^1 \psi(x)\frac{sin(2\pi\sigma(t-x)/4)}{\pi(t-x)} dx = \lambda \psi(t)$</p>
<p>because it is neatly parametered by the time-bandwidth product $\sigma = (2\tau)(2W)$.</p>
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<div class="highlight"><pre><span class="k">def</span> <span class="nf">kernel</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">sigma</span><span class="o">=</span><span class="mi">1</span><span class="p">):</span>
<span class="s">'convenient function to compute kernel of eigenvalue problem'</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">asanyarray</span><span class="p">(</span><span class="n">x</span><span class="p">)</span>
<span class="n">y</span> <span class="o">=</span> <span class="n">pi</span><span class="o">*</span><span class="n">where</span><span class="p">(</span><span class="n">x</span> <span class="o">==</span> <span class="mi">0</span><span class="p">,</span><span class="mf">1.0e-20</span><span class="p">,</span> <span class="n">x</span><span class="p">)</span>
<span class="k">return</span> <span class="n">sin</span><span class="p">(</span><span class="n">sigma</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">y</span><span class="p">)</span><span class="o">/</span><span class="n">y</span>
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<p>Now, we are ready to setup the eigenvalues and see how they change with the time-bandwidth product.</p>
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<div class="highlight"><pre><span class="n">nstep</span><span class="o">=</span><span class="mi">100</span> <span class="c"># quick and dirty integral quantization</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">nstep</span><span class="p">)</span> <span class="c"># quantization of time</span>
<span class="n">dt</span> <span class="o">=</span> <span class="n">diff</span><span class="p">(</span><span class="n">t</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span> <span class="c"># differential step size</span>
<span class="k">def</span> <span class="nf">eigv</span><span class="p">(</span><span class="n">sigma</span><span class="p">):</span>
<span class="k">return</span> <span class="n">eigvalsh</span><span class="p">(</span><span class="n">kernel</span><span class="p">(</span><span class="n">t</span><span class="o">-</span><span class="n">t</span><span class="p">[:,</span><span class="bp">None</span><span class="p">],</span><span class="n">sigma</span><span class="p">))</span><span class="o">.</span><span class="n">max</span><span class="p">()</span> <span class="c"># compute max eigenvalue</span>
<span class="n">sigma</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="mf">0.01</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">15</span><span class="p">)</span> <span class="c"># range of time-bandwidth products to consider</span>
<span class="n">fig</span><span class="o">=</span><span class="n">figure</span><span class="p">()</span>
<span class="n">ax</span><span class="o">=</span><span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">sigma</span><span class="p">,</span> <span class="n">dt</span><span class="o">*</span><span class="n">array</span><span class="p">([</span><span class="n">eigv</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="n">sigma</span><span class="p">]),</span><span class="s">'-o'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'time-bandwidth product $\sigma$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'max eigenvalue'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">ymax</span><span class="o">=</span><span class="mf">1.01</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">grid</span><span class="p">()</span>
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<p>Keep in mind that this cannot actually <em>achieve</em> a maximum of one since that would mean that it <em>is</em> possible to have both time/band-limited functions (which is impossible). But, note that if you numerically achieve this than it has to do with the crude approximation to the integral that we are using.</p>
<p>The important part of this graph is how steep the curve is. Namely, if we have a time-bandwidth product of $3$ or more, then we have already found an eigenfunction (or, vector, in our case) that is compressed <em>almost</em> entirely into the interval $[-1,1]$. Now let's look at the corresponding eigenvector.</p>
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<div class="highlight"><pre><span class="n">sigma</span><span class="o">=</span><span class="mi">3</span>
<span class="n">w</span><span class="p">,</span><span class="n">v</span><span class="o">=</span><span class="n">eigh</span><span class="p">(</span><span class="n">kernel</span><span class="p">(</span><span class="n">t</span><span class="o">-</span><span class="n">t</span><span class="p">[:,</span><span class="bp">None</span><span class="p">],</span><span class="n">sigma</span><span class="p">))</span>
<span class="n">maxv</span><span class="o">=</span><span class="n">v</span><span class="p">[:,</span> <span class="n">w</span><span class="o">.</span><span class="n">argmax</span><span class="p">()]</span>
<span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">maxv</span><span class="p">)</span>
<span class="n">xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'Eigenvector corresponding to e-value=</span><span class="si">%2.4e</span><span class="s">;$\sigma$=</span><span class="si">%3.2f</span><span class="s">'</span><span class="o">%</span><span class="p">(</span><span class="n">w</span><span class="o">.</span><span class="n">max</span><span class="p">()</span><span class="o">*</span><span class="n">dt</span><span class="p">,</span><span class="n">sigma</span><span class="p">))</span>
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<pre><matplotlib.text.Text at 0x30d95d0></pre>
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<p>In fact, since the angular prolate spheroidal wave functions are in <code>scipy.special</code>, we can check our crude result against a better numerical solution</p>
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<div class="highlight"><pre><span class="kn">import</span> <span class="nn">scipy.special</span>
<span class="c"># quick definition to clean up the function signature and normalize</span>
<span class="k">def</span> <span class="nf">pro</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">k</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span><span class="n">sigma</span><span class="o">=</span><span class="mi">3</span><span class="p">):</span>
<span class="s">'normalized prolate angular spherioidal wave function wrapper'</span>
<span class="n">tmp</span><span class="o">=</span> <span class="n">scipy</span><span class="o">.</span><span class="n">special</span><span class="o">.</span><span class="n">pro_ang1</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="n">k</span><span class="p">,</span><span class="n">pi</span><span class="o">/</span><span class="mi">2</span><span class="o">*</span><span class="n">sigma</span><span class="p">,</span><span class="n">t</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span> <span class="c">#kth prolate function</span>
<span class="n">den</span><span class="o">=</span><span class="n">linalg</span><span class="o">.</span><span class="n">norm</span><span class="p">(</span><span class="n">tmp</span><span class="p">[</span><span class="n">np</span><span class="o">.</span><span class="n">logical_not</span><span class="p">(</span> <span class="n">np</span><span class="o">.</span><span class="n">isnan</span><span class="p">(</span><span class="n">tmp</span><span class="p">))])</span><span class="c"># drop those pesky NaNs at edges</span>
<span class="k">return</span> <span class="n">tmp</span><span class="o">/</span><span class="n">den</span> <span class="c"># since e-vectors are likewise normalized</span>
<span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">pro</span><span class="p">(</span><span class="n">t</span><span class="p">),</span><span class="s">'b.'</span><span class="p">,</span><span class="n">t</span><span class="p">,</span><span class="n">maxv</span><span class="p">,</span><span class="s">'-g'</span><span class="p">);</span>
<span class="n">xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'Comparison of $0^{th}$ prolate function'</span><span class="p">)</span>
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<pre><matplotlib.text.Text at 0x3349ab0></pre>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>Since these two lines are basically on top of each other, it implies that our extremely crude approximation to this function is actually not bad. You can play with the <code>nstep</code> variable above to see how the discrete time step influences this result.</p>
<p>What does this all mean?</p>
<p>The key fact we have demonstrated here is that when the time-bandwith product is relatively large ($\sigma>3$), then all of the signal energy in our <em>approximately</em> time-limited functions is concentrated in the zero-th prolate function. This means that any approximately time-limited function can be expanded into the basis of prolate functions in the following way:</p>
<p>$ f(t) = \sum_k a_k \psi_k(t) $</p>
<p>where $a_0 >> a_k$ when the time-bandwidth product is bigger than three, $\sigma>3$.</p>
<p>What we have been exploring here is formally the result of the Landau-Pollak theorem which says that if we take $\sigma+1$ prolate functions, given a fixed $\epsilon$, then we can then reconstruct the signal with a mean-squared error ($\mathbb{L}_2$-norm) less than $12\epsilon$. Notwithstanding what the theorem says, we still must control $\epsilon$ here by expanding the time-interval over which we consider the signal. This means that the time-extent, or, duration, of the signal is what is improving the approximation. The prolate functions are a <em>means</em> to getting under this bound. The choice of the duration fixes $\epsilon$.</p>
<p>Let's test this bound in Python by considering our favorite band-limited function: $X(f)=1$ if $|f| < W $ and zero otherwise.</p>
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<div class="prompt input_prompt">In [6]:</div>
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<div class="highlight"><pre><span class="n">W</span><span class="o">=</span><span class="mi">1</span><span class="o">/</span><span class="mi">2</span> <span class="c"># take bandwidth = 1 Hz. This makes the total signal power=1</span>
<span class="c"># compute epsilon for this time-extent</span>
<span class="n">epsilon</span><span class="o">=</span><span class="mi">1</span> <span class="o">-</span> <span class="mi">2</span><span class="o">/</span><span class="n">pi</span><span class="o">*</span><span class="n">scipy</span><span class="o">.</span><span class="n">special</span><span class="o">.</span><span class="n">sici</span><span class="p">(</span> <span class="mi">2</span><span class="o">*</span><span class="n">pi</span> <span class="p">)[</span><span class="mi">0</span><span class="p">]</span> <span class="c"># using sin-integral special function</span>
<span class="n">fig</span><span class="o">=</span><span class="n">figure</span><span class="p">()</span>
<span class="n">ax</span><span class="o">=</span><span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">100</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span><span class="n">t</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hlines</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="o">-</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">)</span>
<span class="n">tt</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">20</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">fill_between</span><span class="p">(</span><span class="n">tt</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span><span class="n">tt</span><span class="p">)</span><span class="o">**</span><span class="mi">2</span><span class="p">,</span><span class="n">facecolor</span><span class="o">=</span><span class="s">'g'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'$x(t)^2$: $\epsilon=</span><span class="si">%2.4f</span><span class="s">$ over shaded region'</span> <span class="o">%</span> <span class="n">epsilon</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">)</span>
</pre></div>
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<p>The theorem says that </p>
<p>$ \int_{-\infty}^\infty | x(t) - \sum_{k=0}^{\sigma+1} a_k \psi_k(t) |^2 dt < 12 \epsilon $</p>
<p>which computes to 0.0972*12 = 1.1664 in our case, which is not saying much since the total energy of the signal was equal to 1 and this means that the error is on the same order as the signal itself! Let's try another band-limited function:</p>
<p>$X(f) = (1+f)$ if $ f \in [-1,0]$</p>
<p>$X(f) = (1-f)$ if $ f \in [0,1]$</p>
<p>and zero otherwise. This corresponds to the following time-domain function</p>
<p>$x(t)=\frac{sin(\pi t)^2}{\pi^2 t^2}$</p>
<p>Let' quickly plot this:</p>
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<div class="highlight"><pre><span class="n">epsilon</span><span class="o">=</span> <span class="mi">2</span><span class="o">/</span><span class="mf">3.</span> <span class="o">+</span> <span class="mi">4</span><span class="o">/</span><span class="mi">3</span><span class="o">/</span><span class="n">pi</span><span class="o">*</span><span class="n">scipy</span><span class="o">.</span><span class="n">special</span><span class="o">.</span><span class="n">sici</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span><span class="o">-</span><span class="mi">8</span><span class="o">/</span><span class="mi">3</span><span class="o">/</span><span class="n">pi</span><span class="o">*</span><span class="n">scipy</span><span class="o">.</span><span class="n">special</span><span class="o">.</span><span class="n">sici</span><span class="p">(</span><span class="mi">4</span><span class="o">*</span><span class="n">pi</span><span class="p">)[</span><span class="mi">0</span><span class="p">]</span> <span class="c"># total energy is 2/3 in this case</span>
<span class="n">fig</span><span class="o">=</span><span class="n">figure</span><span class="p">()</span>
<span class="n">ax</span><span class="o">=</span><span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">100</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span><span class="n">t</span><span class="p">)</span><span class="o">**</span><span class="mi">4</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hlines</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="o">-</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">)</span>
<span class="n">tt</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">20</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">fill_between</span><span class="p">(</span><span class="n">tt</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span><span class="n">tt</span><span class="p">)</span><span class="o">**</span><span class="mi">4</span><span class="p">,</span><span class="n">facecolor</span><span class="o">=</span><span class="s">'g'</span><span class="p">,</span><span class="n">alpha</span><span class="o">=</span><span class="mf">0.5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'$x(t)^2$: $\epsilon=</span><span class="si">%2.4f</span><span class="s">$ over shaded region'</span> <span class="o">%</span> <span class="n">epsilon</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">)</span>
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<pre><matplotlib.text.Text at 0x2ec3e90></pre>
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<p>And in this case, $12\epsilon=0.0236$ which is a lot better, as we would expect since $x(t)$ is much more tightly concentrated in the interval $[-1,1]$.</p>
<p>What have we learned? If the signal is concentrated in the interval, then we don't need a large duration over which to sample it. The Landau-Pollak theorem provides specific criterion for what can expect by using prolate spheroidal wave functions that are constructed to optimize their energies in a fixed interval. The key take-away from all this is that we want large duration-bandwidth products, $ duration >> 1/bandwidth $ so that we can be assured of collecting enough signal energy in the duration.</p>
<p>Under the condition or large time-bandwidth products, let's reconsider our eigenvalue problem but now let's separate the $\tau$ and $W$ terms as in the following:</p>
<p>$ \int_{-\tau}^\tau \phi(x)\frac{sin(2\pi W (t-x)}{\pi(t-x)} dx = \lambda \phi(t)$</p>
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<div class="prompt input_prompt">In [40]:</div>
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<div class="highlight"><pre><span class="k">def</span> <span class="nf">kernel_tau</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">W</span><span class="o">=</span><span class="mi">1</span><span class="p">):</span>
<span class="s">'convenient function to compute kernel of eigenvalue problem'</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">asanyarray</span><span class="p">(</span><span class="n">x</span><span class="p">)</span>
<span class="n">y</span> <span class="o">=</span> <span class="n">pi</span><span class="o">*</span><span class="n">where</span><span class="p">(</span><span class="n">x</span> <span class="o">==</span> <span class="mi">0</span><span class="p">,</span><span class="mf">1.0e-20</span><span class="p">,</span> <span class="n">x</span><span class="p">)</span>
<span class="k">return</span> <span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">W</span><span class="o">*</span><span class="n">y</span><span class="p">)</span><span class="o">/</span><span class="n">y</span>
<span class="n">nstep</span><span class="o">=</span><span class="mi">300</span> <span class="c"># quick and dirty integral quantization</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="n">nstep</span><span class="p">)</span> <span class="c"># quantization of time</span>
<span class="n">tt</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">2</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="n">nstep</span><span class="p">)</span><span class="c"># extend interval</span>
<span class="n">w</span><span class="p">,</span><span class="n">v</span><span class="o">=</span><span class="n">eig</span><span class="p">(</span><span class="n">kernel_tau</span><span class="p">(</span><span class="n">t</span><span class="o">-</span><span class="n">tt</span><span class="p">[:,</span><span class="bp">None</span><span class="p">],</span><span class="mi">5</span><span class="p">))</span>
<span class="n">ii</span> <span class="o">=</span> <span class="n">argsort</span><span class="p">(</span><span class="n">w</span><span class="o">.</span><span class="n">real</span><span class="p">)</span>
<span class="n">maxv</span><span class="o">=</span><span class="n">v</span><span class="p">[:,</span> <span class="n">w</span><span class="o">.</span><span class="n">real</span><span class="o">.</span><span class="n">argmax</span><span class="p">()]</span><span class="o">.</span><span class="n">real</span>
<span class="n">plot</span><span class="p">(</span><span class="n">tt</span><span class="p">,</span><span class="n">maxv</span><span class="p">)</span>
<span class="c">##plot(tt,v[:,ii[-2]].real)</span>
<span class="n">xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'$\phi_{max}(t),\sigma=10*2=20$'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">16</span><span class="p">)</span>
</pre></div>
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YII=
"></img>
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>This turns out to be the <code>sinc</code> function. This means that when the time-bandwidth product is large, the prolate spheroidal Wave Functions devolve into <code>sinc</code> functions in the eigenvalue problem. This means that under these conditions, it is okay to use the Whittaker interpolating functions as we used previously.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>
Summary
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>In this section, at first blush, it may look like we accomplished nothing. We started out investigating why is it that we have some residual error in the reconstruction formula using the Whittaker approximation functions. Then, we recognized that we cannot have signals that are simultaneously time-limited and band-limited. This realization drove us to investigate "approximately" time-limited functions. Through carefully examining the resulting eigenvalue problem, we realized that the Landau-Pollak theorem provides criterion for finite-dimensionality of band-limited signals in terms of approximately time-limited optimal prolate spheroidal wave functions. However, the real profit of these results in our discussion is that we require large time-bandwidth products in order to reduce our approximation errors and under these conditions, we can legitimately use the same Whittaker interpolating functions that we started out with in the first place! Ultimately, if we want to reduce our reconstruction errors, then we just have to sample over a longer duration since capturing the signal energy in the time domain is based on accumulating more of the signal over a longer duration.</p>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<h2>
References
</h2>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<ul>
<li>
<p>This is in the <a href="http://ipython.org/">IPython Notebook format</a> and was converted to HTML using <a href="https://github.com/ipython/nbconvert">nbconvert</a>.</p>
</li>
<li>
<p>See <a href="http://books.google.com/books?id=Re5SAAAAMAAJ">Signal analysis</a> for more detailed mathematical development.</p>
</li>
<li>
<p>The IPython notebook corresponding to this post can be found <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Sampling_Theorem_Part_2.ipynb">here</a>.</p>
</li>
</ul>
</div>Anonymoushttp://www.blogger.com/profile/11211805900015667837noreply@blogger.com0tag:blogger.com,1999:blog-2783842479260003750.post-38421878094462455912012-09-10T17:15:00.001-07:002013-02-21T11:56:54.499-08:00<div class="text_cell_render border-box-sizing rendered_html">
<h1>
Investigating The Sampling Theorem
</h1>
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<p>In this section, we investigate the implications of the sampling theorem. Here is the usual statement of the theorem from wikipedia:</p>
<p><em>"If a function $x(t)$ contains no frequencies higher than B hertz, it is completely determined by giving its ordinates at a series of points spaced 1/(2B) seconds apart."</em></p>
<p>Since a function $x(t)$ is a function from the real line to the real line, there are uncountably many points between any two ordinates, so sampling is a massive reduction of data since it only takes a tiny number of points to completely characterize the function. This is a powerful idea worth exploring. In fact, we have seen this idea of reducing a function to a discrete set of numbers before in Fourier series expansions where (for periodic $x(t)$) </p>
<p>$ a_n = \frac{1}{T} \int^{T}_0 x(t) \exp (-j \omega_n t )dt $</p>
<p>with corresponding reconstruction as:</p>
<p>$ x(t) = \sum_k a_n \exp( j \omega_n t) $</p>
<p>But here we are generating discrete points $a_n$ by integrating over the <strong>entire</strong> function $x(t)$, not just evaluating it at a single point. This means we are collecting information about the entire function to compute a single discrete point $a_n$, whereas with sampling we are just taking individual points in isolation.</p>
<p>Let's come at this the other way: suppose we are given a set of samples $[x_1,x_2,..,x_N]$ and we are then told to reconstruct the function. What would we do? This is the kind of question seldom asked because we typically sample, filter, and then do something else without trying to reconstruct the function from the samples directly.</p>
<p>Returning to our reconstruction challenge, perhaps the most natural thing to do is draw a straight line between each of the points as in linear interpolation. The next block of code takes samples of the $sin$ over a single period and draws a line between sampled ordinates.</p>
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<div class="prompt input_prompt">In [1]:</div>
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<div class="highlight"><pre><span class="n">fig</span><span class="p">,</span><span class="n">ax</span> <span class="o">=</span> <span class="n">subplots</span><span class="p">()</span>
<span class="n">f</span> <span class="o">=</span> <span class="mf">1.0</span> <span class="c"># Hz, signal frequency</span>
<span class="n">fs</span> <span class="o">=</span> <span class="mf">5.0</span> <span class="c"># Hz, sampling rate (ie. >= 2*f) </span>
<span class="n">t</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="o">+</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="p">,</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="p">)</span> <span class="c"># sample interval, symmetric for convenience later</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">t</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">'o-'</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'amplitude'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">show</span><span class="p">()</span>
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"></img>
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<div class="text_cell_render border-box-sizing rendered_html">
<p>In this plot, notice how near the extremes of the $sin$ at $t=1/(4f)$ and $t=3/(4 f)$, we are taking the same density of points since the sampling theorem makes no requirement on <em>where</em> we should sample as long as we sample at a regular intervals. This means that on the up and down slopes of the $sin$, which are obviously linear-looking and where a linear approximation is a good one, we are taking the same density of samples as near the curvy peaks. Here's a bit of code that zooms in to the first peak to illustrate this.</p>
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<div class="prompt input_prompt">In [2]:</div>
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<div class="highlight"><pre><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">x</span><span class="p">,</span><span class="s">'o-'</span><span class="p">)</span>
<span class="n">axis</span><span class="p">(</span> <span class="n">xmin</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="p">(</span><span class="mi">4</span><span class="o">*</span><span class="n">f</span><span class="p">)</span><span class="o">-</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="o">*</span><span class="mi">3</span><span class="p">,</span> <span class="n">xmax</span> <span class="o">=</span> <span class="mi">1</span><span class="o">/</span><span class="p">(</span><span class="mi">4</span><span class="o">*</span><span class="n">f</span><span class="p">)</span><span class="o">+</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="o">*</span><span class="mi">3</span><span class="p">,</span> <span class="n">ymin</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="n">ymax</span> <span class="o">=</span> <span class="mf">1.1</span> <span class="p">)</span>
<span class="n">show</span><span class="p">()</span>
</pre></div>
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<p>To drive this point home (and create some cool matplotlib plots), we can construct the piecewise linear interpolant and compare the quality of the approximation using <code>numpy.piecewise</code>:</p>
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<div class="highlight"><pre><span class="n">interval</span><span class="o">=</span><span class="p">[]</span> <span class="c"># piecewise domains</span>
<span class="n">apprx</span> <span class="o">=</span> <span class="p">[]</span> <span class="c"># line on domains</span>
<span class="c"># build up points *evenly* inside of intervals</span>
<span class="n">tp</span> <span class="o">=</span> <span class="n">hstack</span><span class="p">([</span> <span class="n">linspace</span><span class="p">(</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="p">],</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">],</span><span class="mi">20</span><span class="p">,</span><span class="bp">False</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">t</span><span class="p">)</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span> <span class="p">])</span>
<span class="c"># construct arguments for piecewise2</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">t</span><span class="p">)</span><span class="o">-</span><span class="mi">1</span><span class="p">):</span>
<span class="n">interval</span><span class="o">.</span><span class="n">append</span><span class="p">(</span> <span class="n">np</span><span class="o">.</span><span class="n">logical_and</span><span class="p">(</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o"><=</span> <span class="n">tp</span><span class="p">,</span><span class="n">tp</span> <span class="o"><</span> <span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]))</span>
<span class="n">apprx</span><span class="o">.</span><span class="n">append</span><span class="p">(</span> <span class="p">(</span><span class="n">x</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span><span class="o">-</span><span class="n">x</span><span class="p">[</span><span class="n">i</span><span class="p">])</span><span class="o">/</span><span class="p">(</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span><span class="o">-</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="p">])</span><span class="o">*</span><span class="p">(</span><span class="n">tp</span><span class="p">[</span><span class="n">interval</span><span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">]]</span><span class="o">-</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="p">])</span> <span class="o">+</span> <span class="n">x</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
<span class="n">x_hat</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">piecewise</span><span class="p">(</span><span class="n">tp</span><span class="p">,</span><span class="n">interval</span><span class="p">,</span><span class="n">apprx</span><span class="p">)</span> <span class="c"># piecewise linear approximation</span>
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<p>Now, we can examine the squared errors in the interpolant. The following snippet plots the $sin$ and with the filled-in error of the linear interpolant.</p>
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<div class="highlight"><pre><span class="n">ax1</span> <span class="o">=</span> <span class="n">figure</span><span class="p">()</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">fill_between</span><span class="p">(</span><span class="n">tp</span><span class="p">,</span><span class="n">x_hat</span><span class="p">,</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">tp</span><span class="p">),</span><span class="n">facecolor</span><span class="o">=</span><span class="s">'red'</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'Amplitude'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax2</span> <span class="o">=</span> <span class="n">ax1</span><span class="o">.</span><span class="n">twinx</span><span class="p">()</span>
<span class="n">sqe</span> <span class="o">=</span> <span class="p">(</span> <span class="n">x_hat</span> <span class="o">-</span> <span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">tp</span><span class="p">))</span><span class="o">**</span><span class="mi">2</span>
<span class="n">ax2</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">tp</span><span class="p">,</span> <span class="n">sqe</span><span class="p">,</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">ax2</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">xmin</span><span class="o">=-</span><span class="mi">1</span><span class="p">,</span><span class="n">ymax</span><span class="o">=</span> <span class="n">sqe</span><span class="o">.</span><span class="n">max</span><span class="p">()</span> <span class="p">)</span>
<span class="n">ax2</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'squared error'</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Errors with Piecewise Linear Interpolant'</span><span class="p">)</span>
<span class="n">show</span><span class="p">()</span>
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<p>Note: I urge you to change the $fs$ sampling rate in the code above then rerun this notebook to see how these errors change with more/less sampling points.</p>
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<p>Now, we could pursue this line of reasoning with higher-order polynomials instead of just straight lines, but this would all eventually take us to the same conclusion; namely, that all of these approximations improve as the density of sample points increases, which is the <em>exact</em> opposite of what the sampling theorem says --- there is <em>sparse</em> set of samples points that will retrieve the original function. Furthermore, we observed that the quality of the piecewise linear interpolation is sensitive to <em>where</em> the sample points are taken and the sampling theorem is so powerful that it <em>has no such requirement</em>. </p>
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<h2>
Reconstruction
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<p>Let's look at this another way by examing the Fourier Transform of a signal that is bandlimited and thus certainly satisfies the hypothesis of the sampling theorem:</p>
<p>$ X(f) = 0$ where $ |f|> W $</p>
<p>Now, the inverse Fourier transform of this is the following:</p>
<p>$ x(t) = \int_{-W}^W X(f) e^{j 2 \pi f t} df $</p>
<p>We can take the $X(f)$ and expand it into a Fourier series by pretending that it is periodic with period $2 W$. Thus, we can formally write the following:</p>
<p>$$ X(f) = \sum_k a_k e^{ - j 2 \pi k f/(2 W) } $$</p>
<p>we can compute the coefficients $a_k$ as </p>
<p>$$ a_k = \frac{1}{2 W} \int_{-W}^W X(f) e^{ j 2 \pi k f/(2 W) } df $$</p>
<p>These coefficients bear a striking similarity to the $x(t)$ integral we just computed above. In fact, by lining up terms, we can write:</p>
<p>$$ a_k = \frac{1}{2 W} x \left( t = \frac{k}{2 W} \right) $$</p>
<p>Now, we can write out $X(f)$ in terms of this series and these $a_k$ and then invert the Fourier transform to obtain the following:</p>
<p>$$ x(t) = \int_{-W}^W \sum_k a_k e^{ - j 2 \pi k f/(2 W) } e^{j 2 \pi f t} df $$</p>
<p>substitute for $a_k$</p>
<p>$$ x(t) = \int_{-W}^W \sum_k ( \frac{1}{2 W} x( t = \frac{k}{2 W} ) ) e^{ - j 2 \pi k f/(2 W) } e^{j 2 \pi f t} df $$</p>
<p>switch summation and integration (usually dangerous, but OK here)</p>
<p>$$ x(t) = \sum_k x(t = \frac{k}{2 W}) \frac{1}{2 W} \int_{-W}^W e^{ - j 2 \pi k f/(2 W) +j 2 \pi f t} df $$</p>
<p>which gives finally:</p>
<p>$$ x(t) = \sum_k x(t = \frac{k}{2 W}) \frac{sin(\pi (k-2 t W))} {\pi (k- 2 t W)} $$</p>
<p>And this what we have been seeking! A formula that reconstructs the function from it's samples. Let's try it!</p>
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<p>Note that since our samples are spaced at $t= k/f_s $, we'll use $ W= f_s /2 $ to line things up.</p>
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<div class="prompt input_prompt">In [5]:</div>
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<div class="highlight"><pre><span class="n">t</span> <span class="o">=</span> <span class="n">linspace</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">100</span><span class="p">)</span> <span class="c"># redefine this here for convenience</span>
<span class="n">ts</span> <span class="o">=</span> <span class="n">arange</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="o">+</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="p">,</span><span class="mi">1</span><span class="o">/</span><span class="n">fs</span><span class="p">)</span> <span class="c"># sample points</span>
<span class="n">num_coeffs</span><span class="o">=</span><span class="nb">len</span><span class="p">(</span><span class="n">ts</span><span class="p">)</span>
<span class="n">sm</span><span class="o">=</span><span class="mi">0</span>
<span class="k">for</span> <span class="n">k</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="o">-</span><span class="n">num_coeffs</span><span class="p">,</span><span class="n">num_coeffs</span><span class="p">):</span> <span class="c"># since function is real, need both sides</span>
<span class="n">sm</span><span class="o">+=</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="p">(</span><span class="n">k</span><span class="o">/</span><span class="n">fs</span><span class="p">))</span><span class="o">*</span><span class="n">sinc</span><span class="p">(</span> <span class="n">k</span> <span class="o">-</span> <span class="n">fs</span> <span class="o">*</span> <span class="n">t</span><span class="p">)</span>
<span class="n">close</span><span class="p">(</span><span class="s">'all'</span><span class="p">)</span>
<span class="n">plot</span><span class="p">(</span> <span class="n">t</span><span class="p">,</span><span class="n">sm</span><span class="p">,</span><span class="s">'--'</span><span class="p">,</span><span class="n">t</span><span class="p">,</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">t</span><span class="p">),</span><span class="n">ts</span><span class="p">,</span> <span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">ts</span><span class="p">),</span><span class="s">'o'</span><span class="p">)</span>
<span class="n">title</span><span class="p">(</span><span class="s">'sampling rate=</span><span class="si">%3.2f</span><span class="s"> Hz'</span> <span class="o">%</span> <span class="n">fs</span> <span class="p">)</span>
<span class="n">show</span><span class="p">()</span>
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<p>We can do the same check as we did for the linear interpolant above as</p>
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<div class="prompt input_prompt">In [6]:</div>
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<div class="highlight"><pre><span class="n">ax1</span> <span class="o">=</span> <span class="n">figure</span><span class="p">()</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">fill_between</span><span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">sm</span><span class="p">,</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">t</span><span class="p">),</span><span class="n">facecolor</span><span class="o">=</span><span class="s">'red'</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'Amplitude'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax2</span> <span class="o">=</span> <span class="n">ax1</span><span class="o">.</span><span class="n">twinx</span><span class="p">()</span>
<span class="n">sqe</span> <span class="o">=</span> <span class="p">(</span><span class="n">sm</span> <span class="o">-</span> <span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">f</span><span class="o">*</span><span class="n">t</span><span class="p">))</span><span class="o">**</span><span class="mi">2</span>
<span class="n">ax2</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,</span> <span class="n">sqe</span><span class="p">,</span><span class="s">'r'</span><span class="p">)</span>
<span class="n">ax2</span><span class="o">.</span><span class="n">axis</span><span class="p">(</span><span class="n">xmin</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span><span class="n">ymax</span> <span class="o">=</span> <span class="n">sqe</span><span class="o">.</span><span class="n">max</span><span class="p">())</span>
<span class="n">ax2</span><span class="o">.</span><span class="n">set_ylabel</span><span class="p">(</span><span class="s">'squared error'</span><span class="p">,</span> <span class="n">color</span><span class="o">=</span><span class="s">'r'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">18</span><span class="p">)</span>
<span class="n">ax1</span><span class="o">.</span><span class="n">set_title</span><span class="p">(</span><span class="s">'Errors with sinc Interpolant'</span><span class="p">)</span>
<span class="n">show</span><span class="p">()</span>
</pre></div>
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"></img>
</div>
</div>
</div>
</div>
</div>
<div class="text_cell_render border-box-sizing rendered_html">
<p>These interpolating functions are called the "Whittaker" interpolating functions. Let's examine these functions more closely with the following code</p>
</div>
<div class="cell border-box-sizing code_cell vbox">
<div class="input hbox">
<div class="prompt input_prompt">In [7]:</div>
<div class="input_area box-flex1">
<div class="highlight"><pre><span class="n">fig</span> <span class="o">=</span> <span class="n">figure</span><span class="p">()</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span> <span class="c"># create axis handle</span>
<span class="n">k</span><span class="o">=</span><span class="mi">0</span>
<span class="n">fs</span><span class="o">=</span><span class="mi">2</span> <span class="c"># makes this plot easier to read</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span> <span class="p">(</span><span class="n">t</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span> <span class="n">k</span> <span class="o">-</span> <span class="n">fs</span> <span class="o">*</span> <span class="n">t</span><span class="p">),</span>
<span class="n">t</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span> <span class="n">k</span><span class="o">+</span><span class="mi">1</span> <span class="o">-</span> <span class="n">fs</span> <span class="o">*</span> <span class="n">t</span><span class="p">),</span><span class="s">'--'</span><span class="p">,</span><span class="n">k</span><span class="o">/</span><span class="n">fs</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="s">'o'</span><span class="p">,(</span><span class="n">k</span><span class="p">)</span><span class="o">/</span><span class="n">fs</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="s">'o'</span><span class="p">,</span>
<span class="n">t</span><span class="p">,</span><span class="n">sinc</span><span class="p">(</span> <span class="n">k</span><span class="o">-</span><span class="mi">1</span> <span class="o">-</span> <span class="n">fs</span> <span class="o">*</span> <span class="n">t</span><span class="p">),</span><span class="s">'--'</span><span class="p">,</span><span class="n">k</span><span class="o">/</span><span class="n">fs</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="s">'o'</span><span class="p">,(</span><span class="o">-</span><span class="n">k</span><span class="p">)</span><span class="o">/</span><span class="n">fs</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="s">'o'</span>
<span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">hlines</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">vlines</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="o">-.</span><span class="mi">2</span><span class="p">,</span><span class="mi">1</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span><span class="s">'sample value goes here'</span><span class="p">,</span>
<span class="n">xy</span><span class="o">=</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">1</span><span class="p">),</span>
<span class="n">xytext</span><span class="o">=</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="o">+.</span><span class="mi">1</span><span class="p">,</span><span class="mf">1.1</span><span class="p">),</span>
<span class="n">arrowprops</span><span class="o">=</span><span class="p">{</span><span class="s">'facecolor'</span><span class="p">:</span><span class="s">'red'</span><span class="p">,</span><span class="s">'shrink'</span><span class="p">:</span><span class="mf">0.05</span><span class="p">},</span>
<span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">annotate</span><span class="p">(</span><span class="s">'no interference here'</span><span class="p">,</span>
<span class="n">xy</span><span class="o">=</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span><span class="mi">0</span><span class="p">),</span>
<span class="n">xytext</span><span class="o">=</span><span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="o">+.</span><span class="mi">1</span><span class="p">,</span><span class="mf">0.5</span><span class="p">),</span>
<span class="n">arrowprops</span><span class="o">=</span><span class="p">{</span><span class="s">'facecolor'</span><span class="p">:</span><span class="s">'green'</span><span class="p">,</span><span class="s">'shrink'</span><span class="p">:</span><span class="mf">0.05</span><span class="p">},</span>
<span class="p">)</span>
<span class="n">show</span><span class="p">()</span>
</pre></div>
</div>
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<p>The vertical line in the previous plot shows that where one function has a peak, the other function has a zero. This is why when you put samples at each of the peaks, they match the sampled function exactly at those points. In between those points, the crown shape of the functions fills in the missing values. Furthermore, as the figure above shows, there is no interference between the functions sitting on each of the interpolating functions because the peak of one is perfectly aligned with the zero of the others (dotted lines). Thus, the sampling theorem says that the filled-in values are drawn from the curvature of the sinc functions, not straight lines as we investigated earlier. </p>
<p>As an illustration, the following code shows how the individual Whittaker functions(dashed lines) are assembled into the final approxmation (black-line) using the given samples (blue-dots). I urge you to play with the sampling rate to see what happens. Note the heavy use of <code>numpy</code> broadcasting in this code instead of the multiple loops we used earlier.</p>
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<div class="highlight"><pre><span class="n">fs</span><span class="o">=</span><span class="mf">5.0</span> <span class="c"># sampling rate</span>
<span class="n">k</span><span class="o">=</span><span class="n">array</span><span class="p">(</span><span class="nb">sorted</span><span class="p">(</span><span class="nb">set</span><span class="p">((</span><span class="n">t</span><span class="o">*</span><span class="n">fs</span><span class="p">)</span><span class="o">.</span><span class="n">astype</span><span class="p">(</span><span class="nb">int</span><span class="p">))))</span> <span class="c"># sorted coefficient list</span>
<span class="n">fig</span><span class="o">=</span><span class="n">figure</span><span class="p">()</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">fig</span><span class="o">.</span><span class="n">add_subplot</span><span class="p">(</span><span class="mi">111</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">t</span><span class="p">,(</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="p">(</span><span class="n">k</span><span class="p">[:,</span><span class="bp">None</span><span class="p">]</span><span class="o">/</span><span class="n">fs</span><span class="p">))</span><span class="o">*</span><span class="n">sinc</span><span class="p">(</span><span class="n">k</span><span class="p">[:,</span><span class="bp">None</span><span class="p">]</span><span class="o">-</span><span class="n">fs</span><span class="o">*</span><span class="n">t</span><span class="p">))</span><span class="o">.</span><span class="n">T</span><span class="p">,</span><span class="s">'--'</span><span class="p">,</span> <span class="c"># individual whittaker functions</span>
<span class="n">t</span><span class="p">,(</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="p">(</span><span class="n">k</span><span class="p">[:,</span><span class="bp">None</span><span class="p">]</span><span class="o">/</span><span class="n">fs</span><span class="p">))</span><span class="o">*</span><span class="n">sinc</span><span class="p">(</span><span class="n">k</span><span class="p">[:,</span><span class="bp">None</span><span class="p">]</span><span class="o">-</span><span class="n">fs</span><span class="o">*</span><span class="n">t</span><span class="p">))</span><span class="o">.</span><span class="n">sum</span><span class="p">(</span><span class="n">axis</span><span class="o">=</span><span class="mi">0</span><span class="p">),</span><span class="s">'k-'</span><span class="p">,</span> <span class="c"># whittaker interpolant</span>
<span class="n">k</span><span class="o">/</span><span class="n">fs</span><span class="p">,</span><span class="n">sin</span><span class="p">(</span><span class="mi">2</span><span class="o">*</span><span class="n">pi</span><span class="o">*</span><span class="n">k</span><span class="o">/</span><span class="n">fs</span><span class="p">),</span><span class="s">'ob'</span><span class="p">)</span><span class="c"># samples</span>
<span class="n">ax</span><span class="o">.</span><span class="n">set_xlabel</span><span class="p">(</span><span class="s">'time'</span><span class="p">,</span><span class="n">fontsize</span><span class="o">=</span><span class="mi">14</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">axis</span><span class="p">((</span><span class="o">-</span><span class="mf">1.1</span><span class="p">,</span><span class="mf">1.1</span><span class="p">,</span><span class="o">-</span><span class="mf">1.1</span><span class="p">,</span><span class="mf">1.1</span><span class="p">));</span>
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<p>However, if you've been following carefully, you should be somewhat uncomfortable with the second to the last plot that shows the errors in the Whittaker interpolation. Namely, <em>why are there any errors</em>? Does not the sampling theorem guarantee exact-ness which should mean no error at all? It turns out that answering this question takes us further into the implications of the sampling theorem, but that is the topic of our next post.</p>
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Summary
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<p>In this section, we started our investigation of the famous sampling theorem that is the bedrock of the entire field of signal processing and we asked if we could reverse-engineer the consquences of the sampling theorem by reconstructing a sampled function from its discrete samples. This led us to consider the famous <em>Whittaker interpolator</em>, whose proof we sketched here. However, after all this work, we came to a disturbing conclusion regarding the exact-ness of the sampling theorem that we will investigate in a subsequent posting. In the meantime, I urge you to start at the top of notebook and play with the sampling frequency, and maybe even the sampled function and see what else you can discover about the sampling theorem.</p>
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References
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<p>This is in the <a href="http://ipython.org/">IPython Notebook format</a> and was converted to HTML using <a href="https://github.com/ipython/nbconvert">nbconvert</a>.</p>
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<p>See <a href="http://books.google.com/books?id=Re5SAAAAMAAJ">Signal analysis</a> for more detailed mathematical development.</p>
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<p>The IPython Notebook corresponding to this post can be found <a href="https://github.com/unpingco/Python-for-Signal-Processing/blob/master/Sampling_Theorem.ipynb">here</a>.</p>
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