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<title>PRML 第4章 演習 4.1-4.10</title>
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<h1 class="title">PRML 第4章 演習 4.1-4.10</h1>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#sec-1">PRML 第4章 演習 4.1-4.10</a>
<ul>
<li><a href="#sec-1-1">4.1</a></li>
<li><a href="#sec-1-2">4.2</a></li>
<li><a href="#sec-1-3">4.3</a></li>
<li><a href="#sec-1-4">4.4</a></li>
<li><a href="#sec-1-5">4.5</a></li>
<li><a href="#sec-1-6">4.6</a></li>
<li><a href="#sec-1-7">4.7</a></li>
<li><a href="#sec-1-8">4.8</a></li>
<li><a href="#sec-1-9"><span class="done DONE">DONE</span> 4.9 [www] 多クラス分類問題の生成モデルによる最尤解</a></li>
<li><a href="#sec-1-10">4.10</a></li>
</ul>
</li>
</ul>
</div>
</div>
\begin{align*}
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<div id="outline-container-sec-1" class="outline-2">
<h2 id="sec-1">PRML 第4章 演習 4.1-4.10</h2>
<div class="outline-text-2" id="text-1">
</div><div id="outline-container-sec-1-1" class="outline-3">
<h3 id="sec-1-1">4.1</h3>
</div>
<div id="outline-container-sec-1-2" class="outline-3">
<h3 id="sec-1-2">4.2</h3>
</div>
<div id="outline-container-sec-1-3" class="outline-3">
<h3 id="sec-1-3">4.3</h3>
</div>
<div id="outline-container-sec-1-4" class="outline-3">
<h3 id="sec-1-4">4.4</h3>
</div>
<div id="outline-container-sec-1-5" class="outline-3">
<h3 id="sec-1-5">4.5</h3>
</div>
<div id="outline-container-sec-1-6" class="outline-3">
<h3 id="sec-1-6">4.6</h3>
</div>
<div id="outline-container-sec-1-7" class="outline-3">
<h3 id="sec-1-7">4.7</h3>
</div>
<div id="outline-container-sec-1-8" class="outline-3">
<h3 id="sec-1-8">4.8</h3>
</div>
<div id="outline-container-sec-1-9" class="outline-3">
<h3 id="sec-1-9"><span class="done DONE">DONE</span> 4.9 [www] 多クラス分類問題の生成モデルによる最尤解</h3>
<div class="outline-text-3" id="text-1-9">
<p>
尤度関数<br />
</p>
\begin{align*}
p(\{\bphi_n,\t_n\}|\{\pi_k\})
= & \prod_{n=1}^N \prod_{k=1}^K [p(C_k) p(\bphi_n|C_k)]^{t_{nk}} \\
= & \prod_{n=1}^N \prod_{k=1}^K [\pi_k p(\bphi_n|C_k)]^{t_{nk}} \\
\ln p(\{\bphi_n,\t_n\}|\{\pi_k\})
= & \sum_{n=1}^N \sum_{k=1}^K t_{nk} [\ln \pi_k + \ln p(\bphi_n|C_k)] \\
\end{align*}
<p>
\(\sum_k \pi_k = 1\)という制約のもとで対数尤度関数を最大化する\(\pi_k\)を求めるため、<br />
ラグランジュの未定係数法を用いる。<br />
</p>
\begin{align*}
L(\{\pi_k\},\lambda) = & \ln p(\{\bphi_n,\t_n\}|\{\pi_k\})
+ \lambda \l( \sum_k \pi_k - 1 \r) \\
\end{align*}
\begin{align*}
& \p{}{\pi_k} L(\{\pi_k\},\lambda) \\
= & \sum_{n=1}^N \sum_{j=1}^K t_{nj} \p{}{\pi_k} \ln \pi_j
+ \p{}{\pi_k} \lambda \l( \sum_k \pi_k - 1 \r) \\
= & \sum_{n=1}^N \sum_{j=1}^K t_{nj} I_{jk} \f{1}{\pi_j} + \lambda \\
= & \f{1}{\pi_k} \sum_{n=1}^N t_{nk} + \lambda \\
= & \f{1}{\pi_k} N_k + \lambda \\
\end{align*}
\begin{align*}
0 = & \f{1}{\pi_k} N_k + \lambda \\
- \lambda \pi_k = & N_k \\
\end{align*}
\begin{align*}
- \lambda \sum_k \pi_k = & \sum_k N_k \\
- \lambda = & N \\
\end{align*}
\begin{align*}
\pi_k = & \f{N_k}{N} \\
\end{align*}
</div>
</div>
<div id="outline-container-sec-1-10" class="outline-3">
<h3 id="sec-1-10">4.10</h3>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="creator"><a href="http://www.gnu.org/software/emacs/">Emacs</a> 24.4.1 (<a href="http://orgmode.org">Org</a> mode 8.2.10)</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
</div>
</body>
</html>