prml演習9.26の解答
PRML演習9.26の解答
\[
\newcommand\l{\left}
\newcommand\r{\right}
\newcommand\cmt[1]{\class{Cmt}{\mbox{#1}}}
\newcommand\b[1]{\class{Bold}{\mathrm{#1}}}
\newcommand\bx{\b{x}}
\newcommand\bmu{\b{\mu}}
\newcommand\gammaold[1]{\gamma^{old}(z_{#1k})}
\newcommand\gammanew[1]{\gamma^{new}(z_{#1k})}
\newcommand\Nold{N^{old}_k}
\newcommand\Nnew{N^{new}_k}
\newcommand\bmuold{\bmu^{old}_k}
\newcommand\bmunew{\bmu^{new}_k}
\]
(9.18)より
\[
\begin{align}
\Nnew &= \gammaold{1}+\gammaold{2}+\cdots+\gammanew{m}+\cdots+\gammaold{N} \\
&=\sum_{n=1}^N \gammaold{n} - \gammaold{m} + \gammanew{m} \\
&=\Nold-\gammaold{m}+\gammanew{m} \tag{9.79}
\end{align}
\]
を得る。(9.17)より
\[
\begin{align}
\bmunew &= {1 \over \Nnew}\l\{\gammaold{1}\bx_1+\gammaold{2}\bx_2+\cdots+\gammanew{m}\bx_m+\cdots+\gammaold{N}\bx_N \r\} \\
&={1 \over \Nnew}\l\{\sum_{n=1}^N\gammaold{n}\bx_n-\gammaold{m}\bx_m+\gammanew{m}\bx_m\r\} \\
&={1 \over \Nnew}\l\{\Nold\bmuold-\gammaold{m}\bx_m+\gammanew{m}\bx_m\r\} \\
&={1 \over \Nnew}\big[\underbrace{\l\{\Nnew+\gammaold{m}-\gammanew{m}\r\}}_{(\because\ (9.79))}\bmuold-\gammaold{m}\bx_m+\gammanew{m}\bx_m\big] \\
&={1 \over \Nnew}\big[\Nnew\bmuold-\{\gammanew{m}-\gammaold{m}\}\bmuold+\{\gammanew{m}-\gammaold{m}\}\bx_m \big] \\
&=\bmuold+{1\over\Nnew}\{\gammanew{m}-\gammaold{m}\}(\bx_m-\bmuold) \tag{9.78}\\
\end{align}
\]
を得る。
\(
\begin{align}
\end{align}
\)
prml演習9.26の解答.txt · 最終更新: 2018/01/25 16:40 by ma
