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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

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