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prml演習5.39の解答

PRML演習5.39の解答

\[ \newcommand\l{\left} \newcommand\r{\right} \newcommand\cmt[1]{\class{Cmt}{\mbox{#1}}} \newcommand\b[1]{\class{Bold}{\mathrm{#1}}} \newcommand\bA{\b{A}} \newcommand\bw{\b{w}} \newcommand\bx{\b{x}} \newcommand\bI{\b{I}} \newcommand\N{{\cal N}} \newcommand\D{{\cal D}} \newcommand\T{\mathrm T} \]


\[ \begin{align} p(\D\mid\alpha,\beta) = \int p(\D\mid\bw,\beta)p(\bw\mid\alpha)\,d\bw \tag{5.174} \end{align} \] において、\(f(\bw)=p(\D\mid\bw,\beta)p(\bw\mid\alpha),\ Z=p(\D\mid\alpha,\beta) \) として、(4.135)を適用すると \[ Z\simeq f(\bw_{MAP})\frac{(2\pi)^{W\over 2}}{|\bA|^{1 \over 2}},\ \ \ (W\ は\ \bw\ の次元) \] となる。ここで \[ \begin{align} f(\bw_{MAP}) &= p(\D\mid\bw_{MAP},\beta)p(\bw_{MAP}\mid\alpha) \\ &= \prod_{n=1}^N \underset{(\because\ (5.163))}{\N(t_n\mid y(\bx_n,\bw_{MAP}),\beta^{-1})} \underset{(\because\ (5.162))}{\N(\bw_{MAP}\mid\b{0},\alpha^{-1}\bI)} \\ &= \prod_{n=1}^N {1\over (2\pi)^{1/2}}{1\over (\beta^{-1})^{1/2}}\exp\l[-{1\over 2\beta^{-1}}\{t_n-y(\bx_n,\bw_{MAP})\}^2\r] \\ &~~~~~{1\over (2\pi)^{W/2}}{1\over |\alpha^{-1}\bI|^{1/2}}\exp\l\{-{1\over 2}\bw_{MAP}^\T(\alpha^{-1}\bI)^{-1}\bw_{MAP}\r\} \\ &= \prod_{n=1}^N \l({\beta\over 2\pi}\r)^{1/2}\exp\l[-{\beta\over 2}\{t_n-y(\bx_n,\bw_{MAP})\}^2\r] \\ &~~~~~\underset{\cmt{※1}}{\l({\alpha\over 2\pi}\r)^{W\over 2}} \underset{\cmt{※2}}{\exp\l(-{\alpha\over 2}\bw_{MAP}^\T\bw_{MAP}\r)} \end{align} \]
\( \begin{align} \cmt{※1}~~~ |\alpha^{-1}\bI| = \l|\pmatrix{\alpha^{-1}&0\\0&\alpha^{-1}}\r| =\alpha^{-W} \end{align} \)
\( \begin{align} \cmt{※2}~~~ (\alpha^{-1}\bI)^{-1} = \pmatrix{\alpha^{-1}&0\\0&\alpha^{-1}}^{-1} =\pmatrix{\alpha&0\\0&\alpha}=\alpha\bI \end{align} \)
なので \[ \begin{align} \ln p(\D\mid\alpha,\beta) &\simeq \ln f(\bw_{MAP})+{W\over 2}\ln(2\pi)-{1\over 2}\ln|\bA| \\ &= \sum_{n=1}^N\l[{1\over2}\{\ln\beta-\ln(2\pi)\}-{\beta\over2}\{t_n-y(\bx_n,\bw_{MAP})\}^2\r] \\ &~~~~~ + {W\over2}\{\ln\alpha-\ln(2\pi)\}-{\alpha\over2}\bw_{MAP}^\T\bw_{MAP}+{W\over2}\ln(2\pi)-{1\over2}\ln|\bA| \\ &=-\l[{\beta\over2}\sum_{n=1}^N\{t_n-y(\bx_n,\bw_{MAP})\}^2+{\alpha\over2}\bw_{MAP}^\T\bw_{MAP}\r] \\ &~~~~~ -{1\over2}\ln|\bA|+{N\over2}\ln\beta-{N\over2}\ln(2\pi)+{W\over2}\ln\alpha \\ &=-E(\bw_{MAP})-{1\over2}\ln|\bA|+{N\over2}\ln\beta-{N\over2}\ln(2\pi)+{W\over2}\ln\alpha \tag{5.175} \end{align} \] を得る。ただし \[ \begin{align} E(\bw_{MAP})={\beta\over2}\sum_{n=1}^N\{t_n-y(\bx_n,\bw_{MAP})\}^2+{\alpha\over2}\bw_{MAP}^\T\bw_{MAP} \tag{5.176} \end{align} \] とする。

prml演習5.39の解答.txt · 最終更新: 2018/01/14 19:29 by ma

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