\[
\newcommand\l{\left}
\newcommand\r{\right}
\newcommand\cmt[1]{\class{Cmt}{\mbox{#1}}}
\newcommand\cause[1]{\class{Tiny}{(\because #1)}}
\newcommand\b[1]{\class{Bold}{\mathrm{#1}}}
\newcommand\bt{\b{t}}
\newcommand\bw{\b{w}}
\newcommand\bxi{\b{\xi}}
\newcommand\bphi{\b{\phi}}
\newcommand\bm{\b{m}}
\newcommand\bS{\b{S}}
\newcommand\pdiff[2]{\frac{\partial #1}{\partial #2}}
\]
(10.159) より
\[
{\cal L}(\bxi) = \ln \int h(\bw,\bxi)p(\bw)\,d\bw
\]
10.6.1節のはじめに載っているとおり \(p(\bw)\) は \(h(\bw,\bxi)~(10.153)\) の共役事前分布 \( {\cal N}(\bw \mid \bm_0, \bS_0)~(4.140) \) とし、また \(h(\bw,\bxi)~(10.153)\) を用いて
\[
\small
\begin{align}
& h(\bw,\bxi)p(\bw) = \l[ \prod_{n=1}^N \sigma(\xi_n)\exp\{\bw^T\bphi_n t_n - (\bw^T\bphi_n+\xi_n)/2
-\lambda(\xi_n)([\bw^T\bphi_n]^2-\xi_n^2) \} \r] {\cal N}(\bw \mid \bm_0, \bS_0) \\
&~~~= \l[ \prod_{n=1}^N \sigma(\xi_n)\exp\{\bw^T\bphi_n t_n - (\bw^T\bphi_n+\xi_n)/2
-\lambda(\xi_n)([\bw^T\bphi_n]^2-\xi_n^2) \} \r]
\frac{1}{(2\pi)^{D/2}} \frac{1}{|\bS_0|^{1/2}}
\exp\{-\frac{1}{2}(\bw-\bm_0)^T\bS_0^{-1}(\bw-\bm_0) \} \\
&~~~= \l(\prod_{n=1}^N \sigma(\xi_n)\r) \frac{1}{(2\pi)^{D/2}} \frac{1}{|\bS_0|^{1/2}}
\exp \underbrace{ \l[
\sum_{n=1}^N \l\{\bw^T\bphi_n t_n - (\bw^T\bphi_n+\xi_n)/2 -\lambda(\xi_n)([\bw^T\bphi_n]^2-\xi_n^2) \r\}
- \frac{1}{2} (\bw-\bm_0)^T\bS_0^{-1}(\bw-\bm_0)
\r] }_{(1)} \\
\end{align}
\]
指数の中を整理する
\[
\begin{align}
&(1) = -\frac{1}{2}\bw^T(2\sum_{n=1}^N\lambda(\xi_n)\bphi_n\bphi_n^T+\bS_0^{-1})\bw
+\bw^T\{\sum_{n=1}^N(\bphi_n t_n - \frac{\bphi_n}{2})+\bS_0^{-1}\bm_0\} + C \\
& C = \sum_{n=1}^N(-\frac{\xi_n}{2}+\lambda(\xi_n)\xi_n^2)-\frac{1}{2}\bm_0^T\bS_0^{-1}\bm_0
\end{align}
\]
これを平方完成する。
\(-\frac{1}{2}(\bw-\bm_N)^T\bS_N^{-1}(\bw-\bm_N)
= -\frac{1}{2}\bw^T\bS_N^{-1}\bw+\bw^T\bS_N^{-1}\bm_N-\frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N \) と見比べて
\[
\begin{align}
\bS_N^{-1} &= 2\sum_{n=1}^N\lambda(\xi_n)\bphi_n\bphi_n^T + \bS_0^{-1} \\
\bm_N &= \bS_N\{\sum_{n=1}^N(\bphi_n t_n - \frac{\bphi_n}{2}) + \bS_0^{-1}\bm_0 \}
\end{align}
\]
を得る。これより
\[
(1) = -\frac{1}{2}(\bw - \bm_N)^T\bS_N^{-1}(\bw - \bm_N)
+ \frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N + C
\]
となる。よって
\[
\small
\begin{align}
h(\bw,\bxi)p(\bw)
&= \l(\prod_{n=1}^N \sigma(\xi_n)\r)
\frac{1}{(2\pi)^{D/2}} \frac{1}{|\bS_0|^{1/2}}
\exp \l(\frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N+C \r)
\exp \l\{\frac{1}{2}(\bw-\bm_N)^T\bS_N^{-1}(\bw-\bm_N) \r\} \\
&= \l(\prod_{n=1}^N \sigma(\xi_n)\r)
\frac{1}{(2\pi)^{D/2}} \frac{1}{|\bS_0|^{1/2}}
\exp \l(\frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N+C \r)
(2\pi)^{D/2} |\bS_N|^{1/2} {\cal N}(\bw \mid \bm_N, \bS_N) \\
&= \l(\prod_{n=1}^N \sigma(\xi_n)\r)
\l(\frac{|\bS_N|}{|\bS_0|}\r)^{\frac{1}{2}}
\exp \l(\frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N+C \r)
{\cal N}(\bw \mid \bm_N, \bS_N)
\end{align}
\]
となる。よって
\[
\int h(\bw,\bxi)p(\bw) \,d\bw
= \l(\prod_{n=1}^N \sigma(\xi_n)\r)
\l(\frac{|\bS_N|}{|\bS_0|}\r)^{\frac{1}{2}}
\exp \l(\frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N+C \r)
\]
となる。よって
\[
\begin{align}
{\cal L}(\bxi)
&= \ln \int h(\bw,\bxi)p(\bw)\,d\bw \\
&= \ln \l\{
\l(\prod_{n=1}^N \sigma(\xi_n)\r)
\l(\frac{|\bS_N|}{|\bS_0|}\r)^{\frac{1}{2}}
\exp \l(\frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N+C \r) \r\} \\
&= \sum_{n=1}^N \ln \sigma(\xi_n) + \frac{1}{2}\ln\frac{|\bS_N|}{|\bS_0|}
+ \frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N + C \\
&= \sum_{n=1}^N \ln \sigma(\xi_n) + \frac{1}{2}\ln\frac{|\bS_N|}{|\bS_0|}
+ \frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N
+ \sum_{n=1}^N\l(-\frac{\xi_n}{2} + \lambda(\xi_n)\xi_n^2 \r)
- \frac{1}{2} \bm_0^T\bS_0^{-1}\bm_0 \\
&= \frac{1}{2}\ln\frac{|\bS_N|}{|\bS_0|}
+ \frac{1}{2}\bm_N^T\bS_N^{-1}\bm_N
- \frac{1}{2} \bm_0^T\bS_0^{-1}\bm_0
+ \sum_{n=1}^N \l\{\ln\sigma(\xi_n)-\frac{\xi_n}{2}+\lambda(\xi_n)\xi_n^2\r\} \tag{10.164}
\end{align}
\]
を得る。