\[
\newcommand\l{\left}
\newcommand\r{\right}
\newcommand\cmt[1]{\class{Cmt}{\mbox{#1}}}
\newcommand\cause[1]{\class{Tiny}{(\because #1)}}
\newcommand\b[1]{\class{Bold}{\mathrm{#1}}}
\newcommand\bZ{\b{Z}}
\newcommand\bz[1]{\b{z}^{(#1)}}
\newcommand\fhat{\hat f}
\newcommand\E{\mathbb E}
\newcommand\var{\mathrm{var}}
\]
\(\bZ=(\bz{1},\bz{2},\cdots,\bz{L}) \) とすると、
\[
\begin{align}
\E_{\bZ}[\fhat]
&= \int \fhat p(\bZ) \,d\bZ
= \int \underset{\cmt{※1}}{\l( {1 \over L} \sum_{l=1}^L f(\bz{l}) \r)}
\underset{\cmt{※2}}{p(\bz{1})p(\bz{2})\cdots}
\underset{\cmt{※3}}{\,d\bz{1}d\bz{2}\cdots} \\
&= {1 \over L}\l\{ \int f(\bz{1})p(\bz{1})p(\bz{2})\cdots\,d\bz{1}d\bz{2}\cdots
+\int f(\bz{2})p(\bz{1})p(\bz{2})\cdots\,d\bz{1}d\bz{2}\cdots
+\cdots \r\} \\
&= {1 \over L}\l\{ \int f(\bz{1})p(\bz{1})\,d\bz{1}
+\int f(\bz{2})p(\bz{2})\,d\bz{2}
+\cdots \r\} \\
&= {1 \over L}\sum_{l=1}^L \int f(\bz{l})p(\bz{l})\,d\bz{l}
= {1 \over L}\sum_{l=1}^L \E_{\bz{l}}[f] = \underset{\cmt{※4}}{\E_{\b{z}}[f]}
\end{align}
\]
となり、\(\fhat\) の平均は、\(f\) の平均と同じであることが確認できる。
次に分散について
\[
\begin{align}
\var[\fhat]
&= \E[(\fhat-\E[\fhat])^2] \\
&= \E[\fhat^2-2\fhat\E[\fhat]+(\E[\fhat])^2] \\
&= \E[\fhat^2] - 2\E[\fhat]\E[\fhat] + (\E[\fhat])^2 \\
&= \E[\fhat^2] - (\E[\fhat])^2 \\
&= \E[\fhat^2] - (\E[f])^2~~~(\because 上記\ \E[\fhat] = \E[f] より)
\end{align}
\]
ここで
\[
\begin{align}
\E[\fhat^2]
&= \int \fhat^2 p(\bZ)\,d\bZ \\
&= \int \l( {1 \over L}\sum_{l=1}^L f(\bz{l}) \r)^2 p(\bz{1})p(\bz{2})\cdots\,d\bz{1}d\bz{2}\cdots \\
&= {1 \over L^2}
\int
\l\{\sum_{l=1}^L(f(\bz{l}))^2
+ \sum_{m \ne n}\sum_n f(\bz{m})f(\bz{n}) \r\}
p(\bz{1})p(\bz{2})\cdots\,d\bz{1}d\bz{2}\cdots \\
&= {1 \over L^2}
\l\{
\int
\l( \sum_{l=1}^L(f(\bz{l}))^2 \r)
p(\bz{1})p(\bz{2})\cdots\,d\bz{1}d\bz{2}\cdots \r. \\
&~~~~~~~~~~~~~~~~~
\l.
+\int
\l( \sum_{m \ne n}\sum_n f(\bz{m})f(\bz{n}) \r)
p(\bz{1})p(\bz{2})\cdots\,d\bz{1}d\bz{2}\cdots
\r\} \\
&= {1 \over L^2}
\l\{
\sum_{l=1}^L \int \l( f(\bz{l})\r)^2 p(\bz{l})\,d\bz{l}
+\sum_{m \ne n}\sum_n \int \l( f(\bz{m})f(\bz{n})\r)p(\bz{m})p(\bz{n})\,d\bz{m}d\bz{n}
\r\} \\
&= {1 \over L^2}
\l\{
\sum_{l=1}^L \int \l( f(\bz{l})\r)^2 p(\bz{l})\,d\bz{l}
+\sum_{m \ne n}\sum_n \int f(\bz{m})p(\bz{m})\,d\bz{m} \int f(\bz{n})p(\bz{n})\,d\bz{n}
\r\} \\
&= {1 \over L^2}
\l\{ L\E[f^2]+L(L-1)(\E[f])^2 \r\}~~~(\because \bz{l},\bz{m},\bz{n}は\b{z}と同分布なので) \\
&= {1 \over L}\,\E[f^2] + {L-1 \over L}\,(\E[f])^2
\end{align}
\]
これより
\[
\begin{align}
\var[\fhat]
&= \E[\fhat^2] - (\E[f])^2 \\
&= {1 \over L}\E[f^2] - {L-1 \over L}(\E[f])^2 - (\E[f])^2 \\
&= {1 \over L}\l\{\E[f^2]-(\E[f])^2\r\} \\
&= {1 \over L}\E[(f-\E[f])^2] \tag{11.3}
\end{align}
\]
を得る。