\[ \newcommand\l{\left} \newcommand\r{\right} \newcommand\cmt[1]{\class{Cmt}{\mbox{#1}}} \newcommand\cause[1]{\class{Tiny}{(\because #1)}} \newcommand\b[1]{\class{Bold}{\mathrm{#1}}} \newcommand\bZ{\b{Z}} \newcommand\bz[1]{\b{z}^{(#1)}} \newcommand\fhat{\hat f} \newcommand\E{\mathbb E} \newcommand\var{\mathrm{var}} \]

\(\bZ=(\bz{1},\bz{2},\cdots,\bz{L}) \) 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。