PRML 第1章 1.5(基本)

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$$
\begin{array}{l}
(1.38) & \mathrm{var}[f] = \mathbb{E} \bigl[ (f(x) - \mathbb{E}[f(x)])^2 \bigr]\\[1em]
(1.39) & \mathrm{var}[f] = \mathbb{E}[f(x)^2] - \mathbb{E}[f(x)]^2
\end{array}
$$

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解答

定義(1.38)と$${\mathbb{E}}$$の線形性により，

$$
\begin{array}{lll}
\mathrm{var}[f] &=& \mathbb{E} \Bigl[ (f(x) - \mathbb{E}[f(x)])^2 \Bigr]\\[1em]
&=& \mathbb{E} \Bigl[ f(x)^2 - 2f(x)\mathbb{E}[f(x)] + \mathbb{E}[f(x)]^2 \Bigr]\\[1em]
&=& \mathbb{E} \Bigl[ f(x)^2 \Bigr] - \mathbb{E} \Bigl[ 2f(x)\mathbb{E}[f(x)] \Bigr] + \mathbb{E} \Bigl[ \mathbb{E}[f(x)]^2 \Bigr]\\[1em]
&=& \mathbb{E}[f(x)^2] - 2\mathbb{E}[f(x)]\mathbb{E}[f(x)] + \mathbb{E}[f(x)]^2\\[1em]
&=& \mathbb{E}[f(x)^2] - \mathbb{E}[f(x)]^2
\end{array}
$$

となり，var$${[f(x)]}$$は(1.39)を満たす．

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参考

$${\mathbb{E}}$$の線形性

●離散分布

$$
\begin{array}{ll}
& \mathbb{E}[f(x) + g(x)] = {\displaystyle \sum_x} p(x) (f(x) + g(x))\\[1em]
&\quad = {\displaystyle \sum_x} p(x)f(x) + {\displaystyle \sum_x} p(x)g(x)
= \mathbb{E}[f(x)] + \mathbb{E}[g(x)]\\[1.5em]
& \mathbb{E}[kf(x)] = {\displaystyle \sum_x} p(x) (kf(x))
= k {\displaystyle \sum_x} p(x)f(x)
= k \mathbb{E}[f(x)]
\end{array}
$$

●連続分布

$$
\begin{array}{ll}
& \mathbb{E}[f(x) + g(x)] = {\displaystyle \int} p(x) (f(x) + g(x)) dx\\[1em]
&\quad = {\displaystyle \int} p(x)f(x) dx + {\displaystyle \int} p(x)g(x) dx = \mathbb{E}[f(x)] + \mathbb{E}[g(x)]\\[1.5em]
& \mathbb{E}[kf(x)] = {\displaystyle \int} p(x) (kf(x)) dx = k {\displaystyle \int} p(x)f(x) dx = k \mathbb{E}[f(x)]
\end{array}
$$