PRML 第1章 1.13(基本)

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$$
\begin{array}{ll}
(1.56) & \sigma_{\scriptscriptstyle\mathrm{ML}}^2 = \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} (x_n - \mu_{\scriptscriptstyle\mathrm{ML}})^2
\end{array}
$$

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解答

(1.56)の$${\mu_{\scriptscriptstyle\mathrm{ML}}}$$を$${\mu}$$で置き換えて期待値を計算すると，

$$
\begin{array}{lll}
\mathbb{E} \left[ \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} (x_n - \mu)^2 \right]
&=& \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} \left\{ \mathbb{E}[x_n^2] - 2\mu \mathbb{E}[x_n] + \mu^2 \right\}\\[1.5em]
&=& \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} \left\{ (\mu^2 + \sigma^2) - 2\mu \cdot \mu + \mu^2 \right\}\\[1.5em]
&=& \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} \sigma^2\\[1.5em]
&=& \sigma^2
\end{array}
$$

となる．