PRML 第1章 1.8(標準)

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$$
\begin{array}{ll}
(1.46) & \mathcal{N}(x|\mu,\sigma^2) = \dfrac{1}{(2\pi\sigma^2)^{1/2}} \exp \left\{ -\dfrac{1}{2\sigma^2}(x - \mu)^2 \right\}\\[1em]
(1.49) & \mathbb{E}[x] = {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)~xdx = \mu\\[1em]
(1.50) & \mathbb{E}[x^2] = {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)~x^2dx = \mu^2 + \sigma^2\\[1em]
(1.51) & {\rm var}[x] = \mathbb{E}[x^2] - \mathbb{E}[x]^2 = \sigma^2\\[1.5em]
(1.127) & {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2) dx = 1
\end{array}
$$

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解答

　$${y = x - \mu}$$と変数変換して(1.49)の積分を実行すると，

$$
\begin{array}{lll}
\mathbb{E}[x] &=& {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)~xdx\\[1em]
&=& \dfrac{1}{(2\pi\sigma^2)^{1/2}} {\displaystyle \int_{-\infty}^\infty} x\exp \left[ -\dfrac{1}{2\sigma^2}(x - \mu)^2 \right] dx\\[1em]
&=& \dfrac{1}{(2\pi\sigma^2)^{1/2}} {\displaystyle \int_{-\infty}^\infty} (y + \mu)\exp \left( -\dfrac{1}{2\sigma^2}y^2 \right) dy\\[1em]
&=& \dfrac{1}{(2\pi\sigma^2)^{1/2}} \left\{ \left[ -\sigma^2 \exp\left( -\dfrac{1}{2\sigma^2}y^2 \right) \right]_{-\infty}^\infty + \mu {\displaystyle \int_{-\infty}^\infty} \exp \left( -\dfrac{1}{2\sigma^2}y^2 \right) dy \right\}\\[1em]
&=& \dfrac{\mu}{(2\pi\sigma^2)^{1/2}} {\displaystyle \int_{-\infty}^\infty} \exp \left( -\dfrac{1}{2\sigma^2}y^2 \right) dy\\[1em]
&=& \mu
\end{array}
$$

を得る．また，規格化条件(1.127)の両辺に$${(2\pi\sigma^2)^{1/2}}$$をかけて$${\sigma^2}$$で微分すると，

$$
\begin{array}{lll}
&& \dfrac{1}{2\sigma^4} {\displaystyle \int_{-\infty}^\infty} (x-\mu)^2 \exp \left[ -\dfrac{1}{2\sigma^2}(x - \mu)^2 \right] dx = \dfrac12 (2\pi\sigma^2)^{-1/2} \cdot 2\pi\\[1em]
&\Longleftrightarrow& {\displaystyle \int_{-\infty}^\infty} (x^2 - 2\mu x + \mu^2)~\mathcal{N}(x|\mu,\sigma^2)~dx = \sigma^2\\[1em]
&\Longleftrightarrow& {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)~x^2 dx - 2\mu {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)~xdx + \mu^2 {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)dx = \sigma^2\\[1em]
&\Longleftrightarrow& {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)~x^2 dx - 2\mu^2 + \mu^2 = \sigma^2\\[1em]
&\Longleftrightarrow& {\displaystyle \int_{-\infty}^\infty} \mathcal{N}(x|\mu,\sigma^2)~x^2 dx = \mu^2 + \sigma^2
\end{array}
$$

したがって，

$$
\mathrm{var}[x] = \mathbb{E}[x^2] - \mathbb{E}[x]^2 = (\mu^2 + \sigma^2) - (\mu)^2 = \sigma^2
$$

となる．