連続分布_（単変量）正規分布

（再掲）
$X \sim \N(\mu, \sigma^2)$である時、その確率密度関数$f(x)$は、
$$\begin{aligned} f(x) = \dfrac{1}{\sqrt{2 \pi \sigma^2}} \exp[- \dfrac{(x – \mu)^2}{2 \sigma^2}] ~~ (- \infty \lt x \lt \infty) \end{aligned}$$であり、その特性値は、
$$\begin{aligned} &{\small ①平均：}E[X] = \mu \\ &{\small ②分散：}V[X] = \sigma^2 \\ &{\small ③積率母関数：}M^X(\theta) = \exp [\mu \theta + \dfrac{1}{2}\sigma^2 \theta^2] \end{aligned}$$となる。

（①平均$E[X]$の証明）
$$\begin{aligned} E[X] &= \int_{- \infty}^{\infty} x f(x) dx \\[10px] &= \int_{- \infty}^{\infty} \{ (x – \mu) + \mu \} f(x) dx \\ &{\scriptsize (無理やり(x – \mu)の項を作り出した)} \\[10px] &= \int_{- \infty}^{\infty} (x – \mu) f(x) dx + \mu \int_{- \infty}^{\infty} f(x) dx \\[10px] &= \dfrac{1}{\sqrt{2 \pi \sigma^2}} \int_{- \infty}^{\infty} (x – \mu) \cdot \exp [- \dfrac{(x – \mu)^2}{2 \sigma^2}] dx + \mu \\ &{\scriptsize (\int_{- \infty}^{\infty} f(x) dx = 1より)} \\[10px] &= \dfrac{1}{\sqrt{2 \pi }} \int_{- \infty}^{\infty} (\dfrac{x – \mu}{\sigma}) \cdot \exp [- \dfrac{1}{2} (\dfrac{x – \mu}{\sigma})^2] dx + \mu \\[10px] &= \dfrac{1}{\sqrt{2 \pi }} \int_{- \infty}^{\infty} z \cdot \exp (- \dfrac{z^2}{2}) \cdot \sigma dz + \mu \\ &{\scriptsize (\dfrac{x – \mu}{\sigma} = zとおいた)} \\[10px] &= \dfrac{\sigma}{\sqrt{2 \pi}} \left[ – \exp(- \dfrac{z^2}{2}) \right]_{- \infty}^{\infty} + \mu \\[10px] &= \mu \end{aligned}$$

（②分散$V[X]$の証明）
$$\begin{aligned} V[X] &= \int_{- \infty}^{\infty} (x – E[X])^2 f(x) dx \\[10px] &= \dfrac{1}{\sqrt{2 \pi \sigma^2}} \int_{- \infty}^{\infty} (x – \mu)^2 \cdot \exp[- \dfrac{(x – \mu)^2}{2 \sigma^2}] dx \\[10px] &= \dfrac{\sigma}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} (\dfrac{x – \mu}{\sigma})^2 \cdot \exp [- \dfrac{1}{2} (\dfrac{x – \mu}{\sigma})^2] dx \\[10px] &= \dfrac{\sigma}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} z^2 \cdot \exp (- \dfrac{z^2}{2}) \cdot \sigma dz \\ &{\scriptsize (\dfrac{x – \mu}{\sigma} = zとおいた)} \\[10px] &= \dfrac{\sigma^2}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} z^2 \cdot \exp (- \dfrac{z^2}{2}) dz \\[10px] &= \dfrac{\sigma^2}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} (-z) \cdot (\exp (- \dfrac{z^2}{2}))^{\prime} dz \\[10px] &= \dfrac{\sigma^2}{\sqrt{2 \pi}} \left\{ \left[ (-z) \cdot \exp(- \dfrac{z^2}{2}) \right]_{-\infty}^{\infty} – \int_{-\infty}^{\infty} (-1) \cdot \exp(- \dfrac{z^2}{2}) dz \right\} \\ &{\scriptsize (部分積分を用いた)} \\[10px] &= \dfrac{\sigma^2}{\sqrt{2 \pi}} \left\{ 0 + \int_{-\infty}^{\infty} \exp(- \dfrac{z^2}{2}) dz \right\} \\[10px] &= \sigma^2 \\ &{\scriptsize (ガウス積分より\int_{-\infty}^{\infty} \exp(- \dfrac{z^2}{2}) dz = \sqrt{2 \pi})} \end{aligned}$$

（③積率母関数$M^X(\theta)$の証明）
$$\begin{aligned} M^X(\theta) &= E[e^{\theta X}] \\[10px] &= \int_{-\infty}^{\infty} e^{\theta x} f(x) dx \\[10px] &= \dfrac{1}{\sqrt{2 \pi \sigma^2}} \int_{- \infty}^{\infty} e^{\theta x} \cdot \exp[- \dfrac{(x – \mu)^2}{2 \sigma^2}] dx \\[10px] &= \dfrac{1}{\sqrt{2 \pi \sigma^2}} \int_{- \infty}^{\infty} \exp [- \dfrac{1}{2 \sigma^2} \{ (x – \mu)^2 – 2 \sigma^2 \cdot \theta x \}] dx \\[10px] &= \dfrac{1}{\sqrt{2 \pi \sigma^2}} \int_{- \infty}^{\infty} \exp [- \dfrac{1}{2 \sigma^2} \{ [x – (\mu + \sigma^2 \theta)]^2 – \sigma^4 \theta^2 – 2 \mu \sigma^2 \theta \}] dx \\[10px] &= \dfrac{1}{\sqrt{2 \pi \sigma^2}} \int_{- \infty}^{\infty} \exp [- \dfrac{[x – (\mu + \sigma^2 \theta)]^2}{2 \sigma^2} + \dfrac{\sigma^2 \theta^2}{2} + \mu \theta] dx \\[10px] &= \exp [\mu \theta + \dfrac{1}{2} \sigma^2 \theta^2] \cdot \int_{- \infty}^{\infty} \boldsymbol{\dfrac{1}{\sqrt{2 \pi \sigma^2}} \exp [- \dfrac{[x – (\mu + \sigma^2 \theta)]^2}{2 \sigma^2}] } dx \\[10px] &= \exp [\mu \theta + \dfrac{1}{2}\sigma^2 \theta^2] \\ &{\scriptsize (太字部分は確率密度関数の形なので全区間で積分すると1になる)} \end{aligned}$$