正規分布
Gauss分布とも言う.確率変数\(X\)が平均\(\mu\),分散\(\sigma^2\)の正規分布(normal distribution)に従うとは,確率変数\(X\)の確率密度関数が, \[ f_X(x:\mu,\sigma^2) = \frac{1}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right), (x \in \overline{\mathbb{R}}) \] で与えられることであり,\(\mathcal{N}(\mu,\sigma^2)\)と表す.
平均
\[ E[X] = \int^{\infty}_{-\infty} \frac{x}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right) dx \]
\(y=x-\mu\)と変数変換する. \[ \begin{align} \int ^{\infty} _{-\infty} \frac{x}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right) dx &= \int ^{\infty} _{-\infty} \frac{y+\mu}{\sqrt{2\pi}\sigma} \exp\left( -\frac{{y}^2}{2\sigma^2}\right) dy \\ &= \frac{1}{\sqrt{2\pi}\sigma}\left( \int ^{\infty} _{-\infty} y \exp\left( -\frac{{y}^2}{2\sigma^2}\right) dy + \int ^{\infty} _{-\infty} \mu \exp\left( -\frac{{y}^2}{2\sigma^2}\right) dy\right) \end{align} \]
右辺第1項は\(z=\frac{y}{\sqrt{2}\sigma}\)と変数変換すると,\(\frac{dz}{dy} = \frac{1}{\sqrt{2}\sigma}\)から, \[ \begin{align} \int ^{\infty} _{-\infty} y \exp\left( -\frac{{y}^2}{2\sigma^2}\right) &= \int ^{\infty} _{-\infty} \sqrt{2}\sigma z \exp\left( -z^2 \right) \cdot \sqrt{2}\sigma \cdot dz \\ &= 0 \end{align} \] 最終式はガウス積分より得られる.
右辺第2項も同様に変数変換して, \[ \begin{align} \int ^{\infty} _{-\infty} \mu \exp\left( -\frac{{y}^2}{2\sigma^2}\right) &= \mu \int ^{\infty} _{-\infty}\exp\left( -z^2 \right) \cdot \sqrt{2}\sigma \cdot dz \\ &= \mu \sigma \sqrt{2\pi} \end{align} \] から, \[ E[X] = \int^{\infty} _{-\infty} \frac{x}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right) dx = \frac{1}{\sqrt{2\pi}\sigma} \cdot \mu \sigma \sqrt{2\pi} = \mu \]
分散
\[ E[(X-\mu)^2] = \int^{\infty}_{-\infty} \frac{(x-\mu)^2}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right) dx \] \(y=x-\mu\)と変数変換すると, \[ \begin{align} \int ^{\infty} _{-\infty} \frac{(x-\mu)^2}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right) dx &= \int ^{\infty} _{-\infty} \frac{y^2}{\sqrt{2\pi}\sigma} \exp\left( -\frac{y^2}{2\sigma^2}\right) dy \\ &= \frac{1}{\sqrt{2\pi}\sigma} \frac{1}{2} \sqrt{\pi \cdot 8\sigma^6} \\ &= \sigma^2 \end{align} \]
積率母関数
\[ \begin{align} M_X(t) &= E[e^{tX}] \\ &= \int^{\infty}_{-\infty}\exp(tx)\frac{1}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right)dx \\ \end{align} \] まず,\(\displaystyle z=\frac{x-\mu}{\sqrt{2}\sigma}\)と変数変換する. \[ \begin{align} \int ^{\infty} _{-\infty}\exp(tx)\frac{1}{\sqrt{2\pi}\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2}\right)dx &= \int ^{\infty} _{-\infty}\exp(t(\sqrt{2}\sigma z+\mu))\frac{1}{\sqrt{2\pi}\sigma} \exp\left( -z^2\right)\sqrt{2} \sigma dz \\ &= \frac{e^{\mu t}}{\sqrt{\pi}} \int ^{\infty} _{-\infty} \exp(\sqrt{2} \sigma tz) \exp( -z^2 )dz \\ &= \frac{e^{\mu t}}{\sqrt{\pi}} \int ^{\infty} _{-\infty} \exp( -z^2+\sqrt{2} \sigma tz)dz \\ &= \frac{e^{\mu t}}{\sqrt{\pi}} \int ^{\infty} _{-\infty} \exp\left( -z^2+ 2 \left( \frac{\sigma tz}{\sqrt{2}} \right) \right) dz \\ &= \frac{e^{\mu t}}{\sqrt{\pi}} \int ^{\infty} _{-\infty} \exp\left( - \left(z^2 - 2 \left( \frac{\sigma tz}{\sqrt{2}} \right) + \left( \frac{\sigma t}{\sqrt{2}} \right)^2 \right) + \left( \frac{\sigma t}{\sqrt{2}} \right)^2 \right) dz \\ &= \frac{e^{\mu t}}{\sqrt{\pi}} \int ^{\infty} _{-\infty} \exp\left( -\left(z + \frac{\sigma t}{\sqrt{2}} \right)^2 + \left( \frac{\sigma t}{\sqrt{2}} \right)^2 \right) dz \\ &= \frac{e^{\mu t + \left( \frac{\sigma t}{\sqrt{2}} \right)^2 }}{\sqrt{\pi}} \int ^{\infty} _{-\infty} \exp\left( -\left(z + \frac{\sigma t}{\sqrt{2}} \right)^2 \right) dz \\ &= e^{\mu t + \left( \frac{\sigma t}{\sqrt{2}} \right)^2} \\ &= e^{\mu t + \frac{\sigma^2 t^2}{2}} \end{align} \]
k次モーメント
積率母関数を微分することに原点周りのk次モーメントを得る. \[ \left. \frac{d}{dt} M_X(t) \right| _{t=0} = \left. \frac{d}{dt} e^{\mu t + \frac{\sigma^2 t^2}{2}} \right| _{t=0} = \left. \left( \mu + \sigma^2 t \right) \left( e^{\mu t + \frac{\sigma^2 t^2}{2}} \right) \right| _{t=0} = \mu \] \[ \begin{align} \left. \frac{d^2}{dt^2} M_X(t) \right| _{t=0} &= \left. \frac{d^2}{dt^2} e^{\mu t + \frac{\sigma^2 t^2}{2}} \right| _{t=0} \\ &= \left. \frac{d}{dt} \left( \mu + \sigma^2 t \right) \left( e^{\mu t + \frac{\sigma^2 t^2}{2}} \right) \right| _{t=0} \\ &= \left. \sigma^2 \left( e^{\mu t + \frac{\sigma^2 t^2}{2}} \right) + ( \mu + \sigma^2 t )^2 \left( e^{\mu t + \frac{\sigma^2 t^2}{2}} \right) \right| _{t=0} \\ &= \sigma^2 + \mu^2 \end{align} \]
特性関数
\[ \varphi_X(t) = E[e^{itX}] = e^{i \mu t - \left( \frac{\sigma t}{\sqrt{2}} \right)^2} \]