正規分布(normal distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)

正規分布

$$\begin{array}{rcl} N(\mu, \sigma^2)&=&\frac{1}{\sqrt{2\pi \sigma^2}}\mathrm{e}^{\frac{-(x-\mu)^2}{2\sigma^2}} \end{array}$$

積率母凾数

$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \int_{-\infty}^{\infty}(\mathrm{e}^{tx})\frac{1}{\sqrt{2\pi \sigma^2}}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}(\mathrm{e}^{tx})\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}+tx} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu)^2+(2\sigma^2)(tx)}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t}{2\sigma^2}-\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t+2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x \,\dotso\,a^2-2ab+b^2-2ac+2bc+c^2=(a-b-c)^2\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}+(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}} \displaystyle \int_{-\infty}^{\infty} \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}} \displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \mathrm{d}x\\ &=&\displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \frac{1}{\sqrt{2\pi \sigma^2}} \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}}\mathrm{d}x \\ &=&\displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\,\dotso\,\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}}\mathrm{d}x = N(\mu+\sigma^2t,\sigma^2)の総和=1\\ \end{array}$$

期待値・分散

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }\left(\mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\right) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}(\mu+\sigma^2t) \right\}|_{t=0}\\ &=&\displaystyle \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})}(\mu+\sigma^20) \\ &=&\displaystyle \mathrm{e}^{0}(\mu+0) \\ &=&\displaystyle \mu \\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }\left(\mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\right) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }\left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}(\mu+\sigma^2t)\right) \right\}|_{t=0}\\ &=&\displaystyle \left[ \displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \right)\right\} \left( \mu+\sigma^2t \right) \displaystyle + \left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \right) \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \left( \mu+\sigma^2t \right) \right\} \right]|_{t=0}\\ &=&\displaystyle \left[ \displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \left( \mu+\sigma^2t \right)^2 \displaystyle + \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \sigma^2 \displaystyle \right]|_{t=0}\\ &=&\displaystyle \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})} \left( \mu+\sigma^20 \right)^2 \displaystyle + \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})} \sigma^2\\ &=&\displaystyle \mu^2+\sigma^2\\ \end{array}$$ $$\begin{array}{rcl} V[X]&=&E[X^2]-E[X]^2\\ &=&\mu^2+\sigma^2-\mu^2\\ &=&\sigma^2 \end{array}$$