正規分布

$$\begin{align*} \newcommand{\mat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{\f}[2]{\frac{#1}{#2}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\d}[2]{\frac{{\rm d}#1}{{\rm d}#2}} \newcommand{\T}{\mathsf{T}} \newcommand{\(}{\left(} \newcommand{\)}{\right)} \newcommand{\{}{\left\{} \newcommand{\}}{\right\}} \newcommand{\[}{\left[} \newcommand{\]}{\right]} \newcommand{\dis}{\displaystyle} \newcommand{\eq}[1]{{\rm Eq}(\ref{#1})} \newcommand{\n}{\notag\\} \newcommand{\t}{\ \ \ \ } \newcommand{\tt}{\t\t\t\t} \newcommand{\argmax}{\mathop{\rm arg\, max}\limits} \newcommand{\argmin}{\mathop{\rm arg\, min}\limits} \def\l<#1>{\left\langle #1 \right\rangle} \def\us#1_#2{\underset{#2}{#1}} \def\os#1^#2{\overset{#2}{#1}} \newcommand{\case}[1]{\{ \begin{array}{ll} #1 \end{array} \right.} \newcommand{\s}[1]{{\scriptstyle #1}} \definecolor{myblack}{rgb}{0.27,0.27,0.27} \definecolor{myred}{rgb}{0.78,0.24,0.18} \definecolor{myblue}{rgb}{0.0,0.443,0.737} \definecolor{myyellow}{rgb}{1.0,0.82,0.165} \definecolor{mygreen}{rgb}{0.24,0.47,0.44} \newcommand{\c}[2]{\textcolor{#1}{#2}} \newcommand{\ub}[2]{\underbrace{#1}_{#2}} \end{align*}$$

積率母関数

　確率変数 $Z \sim \mathcal{N}(0, 1)$ を考えて、

$$\begin{align*} M_{Z}(t) &= E[e^{tZ}] \n & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{t z} \cdot \exp \left[-\frac{z^{2}}{2}\right] d z \n & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp \left[-\frac{1}{2}\left(z^{2}-2 t z\right)\right] d z \n & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp \[ -\f{1}{2} \{(z – t)^2 – t^2 \} \] dz \n &= \exp\[ \f{t^2}{2} \] \cdot \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp \[ -\f{(z – t)^2}{2} \] dz \n &= \exp \[ \f{t^2}{2} \]. \end{align*}$$

$X = \sigma Z + \mu$ であるから、

$$\begin{align*} M_X(t) &= E[e^{tX}] \n &= E[e^{t(\sigma Z + \mu)}] \n &= E[e^{t \mu}] \cdot E[e^{t\sigma Z}] \n &= \exp\[\mu t + \f{\sigma^2}{2}t^2 \]. \end{align*}$$

期待値

　確率変数 $Z \sim \mathcal{N}(0, 1)$ を考えて、$M_Z(t) = e^{\f{t^2}{2}}$ より、

$$\begin{align*} E[Z] &= \left. \d{}{t} M_Z(t) \right|_{t=0} \n &= \left. t e^{\f{t^2}{2}} \right|_{t=0} \n &= 0. \end{align*}$$

$X = \sigma Z + \mu$ であるから、

$$\begin{align*} E[X] &= \sigma E[Z] + \mu \n &= \mu. \end{align*}$$

分散

　確率変数 $Z \sim \mathcal{N}(0, 1)$ を考えて、$M_Z(t) = e^{\f{t^2}{2}}$ より、

$$\begin{align*} E[Z^2] &= \left. \f{{\rm d}^2}{{\rm d}t^2} M_Z(t) \right|_{t=0} \n &= \left. \d{}{t} \( t e^{\f{t^2}{2}} \) \right|_{t=0} \n &= \left. \( e^{\f{t^2}{2}} + t^2 e^{\f{t^2}{2}}\) \right|_{t=0} \n &= 1. \end{align*}$$

よって、

$$\begin{align*} V[Z] &= E[Z^2] – E[Z]^2 \n &= 1. \end{align*}$$

$X = \sigma Z + \mu$ であるから、

$$\begin{align*} V[X] &= V[\sigma Z + \mu] \n &= \sigma^2 V[Z] \n &= \sigma^2. \end{align*}$$