第2講確率分布

正規分布

X \sim N (μ, σ)

$\begin{equation} X \sim N(\mu, \sigma) \end{equation}$

f (x) = \frac{1}{\sqrt{2 π σ^{2}}} \exp [- \frac{(x - μ)^{2}}{2 σ^{2}}] (- \infty < x < \infty)

$\begin{equation} f(x) = \frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left[-\frac{(x-\mu)^{2}}{2\sigma^{2}}\right]~~~~(-\infty < x < \infty) \end{equation}$

E [X] = μ

$\begin{equation} E[X] = \mu \end{equation}$

V [X] = σ^{2}

$\begin{equation} V[X] = \sigma^2 \end{equation}$

M_{X} (t) = \exp [μ t + \frac{σ^{2}}{2} t^{2}]

$\begin{equation} M_{X}(t) = \exp\left[\mu\,t + \frac{\sigma^{2}}{2}t^{2}\right] \end{equation}$

\begin{aligned} E [(X - μ)^{k}] & = {\begin{cases} 0 & k = 2 m + 1 (m = 1, 2, \dots) \\ (k - 1)!! σ^{k} & k = 2 m (m = 1, 2, \dots) \end{cases} \\ = {\begin{cases} 0 & k = 2 m + 1 (m = 1, 2, \dots) \\ \frac{(2 m)!}{2^{m} m!} (σ^{2})^{m} & k = 2 m (m = 1, 2, \dots) \end{cases} \\ = {\begin{cases} 0 & k = 2 m + 1 (m = 1, 2, \dots) \\ \frac{Γ (m + \frac{1}{2})}{Γ (\frac{1}{2})} 2^{m} (σ^{2})^{m} & k = 2 m (m = 1, 2, \dots) \end{cases} \end{aligned}

$\begin{align} E[(X - \mu)^k] &= \begin{cases} 0 & k = 2m+1~~(m=1, 2, \ldots) \\ (k - 1)!! \sigma^k & k = 2m~~(m=1, 2, \ldots) \end{cases} \\ &= \begin{cases} 0 & k = 2m+1~~(m=1, 2, \ldots) \\ \dfrac{(2m)!}{2^m\,m!} (\sigma^2)^m & k = 2m~~(m=1, 2, \ldots) \end{cases} \\ &= \begin{cases} 0 & k = 2m+1~~(m=1, 2, \ldots) \\ \dfrac{\varGamma\left(m+\frac{1}{2}\right)}{\varGamma\left(\frac{1}{2}\right)}2^m (\sigma^2)^m & k = 2m~~(m=1, 2, \ldots) \end{cases} \end{align}$

再生性

$X \sim N(\mu_{X},~\sigma_{X}),~Y \sim N(\mu_{Y},~\sigma_{Y})$ のときは，

M_{X + Y} (t) = \exp [(μ_{X} + μ_{Y}) t + \frac{{σ_{X}}^{2} + {σ_{Y}}^{2}}{2} t^{2}]

$\begin{equation} M_{X+Y}(t) = \exp\left[(\mu_{X}+\mu_{Y})t+\frac{{\sigma_{X}}^{2}+{\sigma_{Y}}^{2}}{2}t^{2}\right] \end{equation}$

よって，

X + Y \sim N (μ_{X} + μ_{Y}, σ_{X}^{2} + σ_{Y}^{2})

$\begin{equation} X + Y \sim N(\mu_{X} + \mu_{Y}, \sigma_{X}^{2} + \sigma_{Y}^{2}) \end{equation}$

補足1: $E[X]$ の計算

f_{X} (x) = \frac{1}{\sqrt{2 π σ^{2}}} \exp [- \frac{(x - μ)^{2}}{2 σ^{2}}]

$\begin{equation} f_X(x) = \frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left[-\frac{(x-\mu)^{2}}{2\sigma^{2}}\right] \end{equation}$

ここで，

\begin{aligned} y = \frac{x - μ}{σ} \end{aligned}

$\begin{align} y = \frac{x - \mu}{\sigma} \end{align}$

とおくと，

x = σ y + μ, \frac{d x}{d y} = σ

$x = \sigma y + \mu,~\dfrac{dx}{dy} = \sigma$ であるから，

\begin{aligned} f_{Y} (y) & = f_{X} (σ y + μ) σ \\ = \frac{1}{\sqrt{2 π σ^{2}}} \exp [- \frac{y^{2}}{2}] \cdot σ \\ = \frac{1}{\sqrt{2 π}} \exp [- \frac{y^{2}}{2}] \end{aligned}

$\begin{align} f_Y(y) &= f_X(\sigma y + \mu)\,\sigma \\ &= \frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left[-\frac{y^2}{2}\right]\cdot\sigma \\ &= \frac{1}{\sqrt{2\pi}}\exp\left[-\frac{y^2}{2}\right] \\ \end{align}$

\begin{aligned} E [Y] & = \int_{- \infty}^{\infty} y \frac{1}{\sqrt{2 π}} \exp [- \frac{y^{2}}{2}] d y \\ = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} y \exp [- \frac{y^{2}}{2}] d y \\ = \frac{1}{\sqrt{2 π}} {[- e^{- \frac{y^{2}}{2}}]}_{- \infty}^{\infty} \\ = 0 \end{aligned}

$\begin{align} E[Y] &= \int_{-\infty}^{\infty}y\,\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{y^2}{2}\right]\,dy \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}y\,\exp\left[-\frac{y^2}{2}\right]\,dy \\ &= \frac{1}{\sqrt{2\pi}}\left[-e^{-\frac{y^2}{2}}\right]_{-\infty}^{\infty} \\ &= 0 \end{align}$

よって，

E [X] = E [σ Y + μ] = σ E [Y] + μ = μ

$\begin{equation} E[X] = E[\sigma Y + \mu] = \sigma E[Y] + \mu = \mu \end{equation}$

補足2: $V[X]$ の計算

\begin{aligned} E [Y^{2}] & = \int_{- \infty}^{\infty} y^{2} \frac{1}{\sqrt{2 π}} \exp [- \frac{y^{2}}{2}] d y \\ = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} y \cdot y \exp [- \frac{y^{2}}{2}] d y \\ = \frac{1}{\sqrt{2 π}} {{[- y e^{- \frac{y^{2}}{2}}]}_{- \infty}^{\infty} + \int_{- \infty}^{\infty} e^{- \frac{y^{2}}{2}} d y} \\ = \frac{1}{\sqrt{2 π}} {0 + \sqrt{2 π}} \\ = 1 \end{aligned}

$\begin{align} E[Y^2] &= \int_{-\infty}^{\infty}y^2\,\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{y^2}{2}\right]\,dy \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}y \cdot y\,\exp\left[-\frac{y^2}{2}\right]\,dy \\ &= \frac{1}{\sqrt{2\pi}}\left\{\left[-y\,e^{-\frac{y^2}{2}}\right]_{-\infty}^{\infty} + \int_{-\infty}^{\infty}e^{-\frac{y^2}{2}}\,dy\right\} \\ &= \frac{1}{\sqrt{2\pi}}\left\{0 + \sqrt{2\pi}\right\} \\ &= 1 \end{align}$

V [Y] = E [Y^{2}] - (E [Y])^{2} = 1 - 0 = 1

$\begin{equation} V[Y] = E[Y^2] - (E[Y])^2 = 1 - 0 = 1 \end{equation}$

よって，

V [X] = V [σ Y + μ] = σ^{2} V [Y] = σ^{2}

$\begin{equation} V[X] = V[\sigma Y + \mu] = \sigma^2 V[Y] = \sigma^2 \end{equation}$

補足3: $M_X(t)$ の計算

\begin{aligned} M_{X} (t) & = \int_{- \infty}^{\infty} e^{t x} \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} x^{2}} d x \\ = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} x^{2} + t x} d x \\ = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} (x^{2} - 2 t x + t^{2} - t^{2})} d x \\ = e^{\frac{1}{2} t^{2}} \cdot \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} (x^{2} - 2 t x + t^{2})} d x \\ = e^{\frac{1}{2} t^{2}} \cdot \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} (x - t)^{2}} d x \\ = e^{\frac{1}{2} t^{2}} \end{aligned}

$\begin{align} M_X(t) &= \int_{-\infty}^{\infty}e^{tx}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\,dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}x^2 + tx}\,dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\left(x^2 - 2 t x + t^2 - t^2\right)}\,dx \\ &= e^{\frac{1}{2}t^2} \cdot \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\left(x^2 - 2 t x + t^2\right)}\,dx \\ &= e^{\frac{1}{2}t^2} \cdot \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x - t)^2}\,dx \\ &= e^{\frac{1}{2}t^2} \end{align}$

補足4: 二重階乗

$k \in N$ のとき，

k!! = {\begin{cases} 2^{m} m! & k = 2 m (m = 1, 2, \dots) \\ \frac{(2 m)!}{2^{m} m!} & k = 2 m - 1 (m = 1, 2, \dots) \end{cases}

$\begin{equation} k!! = \begin{cases} 2^m\,m! & k = 2m~~(m = 1, 2, \ldots) \\ \dfrac{(2m)!}{2^m\,m!} & k = 2m-1~~(m = 1, 2, \ldots) \end{cases} \end{equation}$

第2講 確率分布

正規分布

再生性

補足1: E[X]E[X] の計算

補足2: V[X]V[X] の計算

補足3: MX(t)M_X(t) の計算

補足4: 二重階乗

第2講確率分布

補足1: $E[X]$ の計算

補足2: $V[X]$ の計算

補足3: $M_X(t)$ の計算