第2講確率分布

ベータ分布

独立に $U(0, 1)$ に従う $\alpha + \beta - 1$ 個の確率変数を値の昇順に並べ変えたとき，小さい方から $i$ 番目の確率変数 $X$ はベータ分布 $Be(\alpha, \beta)$ に従う．

X \sim B e (α, β)

$\begin{equation} X \sim Be(\alpha, \beta) \end{equation}$

α > 0, β > 0

$\alpha > 0, \beta > 0$ とする．

f (x) = {\begin{cases} \frac{1}{B (α, β)} x^{α - 1} (1 - x)^{β - 1} & (0 < x < 1) \\ 0 & (それ以外) \end{cases}

$\begin{equation} f(x) = \begin{cases}\ds \frac{1}{B(\alpha, \beta)}x^{\alpha-1}(1-x)^{\beta-1} & (0 < x < 1) \\ 0 & (\text{それ以外}) \end{cases} \end{equation}$

E [X] = \frac{α}{α + β}

$\begin{equation} E[X] = \frac{\alpha}{\alpha + \beta} \end{equation}$

V [X] = \frac{α β}{(α + β)^{2} (α + β + 1)}

$\begin{equation} V[X] = \frac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha + \beta + 1)} \end{equation}$

M o d e [X] = \frac{α - 1}{α + β - 2}

$\begin{equation} Mode[X] = \frac{\alpha - 1}{\alpha + \beta - 2} \end{equation}$

ガンマ関数とベータ関数の関係

B (α, β) = \frac{Γ (α) Γ (β)}{Γ (α + β)}

$\begin{equation} B(\alpha,\beta) = \frac{\varGamma(\alpha)\varGamma(\beta)}{\varGamma(\alpha + \beta)} \end{equation}$

X \sim G a (a, λ), Y \sim G a (b, λ)

$X \sim Ga(a, \lambda),~Y \sim Ga(b, \lambda)$ で独立のとき，

\frac{X}{X + Y} \sim B e (a, b)

$\begin{equation} \frac{X}{X + Y} \sim Be(a,b) \end{equation}$

補足1: $E[X]$ の計算

\begin{aligned} E [X] & = \int_{0}^{1} x \frac{x^{α - 1} (1 - x)^{β - 1}}{B (α, β)} d x \\ = \frac{1}{B (α, β)} \int_{0}^{1} x^{(α + 1) - 1} (1 - x)^{β - 1} d x \\ = \frac{1}{B (α, β)} B (α + 1, β) \\ = \frac{Γ (α + β)}{Γ (α) Γ (β)} \cdot \frac{Γ (α + 1) Γ (β)}{Γ (α + 1 + β)} \\ = \frac{α}{α + β} \end{aligned}

$\begin{align} E[X] &= \int_{0}^{1}x\,\frac{x^{\alpha - 1}(1 - x)^{\beta-1}}{B(\alpha, \beta)}\,dx \\ &= \frac{1}{B(\alpha, \beta)}\int_{0}^{1}x^{(\alpha + 1) - 1}(1 - x)^{\beta-1}\,dx \\ &= \frac{1}{B(\alpha, \beta)}B(\alpha + 1, \beta) \\ &= \frac{\varGamma(\alpha + \beta)}{\varGamma(\alpha)\,\varGamma(\beta)}\cdot\frac{\varGamma(\alpha + 1)\,\varGamma(\beta)}{\varGamma(\alpha + 1 + \beta)} \\ &= \frac{\alpha}{\alpha + \beta} \end{align}$

補足2: $V[X]$ の計算

\begin{aligned} E [X^{2}] & = \int_{0}^{1} x^{2} \frac{x^{α - 1} (1 - x)^{β - 1}}{B (α, β)} d x \\ = \frac{1}{B (α, β)} \int_{0}^{1} x^{(α + 2) - 1} (1 - x)^{β - 1} d x \\ = \frac{1}{B (α, β)} B (α + 2, β) \\ = \frac{Γ (α + β)}{Γ (α) Γ (β)} \cdot \frac{Γ (α + 2) Γ (β)}{Γ (α + 2 + β)} \\ = \frac{(α + 1) α}{(α + β) (α + β + 1)} \end{aligned}

$\begin{align} E[X^2] &= \int_{0}^{1}x^2\,\frac{x^{\alpha - 1}(1 - x)^{\beta-1}}{B(\alpha, \beta)}\,dx \\ &= \frac{1}{B(\alpha, \beta)}\int_{0}^{1}x^{(\alpha + 2) - 1}(1 - x)^{\beta-1}\,dx \\ &= \frac{1}{B(\alpha, \beta)}B(\alpha + 2, \beta) \\ &= \frac{\varGamma(\alpha + \beta)}{\varGamma(\alpha)\,\varGamma(\beta)}\cdot\frac{\varGamma(\alpha + 2)\,\varGamma(\beta)}{\varGamma(\alpha + 2 + \beta)} \\ &= \frac{(\alpha + 1)\alpha}{(\alpha + \beta)(\alpha + \beta + 1)} \end{align}$

\begin{aligned} V [X] & = E [X^{2}] - (E [X])^{2} \\ = \frac{(α + 1) α}{(α + β) (α + β + 1)} - {(\frac{α}{α + β})}^{2} \\ = \frac{(α + 1) α}{(α + β) (α + β + 1)} - \frac{α^{2}}{(α + β)^{2}} \\ = \frac{(α + β) (α + 1) α}{(α + β)^{2} (α + β + 1)} - \frac{α^{2} (α + β + 1)}{(α + β)^{2} (α + β + 1)} \\ = \frac{{α^{3} + (β + 1) α^{2} + α β} - {α^{3} + (β + 1) α^{2})}{(α + β)^{2} (α + β + 1)} \\ = \frac{α β}{(α + β)^{2} (α + β + 1)} \end{aligned}

$\begin{align} V[X] &= E[X^2] - (E[X])^2 \\ &= \frac{(\alpha + 1)\alpha}{(\alpha + \beta)(\alpha + \beta + 1)} - \left(\frac{\alpha}{\alpha + \beta}\right)^2 \\ &= \frac{(\alpha + 1)\alpha}{(\alpha + \beta)(\alpha + \beta + 1)} - \frac{\alpha^2}{(\alpha + \beta)^2} \\ &= \frac{(\alpha + \beta)(\alpha + 1)\alpha}{(\alpha + \beta)^2(\alpha + \beta + 1)} - \frac{\alpha^2(\alpha + \beta + 1)}{(\alpha + \beta)^2(\alpha + \beta + 1)} \\ &= \frac{\{\alpha^3 + (\beta + 1)\alpha^2 + \alpha\beta\} - \{\alpha^3 + (\beta + 1)\alpha^2)}{(\alpha + \beta)^2(\alpha + \beta + 1)} \\ &= \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)} \\ \end{align}$

第2講 確率分布

ベータ分布

ガンマ関数とベータ関数の関係

補足1: E[X]E[X] の計算

補足2: V[X]V[X] の計算

第2講確率分布

補足1: $E[X]$ の計算

補足2: $V[X]$ の計算