ベータ分布の期待値・分散と最頻値の導出

（i）期待値
期待値の定義式 $E \left(X\right)=\int_{-\infty}^{\infty}{x \cdot f \left(x\right)dx}$ より、 \begin{align} E \left(X\right)&=\int_{-\infty}^{0}{x \cdot 0dx}+\int_{0}^{1}{x \cdot f \left(x\right)dx}+\int_{1}^{\infty}{x \cdot 0dx}\\ &=\int_{0}^{1}{x \cdot f \left(x\right)dx}\\ &=\int_{0}^{1}{x \cdot \frac{x^{\alpha-1} \left(1-x\right)^{\beta-1}}{B \left(\alpha,\beta\right)}dx}\\ &=\frac{1}{B \left(\alpha,\beta\right)}\int_{0}^{1}{x^\alpha \left(1-x\right)^{\beta-1}dx}\\ &=\frac{1}{B \left(\alpha,\beta\right)}\int_{0}^{1}{x^{ \left(a+1\right)-1} \left(1-x\right)^{\beta-1}dx}\\ \end{align} ベータ関数の定義式 $B \left(\alpha+1,\beta\right)=\int_{0}^{1}{x^{ \left(a+1\right)-1} \left(1-x\right)^{\beta-1}dx}$ より、 \begin{align} E \left(X\right)=\frac{B \left(\alpha+1,\beta\right)}{B \left(\alpha,\beta\right)}\\ \end{align} ベータ関数の性質より、 \begin{align} B \left(\alpha+1,\beta\right)=\frac{\alpha}{\alpha+\beta}B \left(\alpha,\beta\right)\Leftrightarrow\frac{B \left(\alpha+1,\beta\right)}{B \left(\alpha,\beta\right)}=\frac{\alpha}{\alpha+\beta} \end{align} したがって、 \begin{align} E \left(X\right)=\frac{\alpha}{\alpha+\beta} \end{align} $\blacksquare$

（ii）分散
2乗の期待値の定義式 $E \left(X^2\right)=\int_{-\infty}^{\infty}{x^2 \cdot f \left(x\right)dx}$ より、 \begin{align} E \left(X^2\right)&=\int_{-\infty}^{0}{x^2 \cdot 0dx}+\int_{0}^{1}{x^2 \cdot f \left(x\right)dx}+\int_{1}^{\infty}{x^2 \cdot 0dx}\\ &=\int_{0}^{1}{x^2 \cdot f \left(x\right)dx}\\ &=\int_{0}^{1}{x^2 \cdot \frac{x^{\alpha-1} \left(1-x\right)^{\beta-1}}{B \left(\alpha,\beta\right)}dx}\\ &=\frac{1}{B \left(\alpha,\beta\right)}\int_{0}^{1}{x^{\alpha+1} \left(1-x\right)^{\beta-1}dx}\\ &=\frac{1}{B \left(\alpha,\beta\right)}\int_{0}^{1}{x^{ \left(\alpha+2\right)-1} \left(1-x\right)^{\beta-1}dx} \end{align} ベータ関数の定義式 $B \left(\alpha+2,\beta\right)=\int_{0}^{1}{x^{ \left(a+2\right)-1} \left(1-x\right)^{\beta-1}dx}$ より、 \begin{align} E \left(X^2\right)=\frac{B \left(\alpha+2,\beta\right)}{B \left(\alpha,\beta\right)}\\ \end{align} ベータ関数の性質より、 \begin{gather} B \left(\alpha+2,\beta\right)=\frac{\alpha \left(\alpha+1\right)}{ \left(\alpha+\beta\right) \left(\alpha+\beta+1\right)}B \left(\alpha,\beta\right)\\ \Leftrightarrow\frac{B \left(\alpha+2,\beta\right)}{B \left(\alpha,\beta\right)}=\frac{\alpha \left(\alpha+1\right)}{ \left(\alpha+\beta\right) \left(\alpha+\beta+1\right)} \end{gather} したがって、 \begin{align} E \left(X^2\right)=\frac{\alpha \left(\alpha+1\right)}{ \left(\alpha+\beta\right) \left(\alpha+\beta+1\right)} \end{align} 分散の公式 $V \left(X\right)=E \left(X^2\right)- \left\{E \left(X\right)\right\}^2$ より、 \begin{align} V \left(X\right)&=\frac{\alpha \left(\alpha+1\right)}{ \left(\alpha+\beta\right) \left(\alpha+\beta+1\right)}-\frac{\alpha^2}{ \left(\alpha+\beta\right)^2}\\ &=\frac{\alpha \left(\alpha+1\right) \left(\alpha+\beta\right)-\alpha^2 \left(\alpha+\beta+1\right)}{ \left(\alpha+\beta\right)^2 \left(\alpha+\beta+1\right)}\\ &=\frac{\alpha^3+\alpha^2\beta+\alpha^2+\alpha\beta-\alpha^3-\alpha^2\beta-\alpha^2}{ \left(\alpha+\beta\right)^2 \left(\alpha+\beta+1\right)}\\ &=\frac{\alpha\beta}{ \left(\alpha+\beta\right)^2 \left(\alpha+\beta+1\right)} \end{align} $\blacksquare$

ベータ分布の確率密度関数は、 \begin{align} f \left(x\right)=\frac{1}{B \left(\alpha,\beta\right)}x^{\alpha-1} \left(1-x\right)^{\beta-1} \end{align} 1階微分を求めると、積の微分公式より、 \begin{align} f^\prime \left(x\right)&=\frac{1}{B \left(\alpha,\beta\right)} \left\{ \left(1-x\right)^{\beta-1} \cdot \frac{d}{dx} \left(x^{\alpha-1}\right)-x^{\alpha-1} \cdot \frac{d}{dx} \left(1-x\right)^{\beta-1}\right\}\\ &=\frac{1}{B \left(\alpha,\beta\right)} \left\{ \left(\alpha-1\right)x^{\alpha-2} \cdot \left(1-x\right)^{\beta-1}+x^{\alpha-1} \cdot \left(\beta-1\right) \left(1-x\right)^{\beta-2} \cdot \frac{d}{dx} \left(1-x\right)\right\}\\ &=\frac{1}{B \left(\alpha,\beta\right)} \left\{ \left(\alpha-1\right)x^{\alpha-2} \cdot \left(1-x\right)^{\beta-1}-x^{\alpha-1} \cdot \left(\beta-1\right) \left(1-x\right)^{\beta-2}\right\}\\ &=\frac{1}{B \left(\alpha,\beta\right)}x^{\alpha-2} \left(1-x\right)^{\beta-2} \left\{ \left(\alpha-1\right) \left(1-x\right)- \left(\beta-1\right)x\right\}\\ &=\frac{1}{B \left(\alpha,\beta\right)}x^{\alpha-2} \left(1-x\right)^{\beta-2} \left\{ \left(\alpha-1\right)- \left(\alpha+\beta-2\right)x\right\}\\ \end{align} 極値 $f^\prime \left(x\right)=0$ を求めると、 \begin{align} x=0 \quad x=1 \quad x=\frac{\alpha-1}{\alpha+\beta-2} \end{align} 増減表は、 \begin{array}{c|ccccc} x & 0 & \cdots & \displaystyle\frac{\alpha-1}{\alpha+\beta-2} & \cdots & 1 \\ \hline f^\prime \left(x\right) & 0 & + & 0 & - & 0 \\ \hline f \left(x\right) & 0 & \nearrow & f \left(\frac{\alpha-1}{\alpha+\beta-2}\right) & \searrow & 0 \\ \end{array} したがって、$1 \lt \alpha,1 \lt \beta$ のとき、$0 \lt x \lt 1$ の範囲で、 \begin{align} x=\frac{\alpha-1}{\alpha+\beta-2} \end{align} が極大、かつ最大である。 $\blacksquare$