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 $E\left(X\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\cdot f\left(x\right)dx$ より、 $$\begin{array}{rl}E\left(X\right)& ={\int }_{-\mathrm{\infty }}^{0}x\cdot 0dx+{\int }_0^{1}x\cdot f\left(x\right)dx+{\int }_1^{\mathrm{\infty }}x\cdot 0dx\\ & ={\int }_0^{1}x\cdot f\left(x\right)dx\\ & ={\int }_0^{1}x\cdot \frac{x^{\alpha -1}{(1-x)}^{\beta -1}}{B(\alpha ,\beta )}dx\\ & =\frac{1}{B(\alpha ,\beta )}{\int }_0^1{x}^{\alpha }{(1-x)}^{\beta -1}dx\\ & =\frac{1}{B(\alpha ,\beta )}{\int }_0^1{x}^{(a+1)-1}{(1-x)}^{\beta -1}dx\end{array}$$ ベータ関数の定義式 $B(\alpha +1,\beta )={\int }_0^1{x}^{(a+1)-1}{(1-x)}^{\beta -1}dx$ より、 $$\begin{array}{r}E\left(X\right)=\frac{B(\alpha +1,\beta )}{B(\alpha ,\beta )}\end{array}$$ ベータ関数の性質より、 $$\begin{array}{r}B(\alpha +1,\beta )=\frac{\alpha }{\alpha +\beta }B(\alpha ,\beta )\iff \frac{B(\alpha +1,\beta )}{B(\alpha ,\beta )}=\frac{\alpha }{\alpha +\beta }\end{array}$$ したがって、 $$\begin{array}{r}E\left(X\right)=\frac{\alpha }{\alpha +\beta }\end{array}$$ $◼$

（ii）分散
2乗の期待値の定義式 $E\left(X^2\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2\cdot f\left(x\right)dx$ より、 $$\begin{array}{rl}E\left(X^2\right)& ={\int }_{-\mathrm{\infty }}^0{x}^2\cdot 0dx+{\int }_0^1{x}^2\cdot f\left(x\right)dx+{\int }_1^{\mathrm{\infty }}{x}^2\cdot 0dx\\ & ={\int }_0^1{x}^2\cdot f\left(x\right)dx\\ & ={\int }_0^1{x}^2\cdot \frac{x^{\alpha -1}{(1-x)}^{\beta -1}}{B(\alpha ,\beta )}dx\\ & =\frac{1}{B(\alpha ,\beta )}{\int }_0^1{x}^{\alpha +1}{(1-x)}^{\beta -1}dx\\ & =\frac{1}{B(\alpha ,\beta )}{\int }_0^1{x}^{(\alpha +2)-1}{(1-x)}^{\beta -1}dx\end{array}$$ ベータ関数の定義式 $B(\alpha +2,\beta )={\int }_0^1{x}^{(a+2)-1}{(1-x)}^{\beta -1}dx$ より、 $$\begin{array}{r}E\left(X^2\right)=\frac{B(\alpha +2,\beta )}{B(\alpha ,\beta )}\end{array}$$ ベータ関数の性質より、 $$\begin{array}{c}B(\alpha +2,\beta )=\frac{\alpha (\alpha +1)}{(\alpha +\beta )(\alpha +\beta +1)}B(\alpha ,\beta )\\ \iff \frac{B(\alpha +2,\beta )}{B(\alpha ,\beta )}=\frac{\alpha (\alpha +1)}{(\alpha +\beta )(\alpha +\beta +1)}\end{array}$$ したがって、 $$\begin{array}{r}E\left(X^2\right)=\frac{\alpha (\alpha +1)}{(\alpha +\beta )(\alpha +\beta +1)}\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =\frac{\alpha (\alpha +1)}{(\alpha +\beta )(\alpha +\beta +1)}-\frac{{\alpha }^2}{{(\alpha +\beta )}^2}\\ & =\frac{\alpha (\alpha +1)(\alpha +\beta )-{\alpha }^2(\alpha +\beta +1)}{{(\alpha +\beta )}^2(\alpha +\beta +1)}\\ & =\frac{{\alpha }^3+{\alpha }^2\beta +{\alpha }^2+\alpha \beta -{\alpha }^3-{\alpha }^2\beta -{\alpha }^2}{{(\alpha +\beta )}^2(\alpha +\beta +1)}\\ & =\frac{\alpha \beta }{{(\alpha +\beta )}^2(\alpha +\beta +1)}\end{array}$$ $◼$

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