ワイブル分布の期待値・分散の導出

（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、 $$\begin{array}{rl}E\left(X\right)& ={\int }_{-\mathrm{\infty }}^{0}x\cdot 0dx+{\int }_0^{\mathrm{\infty }}x\cdot f\left(x\right)dx\\ & ={\int }_0^{\mathrm{\infty }}x\cdot \frac{b{x}^{b-1}}{a^b}{e}^{-{\left(\frac{x}{a}\right)}^b}dx\\ & =\frac{b}{a^b}{\int }_0^{\mathrm{\infty }}{x}^b{e}^{-{\left(\frac{x}{a}\right)}^b}dx\end{array}$$ ここで、以下のように変数変換すると、 $$\begin{array}{c}t={\left(\frac{x}{a}\right)}^b\iff x=a{t}^{\frac{1}{b}}\\ \frac{dx}{dt}=\frac{a}{b}{t}^{\frac{1}{b}-1}\Rightarrow dx=\frac{a}{b}{t}^{\frac{1}{b}-1}dt\\ x:0\to \mathrm{\infty }\Rightarrow t:0\to \mathrm{\infty }\end{array}$$ となるので、 置換積分法により、 $$\begin{array}{rl}E\left(X\right)& =\frac{b}{a^b}{\int }_0^{\mathrm{\infty }}{\left(a{t}^{\frac{1}{b}}\right)}^b{e}^{-t}\cdot \frac{a}{b}{t}^{\frac{1}{b}-1}dt\\ & =\frac{b}{a^b}\cdot \frac{a}{b}\cdot a^b{\int }_0^{\mathrm{\infty }}{t}^{\frac{b+1}{b}-1}{e}^{-t}dt\\ & =a{\int }_0^{\mathrm{\infty }}{t}^{\left(\frac{b+1}{b}\right)-1}{e}^{-t}dt\end{array}$$ ガンマ関数の定義式 $\mathrm{\Gamma }\left(\alpha \right)={\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}\cdot e^{-x}dx$ より、 $$\begin{array}{r}\mathrm{\Gamma }\left(\frac{b+1}{b}\right)={\int }_0^{\mathrm{\infty }}{t}^{\left(\frac{b+1}{b}\right)-1}{e}^{-t}dt\end{array}$$ よって、 $$\begin{array}{r}E\left(X\right)=a\mathrm{\Gamma }\left(\frac{b+1}{b}\right)\end{array}$$ $◼$

（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、 $$\begin{array}{rl}E\left(X^2\right)& ={\int }_{-\mathrm{\infty }}^0{x}^2\cdot 0dx+{\int }_0^{\mathrm{\infty }}{x}^2\cdot f\left(x\right)dx\\ & ={\int }_0^{\mathrm{\infty }}{x}^2\cdot \frac{b{x}^{b-1}}{a^b}{e}^{-{\left(\frac{x}{a}\right)}^b}dx\\ & =\frac{b}{a^b}{\int }_0^{\mathrm{\infty }}{x}^{b+1}{e}^{-{\left(\frac{x}{a}\right)}^b}dx\end{array}$$ （i）と同様の変数変換を行うと、置換積分法により、 $$\begin{array}{rl}E\left(X^2\right)& =\frac{b}{a^b}{\int }_0^{\mathrm{\infty }}{\left(a{t}^{\frac{1}{b}}\right)}^{b+1}{e}^{-t}\cdot \frac{a}{b}{t}^{\frac{1}{b}-1}dt\\ & =\frac{b}{a^b}\cdot \frac{a}{b}\cdot a^{b+1}{\int }_0^{\mathrm{\infty }}{t}^{\frac{b+2}{b}-1}{e}^{-t}dt\\ & =a^2{\int }_0^{\mathrm{\infty }}{t}^{\left(\frac{b+2}{b}\right)-1}{e}^{-t}dt\end{array}$$ ガンマ関数の定義式 $\mathrm{\Gamma }\left(\alpha \right)={\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}\cdot e^{-x}dx$ より、 $$\begin{array}{r}\mathrm{\Gamma }\left(\frac{b+2}{b}\right)={\int }_0^{\mathrm{\infty }}{t}^{\left(\frac{b+2}{b}\right)-1}{e}^{-t}dt\end{array}$$ よって、 $$\begin{array}{r}E\left(X^2\right)=a^2\mathrm{\Gamma }\left(\frac{b+2}{b}\right)\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =a^2\mathrm{\Gamma }\left(\frac{b+2}{b}\right)-a^2{\left\{\mathrm{\Gamma }\left(\frac{b+1}{b}\right)\right\}}^2\\ & =a^2[\mathrm{\Gamma }\left(\frac{b+2}{b}\right)-{\left\{\mathrm{\Gamma }\left(\frac{b+1}{b}\right)\right\}}^2]\end{array}$$ $◼$