連続一様分布の期待値・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、 $$\begin{array}{rl}E\left(X\right)& =\frac{1}{b-a}{\int }_a^{b}x\cdot dx\\ & =\frac{1}{b-a}{\left[\frac{1}{2}{x}^2\right]}_a^b\\ & =\frac{1}{b-a}(\frac{b^2}{2}-\frac{a^2}{2})\\ & =\frac{1}{b-a}\cdot \frac{(b+a)(b-a)}{2}\\ & =\frac{a+b}{2}\end{array}$$ $◼$

（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、 $$\begin{array}{rl}E\left(X^2\right)& =\frac{1}{b-a}{\int }_a^b{x}^{2}dx\\ & =\frac{1}{b-a}{\left[\frac{1}{3}{x}^3\right]}_a^b\\ & =\frac{1}{b-a}(\frac{b^3}{3}-\frac{a^3}{3})\\ & =\frac{1}{b-a}\cdot \frac{(b-a)(b^2+ab+a^2)}{3}\\ & =\frac{a^2+ab+b^2}{3}\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =\frac{a^2+ab+b^2}{3}-{\left(\frac{a+b}{2}\right)}^2\\ & =\frac{a^2+ab+b^2}{3}-\frac{a^2+2ab+b^2}{4}\\ & =\frac{a^2-2ab+b^2}{12}\\ & =\frac{{(b-a)}^2}{12}\end{array}$$ $◼$

（i）期待値
連続一様分布のモーメント母関数の公式より、 $$\begin{array}{r}{M}_X\left(\theta \right)=\frac{e^{\theta b}-e^{\theta a}}{\theta (b-a)}\end{array}$$ モーメント母関数の1階微分を求めると、商の微分公式と合成関数の微分法より、 $$\begin{array}{rl}{M}_X^{\left(1\right)}\left(\theta \right)& =\frac{e^{\theta b}-e^{\theta a}}{b-a}\cdot \frac{d}{d\theta }\left(\frac{1}{\theta }\right)+\frac{1}{\theta (b-a)}\cdot \frac{d}{d\theta }(e^{\theta b}-e^{\theta a})\\ & =-\frac{e^{\theta b}-e^{\theta a}}{{\theta }^2(b-a)}+\frac{b{e}^{\theta b}-a{e}^{\theta a}}{\theta (b-a)}\\ & =-\frac{e^{\theta b}-e^{\theta a}}{{\theta }^2(b-a)}+\frac{b\theta e^{\theta b}-e^{\theta b}-a\theta e^{\theta a}+e^{\theta a}}{{\theta }^2(b-a)}\\ & =\frac{e^{\theta b}(b\theta -1)-e^{\theta a}(a\theta -1)}{{\theta }^2(b-a)}\end{array}$$ 1次モーメントとモーメント母関数の関係 $M_X^{\left(1\right)}\left(0\right)=E\left(X\right)$ より、 $$\begin{array}{rl}E\left(X\right)& =\frac{e^0(0-1)-e^0(0-1)}{0^2(b-a)}\\ & =\frac{-1+1}{0^2(b-a)}\\ & =\frac{0}{0}\end{array}$$ ここで、 $$\begin{array}{c}f\left(\theta \right)={\theta }^2(b-a)\\ g\left(\theta \right)=g_1\left(\theta \right)-g_2\left(\theta \right)\\ g_1\left(\theta \right)=e^{\theta b}(b\theta -1)g_2\left(\theta \right)=e^{\theta a}(a\theta -1)\end{array}$$ とすると、 $$\begin{array}{c}{f}^{\prime }\left(\theta \right)=2\theta (b-a)\\ f^{\prime \prime }\left(\theta \right)=2(b-a)\end{array}$$ $$\begin{array}{rl}{g}_1^{\prime }\left(\theta \right)& =(b\theta -1)\cdot \frac{d}{d\theta }{e}^{\theta b}+e^{\theta b}\cdot \frac{d}{d\theta }(b\theta -1)\\ & =b{e}^{\theta b}(b\theta -1)+b{e}^{\theta b}\\ & =b{e}^{\theta b}(b\theta -1+1)\\ & =b^2\cdot \theta e^{\theta b}\end{array}$$ $$\begin{array}{rl}{g}_1^{\prime \prime }\left(\theta \right)& =b^2\theta \cdot \frac{d}{d\theta }{e}^{\theta b}+b^2{e}^{\theta b}\cdot \frac{d\theta }{d\theta }\\ & =b^2\theta \cdot b{e}^{\theta b}+b^2{e}^{\theta b}\\ & =b^2{e}^{\theta b}(b\theta +1)\end{array}$$ 同様に、 $$\begin{array}{c}{g}_1^{\prime \prime }\left(\theta \right)=a^2{e}^{\theta a}(a\theta +1)\end{array}$$ したがって、 $$\begin{array}{c}{g}^{\prime \prime }\left(\theta \right)=b^2{e}^{\theta b}(b\theta +1)-a^2{e}^{\theta a}(a\theta +1)\end{array}$$ $f^{\prime \prime }\left(\theta \right),g^{\prime \prime }\left(\theta \right)$ の $\theta \to 0$ のときの極限を取ると、 $$\begin{array}{c}\underset{\theta \to 0}{lim}{f}^{\prime \prime }\left(\theta \right)=2(b-a)\\ \underset{\theta \to 0}{lim}{g}^{\prime \prime }\left(\theta \right)=b^2{e}^0(0+1)-a^2{e}^0(0+1)=b^2-a^2\end{array}$$ したがって、 $$\begin{array}{r}\underset{\theta \to 0}{lim}{f}^{\prime \prime }\left(\theta \right)\ne 0\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{f\left(\theta \right)}=\frac{0}{0}\end{array}$$ よって適用条件を満たすので、ロピタルの定理により、 $$\begin{array}{rl}E\left(X\right)& =\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{f\left(\theta \right)}\\ & =\underset{\theta \to 0}{lim}\frac{g^{\prime \prime }\left(\theta \right)}{f^{\prime \prime }\left(\theta \right)}\\ & =\frac{b^2-a^2}{2(b-a)}\\ & =\frac{(b+a)(b-a)}{2(b-a)}\\ & =\frac{b+a}{2}\end{array}$$ $◼$

（ii）分散
$$\begin{array}{r}{M}_X^{\left(1\right)}\left(\theta \right)=\frac{e^{\theta b}(b\theta -1)}{{\theta }^2(b-a)}-\frac{e^{\theta a}(a\theta -1)}{{\theta }^2(b-a)}\end{array}$$ ここで、 $$\begin{array}{c}{h}_1\left(\theta \right)=\frac{e^{\theta b}(b\theta -1)}{{\theta }^2(b-a)}{h}_2\left(\theta \right)=\frac{e^{\theta a}(a\theta -1)}{{\theta }^2(b-a)}\end{array}$$ として微分すると、 $$\begin{array}{rl}{h}_1^{\prime }\left(\theta \right)& =\frac{e^{\theta b}(b\theta -1)}{b-a}\cdot \frac{d}{d\theta }\left(\frac{1}{{\theta }^2}\right)+\frac{1}{{\theta }^2(b-a)}\cdot \frac{d}{d\theta }\left\{e^{\theta b}(b\theta -1)\right\}\\ & =-\frac{2e^{\theta b}(b\theta -1)}{{\theta }^3(b-a)}+\frac{1}{{\theta }^2(b-a)}\{e^{\theta b}\cdot \frac{d}{d\theta }(b\theta -1)+(b\theta -1)\cdot \frac{d}{d\theta }{e}^{\theta b}\}\\ & =-\frac{2e^{\theta b}(b\theta -1)}{{\theta }^3(b-a)}+\frac{b^2\theta e^{\theta b}}{{\theta }^2(b-a)}\\ & =\frac{b^2{\theta }^2{e}^{\theta b}-2e^{\theta b}(b\theta -1)}{{\theta }^3(b-a)}\\ & =\frac{(b^2{\theta }^2-2b\theta +2)e^{\theta b}}{{\theta }^3(b-a)}\end{array}$$ 同様に、 $$\begin{array}{rl}{h}_2^{\prime }\left(\theta \right)& =\frac{(a^2{\theta }^2-2a\theta +2)e^{\theta a}}{{\theta }^3(b-a)}\end{array}$$ したがって、モーメント母関数の2階微分は、 $$\begin{array}{r}{M}_X^{\left(2\right)}\left(\theta \right)=\frac{(b^2{\theta }^2-2b\theta +2)e^{\theta b}-(a^2{\theta }^2-2a\theta +2)e^{\theta a}}{{\theta }^3(b-a)}\end{array}$$ 2次モーメントとモーメント母関数の関係 $M_X^{\left(2\right)}\left(0\right)=E\left(X^2\right)$ より、 $$\begin{array}{rl}E\left(X^2\right)& =\frac{(b^2\cdot 0^2-2b\cdot 0+2)e^0-(a^2\cdot 0^2-2a\cdot 0+2)e^0}{0^3(b-a)}\\ & =\frac{2-2}{0^3(b-a)}\\ & =\frac{0}{0}\end{array}$$ ここで、 $$\begin{array}{c}f\left(\theta \right)={\theta }^3(b-a)\\ g\left(\theta \right)=g_1\left(\theta \right)-g_2\left(\theta \right)\\ g_1\left(\theta \right)=(b^2{\theta }^2-2b\theta +2)e^{\theta b}{g}_2\left(\theta \right)=(a^2{\theta }^2-2a\theta +2)e^{\theta a}\end{array}$$ とすると、 $$\begin{array}{c}{f}^{\prime }\left(\theta \right)=3{\theta }^2(b-a)\\ f^{\prime \prime }\left(\theta \right)=6\theta (b-a)\\ f^{\prime \prime \prime }\left(\theta \right)=6(b-a)\end{array}$$ $$\begin{array}{rl}{g}_1^{\prime }\left(\theta \right)& =e^{\theta b}\cdot \frac{d}{d\theta }(b^2{\theta }^2-2b\theta +2)+(b^2{\theta }^2-2b\theta +2)\cdot \frac{d}{d\theta }{e}^{\theta b}\\ & =(2b^2\theta -2b)e^{\theta b}+(b^3{\theta }^2-2b^2\theta +2b)e^{\theta b}\\ & =b^3{\theta }^2{e}^{\theta b}\end{array}$$ $$\begin{array}{rl}{g}_1^{\prime \prime }\left(\theta \right)& =b^3{\theta }^2\cdot \frac{d}{d\theta }{e}^{\theta b}+b^3{e}^{\theta b}\cdot \frac{d}{d\theta }{\theta }^2\\ & =b^4{\theta }^2{e}^{\theta b}+2b^3\theta e^{\theta b}\\ & =b^3(b{\theta }^2+2\theta )e^{\theta b}\end{array}$$ $$\begin{array}{rl}{g}_1^{\prime \prime \prime }\left(\theta \right)& =b^3(b{\theta }^2+2\theta )\cdot \frac{d}{d\theta }{e}^{\theta b}+b^3{e}^{\theta b}\cdot \frac{d}{d\theta }(b{\theta }^2+2\theta )\\ & =b^4(b{\theta }^2+2\theta )e^{\theta b}+b^3{e}^{\theta b}(2b\theta +2)\\ & =b^3(b^2{\theta }^2+4b\theta +2)e^{\theta b}\end{array}$$ 同様に、 $$\begin{array}{rl}{g}_2^{\prime \prime \prime }\left(\theta \right)& =a^3(a^2{\theta }^2+4a\theta +2)e^{\theta a}\end{array}$$ したがって、 $$\begin{array}{c}{g}^{\prime \prime \prime }\left(\theta \right)=b^3(b^2{\theta }^2+4b\theta +2)e^{\theta b}-a^3(a^2{\theta }^2+4a\theta +2)e^{\theta a}\end{array}$$ $f^{\prime \prime }\left(\theta \right),g^{\prime \prime }\left(\theta \right)$ の $\theta \to 0$ のときの極限を取ると、 $$\begin{array}{r}\underset{\theta \to 0}{lim}{f}^{\prime \prime \prime }\left(\theta \right)=6(b-a)\end{array}$$ $$\begin{array}{rl}\underset{\theta \to 0}{lim}{g}^{\prime \prime \prime }\left(\theta \right)& =b^3(b^2\cdot 0^2+4b\cdot 0+2)e^0-a^3(a^2\cdot 0^2+4a\cdot 0+2)e^0\\ & =2(b^3-a^3)\end{array}$$ したがって、 $$\begin{array}{r}\underset{\theta \to 0}{lim}{f}^{\prime \prime \prime }\left(\theta \right)\ne 0\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{f\left(\theta \right)}=\frac{0}{0}\end{array}$$ よって適用条件を満たすので、ロピタルの定理により、 $$\begin{array}{rl}E\left(X\right)& =\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{f\left(\theta \right)}\\ & =\underset{\theta \to 0}{lim}\frac{g^{\prime \prime \prime }\left(\theta \right)}{f^{\prime \prime \prime }\left(\theta \right)}\\ & =\frac{2(b^3-a^3)}{6(b-a)}\\ & =\frac{(b^2+ab+a^2)(b-a)}{3(b-a)}\\ & =\frac{b^2+ab+a^2}{3}\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =\frac{a^2+ab+b^2}{3}-{\left(\frac{a+b}{2}\right)}^2\\ & =\frac{a^2+ab+b^2}{3}-\frac{a^2+2ab+b^2}{4}\\ & =\frac{a^2-2ab+b^2}{12}\\ & =\frac{{(b-a)}^2}{12}\end{array}$$ $◼$