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 $E \left(X\right)=\int_{-\infty}^{\infty}{x \cdot f \left(x\right)dx}$ より、 \begin{align} E \left(X\right)&=\frac{1}{b-a}\int_{a}^{b}{x \cdot d x}\\ &=\frac{1}{b-a} \left[\frac{1}{2}x^2\right]_a^b\\ &=\frac{1}{b-a} \left(\frac{b^2}{2}-\frac{a^2}{2}\right)\\ &=\frac{1}{b-a} \cdot \frac{ \left(b+a\right) \left(b-a\right)}{2}\\ &=\frac{a+b}{2} \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)&=\frac{1}{b-a}\int_{a}^{b}{x^2dx}\\ &=\frac{1}{b-a} \left[\frac{1}{3}x^3\right]_a^b\\ &=\frac{1}{b-a} \left(\frac{b^3}{3}-\frac{a^3}{3}\right)\\ &=\frac{1}{b-a} \cdot \frac{ \left(b-a\right) \left(b^2+ab+a^2\right)}{3}\\ &=\frac{a^2+ab+b^2}{3} \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{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{ \left(b-a\right)^2}{12} \end{align} $\blacksquare$

（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{align} M_X \left(\theta\right)=\frac{e^{\theta b}-e^{\theta a}}{\theta \left(b-a\right)} \end{align} 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{align} 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 \left(b-a\right)} \cdot \frac{d}{d\theta} \left(e^{\theta b}-e^{\theta a}\right)\\ &=-\frac{e^{\theta b}-e^{\theta a}}{\theta^2 \left(b-a\right)}+\frac{be^{\theta b}-ae^{\theta a}}{\theta \left(b-a\right)}\\ &=-\frac{e^{\theta b}-e^{\theta a}}{\theta^2 \left(b-a\right)}+\frac{b\theta e^{\theta b}-e^{\theta b}-a\theta e^{\theta a}+e^{\theta a}}{\theta^2 \left(b-a\right)}\\ &=\frac{e^{\theta b} \left(b\theta-1\right)-e^{\theta a} \left(a\theta-1\right)}{\theta^2 \left(b-a\right)} \end{align} 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{align} E \left(X\right)&=\frac{e^0 \left(0-1\right)-e^0 \left(0-1\right)}{0^2 \left(b-a\right)}\\ &=\frac{-1+1}{0^2 \left(b-a\right)}\\ &=\frac{0}{0} \end{align} ここで、 \begin{gather} f \left(\theta\right)=\theta^2 \left(b-a\right)\\ g \left(\theta\right)=g_1 \left(\theta\right)-g_2 \left(\theta\right)\\ g_1 \left(\theta\right)=e^{\theta b} \left(b\theta-1\right) \quad g_2 \left(\theta\right)=e^{\theta a} \left(a\theta-1\right) \end{gather} とすると、 \begin{gather} f^\prime \left(\theta\right)=2\theta \left(b-a\right)\\ f^{\prime\prime} \left(\theta\right)=2 \left(b-a\right) \end{gather} \begin{align} g_1^\prime \left(\theta\right)&= \left(b\theta-1\right) \cdot \frac{d}{d\theta}e^{\theta b}+e^{\theta b} \cdot \frac{d}{d\theta} \left(b\theta-1\right)\\ &=be^{\theta b} \left(b\theta-1\right)+be^{\theta b}\\ &=be^{\theta b} \left(b\theta-1+1\right)\\ &=b^2 \cdot \theta e^{\theta b} \end{align} \begin{align} g_1^{\prime\prime} \left(\theta\right)&=b^2\theta \cdot \frac{d}{d\theta}e^{\theta b}+b^2e^{\theta b} \cdot \frac{d\theta}{d\theta}\\ &=b^2\theta \cdot be^{\theta b}+b^2e^{\theta b}\\ &=b^2e^{\theta b} \left(b\theta+1\right)\\ \end{align} 同様に、 \begin{gather} g_1^{\prime\prime} \left(\theta\right)=a^2e^{\theta a} \left(a\theta+1\right) \end{gather} したがって、 \begin{gather} g^{\prime\prime} \left(\theta\right)=b^2e^{\theta b} \left(b\theta+1\right)-a^2e^{\theta a} \left(a\theta+1\right) \end{gather} $f^{\prime\prime} \left(\theta\right),g^{\prime\prime} \left(\theta\right)$ の $\theta\rightarrow0$ のときの極限を取ると、 \begin{gather} \lim_{\theta\rightarrow0}{f^{\prime\prime} \left(\theta\right)}=2 \left(b-a\right)\\ \lim_{\theta\rightarrow0}{g^{\prime\prime} \left(\theta\right)}=b^2e^0 \left(0+1\right)-a^2e^0 \left(0+1\right)=b^2-a^2 \end{gather} したがって、 \begin{align} \lim_{\theta\rightarrow0}{f^{\prime\prime} \left(\theta\right)} \neq 0 \quad \lim_{\theta\rightarrow0}{\frac{g \left(\theta\right)}{f \left(\theta\right)}}=\frac{0}{0} \end{align} よって適用条件を満たすので、ロピタルの定理により、 \begin{align} E \left(X\right)&=\lim_{\theta\rightarrow0}{\frac{g \left(\theta\right)}{f \left(\theta\right)}}\\ &=\lim_{\theta\rightarrow0}{\frac{g^{\prime\prime} \left(\theta\right)}{f^{\prime\prime} \left(\theta\right)}}\\ &=\frac{b^2-a^2}{2 \left(b-a\right)}\\ &=\frac{ \left(b+a\right) \left(b-a\right)}{2 \left(b-a\right)}\\ &=\frac{b+a}{2} \end{align} $\blacksquare$

（ii）分散
\begin{align} M_X^{ \left(1\right)} \left(\theta\right)=\frac{e^{\theta b} \left(b\theta-1\right)}{\theta^2 \left(b-a\right)}-\frac{e^{\theta a} \left(a\theta-1\right)}{\theta^2 \left(b-a\right)} \end{align} ここで、 \begin{gather} h_1 \left(\theta\right)=\frac{e^{\theta b} \left(b\theta-1\right)}{\theta^2 \left(b-a\right)} \quad h_2 \left(\theta\right)=\frac{e^{\theta a} \left(a\theta-1\right)}{\theta^2 \left(b-a\right)} \end{gather} として微分すると、 \begin{align} h_1^\prime \left(\theta\right)&=\frac{e^{\theta b} \left(b\theta-1\right)}{b-a} \cdot \frac{d}{d\theta} \left(\frac{1}{\theta^2}\right)+\frac{1}{\theta^2 \left(b-a\right)} \cdot \frac{d}{d\theta} \left\{e^{\theta b} \left(b\theta-1\right)\right\}\\ &=-\frac{2e^{\theta b} \left(b\theta-1\right)}{\theta^3 \left(b-a\right)}+\frac{1}{\theta^2 \left(b-a\right)} \left\{e^{\theta b} \cdot \frac{d}{d\theta} \left(b\theta-1\right)+ \left(b\theta-1\right) \cdot \frac{d}{d\theta}e^{\theta b}\right\}\\ &=-\frac{2e^{\theta b} \left(b\theta-1\right)}{\theta^3 \left(b-a\right)}+\frac{b^2\theta e^{\theta b}}{\theta^2 \left(b-a\right)}\\ &=\frac{b^2\theta^2e^{\theta b}-2e^{\theta b} \left(b\theta-1\right)}{\theta^3 \left(b-a\right)}\\ &=\frac{ \left(b^2\theta^2-2b\theta+2\right)e^{\theta b}}{\theta^3 \left(b-a\right)} \end{align} 同様に、 \begin{align} h_2^\prime \left(\theta\right)&=\frac{ \left(a^2\theta^2-2a\theta+2\right)e^{\theta a}}{\theta^3 \left(b-a\right)}\\ \end{align} したがって、モーメント母関数の2階微分は、 \begin{align} M_X^{ \left(2\right)} \left(\theta\right)=\frac{ \left(b^2\theta^2-2b\theta+2\right)e^{\theta b}- \left(a^2\theta^2-2a\theta+2\right)e^{\theta a}}{\theta^3 \left(b-a\right)} \end{align} 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{align} E \left(X^2\right)&=\frac{ \left(b^2 \cdot 0^2-2b \cdot 0+2\right)e^0- \left(a^2 \cdot 0^2-2a \cdot 0+2\right)e^0}{0^3 \left(b-a\right)}\\ &=\frac{2-2}{0^3 \left(b-a\right)}\\ &=\frac{0}{0} \end{align} ここで、 \begin{gather} f \left(\theta\right)=\theta^3 \left(b-a\right)\\ g \left(\theta\right)=g_1 \left(\theta\right)-g_2 \left(\theta\right)\\ g_1 \left(\theta\right)= \left(b^2\theta^2-2b\theta+2\right)e^{\theta b} \quad g_2 \left(\theta\right)= \left(a^2\theta^2-2a\theta+2\right)e^{\theta a} \end{gather} とすると、 \begin{gather} f^\prime \left(\theta\right)=3\theta^2 \left(b-a\right)\\ f^{\prime\prime} \left(\theta\right)=6\theta \left(b-a\right)\\ f^{\prime\prime\prime} \left(\theta\right)=6 \left(b-a\right) \end{gather} \begin{align} g_1^\prime \left(\theta\right)&=e^{\theta b} \cdot \frac{d}{d\theta} \left(b^2\theta^2-2b\theta+2\right)+ \left(b^2\theta^2-2b\theta+2\right) \cdot \frac{d}{d\theta}e^{\theta b}\\ &= \left(2b^2\theta-2b\right)e^{\theta b}+ \left(b^3\theta^2-2b^2\theta+2b\right)e^{\theta b}\\ &=b^3\theta^2e^{\theta b} \end{align} \begin{align} g_1^{\prime\prime} \left(\theta\right)&=b^3\theta^2 \cdot \frac{d}{d\theta}e^{\theta b}+b^3e^{\theta b} \cdot \frac{d}{d\theta}\theta^2\\ &=b^4\theta^2e^{\theta b}+2b^3\theta e^{\theta b}\\ &=b^3 \left(b\theta^2+2\theta\right)e^{\theta b} \end{align} \begin{align} g_1^{\prime\prime\prime} \left(\theta\right)&=b^3 \left(b\theta^2+2\theta\right) \cdot \frac{d}{d\theta}e^{\theta b}+b^3e^{\theta b} \cdot \frac{d}{d\theta} \left(b\theta^2+2\theta\right)\\ &=b^4 \left(b\theta^2+2\theta\right)e^{\theta b}+b^3e^{\theta b} \left(2b\theta+2\right)\\ &=b^3 \left(b^2\theta^2+4b\theta+2\right)e^{\theta b} \end{align} 同様に、 \begin{align} g_2^{\prime\prime\prime} \left(\theta\right)&=a^3 \left(a^2\theta^2+4a\theta+2\right)e^{\theta a}\\ \end{align} したがって、 \begin{gather} g^{\prime\prime\prime} \left(\theta\right)=b^3 \left(b^2\theta^2+4b\theta+2\right)e^{\theta b}-a^3 \left(a^2\theta^2+4a\theta+2\right)e^{\theta a} \end{gather} $f^{\prime\prime} \left(\theta\right),g^{\prime\prime} \left(\theta\right)$ の $\theta\rightarrow0$ のときの極限を取ると、 \begin{align} \lim_{\theta\rightarrow0}{f^{\prime\prime\prime} \left(\theta\right)}=6 \left(b-a\right) \end{align} \begin{align} \lim_{\theta\rightarrow0}{g^{\prime\prime\prime} \left(\theta\right)}&=b^3 \left(b^2 \cdot 0^2+4b \cdot 0+2\right)e^0-a^3 \left(a^2 \cdot 0^2+4a \cdot 0+2\right)e^0\\ &=2 \left(b^3-a^3\right) \end{align} したがって、 \begin{align} \lim_{\theta\rightarrow0}{f^{\prime\prime\prime} \left(\theta\right)} \neq 0 \quad \lim_{\theta\rightarrow0}{\frac{g \left(\theta\right)}{f \left(\theta\right)}}=\frac{0}{0} \end{align} よって適用条件を満たすので、ロピタルの定理により、 \begin{align} E \left(X\right)&=\lim_{\theta\rightarrow0}{\frac{g \left(\theta\right)}{f \left(\theta\right)}}\\ &=\lim_{\theta\rightarrow0}{\frac{g^{\prime\prime\prime} \left(\theta\right)}{f^{\prime\prime\prime} \left(\theta\right)}}\\ &=\frac{2 \left(b^3-a^3\right)}{6 \left(b-a\right)}\\ &=\frac{ \left(b^2+ab+a^2\right) \left(b-a\right)}{3 \left(b-a\right)}\\ &=\frac{b^2+ab+a^2}{3} \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{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{ \left(b-a\right)^2}{12} \end{align} $\blacksquare$