連続型確率変数(continuous random variable) の一様分布(uniform distribution)の分散(variance)

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$

積率母凾数の二階微分

$$\begin{array}{rcl} \displaystyle M_X^{(2)} &=& \displaystyle \frac{\mathrm{d}^2}{\mathrm{d}t^2}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform.html}{\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (t^{-2})'\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\}' \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[\left\{(tb-1)\mathrm{e}^{tb}\right\}'-\left\{(ta-1)\mathrm{e}^{ta}\right\}'\right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ (tb-1)'\mathrm{e}^{tb}+(tb-1)(\mathrm{e}^{tb})' \right\} -\left\{ (ta-1)'\mathrm{e}^{ta}+(ta-1)(\mathrm{e}^{ta})' \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[\frac{ \displaystyle -2\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +t\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right]}{t^3} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle -2(tb-1)\mathrm{e}^{tb}+2(ta-1)\mathrm{e}^{ta} \displaystyle +tb\mathrm{e}^{tb}+tb(tb-1)\mathrm{e}^{tb} \displaystyle -ta\mathrm{e}^{ta}-ta(ta-1)\mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle \left\{-2(tb-1)+tb+tb(tb-1)\right\} \mathrm{e}^{tb} \displaystyle +\left\{2(ta-1)-ta-ta(ta-1)\right\} \mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} \displaystyle -((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\}\\ &=& \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)}\\ \end{array}$$

原点周りの二次モーメント

$$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)} \displaystyle \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} - ((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}+\frac{(tb)^3}{3!}\right) \displaystyle -((ta-1)^2+1) \left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}+\frac{(ta)^3}{3!}\right) \displaystyle \right\} \displaystyle \right]\\ && \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb} (マクローリン展開), t^3が分母にあるのでt^4の項以上は分子にtが残ることになるのでt^3の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle (t^2b^2-2tb+2) \left(1+tb+t^2\frac{b^2}{2}+t^3\frac{b^3}{6}\right) \displaystyle -(t^2a^2-2ta+2) \left(1+ta+t^2\frac{a^2}{2}+t^3\frac{a^3}{6}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left[ \left\{ \left( t^2b^2 +t^3b^3 +t^4\frac{b^4}{2} +t^5\frac{b^5}{6} \right) +\left( -2tb -2t^2b^2 -t^3b^3 -t^4\frac{b^4}{3} \right) +\left(2+2tb + t^2b^2 +t^3\frac{b^3}{3} \right) \right\} -\left\{ \left( t^2a^2 +t^3a^3 +t^4\frac{a^4}{2} +t^5\frac{a^5}{6} \right) +\left( -2ta -2t^2a^2 -t^3a^3 -t^4\frac{a^4}{3} \right) +\left(2+2ta + t^2a^2 +t^3\frac{a^3}{3} \right) \right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \left( 2+t^3\frac{b^3}{3}+t^4\frac{b^4}{6}+t^5\frac{b^5}{6} \right) -\left( 2+t^3\frac{a^3}{3}+t^4\frac{a^4}{6}+t^5\frac{a^5}{6} \right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{b^3-a^3}{3} +t^4\frac{b^4-a^4}{6} +t^5\frac{b^5-a^5}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{(b-a)(b^2+ab+a^2)}{3} +t^4\frac{(b-a)(b+a)(a^2+b^2)}{6} +t^5\frac{(b-a)\frac{b^5-a^5}{b-a}}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{b^2+ab+a^2}{3} +t \frac{(b+a)(a^2+b^2)}{6} +t^2\frac{\left(\frac{b^5-a^5}{b-a}\right)}{6} \displaystyle \right\}\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}=\frac{a^2+ab+b^2}{3} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$

分散(二次の中心モーメント)

$$\begin{array}{rcl} \displaystyle V[X] &=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\left(\href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{b+a}{2}}\right)^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\frac{b^2+2ab+a^2}{4}\\ &=&\displaystyle \frac{4(b^2+ab+a^2)-3(b^2+2ab+a^2)}{12}\\ &=&\displaystyle \frac{4b^2+4ab+4a^2-3b^2-6ab-3a^2}{12}\\ &=&\displaystyle \frac{b^2-2ab+a^2}{12}\\ &=&\displaystyle \frac{(b-a)^2}{12}\\ \end{array}$$