連続型確率変数(continuous random variable) の一様分布(uniform distribution)の期待値(expected value)

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

積率母凾数の一階微分

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

原点周りの一次モーメント=期待値

$$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(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^2(b-a)}\left\{ \displaystyle (tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}\right) \displaystyle -(ta-1)\left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}\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!}+\dotsb} (マクローリン展開), t^2が分母にあるのでt^3の項以上は分子にtが残ることになるのでt^2の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(1+tb+t^2\frac{b^2}{2}\right) \displaystyle -(ta-1)\left(1+ta+t^2\frac{a^2}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left[ \displaystyle \left\{\left(tb+t^2b^2+t^3\frac{b^3}{2}\right)-\left(1+tb+t^2\frac{b^2}{2}\right)\right\} \displaystyle -\left\{\left(ta+t^2a^2+t^3\frac{a^3}{2}\right)-\left(1+ta+t^2\frac{a^2}{2}\right)\right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t(b-b)+t^2(b^2-\frac{b^2}{2})+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t(a-a)+t^2(a^2-\frac{a^2}{2})+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t^2\frac{b^2}{2}+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t^2\frac{a^2}{2}+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{1}{t^2(b-a)}\left( \displaystyle t^2\frac{b^2-a^2}{2} \displaystyle +t^3\frac{b^3-a^3}{2} \displaystyle \right) \displaystyle \right\}\\ &=&\displaystyle \lim_{t \to 0} \left[\frac{1}{t^2(b-a)}\left\{ \displaystyle t^2 \frac{ (b-a)(b+a) }{2} \displaystyle +t^3 \frac{ (b-a)(b^2+ab+a^2) }{2} \displaystyle \right\}\right] \,\dotso\,b^2-a^2=(b-a)(b+a),\,b^3-a^3=(b-a)(b^2+ab+a^2)\\ &=&\displaystyle \lim_{t \to 0} \displaystyle \left\{ \displaystyle \frac{ (b+a) }{2} \displaystyle +t \frac{ (b^2+ab+a^2) }{2} \displaystyle \right\}\\ &=&\displaystyle \frac{b+a}{2}=\frac{a+b}{2} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$