離散型確率変数(discrete 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/uniform-distribution.html}{\frac{1}{n}\frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{n} \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{n} \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ (\mathrm{e}^{t})'(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)'(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)((\mathrm{e}^{t}-1)^{-1})' \right\}\\ &&\,\dotso\,(uvw)'=u'(vw)+u(vw)'=u'(vw)+u(v'w+vw')=u'vw+uv'w+uvw'\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(n\mathrm{e}^{nt})(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(-\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-2}) \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}(n\mathrm{e}^{nt})}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{-(\mathrm{e}^{t}-1)^2} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{ \frac{(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} + \frac{(n\mathrm{e}^{nt})}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{-(\mathrm{e}^{t}-1)^2} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1) + (n\mathrm{e}^{nt})(\mathrm{e}^{t}-1) - \mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{(\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{nt}-\mathrm{e}^{t}+1) + (n\mathrm{e}^{nt}\mathrm{e}^{t}-n\mathrm{e}^{nt}) - (\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{t})}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{nt}-\mathrm{e}^{t}+1 + n\mathrm{e}^{nt}\mathrm{e}^{t}-n\mathrm{e}^{nt} - \mathrm{e}^{nt}\mathrm{e}^{t}+\mathrm{e}^{t}}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}(\mathrm{e}^{t} - 1 +n\mathrm{e}^{t} -n-\mathrm{e}^{t}) +1}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}(n\mathrm{e}^{t}-(n+1))+1}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2}\\ \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\{ \frac{\mathrm{e}^{t}}{n} \displaystyle \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2} \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\{ \frac{\mathrm{e}^{t}}{n} \displaystyle \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{\mathrm{e}^{2t}-2\mathrm{e}^{t}+1} \right\}\\ &=&\displaystyle \lim_{t \to 0} \left\{\frac{\left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}\right)}{n} \frac{n\left(\frac{((n+1)t)^0}{0!}+\frac{((n+1)t)^1}{1!}+\frac{((n+1)t)^2}{2!}+\frac{((n+1)t)^3}{3!}\right) -(n+1)\left(\frac{(nt)^0}{0!}+\frac{(nt)^1}{1!}+\frac{(nt)^2}{2!}+\frac{(nt)^3}{3!}\right) +1}{ \left(\frac{(2t)^0}{0!}+\frac{(2t)^1}{1!}+\frac{(2t)^2}{2!}+\frac{(2t)^3}{3!}\right) -2 \left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}\right) +1} \right\}\\ && \displaystyle \,\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), ひとまずt^3の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{ \left(1+(2t)+\frac{(2t)^2}{2}+\frac{(2t)^3}{6}\right) -2 \left(1+(t)+\frac{t^2}{2}+\frac{t^3}{6}\right) +1} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{ 1+2t+2t^2+\frac{4}{3}t^3 -2-2t-t^2-\frac{1}{3}t^3 +1} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{t^2(1+t)} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{n+n(n+1)t+\frac{n(n+1)^2}{2}t^2+\frac{n(n+1)^3}{6}t^3 -(n+1)-(n+1)nt-\frac{(n+1)n^2}{2}t^2-\frac{(n+1)n^3}{6}t^3 +1\right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{(n-(n+1)+1) +(n(n+1)-(n+1)n)t +(\frac{n(n+1)^2}{2}-\frac{(n+1)n^2}{2})t^2 +(\frac{n(n+1)^3}{6}-\frac{(n+1)n^3}{6})t^3 \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{(0) +(0)t +(\frac{n+1}{2})nt^2 +(\frac{(n+1)(2n+1)}{6})nt^3 \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} nt^2\left(\frac{n+1}{2}+\frac{(n+1)(2n+1)}{6}t\right) \right\}\\ &&\,\dotso\,分母はt^2の項からが残っている．t^3以上の項はtが残るのでマクローリン展開はt^3で十分となる．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right) \left(\frac{n+1}{2}+\frac{(n+1)(2n+1)}{6}t\right) \right\}\\ &=&\displaystyle \frac{n+1}{2}\\ &&\,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$