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)
$$\begin{array}{rcl} \displaystyle V[X] &=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\left(\href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{b+a}{2}}\right)^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\frac{b^2+2ab+a^2}{4}\\ &=&\displaystyle \frac{4(b^2+ab+a^2)-3(b^2+2ab+a^2)}{12}\\ &=&\displaystyle \frac{4b^2+4ab+4a^2-3b^2-6ab-3a^2}{12}\\ &=&\displaystyle \frac{b^2-2ab+a^2}{12}\\ &=&\displaystyle \frac{(b-a)^2}{12}\\ \end{array}$$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}$$連続型確率変数(continuous random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)
$$f_X(x) =
\begin{cases}
\displaystyle \frac{1}{b-a} & \quad \left\{ a\leq x \leq b\right\}\\
\displaystyle 0 & \quad \left\{ x\lt a,\,b\lt x\right\}\\
\end{cases}
$$
$$\begin{array}{rcl}
\displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\
&=&\displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{tx}f_X(x)\mathrm{d}x\\
&=&\displaystyle \int_{a}^{b}\mathrm{e}^{tx}\left(\frac{1}{b-a}\right)\mathrm{d}x\\
&=&\displaystyle \frac{1}{b-a}\int_{a}^{b}\mathrm{e}^{tx}\mathrm{d}x\\
&=&\displaystyle \frac{1}{b-a}\int_{ta}^{tb}\mathrm{e}^{s}\frac{1}{t}\mathrm{d}s\,\dotso\,s=tx,\,\frac{\mathrm{d}s}{\mathrm{d}x}=t,\, \mathrm{d}x=\frac{1}{t}\mathrm{d}s,\,a\to ta,\,b\to tb\\
&=&\displaystyle \frac{1}{b-a}\frac{1}{t}\int_{ta}^{tb}\mathrm{e}^{s}\mathrm{d}s\\
&=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{s}\right]_{ta}^{tb}\\
&=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{tb}-\mathrm{e}^{ta}\right]\\
&=&\displaystyle \frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}\\
\end{array}$$
離散型確率変数(discrete 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/uniform-distribution.html}{\frac{1}{n}\frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)}} \displaystyle \right\}\\ &=&\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/discrete-random-variable-uniform.html}{\frac{\mathrm{e}^{t}}{n}\frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2}} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t} \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t} (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ (\mathrm{e}^{t})'(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)'(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left((\mathrm{e}^{t}-1)^{-2}\right)' \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}((n\mathrm{e}^{(n+1)t})'-((n+1)\mathrm{e}^{nt})'+(1)')(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt})+0)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{ (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt}))(\mathrm{e}^{t}-1)^{-2} +(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1) +((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt}))(\mathrm{e}^{t}-1) -2\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1) }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n\mathrm{e}^{(n+2)t}-(n+1)\mathrm{e}^{(n+1)t}+\mathrm{e}^{t})-(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1) +((n(n+1)\mathrm{e}^{(n+2)t})-(n(n+1)\mathrm{e}^{(n+1)t}))-((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt})) -(2n\mathrm{e}^{(n+2)t}-2(n+1)\mathrm{e}^{(n+1)t}+2\mathrm{e}^{t}) }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n\mathrm{e}^{(n+2)t}-(n+1)\mathrm{e}^{(n+1)t}+\mathrm{e}^{t}-n\mathrm{e}^{(n+1)t}+(n+1)\mathrm{e}^{nt}-1 +n(n+1)\mathrm{e}^{(n+2)t}-n(n+1)\mathrm{e}^{(n+1)t}-n(n+1)\mathrm{e}^{(n+1)t}+n(n+1)\mathrm{e}^{nt} -2n\mathrm{e}^{(n+2)t}+2(n+1)\mathrm{e}^{(n+1)t}-2\mathrm{e}^{t} }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n+n(n+1)-2n)\mathrm{e}^{(n+2)t} +(-(n+1)-n-n(n+1)-n(n+1)+2(n+1))\mathrm{e}^{(n+1)t} +((n+1)+n(n+1))\mathrm{e}^{nt} +(1-2)\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\}\\ \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[ \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\} \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} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{ \mathrm{e}^{3t} -3\mathrm{e}^{2t} +3\mathrm{e}^{t} -1} \right\} \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!}+\frac{t^4}{4!}\right)}{n} \left\{\frac{ n^2\left( \frac{((n+2)t)^0}{0!}+\frac{((n+2)t)^1}{1!}+\frac{((n+2)t)^2}{2!}+\frac{((n+2)t)^3}{3!}+\frac{((n+2)t)^4}{4!} \right) -(2n^2+2n-1)\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!}+\frac{((n+1)t)^4}{4!} \right) +(n^2+2n+1)\left( \frac{(nt)^0}{0!}+\frac{(nt)^1}{1!}+\frac{(nt)^2}{2!}+\frac{(nt)^3}{3!}+\frac{(nt)^4}{4!} \right) -\left( \frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!} \right) -1}{ \left( \frac{(3t)^0}{0!}+\frac{(3t)^1}{1!}+\frac{(3t)^2}{2!}+\frac{(3t)^3}{3!}+\frac{(3t)^4}{4!} \right) -3\left( \frac{(2t)^0}{0!}+\frac{(2t)^1}{1!}+\frac{(2t)^2}{2!}+\frac{(2t)^3}{3!}+\frac{(2t)^4}{4!} \right) +3\left( \frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!} \right) -1} \right\} \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!}+\frac{x^4}{4!}+\dotsb} (マクローリン展開), ひとまず4乗の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{n} \left\{\frac{ n^2\left( 1 +(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left( 1 +(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1)\left( 1 +nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) -\left( 1 +t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1}{ 1 +3t +\frac{9}{2}t^2 +\frac{27}{6}t^3 +\frac{81}{24}t^4 -3 -6t -\frac{12}{2}t^2 -\frac{24}{6}t^3 -\frac{48}{24}t^4 +3 +3t +\frac{3}{2}t^2 +\frac{3}{6}t^3 +\frac{3}{24}t^4 -1} \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{n} \left\{\frac{ n^2 \left(1+(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left(1+(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1) \left(1+nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) - \left(1+t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1}{t^3+\frac{3}{2}t^4} \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ n^2 \left(1+(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left(1+(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1) \left(1+nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) - \left(1+t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ n^2 +n^2(n+2)t +\frac{n^2(n+2)^2}{2}t^2 +\frac{n^2(n+2)^3}{6}t^3 +\frac{n^2(n+2)^4}{24}t^4 -(2n^2+2n-1) -(2n^2+2n-1)(n+1)t -\frac{(2n^2+2n-1)(n+1)^2}{2}t^2 -\frac{(2n^2+2n-1)(n+1)^3}{6}t^3 +\frac{(2n^2+2n-1)(n+1)^4}{24}t^4 +(n^2+2n+1) +(n^2+2n+1)nt +\frac{(n^2+2n+1)n^2}{2}t^2 +\frac{(n^2+2n+1)n^3}{6}t^3 +\frac{(n^2+2n+1)n^4}{24}t^4 -1 -t -\frac{1}{2}t^2 -\frac{1}{6}t^3 +\frac{1}{24}t^4 -1 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ (n^2 -(2n^2+2n-1) +(n^2+2n+1) -1 -1) +(n^2(n+2) -(2n^2+2n-1)(n+1) +(n^2+2n+1)n -1 )t +(\frac{n^2(n+2)^2}{2} -\frac{(2n^2+2n-1)(n+1)^2}{2} +\frac{(n^2+2n+1)n^2}{2} -\frac{1}{2} )t^2 +(\frac{n^2(n+2)^3}{6} -\frac{(2n^2+2n-1)(n+1)^3}{6} +\frac{(n^2+2n+1)n^3}{6} -\frac{1}{6} )t^3 +(\frac{n^2(n+2)^4}{24} -\frac{(2n^2+2n-1)(n+1)^4}{24} +\frac{(n^2+2n+1)n^4}{24} -\frac{1}{24} )t^4 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ (0) +(0)t +(0)t^2 +\frac{n(n+1)(2n+1)}{6}t^3 +\frac{n(n+1)(3n^2+5n+1)}{12}t^4 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} nt^3 \left\{ \frac{(n+1)(2n+1)}{6} +\frac{(n+1)(3n^2+5n+1)}{12}t \right\}\right]\\ &&\,\dotso\,分母はt^3の項からが残っている．t^4以上の項はtが残るのでマクローリン展開はt^4で十分となる\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{1+\frac{3}{2}t} \left\{ \frac{(n+1)(2n+1)}{6} +\frac{(n+1)(3n^2+5n+1)}{12}t \right\}\right]\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}\\ &&\,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$分散
$$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}-\left(\href{https://shikitenkai.blogspot.com/2019/07/discrete-random-variable-uniform.html}{\frac{n+1}{2}}\right)^2\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}\\ &=&\displaystyle \frac{2(n+1)(2n+1)-3(n+1)^2}{12}\\ &=&\displaystyle \frac{(n+1)(2(2n+1)-3(n+1))}{12}\\ &=&\displaystyle \frac{(n+1)(4n+2-3n-3)}{12}\\ &=&\displaystyle \frac{(n+1)(n-1)}{12}\\ &=&\displaystyle \frac{n^2-1}{12}\\ \end{array}$$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}$$離散型確率変数(discrete random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)
$$f_X(x) =
\begin{cases}
\displaystyle \frac{1}{n} & \quad x \in \left\{1,2, \dots ,n\right\}\\
\displaystyle 0 & \quad x \notin \left\{1,2, \dots ,n\right\}
\end{cases}
$$
$$\begin{array}{rcl}
\displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\
&=&\displaystyle \sum_{x=1}^{n}\mathrm{e}^{tx}\left(\frac{1}{n}\right)\\
&=&\displaystyle \frac{1}{n} \sum_{x=1}^{n}\mathrm{e}^{tx}\\
&=&\displaystyle \frac{1}{n} \left(\mathrm{e}^t+\mathrm{e}^{2t}+\dotsb+\mathrm{e}^{nt}\right)\\
&=&\displaystyle \frac{1}{n} \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{\mathrm{e}^{t}-1}
\,\dotso\,a+ar+ar^2+\dotsb+ar^{n-1}=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}\;(r \neq 1)\\
&=&\displaystyle \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{n(\mathrm{e}^{t}-1)}\\
\end{array}$$
正規分布(normal distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)
正規分布
$$\begin{array}{rcl} N(\mu, \sigma^2)&=&\frac{1}{\sqrt{2\pi \sigma^2}}\mathrm{e}^{\frac{-(x-\mu)^2}{2\sigma^2}} \end{array}$$積率母凾数
$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \int_{-\infty}^{\infty}(\mathrm{e}^{tx})\frac{1}{\sqrt{2\pi \sigma^2}}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}(\mathrm{e}^{tx})\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}+tx} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu)^2+(2\sigma^2)(tx)}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t}{2\sigma^2}-\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t+2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x \,\dotso\,a^2-2ab+b^2-2ac+2bc+c^2=(a-b-c)^2\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}+(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}} \displaystyle \int_{-\infty}^{\infty} \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}} \displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \mathrm{d}x\\ &=&\displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \frac{1}{\sqrt{2\pi \sigma^2}} \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}}\mathrm{d}x \\ &=&\displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\,\dotso\,\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}}\mathrm{d}x = N(\mu+\sigma^2t,\sigma^2)の総和=1\\ \end{array}$$期待値・分散
$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }\left(\mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\right) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}(\mu+\sigma^2t) \right\}|_{t=0}\\ &=&\displaystyle \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})}(\mu+\sigma^20) \\ &=&\displaystyle \mathrm{e}^{0}(\mu+0) \\ &=&\displaystyle \mu \\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }\left(\mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\right) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }\left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}(\mu+\sigma^2t)\right) \right\}|_{t=0}\\ &=&\displaystyle \left[ \displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \right)\right\} \left( \mu+\sigma^2t \right) \displaystyle + \left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \right) \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \left( \mu+\sigma^2t \right) \right\} \right]|_{t=0}\\ &=&\displaystyle \left[ \displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \left( \mu+\sigma^2t \right)^2 \displaystyle + \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \sigma^2 \displaystyle \right]|_{t=0}\\ &=&\displaystyle \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})} \left( \mu+\sigma^20 \right)^2 \displaystyle + \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})} \sigma^2\\ &=&\displaystyle \mu^2+\sigma^2\\ \end{array}$$ $$\begin{array}{rcl} V[X]&=&E[X^2]-E[X]^2\\ &=&\mu^2+\sigma^2-\mu^2\\ &=&\sigma^2 \end{array}$$ポアソン分布(Poisson distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)
ポアソン分布
$$\begin{array}{rcl} \displaystyle Po(\lambda)&=&\displaystyle \frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ \end{array}$$積率母凾数
$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{\mathrm{e}^{tx}\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{(\mathrm{e}^{t}\lambda)^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\mathrm{e}^{\mathrm{e}^{t}\lambda} \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{x=0}^{\infty}\frac{a^{x}}{x!}=\mathrm{e}^a}\\ &=&\displaystyle \mathrm{e}^{\mathrm{e}^{t}\lambda-\lambda}=\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)} \end{array}$$期待値・分散
$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1),\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}\\ &=&\displaystyle \left\{ (\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)})(\lambda\mathrm{e}^{t}) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }\mathrm{e}^{x}=\mathrm{e}^{x}\\ &=&\displaystyle \left\{ \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0}\\ &=&\displaystyle \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+0})\\ &=&\displaystyle \lambda\mathrm{e}^0 \,\dotso\,a^0=1\\ &=&\lambda \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0} \,\dotso\,E[X]の展開から．\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\lambda\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1)+t,\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}+1\\ &=&\displaystyle \left\{ (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+1})(\lambda\mathrm{e}^{t}+1) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }C\mathrm{e}^{x}=C\mathrm{e}^{x}\\ &=&\displaystyle (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+1})(\lambda\mathrm{e}^{0}+1)\\ &=&\displaystyle \lambda\mathrm{e}^0(\lambda+1) \,\dotso\,a^0=1\\ &=&\lambda(\lambda+1) \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\lambda(\lambda+1)-\lambda^2\\ &=&\lambda^2+\lambda-\lambda^2\\ &=&\lambda\\ \end{array}$$
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