$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
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
$$\begin{array}{rcl}
\displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\
&=&\displaystyle \frac{(n+1)(2n+1)}{6}-\left(\href{https://shikitenkai.blogspot.com/2019/07/discrete-random-variable-uniform.html}{\frac{n+1}{2}}\right)^2\\
&=&\displaystyle \frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}\\
&=&\displaystyle \frac{2(n+1)(2n+1)-3(n+1)^2}{12}\\
&=&\displaystyle \frac{(n+1)(2(2n+1)-3(n+1))}{12}\\
&=&\displaystyle \frac{(n+1)(4n+2-3n-3)}{12}\\
&=&\displaystyle \frac{(n+1)(n-1)}{12}\\
&=&\displaystyle \frac{n^2-1}{12}\\
\end{array}$$
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