標本平均の母平均まわりの4次モーメント (標本平均の4次の中心(化)モーメント)

標本平均\(\overline{X}\)の母平均\(\mu\)まわりの4次モーメント(=標本平均\(\overline{X}\)の4次の中心(化)モーメント)

$$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^4\right] &=&\mathrm{E}\left[\left\{\left(\frac{1}{n}\sum_{k=1}^{n}X_k\right) - \mu\right\}^4\right] \;\cdots\;\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i \\&=&\mathrm{E}\left[\left\{\left(\frac{1}{n}\sum_{k=1}^{n}X_k\right) - \left(\frac{1}{n}\sum_{k=1}^{n}\mu\right)\right\}^4\right] \;\cdots\;C=\frac{n}{n}C=\frac{1}{n}C\sum_{i=1}^{n}1=\frac{1}{n}\sum_{i=1}^{n}C\;(C:iによらない数，\sumにとって定数) \\&=&\mathrm{E}\left[\left[\frac{1}{n}\left\{\left(\sum_{k=1}^{n}X_k\right)-\left(\sum_{k=1}^{n}\mu\right)\right\}\right]^4\right] \\&=&\mathrm{E}\left[\frac{1}{n^4}\left\{\left(\sum_{k=1}^{n}X_k\right)-\left(\sum_{k=1}^{n}\mu\right)\right\}^4\right] \;\cdots\;(AB)^C=A^CB^C \\&=&\mathrm{E}\left[\frac{1}{n^4}\left\{\sum_{k=1}^{n}\left(X_k-\mu\right)\right\}^4\right] \;\cdots\;\sum_{i=1}^{n}X_i-\sum_{i=1}^{n}Y_i=\sum_{i=1}^{n}\left(X_i-Y_i\right) \end{eqnarray} $$ 総和の指数計算において掛け合わせる添え字の組合せについて考える． $$ \begin{eqnarray} \left(\sum_{k=1}^{n}A_k\right)^4 &=&\left(\sum_{k=1}^{n} A_k \right)\left(\sum_{l=1}^{n}A_l\right)\left(\sum_{m=1}^{n}A_m\right)\left(\sum_{s=1}^{n}A_s\right) \\&=&(A_1+A_2+\cdots+A_k+\cdots+A_n)(A_1+A_2+\cdots+A_l+\cdots+A_n)(A_1+A_2+\cdots+A_m+\cdots+A_n)(A_1+A_2+\cdots+A_s+\cdots+A_n) \\&=&_4\mathrm{P}_0\times\left(\sum_{k=1}^{n} A_k^4 \right)\;\cdots\;4つとも同じ添え字(どれか0個の添え字が異なるケース) \\&&+_4\mathrm{P}_1\times\left(\sum_{k \neq l} A_k^3 A_l\right)\;\cdots\;いずれか3つが同じ添え字(どれか1個の添え字が異なるのケース) \\&&+_4\mathrm{C}_2\times\left(\sum_{k \lt l} A_k^2 A_l^2\right)\;\cdots\;いずれか2つの添え字が同じで残りの2つの添え字同士も同じケース(どれか2個の添え字が異なるのケース(異なった添え字同士も同じ)) \\&&+_4\mathrm{P}_2\times\left(\sum_{k \neq l,m かつ l\lt m} A_k^2 A_l A_m\right)\;\cdots\;いずれか2つの添え字が同じで残りの2つの添え字同士は異なるケース(どれか2個の添え字が異なるのケース(異なった添え字同士は異なる)) \\&&+_4\mathrm{P}_3\times\left(\sum_{k \lt l \lt m \lt s} A_k A_l A_m A_s\right)\;\cdots\;すべての添え字が異なるケース(どれか3個の添え字が異なるのケース) \\&&\;\cdots\;各ケースでの(重複する数 \times 組合せで総和)の和 \\&=&\frac{4!}{(4-0)!}\left(\sum_{k=1}^{n} A_k^4 \right) +\frac{4!}{(4-1)!}\left(\sum_{k \neq l} A_k^3 A_l\right) +\frac{4!}{(4-2)!2!}\left(\sum_{k \lt l} A_k^2 A_l^2\right) +\frac{4!}{(4-2)!}\left(\sum_{k \neq l,m かつ l\lt m} A_k^2 A_l A_m\right) +\frac{4!}{(4-3)!}\left(\sum_{k \lt l \lt m \lt s} A_k A_l A_m A_s\right) \\&=&\frac{4\times3\times2\times1}{4\times3\times2\times1}\left(\sum_{k=1}^{n} A_k^4 \right) +\frac{4\times3\times2\times1}{3\times2\times1}\left(\sum_{k \neq l} A_k^3 A_l\right) +\frac{4\times3\times2\times1}{2\times1\cdot2\times1}\left(\sum_{k \lt l} A_k^2 A_l^2\right) +\frac{4\times3\times2\times1}{2\times1}\left(\sum_{k \neq l,m かつ l\lt m} A_k^2 A_l A_m\right) +\frac{4\times3\times2\times1}{1}\left(\sum_{k \lt l \lt m \lt s} A_k A_l A_m A_s\right) \\&=&1\cdot\left(\sum_{k=1}^{n} A_k^4 \right) +4\cdot\left(\sum_{k \neq l} A_k^3 A_l\right) +6\cdot\left(\sum_{k \lt l} A_k^2 A_l^2\right) +12\cdot\left(\sum_{k \neq l,m かつ l\lt m} A_k^2 A_l A_m\right) +24\cdot\left(\sum_{k \lt l \lt m \lt s} A_k A_l A_m A_s\right) \end{eqnarray} $$ よって， $$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^4\right] &=&\mathrm{E}\left[\frac{1}{n^4}\left\{\sum_{k=1}^{n}\left(X_k-\mu\right)\right\}^4\right] \\&=&\mathrm{E}\left[\frac{1}{n^4}\left\{ \sum_{k=1}^{n} \left(X_k-\mu\right)^4 +4\sum_{k \neq l} \left(X_k-\mu\right)^3\left(X_l-\mu\right) +6\sum_{k \lt l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)^2 +12\sum_{k \neq l,m かつ l\lt m} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\left(X_m-\mu\right) +24\sum_{k \lt l \lt m \lt s} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\left(X_s-\mu\right) \right\}\right] \\&=&\frac{1}{n^4}\mathrm{E}\left[ \sum_{k=1}^{n} \left(X_k-\mu\right)^4 +4\sum_{k \neq l} \left(X_k-\mu\right)^3\left(X_l-\mu\right) +6\sum_{k \lt l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)^2 +12\sum_{k \neq l,m かつ l\lt m} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\left(X_m-\mu\right) +24\sum_{k \lt l \lt m \lt s} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\left(X_s-\mu\right) \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[cX]=c\mathrm{E}[X]} \\&=&\frac{1}{n^4}\left[ \mathrm{E}\left[ \sum_{k=1}^{n} \left(X_k-\mu\right)^4 \right] +\mathrm{E}\left[ 4\sum_{k \neq l} \left(X_k-\mu\right)^3\left(X_l-\mu\right) \right] +\mathrm{E}\left[ 6\sum_{k \lt l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)^2 \right] +\mathrm{E}\left[ 12\sum_{k \neq l,m かつ l\lt m} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\left(X_m-\mu\right) \right] +\mathrm{E}\left[ 24\sum_{k \lt l \lt m \lt s} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\left(X_s-\mu\right) \right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X+Y]=\mathrm{E}[X]+\mathrm{E}[Y]} \\&=&\frac{1}{n^4}\left[ \mathrm{E}\left[ \sum_{k=1}^{n} \left(X_k-\mu\right)^4 \right] +4\mathrm{E}\left[ \sum_{k \neq l} \left(X_k-\mu\right)^3\left(X_l-\mu\right) \right] +6\mathrm{E}\left[ \sum_{k \lt l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)^2 \right] +12\mathrm{E}\left[ \sum_{k \neq l,m かつ l\lt m} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\left(X_m-\mu\right) \right] +24\mathrm{E}\left[ \sum_{k \lt l \lt m \lt s} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\left(X_s-\mu\right) \right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[cX]=c\mathrm{E}[X]} \\&=&\frac{1}{n^4}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^4 \right] +4\sum_{k \neq l} \mathrm{E}\left[\left(X_k-\mu\right)^3\left(X_l-\mu\right) \right] +6\sum_{k \lt l} \mathrm{E}\left[\left(X_k-\mu\right)^2\left(X_l-\mu\right)^2 \right] +12\sum_{k \neq l,m かつ l\lt m} \mathrm{E}\left[\left(X_k-\mu\right)^2\left(X_l-\mu\right)\left(X_m-\mu\right) \right] +24\sum_{k \lt l \lt m \lt s} \mathrm{E}\left[\left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\left(X_s-\mu\right) \right] \right] \\&&\;\cdots\;\mathrm{E}\left[\sum_{i=1}^n A_i\right]=\mathrm{E}\left[A_1+A_2+\cdots+A_i+\cdots+A_n\right]=\mathrm{E}\left[A_1\right]+\mathrm{E}\left[A_2\right]+\cdots+\mathrm{E}\left[A_i\right]+\cdots+\mathrm{E}\left[A_n\right]=\sum_{i=1}^n \mathrm{E}\left[A_i\right] \\&=&\frac{1}{n^4}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^4 \right] +4\sum_{k \neq l} \mathrm{E}\left[\left(X_k-\mu\right)^3\right]\mathrm{E}\left[\left(X_l-\mu\right) \right] +6\sum_{k \lt l} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[\left(X_l-\mu\right)^2 \right] +12\sum_{k \neq l,m かつ l\lt m} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[\left(X_l-\mu\right)\right]\mathrm{E}\left[\left(X_m-\mu\right) \right] +24\sum_{k \lt l \lt m \lt s} \mathrm{E}\left[\left(X_k-\mu\right)\right]\mathrm{E}\left[\left(X_l-\mu\right)\right]\mathrm{E}\left[\left(X_m-\mu\right)\right]\mathrm{E}\left[\left(X_s-\mu\right) \right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{X,Yが独立の場合\,\,\mathrm{E}[XY]=\mathrm{E}[X]\mathrm{E}[Y]} \end{eqnarray} $$ 1次の中心(化)モーメントについて考える． $$ \begin{eqnarray} \mathrm{E}\left[X_i-\mu\right] &=& \mathrm{E}\left[X_i\right]-\mathrm{E}\left[\mu\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X-Y]=\mathrm{E}[X]-\mathrm{E}[Y]} \\&=& \mu-\mu \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable.html}{\mathrm{E}[X_i]=\mathrm{E}[X]=\mu},\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[C]=C\;(C定数)} \\&=& 0 \end{eqnarray} $$ これを用いて $$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^4\right] \\&=&\frac{1}{n^4}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^4 \right] +4\sum_{k \neq l} \mathrm{E}\left[\left(X_k-\mu\right)^3\right]\mathrm{E}\left[\left(X_l-\mu\right) \right] +6\sum_{k \lt l} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[\left(X_l-\mu\right)^2 \right] +12\sum_{k \neq l,m かつ l\lt m} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[\left(X_l-\mu\right)\right]\mathrm{E}\left[\left(X_m-\mu\right) \right] +24\sum_{k \lt l \lt m \lt s} \mathrm{E}\left[\left(X_k-\mu\right)\right]\mathrm{E}\left[\left(X_l-\mu\right)\right]\mathrm{E}\left[\left(X_m-\mu\right)\right]\mathrm{E}\left[\left(X_s-\mu\right) \right] \right] \\&=&\frac{1}{n^4}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^4 \right] +4\sum_{k \neq l} \left( \mathrm{E}\left[\left(X_k-\mu\right)^3\right] \cdot 0 \right) +6\sum_{k \lt l} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[\left(X_l-\mu\right)^2 \right] +12\sum_{k \neq l,m かつ l\lt m} \left( \mathrm{E}\left[\left(X_k-\mu\right)^2\right] \cdot 0 \cdot 0 \right) +24\sum_{k \lt l \lt m \lt s} \left( 0 \cdot 0 \cdot 0 \cdot 0 \right) \right] \;\cdots\;\mathrm{E}\left[X_i-\mu\right]=0 \\&=&\frac{1}{n^4}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^4\right] +0 +6\sum_{k \lt l}\mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[\left(X_l-\mu\right)^2 \right] +0 +0 \right] \\&=&\frac{1}{n^4}\left\{ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^4\right] +6\sum_{k \lt l}\mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[\left(X_l-\mu\right)^2 \right] \right\} \\&=&\frac{1}{n^4}\left\{ \sum_{k=1}^{n} \mu_4 +6\sum_{k \lt l} \left(\sigma^2 \cdot \sigma^2\right) \right\} \;\cdots\;\mathrm{E}\left[\left(X_i-\mu\right)^4\right]=\mu_4\;:4次の中心(化)モーメント,\;\mathrm{E}\left[\left(X_i-\mu\right)^2\right]=\sigma^2\;:2次の中心(化)モーメント \\&=&\frac{1}{n^4}\left\{ \sum_{k=1}^{n} \mu_4 +6\sum_{k \lt l} \sigma^4 \right\} \\&=&\frac{1}{n^4}\left\{ \sum_{k=1}^{n} \mu_4 +6\sigma^4\sum_{k \lt l} 1 \right\} \\&=&\frac{1}{n^4}\left\{n\mu_4+6\frac{n(n-1)}{2}\sigma^4\right\} \;\cdots\;\sum_{k \lt l} 1=\frac{n(n-1)}{2}\;(各k=1〜n-1に対してl=k+1からl=nまでの和) \\&=&\frac{1}{n^3}\left(\mu_4+3(n-1)\sigma^4\right) \\&=&\mu_4\left(\overline{X}\right)\;\cdots\;\mu_4\left(\overline{X}\right):標本平均\overline{X}の母平均\muまわりの4次モーメント(4次の中心(化)モーメント) \end{eqnarray} $$

標本平均\(\overline{X}\)の母平均\(\mu\)まわりの4次モーメント(標本平均\(\overline{X}\)の4次の中心(化)モーメント)を尖度\(\beta_2\)で表す

$$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^4\right] &=&\mu_4\left(\overline{X}\right) \\&=&\frac{1}{n^3}\left(\mu_4+3(n-1)\sigma^4\right) \\&=&\frac{1}{n^3}\left(\frac{\beta_2+3}{\sigma^4}+3(n-1)\sigma^4\right) \;\cdots\;\mu_4=\frac{\beta_2+3}{\sigma^4}\;:4次の中心(化)モーメント \\&=&\frac{1}{n^3}\left(\frac{\beta_2+3+3(n-1)}{\sigma^4}\right) \\&=&\frac{1}{n^3}\left(\frac{\beta_2+3+3n-3}{\sigma^4}\right) \\&=&\frac{1}{n^3}\left(\frac{\beta_2+3n}{\sigma^4}\right) \end{eqnarray} $$

標本平均\(\overline{X}\)の尖度\(\beta_2\left(\overline{X}\right)\)

$$ \begin{eqnarray} \href{https://shikitenkai.blogspot.com/2020/08/blog-post_39.html}{\beta_2\left(\overline{X}\right)=\frac{\beta_2}{n}} \end{eqnarray} $$

標本平均の母平均まわりの3次モーメント

標本平均\(\overline{X}\)の母平均\(\mu\)まわりの3次モーメント(=標本平均\(\overline{X}\)の3次の中心(化)モーメント)

$$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] &=&\mathrm{E}\left[\left\{\left(\frac{1}{n}\sum_{k=1}^{n}X_k\right) - \mu\right\}^3\right] \;\cdots\;\overline{X}=\frac{1}{n}\sum_{k=1}^{n}X_k \\&=&\mathrm{E}\left[\left\{\left(\frac{1}{n}\sum_{k=1}^{n}X_k\right) - \left(\frac{1}{n}\sum_{k=1}^{n}\mu\right)\right\}^3\right] \;\cdots\;C=\frac{n}{n}C=\frac{1}{n}C\sum_{k=1}^{n}1=\frac{1}{n}\sum_{k=1}^{n}C\;(C:kによらない数，\sumにとって定数) \\&=&\mathrm{E}\left[\left[\frac{1}{n}\left\{\left(\sum_{k=1}^{n}X_k\right)-\left(\sum_{k=1}^{n}\mu\right)\right\}\right]^3\right] \\&=&\mathrm{E}\left[\frac{1}{n^3}\left\{\left(\sum_{k=1}^{n}X_k\right)-\left(\sum_{k=1}^{n}\mu\right)\right\}^3\right] \;\cdots\;(AB)^C=A^CB^C \\&=&\mathrm{E}\left[\frac{1}{n^3}\left\{\sum_{k=1}^{n}\left(X_k-\mu\right)\right\}^3\right] \;\cdots\;\sum_{k=1}^{n}X_k-\sum_{k=1}^{n}Y_k=\sum_{k=1}^{n}\left(X_k-Y_k\right) \end{eqnarray} $$ 総和の指数計算において掛け合わせる添え字の組合せについて考える． $$ \begin{eqnarray} \left(\sum_{k=1}^{n}A_k\right)^3 &=&\left(\sum_{k=1}^{n} A_k \right)\left(\sum_{l=1}^{n}A_l\right)\left(\sum_{m=1}^{n}A_m\right) \\&=&(A_1+A_2+\cdots+A_k+\cdots+A_n)(A_1+A_2+\cdots+A_l+\cdots+A_n)(A_1+A_2+\cdots+A_m+\cdots+A_n) \\&=&_3\mathrm{P}_0\times\left(\sum_{k=1}^{n} A_k^3 \right)\;\cdots\;3つとも同じ添え字(どれか0個の添え字が異なるケース) \\&&+_3\mathrm{P}_1\times\left(\sum_{k \neq l} A_k^2 A_l\right)\;\cdots\;いずれか2つが同じ添え字(どれか1個の添え字が異なるのケース) \\&&+_3\mathrm{P}_2\times\left(\sum_{k \lt l \lt m} A_k A_l A_m\right)\;\cdots\;すべての添え字が異なるケース(どれか2個の添え字が異なるのケース) \\&&\;\cdots\;各ケースでの(重複する数 \times 組合せで総和)の和 \\&=&\frac{3!}{(3-0)!}\left(\sum_{k=1}^{n} A_k^3 \right) +\frac{3!}{(3-1)!}\left(\sum_{k \neq l} A_k^2 A_l\right) +\frac{3!}{(3-2)!}\left(\sum_{k \lt l \lt m} A_k A_l A_m\right) \\&=&\frac{3\times2\times1}{3\times2\times1}\left(\sum_{k=1}^{n} A_k^3 \right) +\frac{3\times2\times1}{2\times1}\left(\sum_{k \neq l} A_k^2 A_l\right) +\frac{3\times2\times1}{1}\left(\sum_{k \lt l \lt m} A_k A_l A_m\right) \\&=&1\cdot\left(\sum_{k=1}^{n} A_k^3 \right) +3\cdot\left(\sum_{k \neq l} A_k^2 A_l\right) +6\cdot\left(\sum_{k \lt l \lt m} A_k A_l A_m\right) \end{eqnarray} $$ よって， $$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] &=&\mathrm{E}\left[\frac{1}{n^3}\left\{\sum_{k=1}^{n}\left(X_k-\mu\right)\right\}^3\right] \\&=&\mathrm{E}\left[\frac{1}{n^3}\left\{ \sum_{k=1}^{n} \left(X_k-\mu\right)^3 +3\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right) +6\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right) \right\}\right] \\&=&\frac{1}{n^3}\mathrm{E}\left[ \sum_{k=1}^{n} \left(X_k-\mu\right)^3 +3\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right) +6\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right) \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[cX]=c\mathrm{E}[X]} \\&=&\frac{1}{n^3}\left[ \mathrm{E}\left[\sum_{k=1}^{n} \left(X_k-\mu\right)^3\right] +\mathrm{E}\left[3\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\right] +\mathrm{E}\left[6\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X+Y]=\mathrm{E}[X]+\mathrm{E}[Y]} \\&=&\frac{1}{n^3}\left[ \mathrm{E}\left[\sum_{k=1}^{n} \left(X_k-\mu\right)^3\right] +3\mathrm{E}\left[\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\right] +6\mathrm{E}\left[\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[cX]=c\mathrm{E}[X]} \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[ \left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \mathrm{E}\left[ \left(X_k-\mu\right)^2\left(X_l-\mu\right)\right] +6\sum_{k \lt l \lt m} \mathrm{E}\left[ \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\right] \right] \\&&\;\cdots\;\mathrm{E}\left[\sum_{k=1}^n A_k\right]=\mathrm{E}\left[A_1+A_2+\;\cdots\;+A_n\right]=\mathrm{E}\left[A_1\right]+\mathrm{E}\left[A_2\right]+\cdots+\mathrm{E}\left[A_n\right]=\sum_{k=1}^n\mathrm{E}\left[A_k\right] ,\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X+Y]=\mathrm{E}[X]+\mathrm{E}[Y]} \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[X_l-\mu\right] +6\sum_{k \lt l \lt m} \mathrm{E}\left[X_k-\mu\right]\mathrm{E}\left[X_l-\mu\right]\mathrm{E}\left[X_m-\mu\right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{X,Yが独立の場合\,\,\mathrm{E}[XY]=\mathrm{E}[X]\mathrm{E}[Y]} \end{eqnarray} $$ ここで1次の中心(化)モーメントについて考える． $$ \begin{eqnarray} \mathrm{E}\left[X_i-\mu\right] &=& \mathrm{E}\left[X_i\right]-\mathrm{E}\left[\mu\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X-Y]=\mathrm{E}[X]-\mathrm{E}[Y]} \\&=& \mu-\mu \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable.html}{\mathrm{E}[X_i]=\mathrm{E}[X]=\mu},\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[C]=C\;(C定数)} \\&=& 0 \end{eqnarray} $$ これを用いて $$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[X_l-\mu\right] +6\sum_{k \lt l \lt m} \mathrm{E}\left[X_k-\mu\right]\mathrm{E}\left[X_l-\mu\right]\mathrm{E}\left[X_m-\mu\right] \right] \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \left(\mathrm{E}\left[\left(X_k-\mu\right)^2\right]\cdot0\right) +6\sum_{k \lt l \lt m} \left(0\cdot0\cdot0\right) \right] \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +0+0 \right] \\&=&\frac{1}{n^3}\sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] \\&=&\frac{1}{n^3}\sum_{k=1}^{n} \mu_3 \,\cdots\,\href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{\mathrm{E}\left[\left(X-\mu\right)^3\right]=\mu_3\;:3次の中心(化)モーメント} \\&=&\frac{1}{n^3}n\mu_3 \\&=&\frac{\mu_3}{n^2} \\&=&\mu_3\left(\overline{X}\right)\;\cdots\;\mu_3\left(\overline{X}\right):標本平均\overline{X}の母平均\muまわりの3次モーメント(3次の中心(化)モーメント) \end{eqnarray} $$

標本平均\(\overline{X}\)の母平均\(\mu\)まわりの3次モーメント(標本平均\(\overline{X}\)の3次の中心(化)モーメント)を歪度\(\beta_1\)で表す

$$ \begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] &=&\mu_3\left(\overline{X}\right) \\&=&\frac{\mu_3}{n^2} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{\mu_3=\mu_3\left(X\right)=\beta_1\sigma^3\;:3次の中心(化)モーメント} \\&=&\frac{\beta_1 \sigma^3}{n^2} \end{eqnarray} $$

標本平均\(\overline{X}\)の歪度\(\beta_1\left(\overline{X}\right)\)

$$ \begin{eqnarray} \href{https://shikitenkai.blogspot.com/2019/07/blog-post_22.html}{\beta_1\left(\overline{X}\right)=\frac{\beta_1}{\sqrt{n}}} \end{eqnarray} $$

標本平均まわりの3次モーメントの和

“標本平均\(\overline{X}\)まわりの3次モーメントの和”を“母平均\(\mu\)まわりの3次モーメント\(\mu_3\)”で表す

$$\begin{array}{rcl} \displaystyle \sum_{k=1}^{n}E\left[(X_k-\overline{X})^3\right] &=&\displaystyle E\left[\sum_{k=1}^{n}(X_k-\overline{X})^3\right]\\ &&\displaystyle\,\dotso\,E\left[X\right]+E\left[Y\right]=E\left[X+Y\right]\\ &=&\displaystyle E\left[\sum_{k=1}^{n}\left\{ \displaystyle \left( X_k-\mu \right) \displaystyle -\left( \overline{X}-\mu \right) \displaystyle \right\}^3\right]\\ &&\displaystyle\,\dotso\,(A-B)=(A-C)-(B-C)\\ &=&\displaystyle E\left[\sum_{k=1}^{n}\left\{ \displaystyle \left( X_k-\mu \right)^3 \displaystyle -3\left( X_k-\mu \right)^2\left( \overline{X}-\mu \right) \displaystyle +3\left( X_k-\mu \right)\left( \overline{X}-\mu \right)^2 \displaystyle - \left( \overline{X}-\mu \right)^3 \displaystyle \right\}\right]\\ &&\displaystyle\,\dotso\,(A-B)^3=A^3-3A^2B+3AB^2-B^3\\ &=&\displaystyle E\left[ \displaystyle \sum_{k=1}^{n}\left( X_k-\mu \right)^3 \displaystyle -3\left\{ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right\} \displaystyle +3\left\{ \left( \overline{X}-\mu \right)^2 \sum_{k=1}^{n} \left( X_k-\mu \right) \right\} \displaystyle - \left( \overline{X}-\mu \right)^3 \sum_{k=1}^{n}1 \displaystyle \right]\\ &&\displaystyle\,\dotso\,\sum_{k=1}^{n} (X+Y)=\sum_{k=1}^{n} X+\sum_{k=1}^{n} Y\\ &=&\displaystyle E\left[\sum_{k=1}^{n}\left( X_k-\mu \right)^3\right] \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right] \displaystyle +3E\left[ \left( \overline{X}-\mu \right)^2 \sum_{k=1}^{n} \left( X_k-\mu \right) \right] \displaystyle - E\left[\left( \overline{X}-\mu \right)^3 \sum_{k=1}^{n}1\right] \displaystyle \\ &&\displaystyle\,\dotso\,E\left[X+Y\right]=E\left[X\right]+E\left[Y\right]\\ &=&\displaystyle \sum_{k=1}^{n}E\left[\left( X_k-\mu \right)^3\right] \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right] \displaystyle +3E\left[ \left( \overline{X}-\mu \right)^2 \left( \sum_{k=1}^{n}X_k - \sum_{k=1}^{n}\mu \right) \right] \displaystyle - E\left[n\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &&\displaystyle\,\dotso\, \displaystyle E\left[X+Y\right]=E\left[X\right]+E\left[Y\right] ,\quad \sum_{k=1}^{n} (X+Y)=\sum_{k=1}^{n} X+\sum_{k=1}^{n} Y\\ &=&\displaystyle \sum_{k=1}^{n}E\left[\left( X_k-\mu \right)^3\right] \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right] \displaystyle +3E\left[ \left( \overline{X}-\mu \right)^2 \left( n\overline{X} - n\mu \right) \right] \displaystyle - E\left[n\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &&\displaystyle\,\dotso\, \sum_{k=1}^{n}X_k=n\overline{X} ,\quad \sum_{k=1}^{n}\mu=n\mu\\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +3E\left[ n\left( \overline{X}-\mu \right)^3\right] \displaystyle -nE\left[\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &&\displaystyle\,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{E\left[\left( X_k-\mu \right)^3\right]=\mu_3}\\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +3nE\left[ \left( \overline{X}-\mu \right)^3\right] \displaystyle - nE\left[\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +2nE\left[ \left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +2n\frac{\mu_3}{n^2} \displaystyle \\ &&\displaystyle\,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/overlinexmu3.html}{E\left[\left( \overline{X}-\mu \right)^3\right]=\frac{\mu_3}{n^2}}\\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +2\frac{\mu_3}{n} \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k^2-2\mu X_k+\mu^2 \right) \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \left( \sum_{k=1}^{n}X_k^2-2\mu \sum_{k=1}^{n}X_k+\mu^2\sum_{k=1}^{n}1 \right) \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \left( \sum_{k=1}^{n}X_k^2-2\mu n\overline{X}+n\mu^2 \right) \right] \displaystyle \\ &&\displaystyle\,\dotso\, \sum_{k=1}^{n}X_k=n\overline{X} ,\quad \sum_{k=1}^{n}1=n\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\left( \sum_{k=1}^{n}X_k^2- 2\mu n\overline{X}+ n\mu^2 \right) \displaystyle -\mu\left( \sum_{k=1}^{n}X_k^2-2\mu n\overline{X}+ n\mu^2 \right) \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \left( \overline{X}\sum_{k=1}^{n}X_k^2-2n\mu \overline{X}^2+ n\mu^2\overline{X} \right) \displaystyle -\left( \mu\sum_{k=1}^{n}X_k^2-2\mu^2 n\overline{X}+n\mu^3 \right) \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2-2n\mu \overline{X}^2+ n\mu^2\overline{X} \displaystyle -\mu\sum_{k=1}^{n}X_k^2+2\mu^2 n\overline{X}-n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2 \displaystyle -2n\mu \overline{X}^2- \mu \sum_{k=1}^{n}X_k^2 \displaystyle + n\mu^2\overline{X} +2\mu^2 n\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2 \displaystyle - \mu \left(2n \overline{X}^2+\sum_{k=1}^{n}X_k^2\right) \displaystyle + n\mu^2\left( \overline{X} +2 \overline{X}\right) \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2 \displaystyle - \mu\left(2n \overline{X}^2+\sum_{k=1}^{n}X_k^2\right) \displaystyle + 3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &&\displaystyle\,\dotso\, \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/blog-post_41.html}{\sum_{k=1}^{n}X_k^2=\sum_{k=1}^{n}(X_k-\overline{X})^2+n\overline{X}^2}\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\,\left(\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2+n\overline{X}^2\right) \displaystyle - \mu \left(2n \overline{X}^2+\left(\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2+n\overline{X}^2\right)\right) \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle +n\overline{X}^3 \displaystyle -2n\mu \overline{X}^2 \displaystyle -\mu\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle - n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \left( \overline{X}-\mu \right) \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle +n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \left( \overline{X}-\mu \right) \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle \right] \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[\overline{X}-\mu \right] \displaystyle \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3\left(E\left[\overline{X} \right]-\mu\right) \displaystyle \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3\left(\mu-\mu\right) \displaystyle \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[n \displaystyle \left(\overline{X}^3 \displaystyle -3\mu \overline{X}^2 \displaystyle +3\mu^2\overline{X} \displaystyle -\mu^3\right) \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle n\left(\overline{X}-\mu\right)^3 \displaystyle \right] \displaystyle \\ &&\displaystyle\,\dotso\,A^3-3A^2B+3AB^2-B^3=(A-B)^3\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3nE\left[ \displaystyle \left(\overline{X}-\mu\right)^3 \displaystyle \right] \displaystyle \\ &&\displaystyle\,\dotso\,E[cX]=cE[X]\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3n\frac{\mu_3}{n^2} \displaystyle \\ &&\displaystyle\,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/overlinexmu3.html}{E\left[\left( \overline{X}-\mu \right)^3\right]=\frac{\mu_3}{n^2}}\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3\frac{\mu_3}{n} \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} -3\frac{n}{n}\right) \displaystyle \\ &=&\displaystyle \mu_3\frac{n^2-3n+2}{n}\\ &=&\displaystyle \mu_3\frac{(n-1)(n-2)}{n}\\ \end{array}$$

“標本平均\(\overline{X}\)まわりの3次モーメントの和”から“母平均\(\mu\)まわりの3次モーメント\(\mu_3\)”を推定する\(\hat{\mu}_3\)

$$\begin{array}{rcl} \displaystyle \hat{\mu}_3 &=&\displaystyle \frac{n}{(n-1)(n-2)}E\left[\sum_{k=1}^{n}(X_k-\overline{X})^3\right]\\ \end{array}$$

標本平均まわりの2次モーメントの和

標本平均\(\overline{X}\)まわりの2次モーメント

\begin{array}{rclcl} \displaystyle E\left[ \left(X_k -\overline{X}\right)^2 \right] \end{array}

標本平均\(\overline{X}\)まわりの2次モーメントの和

\begin{array}{rclcl} \displaystyle \sum_{k=1}^{n}E\left[ \left(X_k -\overline{X}\right)^2 \right] &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(\left(X_k - \mu\right) - \left(\overline{X} - \mu\right)\right)^2 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(\left(X_k - \mu\right)^2 -2\left(X_k - \mu\right)\left(\overline{X} - \mu\right) + \left(\overline{X} - \mu\right)^2\right) \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k - \mu\right)^2 -2\left(\overline{X} - \mu\right)\sum_{k=1}^{n}\left(X_k - \mu\right) + \left(\overline{X} - \mu\right)^2\sum_{k=1}^{n}1 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k - \mu\right)^2 -2\left(\overline{X} - \mu\right)\sum_{k=1}^{n}\left(X_k - \mu\right) + n\left(\overline{X} - \mu\right)^2 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k - \mu\right)^2\right] +E\left[-2\left(\overline{X} - \mu\right)\sum_{k=1}^{n}\left(X_k - \mu\right) + n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle \sum_{k=1}^{n}E\left[ \left(X_k - \mu\right)^2\right] + E\left[-2\left(\overline{X} - \mu\right)\left(\sum_{k=1}^{n}X_k - \sum_{k=1}^{n}\mu\right) + n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle \sum_{k=1}^{n}V\left[X_k\right] + E\left[-2\left(\overline{X} - \mu\right)\left(n\overline{X} - n\mu\right) + n\left(\overline{X} - \mu\right)^2\right] \,\dots\,\displaystyle\sum_{k=1}^{n}X_k=n\overline{X}\\ &=&\displaystyle \sum_{k=1}^{n}\sigma^2 + E\left[-2\left(\overline{X} - \mu\right)n\left(\overline{X} - \mu\right) + n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle n\sigma^2 + E\left[-n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle n\sigma^2 - nE\left[\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle n\sigma^2 - nV\left[\overline{X}\right] \\ &=&\displaystyle n\sigma^2 - n\frac{\sigma^2}{n} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable_3.html}{V\left[\overline{X}\right]=\frac{\sigma^2}{n}} \\&=&\displaystyle \left(n-1\right)\sigma^2 \\ \end{array}

“標本平均\(\overline{X}\)まわりの2次モーメントの和”から母分散\(\sigma^2\)を推定する\(\hat{\sigma}^2\)

\begin{array}{rclcl} \displaystyle \hat{\sigma}^2 &=&\displaystyle \frac{1}{\left(n-1\right)}\sum_{k=1}^{n}E\left[ \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle \frac{1}{\left(n-1\right)}E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle \frac{1}{\left(n-1\right)}E\left[\sum_{k=1}^{n} \left(X_k^2 \right)-\overline{X}^2 \right] \end{array}