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\(\beta_2\left(\overline{X}\right)\)

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