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\(\beta_1\left(\overline{X}\right)\)

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