標本平均の分布の尖度
標本平均\(\overline{X}\)の尖度\(\beta_2\left(\overline{X}\right)\)
$$ \begin{eqnarray} \beta_2\left(\overline{X}\right) &=&\frac{\mathrm{E}\left[(\overline{X}-\mu)^4\right]}{\mathrm{V}\left[\overline{X}\right]^{\frac{4}{2}}}-3 \;\cdots\; \href{https://shikitenkai.blogspot.com/2020/08/blog-post_27.html}{\beta_2=\beta_2\left(X\right)=\frac{\mathrm{E}\left[(X-\mu)^4\right]}{\mathrm{V}\left[X\right]^{\frac{4}{2}}}-3=\frac{\mu_4}{\sigma^4}-3\;:尖度} \\&=&\frac{\frac{1}{n^3}\left(\mu_4+3(n-1)\sigma^4\right)}{\left(\frac{\sigma^2}{n}\right)^2}-3 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/4-4.html}{\mathrm{E}\left[(\overline{X}-\mu)^4\right]=\frac{1}{n^3}\left(\mu_4+3(n-1)\sigma^4\right) \;:標本平均の母平均まわりの4次モーメント (標本平均の4次の中心(化)モーメント) } \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable_3.html}{\mathrm{V}\left[\overline{X}\right]=\frac{\sigma^2}{n}\;:標本平均の分散} \\&=&\frac{n^2}{n^3}\frac{\mu_4+3(n-1)\sigma^4}{\sigma^4}-3 \\&=&\frac{1}{n}\left(\frac{\mu_4}{\sigma^4}+3(n-1)\right)-3 \\&=&\frac{1}{n}\left(\frac{\mu_4}{\sigma^4}+3n-3\right)-3 \\&=&\frac{\frac{\mu_4}{\sigma^4}-3}{n}+3-3 \\&=&\frac{\frac{\mu_4}{\sigma^4}-3}{n} \\&=&\frac{\beta_2}{n} \;\cdots\;\beta_2=\frac{\mu_4}{\sigma^4}-3 \end{eqnarray} $$ よって標本数\(n\)を増やすことで\(\beta_2\left(\overline{X}\right)\)は\(0\)に近づいていく．水準Lと繰返しRが有る標本平均周りの平方和の式展開
水準と繰返しが有る標本平均周りの平方和の式展開(分解あり)
L:水準, R:繰返し $$\begin{eqnarray} \sum_{n=1}^{N}{(x_{n}-\bar{x})^2} &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x})^2}&&\dots L:水準, R:繰返し, N=LR \\&=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(\bar{x_{l}}-\bar{x})^2} &+\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x_{l}})^2} &\dots \sum_{l=1}^{L}\sum_{r=1}^{R}{(\bar{x_{l}}-\bar{x})^2}:各水準平均と全体平均の 差の平方和, \sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x_{l}})^2}:各値と各水準との差の平方和\\ &=&R\sum_{l=1}^{L}{(\bar{x_{l}}^2-2\bar{x_{l}}\bar{x}+\bar{x}^2)} &+\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}^2-2x_{lr}\bar{x_{l}}+\bar{x_{l}}^2)} &\dots (a-b)^2=a^2-2ab+b^2\\ &=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-R\sum_{l=1}^{L}{2\bar{x_{l}}\bar{x}}+R\sum_{l=1}^{L}{\bar{x}^2} &+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-\sum_{l=1}^{L}\sum_{r=1}^{R}{2x_{lr}\bar{x_{l}}}+\sum_{l=1}^{L}\sum_{r=1}^{R}{\bar{x_{l}}^2} &\dots \sum_{k=1}^{K}{(x_k+y_k)}=\sum_{k=1}^{K}{x_k}+\sum_{k=1}^{K}{y_k}\\ &=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-2R\bar{x}\sum_{l=1}^{L}{\bar{x_{l}}}+R\bar{x}^2\sum_{l=1}^{L}{1} &+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\sum_{l=1}^{L}{(\bar{x_{l}}\sum_{r=1}^{R}{x_{lr}})}+\sum_{l=1}^{L}{(\bar{x_{l}}^2\sum_{r=1}^{R}{1})} &\dots \sum_{k=1}^{K}{Cx_k}=C\sum_{k=1}^{K}{x_k}\\ &=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-2R\bar{x}L\bar{x}+R\bar{x}^2L &+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\sum_{l=1}^{L}{(\bar{x_{l}}R\bar{x_{l}})}+\sum_{l=1}^{L}{\bar{x_{l}}^2R} &\dots \sum_{k=1}^{K}{1}=K, \bar{x}:=\frac{\sum_{k=1}^{K}{x_k}}{K} \rightarrow \sum_{k=1}^{K}{x_k}=K\bar{x}\\ &=&R\sum_{l=1}^{L}{\bar{x_{l}}^2}-2LR\bar{x}^2+LR\bar{x}^2 &+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2R\sum_{l=1}^{L}{\bar{x_{l}}^2}+R\sum_{l=1}^{L}{\bar{x_{l}}^2}\\ &=&\color{red}{R\sum_{l=1}^{L}{\bar{x_{l}}^2}}\color{black}{-LR\bar{x}^2} &+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}\color{red}{-R\sum_{l=1}^{L}{\bar{x_{l}}^2}}\\ &=&-LR\bar{x}^2 &+\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}\\ &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-LR\bar{x}^2\\ &=&\sum_{n=1}^{N}{x_{n}^2}-N\bar{x}^2&\dots LR=N\\ \end{eqnarray}$$水準と繰返しが有る標本平均周りの平方和の式展開(分解なし)
$$\begin{eqnarray} \sum_{n=1}^{N}{(x_{n}-\bar{x})^2} &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x})^2}\dots L:水準, R:繰返し, N=LR\\ &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}^2-2x_{lr}\bar{x}+\bar{x}^2)}\\ &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-\sum_{l=1}^{L}\sum_{r=1}^{R}{2x_{lr}\bar{x}}+\sum_{l=1}^{L}\sum_{r=1}^{R}{\bar{x}^2}\\ &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\bar{x}\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}}+\bar{x}^2\sum_{l=1}^{L}\sum_{r=1}^{R}{1}\\ &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2\bar{x}LR\bar{x}+\bar{x}^2LR\\ &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-2LR\bar{x}^2+LR\bar{x}^2\\ &=&\sum_{l=1}^{L}\sum_{r=1}^{R}{x_{lr}^2}-LR\bar{x}^2\\ &=&\sum_{n=1}^{N}{x_{n}^2}-N\bar{x}^2&\dots LR=N\\ &=&\sum_{n=1}^{N}{x_{n}^2}-N(\frac{\sum_{n=1}^{N}{x_{n}}}{N})^2\\ &=&\sum_{n=1}^{N}{x_{n}^2}-\frac{1}{N}(\sum_{n=1}^{N}{x_{n}})^2\\ &=&\sum_{n=1}^{N}{x_{n}^2}-\frac{1}{N}(N\bar{x})^2\\ &=&\sum_{n=1}^{N}{x_{n}^2}-\frac{1}{N}N^2\bar{x}^2\\ &=&\sum_{n=1}^{N}{x_{n}^2}-N\bar{x}^2\\ \end{eqnarray}$$各水準平均と全体平均の差の平方和と各値と各水準との差の平方和に分解できること
以上より分解有りと無しで同じ値になることから，\(\sum_{n=1}^{N}{(x_{n}-\bar{x})^2}=\sum_{l=1}^{L}\sum_{r=1}^{R}{(x_{lr}-\bar{x})^2}\)は\(\sum_{l=1}^{L}\sum_{r=1}^{R}{(\bar{x_{l}}-\bar{x})^2}\)と\( \sum_{l=1}^{L}\sum_{r=1}^{R}{ (x_{lr}-\bar{x_{l}})^2 } \)に分解できることがわかる．標本平均の分布の歪度
標本平均(\(\overline{X}\))の分布の歪度
$$\begin{array}{rcl} \displaystyle \beta_1(\overline{X}) &=&\displaystyle \frac{E\left[\left(\overline{X}-\mu\right)^3\right]}{\left(V\left[\overline{X}\right]^{\frac{1}{2}}\right)^3}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{\left\{ \left( \frac{\sigma^2}{n} \right) ^{\frac{1}{2}} \right\}^3} &&\displaystyle\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/overlinexmu3.html}{E\left[\left(\overline{X}-\mu\right)^3\right]=\mu_3\left(\overline{X}\right)=\frac{\mu_3}{n^2}} ,\,\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable_3.html}{V\left[\overline{X}\right]=\frac{\sigma^2}{n}}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{\left( \frac{\sigma}{\sqrt{n}} \right)^3}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{ \frac{\sigma^3}{n\sqrt{n}} }\\ &=&\displaystyle \frac{\mu_3}{n^2} \frac{n\sqrt{n}}{\sigma^3}\\ &=&\displaystyle \frac{\mu_3}{\sigma^3} \frac{1}{\sqrt{n}}\\ &=&\displaystyle \frac{\beta_1}{\sqrt{n}} &&\displaystyle\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{\frac{\mu_3}{\sigma^3}=\beta_1}\\ \end{array}$$ よって標本数\(n\)を増やすことで\(\beta_1\left(\overline{X}\right)\)は\(0\)に近づいていく．標本の二乗和
標本分散の式展開
$$\begin{array}{rcl} \displaystyle \sum_{k=1}^{n}(X_k-\overline{X})^2 &=&\displaystyle \sum_{k=1}^{n}(X_k^2-2\overline{X} X_k+\overline{X}^2)\\ &=&\displaystyle \sum_{k=1}^{n}X_k^2-2\overline{X} \sum_{k=1}^{n}X_k+\overline{X}^2\sum_{k=1}^{n}1\\ &=&\displaystyle \sum_{k=1}^{n}X_k^2-2n\overline{X}^2+n\overline{X}^2\\ &&\displaystyle\,\dotso\,\sum_{k=1}^{n}X_k=n\overline{X}\\ &=&\displaystyle \sum_{k=1}^{n}X_k^2-n\overline{X}^2\\ \end{array}$$標本の二乗和
$$\begin{array}{rcl} \displaystyle \sum_{k=1}^{n}X_k^2 &=&\displaystyle \sum_{k=1}^{n}(X_k-\overline{X})^2+n\overline{X}^2\\ \end{array}$$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\(\hat{\mu}_3\)
$$\begin{array}{rcl} \displaystyle \hat{\mu}_3 &=&\displaystyle \frac{n}{(n-1)(n-2)}E\left[\sum_{k=1}^{n}(X_k-\overline{X})^3\right]\\ \end{array}$$不偏分散の期待値
不偏分散の期待値
$$\begin{array}{rclcl} \hat{\sigma}^2 &=& \href{https://shikitenkai.blogspot.com/2019/07/specimen-random-variable.html}{\displaystyle \frac{1}{n-1}\sum_{k=1}^{n}\left( \displaystyle X_k - \overline{X} \displaystyle \right)^2}\,\dotso\,不偏分散(unbiased \, variance)\\ E\left[\hat{\sigma}^2\right] &=&E\left[ \displaystyle\frac{1}{n-1}\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle\frac{1}{n-1}E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle\frac{1}{n-1}\left(n-1\right)\sigma^2 \,\dotso\,\displaystyle \href{https://shikitenkai.blogspot.com/2019/07/overlinex2.html}{E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]=\left(n-1\right)\sigma^2}\\ &=&\displaystyle\sigma^2\\ \end{array}$$標本分散の期待値
$$\begin{array}{rcl}
s^2&=&\displaystyle \frac{1}{n}\sum_{k=1}^{n}\left(
\displaystyle X_k - \overline{X}
\displaystyle \right)^2\,\dotso\,標本分散(sample \, variance)\\
E\left[s^2\right]&=&E\left[ \displaystyle\frac{1}{n}\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\
&=&\displaystyle \frac{1}{n} E\left[ \sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\
&=&\displaystyle \frac{1}{n} \left(n-1\right)\sigma^2 \,\dotso\,\displaystyle \href{https://shikitenkai.blogspot.com/2019/07/overlinex2.html}{E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]=\left(n-1\right)\sigma^2}\\
&=&\displaystyle \frac{n-1}{n}\sigma^2 \\
\end{array}$$
標本確率変数(Specimen random variable) / 標本確率変数の期待値と分散
$$\begin{array}{rcl}
母集団の確率変数&:&X\\
標本確率変数&:&X_k (k=1,2,\dotsc ,n)\\
\end{array}$$
$$\begin{array}{rcl}
\mu&=&E[X]\\
\sigma^2&=&V[X]\\
\end{array}$$
標本確率変数\(X_k\)は母集団の確率変数\(X\)と同じ確率分布(母集団)に従うので，\(X_k\)の期待値・分散は\(X\)の期待値・分散と等しい．
$$\begin{array}{rclcl}
E[X_k]&=&E[X]&=&\mu\\
V[X_k]&=&V[X]&=&\sigma^2\\
\end{array}$$
標本確率変数(Specimen random variable) / 標本平均・標本分散・不偏分散
\(n\)個の標本(試行して得られたもの)に対して求めたものが，標本平均,標本分散，不偏分散となる
$$\begin{array}{rcl} 標本確率変数&:&X_k (k=1,2,\dotsc ,n)\\ \end{array}$$ $$\begin{array}{rclcl} \overline{X}&=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k &\dots&標本平均(sample \, mean)\\ s^2&=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} (X_k - \overline{X})^2 &\dots&標本分散(sample \, variance)\\ \hat{\sigma}^2&=&\displaystyle\frac{1}{n-1}\sum_{k=1}^{n} (X_k - \overline{X})^2 &\dots&不偏分散(unbiased \, variance)\\ \end{array}$$全事象をもとに求めたものが母平均,母分散となる
$$\begin{array}{rclcl} \mu&=&\displaystyle\frac{1}{N}\sum_{k=1}^{N} X_i &\dots&母平均(population \, mean)\\ \sigma^2&=&\displaystyle\frac{1}{N}\sum_{i=1}^{N} (X_i - \mu)^2 &\dots&母分散(population \, variance)\\ \end{array}$$標本確率変数(Specimen random variable) / 標本平均の期待値
$$\begin{array}{rcl}
標本確率変数&:&X_k (k=1,2,\dotsc ,n)\\
標本平均(sample \, mean)&:&\overline{X}=\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k\\
\end{array}$$
$$\begin{array}{rclcl}
\mu&=&E[X]\\
E[X_k]&=&E[X]&=&\mu\\
E[cX]&=&c E[X]\\
\end{array}$$
$$\begin{array}{rclcl}
E[\overline{X}]&=&E[\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k]\\
&=&\displaystyle\frac{1}{n}E[\sum_{k=1}^{n} X_k]\\
&=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} E[X_k]\\
&=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} \mu\\
&=&\displaystyle\frac{1}{n}n\mu\\
&=&\displaystyle\mu\\
\end{array}$$
標本確率変数(Specimen random variable) / 標本平均の分散
$$\begin{array}{rcl}
母集団の確率変数&:&X\\
標本確率変数&:&X_k (k=1,2,\dotso ,n)\\
標本平均(sample \, mean)&:&\overline{X}=\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k\\
\end{array}$$
$$\begin{array}{rcl}
\sigma^2&=&V[X]\,\dotso\,母集団の分散\\
\end{array}$$
$$\begin{array}{rclcl}
V[\overline{X}]&=&V[\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k]\\
&=&\displaystyle\left( \frac{1}{n} \right)^2 V[\sum_{k=1}^{n} X_k]
\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-variance.html}{V[cX]=c^2V[X]}\\
&=&\displaystyle\frac{1}{n^2}\sum_{k=1}^{n} V[X_k]
\;\cdots\;X_k同士が互いに独立\left(\mathrm{Con}\left(X_i, X_j\right)=0\right)であるならV[\sum_{k=1}^{n} X_k]=\sum_{k=1}^{n} V[X_k]
\\&=&\displaystyle\frac{1}{n^2}\sum_{k=1}^{n} \sigma^2
\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable.html}{V[X_k]=V[X]=\sigma^2}\\
&=&\displaystyle\frac{1}{n^2}n\sigma^2\\
&=&\displaystyle\frac{\sigma^2}{n}\\
\end{array}$$
よって標本数\(n\)を増やすことで\(\mathrm{V}[\overline{X}]\)は\(0\)に近づいていく．
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