標本平均まわりの2次モーメントの和

標本平均$\bar{X}$まわりの2次モーメント

$$\begin{array}{r}{\displaystyle E\left[{(X_k-\bar{X})}^2\right]}\end{array}$$

標本平均$\bar{X}$まわりの2次モーメントの和

$$\begin{array}{rcl}{\displaystyle \sum _{k=1}^{n}E\left[{(X_k-\bar{X})}^2\right]}& =& {\displaystyle E\left[\sum _{k=1}^n{(X_k-\bar{X})}^2\right]}\\ & =& {\displaystyle E\left[\sum _{k=1}^n{((X_k-\mu )-(\bar{X}-\mu ))}^2\right]}\\ & =& {\displaystyle E\left[\sum _{k=1}^n({(X_k-\mu )}^2-2(X_k-\mu )(\bar{X}-\mu )+{(\bar{X}-\mu )}^2)\right]}\\ & =& {\displaystyle E[\sum _{k=1}^n{(X_k-\mu )}^2-2(\bar{X}-\mu )\sum _{k=1}^n(X_k-\mu )+{(\bar{X}-\mu )}^2\sum _{k=1}^{n}1]}\\ & =& {\displaystyle E[\sum _{k=1}^n{(X_k-\mu )}^2-2(\bar{X}-\mu )\sum _{k=1}^n(X_k-\mu )+n{(\bar{X}-\mu )}^2]}\\ & =& {\displaystyle E\left[\sum _{k=1}^n{(X_k-\mu )}^2\right]+E[-2(\bar{X}-\mu )\sum _{k=1}^n(X_k-\mu )+n{(\bar{X}-\mu )}^2]}\\ & =& {\displaystyle \sum _{k=1}^{n}E\left[{(X_k-\mu )}^2\right]+E[-2(\bar{X}-\mu )(\sum _{k=1}^n{X}_k-\sum _{k=1}^n\mu )+n{(\bar{X}-\mu )}^2]}\\ & =& {\displaystyle \sum _{k=1}^{n}V\left[X_k\right]+E[-2(\bar{X}-\mu )(n\bar{X}-n\mu )+n{(\bar{X}-\mu )}^2]\dots {\displaystyle \sum _{k=1}^n{X}_k=n\bar{X}}}\\ & =& {\displaystyle \sum _{k=1}^n{\sigma }^2+E[-2(\bar{X}-\mu )n(\bar{X}-\mu )+n{(\bar{X}-\mu )}^2]}\\ & =& {\displaystyle n{\sigma }^2+E[-n{(\bar{X}-\mu )}^2]}\\ & =& {\displaystyle n{\sigma }^2-nE\left[{(\bar{X}-\mu )}^2\right]}\\ & =& {\displaystyle n{\sigma }^2-nV\left[\bar{X}\right]}\\ & =& {\displaystyle n{\sigma }^2-n\frac{{\sigma }^2}{n}\cdots V\left[\bar{X}\right]=\frac{{\sigma }^2}{n}}\\ & =& {\displaystyle (n-1){\sigma }^2}\end{array}$$

“標本平均$\bar{X}$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${\hat{\sigma }}^2$

$$\begin{array}{rcl}{\displaystyle {\hat{\sigma }}^2}& =& {\displaystyle \frac{1}{(n-1)}\sum _{k=1}^{n}E\left[{(X_k-\bar{X})}^2\right]}\\ & =& {\displaystyle \frac{1}{(n-1)}E\left[\sum _{k=1}^n{(X_k-\bar{X})}^2\right]}\\ & =& {\displaystyle \frac{1}{(n-1)}E[\sum _{k=1}^n\left(X_k^2\right)-{\bar{X}}^2]}\end{array}$$