間違いしかありません.コメントにてご指摘いただければ幸いです(気が付いた点を特に断りなく頻繁に書き直していますのでご注意ください).

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}

0 件のコメント:

コメントを投稿