統計検定 1級 過去問 解答/解答例と解説
2015年11月29日 (日) 試験

$$\begin{align} MSE[c S^2] &= E\left[\left(c S^2 - \sigma^2\right)^2\right] \\ &= E\left[(c S^2)^2 - 2c S^2\sigma^2 + \sigma^4\right] \\ &= E\left[(c S^2)^2\right] - E\left[2c S^2\sigma^2\right] + E\left[\sigma^4\right] \\ &= E\left[(c S^2)^2\right] - 2c\sigma^2E\left[S^2\right] + E\left[\sigma^4\right] \\ &= E\left[(c S^2)^2\right] - 2c\sigma^4 + \sigma^4 \\ &= V\left[c S^2\right] + \left(E\left[c S^2\right]\right)^2 - 2c\sigma^4 + \sigma^4 \\ &= c^2V\left[S^2\right] + \left(c E\left[S^2\right]\right)^2 - 2c\sigma^4 + \sigma^4 \\ &= c^2\frac{2\sigma^4}{n-1} + \left(c \sigma^2\right)^2 - 2c\sigma^4 + \sigma^4 \\ &= \left\{\frac{2c^2}{n-1} + c^2 - 2c + 1\right\}\sigma^4 \\ &= \left\{\frac{2c^2}{n-1} + (c - 1)^2\right\}\sigma^4 \\ \end{align}$$

$$\begin{align} \frac{d}{dc}MSE[c S^2] &= \frac{d}{dc}\left\{\frac{2c^2}{n-1} + (c - 1)^2\right\}\sigma^4 \\ &= \left\{\frac{4c}{n-1} + 2(c - 1)\right\}\sigma^4 \\ &= (n - 1)\left\{4c + 2(n - 1)(c - 1)\right\}\sigma^4 \\ &= (n - 1)\left\{4c + 2cn - 2n - 2c + 2\right\}\sigma^4 \\ &= (n - 1)\left\{2(n + 1)c - 2(n - 1)\right\}\sigma^4 \\ &= 2(n - 1)\left\{(n + 1)c - (n - 1)\right\}\sigma^4 \\ &= 0 \end{align}$$

$$\begin{equation} \hat{c} = \frac{n - 1}{n + 1} \end{equation}$$

$$\begin{align} MSE[\hat{c} S^2] &= \left\{\frac{2\hat{c}^2}{n-1} + (\hat{c} - 1)^2\right\}\sigma^4 \\ &= \left\{\frac{2}{n-1}\left(\frac{n - 1}{n + 1}\right)^2 + \left(\frac{n - 1}{n + 1} - 1\right)^2\right\}\sigma^4 \\ &= \left\{\frac{2(n-1)}{(n+1)^2} + \left(\frac{n - 1}{n + 1} - 1\right)^2\right\}\sigma^4 \\ &= \left\{\frac{2(n-1)}{(n+1)^2} + \frac{4}{(n+1)^2}\right\}\sigma^4 \\ &= \frac{2\sigma^4}{(n+1)^2}\left\{(n-1) + 2\right\} \\ &= \frac{2\sigma^4}{n+1} \end{align}$$