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

統計数理問1 [2]

設問の要約

$S^2$ は $\sigma^2$ の不偏推定量であることを示せ．

解答例

\begin{aligned} E [{X_{i}}^{2}] = (E [X_{i}])^{2} + V [X_{i}] = μ^{2} + σ^{2} \end{aligned}

$\begin{align} E[{X_i}^2] = (E[X_i])^2 + V[X_i] = \mu^2 + \sigma^2 \end{align}$

\begin{aligned} E [S^{2}] & = E [\frac{1}{n - 1} \sum_{i = 1}^{n} (X_{i} - \bar{X})^{2}] \\ = \frac{1}{n - 1} E [\sum_{i = 1}^{n} ({X_{i}}^{2} - 2 X_{i} \bar{X} + {\bar{X}}^{2})] \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} (E [{X_{i}}^{2}] - 2 E [X_{i} \bar{X}] + E [{\bar{X}}^{2}]) \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} (E [{X_{i}}^{2}] - 2 E [X_{i} \frac{1}{n} \sum_{j = 1}^{n} X_{j}] + E [{\bar{X}}^{2}]) \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} (E [{X_{i}}^{2}] - \frac{2}{n} E [\sum_{j = 1}^{n} X_{i} X_{j}] + E [{\bar{X}}^{2}]) \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} (E [{X_{i}}^{2}] - \frac{2}{n} E [X_{i} X_{i} + \sum_{j \neq i} X_{i} X_{j}] + E [{\bar{X}}^{2}]) \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} (E [{X_{i}}^{2}] - \frac{2}{n} E [X_{i}^{2} + \sum_{j \neq i} X_{i} X_{j}] + E [{\bar{X}}^{2}]) \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} (E [{X_{i}}^{2}] - \frac{2}{n} E [X_{i}^{2}] - \frac{2}{n} \sum_{j \neq i} E [X_{i} X_{j}] + E [{\bar{X}}^{2}]) \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} (\frac{n - 2}{n} E [{X_{i}}^{2}] - \frac{2}{n} \sum_{j \neq i} E [X_{i}] E [X_{j}] + E [{\bar{X}}^{2}]) \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} {\frac{n - 2}{n} (μ^{2} + σ^{2}) - \frac{2}{n} \sum_{j \neq i} μ^{2} + (μ^{2} + \frac{σ^{2}}{n})} \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} {\frac{n - 2}{n} (μ^{2} + σ^{2}) - \frac{2 (n - 1)}{n} μ^{2} + (μ^{2} + \frac{σ^{2}}{n})} \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} {(\frac{n - 2}{n} - \frac{2 (n - 1)}{n} + 1) μ^{2} + (\frac{n - 2}{n} + \frac{1}{n}) σ^{2}} \\ = \frac{1}{n - 1} \sum_{i = 1}^{n} \frac{n - 1}{n} σ^{2} \\ = \frac{1}{n} \sum_{i = 1}^{n} σ^{2} \\ = σ^{2} \end{aligned}

$\begin{align} E[S^2] &= E\left[\frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar{X})^2\right] \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}({X_i}^2 - 2X_i\bar{X} + \bar{X}^2)\right] \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left(E[{X_i}^2] - 2E[X_i\bar{X}] + E[\bar{X}^2]\right) \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left(E[{X_i}^2] - 2E\left[X_i\frac{1}{n}\sum_{j=1}^{n}X_j\right] + E[\bar{X}^2]\right) \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left(E[{X_i}^2] - \frac{2}{n}E\left[\sum_{j=1}^{n}X_iX_j\right] + E[\bar{X}^2]\right) \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left(E[{X_i}^2] - \frac{2}{n}E\left[X_i X_i + \sum_{j \ne i}X_iX_j\right] + E[\bar{X}^2]\right) \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left(E[{X_i}^2] - \frac{2}{n}E\left[X_i^2 + \sum_{j \ne i}X_iX_j\right] + E[\bar{X}^2]\right) \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left(E[{X_i}^2] - \frac{2}{n}E\left[X_i^2\right] - \frac{2}{n}\sum_{j \ne i}E\left[X_iX_j\right] + E[\bar{X}^2]\right) \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left(\frac{n-2}{n}E[{X_i}^2] - \frac{2}{n}\sum_{j \ne i}E[X_i]E[X_j] + E[\bar{X}^2]\right) \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left\{\frac{n-2}{n}(\mu^2 + \sigma^2) - \frac{2}{n}\sum_{j \ne i}\mu^2 + \left(\mu^2 + \frac{\sigma^2}{n}\right)\right\} \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left\{\frac{n-2}{n}(\mu^2 + \sigma^2) - \frac{2(n-1)}{n}\mu^2 + \left(\mu^2 + \frac{\sigma^2}{n}\right)\right\} \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\left\{\left(\frac{n-2}{n} - \frac{2(n-1)}{n} + 1\right)\mu^2 + \left(\frac{n-2}{n} + \frac{1}{n}\right)\sigma^2\right\} \\ &= \frac{1}{n-1}\sum_{i=1}^{n}\frac{n-1}{n}\sigma^2 \\ &= \frac{1}{n}\sum_{i=1}^{n}\sigma^2 \\ &= \sigma^2 \end{align}$

よって，

E [S^{2}]

$E[S^2]$ は

σ^{2}

$\sigma^2$ の不偏推定量である．

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

統計数理 問1 [2]

設問の要約

解答例

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

統計数理問1 [2]