統計検定 1級過去問解答/解答例と解説
2017年11月26日 (日) 試験

統計数理問1 [1]

設問の要約

$E[\bar{X}]$ と $V[\bar{X}]$ を $\mu$ ， $\sigma^2$ ， $n$ を用いて表わせ
$T^2$ と $S^2$ は $\sigma^2$ の不偏推定量であることを示せ

解答例

\begin{aligned} E [\bar{X}] & = E [\frac{1}{n} \sum_{i = 1}^{n} X_{i}] \\ = \frac{1}{n} \sum_{i = 1}^{n} E [X_{i}] \\ = \frac{1}{n} \cdot n μ \\ = μ \end{aligned}

$\begin{align} E[\bar{X}] &= E\left[\frac{1}{n}\sum_{i=1}^{n}X_i\right] \\ &= \frac{1}{n}\sum_{i=1}^{n}E\left[X_i\right] \\ &= \frac{1}{n} \cdot n \mu \\ &= \mu \end{align}$

\begin{aligned} V [\bar{X}] & = V [\frac{1}{n} \sum_{i = 1}^{n} X_{i}] \\ = \frac{1}{n^{2}} \sum_{i = 1}^{n} V [X_{i}] \\ = \frac{1}{n^{2}} \cdot n σ^{2} \\ = \frac{σ^{2}}{n} \end{aligned}

$\begin{align} V[\bar{X}] &= V\left[\frac{1}{n}\sum_{i=1}^{n}X_i\right] \\ &= \frac{1}{n^2}\sum_{i=1}^{n}V\left[X_i\right] \\ &= \frac{1}{n^2} \cdot n \sigma^2 \\ &= \frac{\sigma^2}{n} \end{align}$

\begin{aligned} E [T^{2}] & = E [\frac{1}{n} \sum_{i = 1}^{n} (X_{i} - μ)^{2}] \\ = \frac{1}{n} \sum_{i = 1}^{n} E [(X_{i} - μ)^{2}] \\ = \frac{1}{n} \sum_{i = 1}^{n} V [X] \\ = \frac{1}{n} \cdot n σ^{2} \\ = σ^{2} \end{aligned}

$\begin{align} E[T^2] &= E\left[\frac{1}{n}\sum_{i=1}^{n}(X_i - \mu)^2\right] \\ &= \frac{1}{n}\sum_{i=1}^{n}E\left[(X_i - \mu)^2\right] \\ &= \frac{1}{n}\sum_{i=1}^{n}V\left[X\right] \\ &= \frac{1}{n} \cdot n \sigma^2 \\ &= \sigma^2 \end{align}$

よって，

T^{2}

$T^2$ は，

σ^{2}

$\sigma^2$ の不偏推定量である．

\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} - \bar{X})^{2}] \\ = \frac{1}{n - 1} E [\sum_{i = 1}^{n} {(X_{i} - μ) - (\bar{X} - μ)}^{2}] \\ = \frac{1}{n - 1} E [\sum_{i = 1}^{n} {(X_{i} - μ)^{2} - 2 (X_{i} - μ) (\bar{X} - μ) + (\bar{X} - μ)^{2}}] \\ = \frac{1}{n - 1} E [\sum_{i = 1}^{n} (X_{i} - μ)^{2} - 2 (\bar{X} - μ) \sum_{i = 1}^{n} (X_{i} - μ) + \sum_{i = 1}^{n} (\bar{X} - μ)^{2}] \\ = \frac{1}{n - 1} E [\sum_{i = 1}^{n} (X_{i} - μ)^{2} - 2 (\bar{X} - μ) \cdot n (\bar{X} - μ) + n (\bar{X} - μ)^{2}] \\ = \frac{1}{n - 1} E [\sum_{i = 1}^{n} (X_{i} - μ)^{2} - \sum_{i = 1}^{n} (\bar{X} - μ)^{2}] \\ = \frac{1}{n - 1} {\sum_{i = 1}^{n} E [(X_{i} - μ)^{2}] - n E [(\bar{X} - μ)^{2}]} \\ = \frac{1}{n - 1} {\sum_{i = 1}^{n} V [X] - n V [\bar{X}]} \\ = \frac{1}{n - 1} {\sum_{i = 1}^{n} σ^{2} - n \frac{σ^{2}}{n}} \\ = \frac{1}{n - 1} {n σ^{2} - σ^{2}} \\ = \frac{1}{n - 1} (n - 1) σ^{2} \\ = σ^{2} \end{aligned}

$\begin{align} E[S^2] &= E\left[\frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar{X})^2\right] \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}(X_i - \bar{X})^2\right] \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}\left\{(X_i - \mu) - (\bar{X} - \mu)\right\}^2\right] \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}\left\{(X_i - \mu)^2 - 2(X_i - \mu)(\bar{X} - \mu) + (\bar{X} - \mu)^2\right\}\right] \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}(X_i - \mu)^2 - 2(\bar{X} - \mu)\sum_{i=1}^{n}(X_i - \mu) + \sum_{i=1}^{n}(\bar{X} - \mu)^2\right] \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}(X_i - \mu)^2 - 2(\bar{X} - \mu) \cdot n(\bar{X} - \mu) + n(\bar{X} - \mu)^2\right] \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}(X_i - \mu)^2 - \sum_{i=1}^{n}(\bar{X} - \mu)^2\right] \\ &= \frac{1}{n-1}\left\{\sum_{i=1}^{n}E\left[(X_i - \mu)^2\right] - nE\left[(\bar{X} - \mu)^2\right]\right\} \\ &= \frac{1}{n-1}\left\{\sum_{i=1}^{n}V[X] - nV[\bar{X}]\right\} \\ &= \frac{1}{n-1}\left\{\sum_{i=1}^{n}\sigma^2 - n\frac{\sigma^2}{n}\right\} \\ &= \frac{1}{n-1}\left\{n\sigma^2 - \sigma^2\right\} \\ &= \frac{1}{n-1}(n - 1)\sigma^2 \\ &= \sigma^2 \end{align}$

よって，

S^{2}

$S^2$ 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 2017年11月26日 (日) 試験

統計数理 問1 [1]

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
2017年11月26日 (日) 試験

統計数理問1 [1]