統計検定 1級過去問解答/解答例と解説
2016年11月27日 (日) 試験

統計数理問3 [1]

設問の要約

次の推定量 $b_0$ と $b_1$ は $\beta$ の不偏推定量でありことを示せ．
$b_{0} = \frac{1}{n} \sum_{i = 1}^{n} \frac{Y_{i}}{x_{i}}$
$\begin{equation} b_0 = \dfrac{1}{n}\sum_{i=1}^{n}\frac{Y_i}{x_i} \end{equation}$

$b_{1} = \frac{\sum_{i = 1}^{n} Y_{i}}{\sum_{i = 1}^{n} x_{i}}$
$\begin{equation} b_1 = \dfrac{\sum_{i=1}^{n}Y_i}{\sum_{i=1}^{n}x_i} \end{equation}$

解答例

\begin{aligned} E [b_{0}] & = E [\frac{1}{n} \sum_{i = 1}^{n} \frac{Y_{i}}{x_{i}}] \\ = E [\frac{1}{n} \sum_{i = 1}^{n} \frac{β x_{i} + ϵ_{i}}{x_{i}}] \\ = E [\frac{1}{n} \sum_{i = 1}^{n} {β + \frac{1}{x_{i}} ϵ_{i}}] \\ = \frac{1}{n} \sum_{i = 1}^{n} E [β] + \frac{1}{n} \sum_{i = 1}^{n} \frac{1}{x_{i}} E [ϵ_{i}] \\ = \frac{1}{n} \sum_{i = 1}^{n} β + \frac{1}{n} \sum_{i = 1}^{n} \frac{1}{x_{i}} \times 0 \\ = β \end{aligned}

$\begin{align} E[b_0] &= E \left[\frac{1}{n}\sum_{i=1}^{n}\frac{Y_i}{x_i}\right] \\ &= E \left[\frac{1}{n}\sum_{i=1}^{n}\frac{\beta x_i + \epsilon_i}{x_i}\right] \\ &= E \left[\frac{1}{n}\sum_{i=1}^{n}\left\{\beta + \frac{1}{x_i}\epsilon_i\right\}\right] \\ &= \frac{1}{n}\sum_{i=1}^{n}E[\beta] + \frac{1}{n}\sum_{i=1}^{n}\frac{1}{x_i}E[\epsilon_i] \\ &= \frac{1}{n}\sum_{i=1}^{n}\beta + \frac{1}{n}\sum_{i=1}^{n}\frac{1}{x_i} \times 0 \\ &= \beta \end{align}$

\begin{aligned} E [b_{1}] & = E [\frac{\sum_{i = 1}^{n} Y_{i}}{\sum_{i = 1}^{n} x_{i}}] \\ = \frac{1}{\sum_{i = 1}^{n} x_{i}} E [\sum_{i = 1}^{n} Y_{i}] \\ = \frac{1}{\sum_{i = 1}^{n} x_{i}} E [\sum_{i = 1}^{n} (β x_{i} + ϵ_{i})] \\ = \frac{1}{\sum_{i = 1}^{n} x_{i}} {β \sum_{i = 1}^{n} x_{i} + \sum_{i = 1}^{n} E [ϵ_{i}]} \\ = \frac{1}{\sum_{i = 1}^{n} x_{i}} {β \sum_{i = 1}^{n} x_{i}} \\ = β \end{aligned}

$\begin{align} E[b_1] &= E \left[\frac{\sum_{i=1}^{n}Y_i}{\sum_{i=1}^{n}x_i}\right] \\ &= \frac{1}{\sum_{i=1}^{n}x_i}E \left[\sum_{i=1}^{n}Y_i\right] \\ &= \frac{1}{\sum_{i=1}^{n}x_i}E \left[\sum_{i=1}^{n}(\beta x_i + \epsilon_i)\right] \\ &= \frac{1}{\sum_{i=1}^{n}x_i}\left\{\beta\sum_{i=1}^{n} x_i + \sum_{i=1}^{n} E[\epsilon_i]\right\} \\ &= \frac{1}{\sum_{i=1}^{n}x_i}\left\{\beta\sum_{i=1}^{n} x_i\right\} \\ &= \beta \end{align}$

統計検定 1級 過去問 解答/解答例と解説 2016年11月27日 (日) 試験

統計数理 問3 [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
2016年11月27日 (日) 試験

統計数理問3 [1]