統計検定 1級 過去問 解答/解答例と解説
2014年11月30日 (日) 試験

$$\begin{equation} X = \begin{pmatrix} 1 & x_{1} & {x_{1}}^2 \\ 1 & x_{2} & {x_{2}}^2 \\ \vdots & \vdots & \vdots \\ 1 & x_{n} & {x_{n}}^2 \end{pmatrix} \end{equation}$$

$$\begin{align} X'X &= \begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_{1} & x_{2} & \cdots & {x_{n}}^2 \\ {x_{1}}^2 & {x_{2}}^2 & \cdots & {x_{n}}^2 \\ \end{pmatrix} \begin{pmatrix} 1 & x_{1} & {x_{1}}^2 \\ 1 & x_{2} & {x_{2}}^2 \\ \vdots & \vdots & \vdots \\ 1 & x_{n} & {x_{n}}^2 \end{pmatrix} \\ &= \begin{pmatrix} n & \sum_{i=1}^{n}x_i & \sum_{i=1}^{n}{x_i}^2 \\ \sum_{i=1}^{n}x_i & \sum_{i=1}^{n}{x_i}^2 & \sum_{i=1}^{n}{x_i}^3 \\ \sum_{i=1}^{n}{x_i}^2 & \sum_{i=1}^{n}{x_i}^3 & \sum_{i=1}^{n}{x_i}^4 \\ \end{pmatrix} \end{align}$$

$$\begin{align} \left|\sigma^2(X'X)^{-1}\right| &= (\sigma^2)^n\left|(X'X)^{-1}\right| \\ &= \frac{(\sigma^2)^n}{|X'X|} \end{align}$$

$$\begin{equation} \sum_{i=1}^{n}x_i = (-1) \times \frac{1-p}{2}n + 0 \times np + 1 \times \frac{1-p}{2}n = 0 \end{equation}$$

$$\begin{align} \sum_{i=1}^{n}{x_i}^2 &= (-1)^2 \times \frac{1-p}{2}n + 0^2 \times np + 1^2 \times \frac{1-p}{2}n \\ &= n(1 - p) \end{align}$$

$$\begin{align} \sum_{i=1}^{n}{x_i}^3 &= (-1)^3 \times \frac{1-p}{2}n + 0^3 \times np + 1^3 \times \frac{1-p}{2}n \\ &= 0 \end{align}$$

$$\begin{align} \sum_{i=1}^{n}{x_i}^4 &= (-1)^4 \times \frac{1-p}{2}n + 0^4 \times np + 1^4 \times \frac{1-p}{2}n \\ &= n(1 - p) \end{align}$$

$$\begin{align} |X'X| &= \begin{vmatrix} n & \sum_{i=1}^{n}x_i & \sum_{i=1}^{n}{x_i}^2 \\ \sum_{i=1}^{n}x_i & \sum_{i=1}^{n}{x_i}^2 & \sum_{i=1}^{n}{x_i}^3 \\ \sum_{i=1}^{n}{x_i}^2 & \sum_{i=1}^{n}{x_i}^3 & \sum_{i=1}^{n}{x_i}^4 \\ \end{vmatrix} \\ &= \begin{vmatrix} n & 0 & n(1-p) \\ 0 & n(1-p) & 0 \\ n(1-p) & 0 & n(1-p) \\ \end{vmatrix} \\ &= \begin{vmatrix} n - n(1-p) & 0 & n(1-p) \\ 0 & n(1-p) & 0 \\ 0 & 0 & n(1-p) \\ \end{vmatrix}~\cdots\cdots~\text{(第1列)} - \text{(第3列)} \\ &= \{n - n(1-p)\} \begin{vmatrix} n(1-p) & 0 \\ 0 & n(1-p) \\ \end{vmatrix} \\ &= \{n - n(1-p)\}\{n^2(1-p)^2 - 0\} \\ &= n^3p(1-p)^2 \end{align}$$

$$\begin{align} \frac{d}{dp}|X'X| &= n^3(1-p)^2 - 2 n^3 p(1-p) \\ &= n^3(1-p)\{(1-p) - 2 p\} \\ &= n^3(1-p)(1 - 3 p) = 0\\ \end{align}$$

$$\begin{equation} \therefore~p = 1, \frac{1}{3} \end{equation}$$

$$\begin{equation} \hat{Y}_0 = \hat{\alpha} + \hat{\beta} x_0 + \hat{\gamma} {x_0}^2 \end{equation}$$

$$\begin{equation} \hat{Y}_0 = \bm{x}'\hat{\bm{\beta}} \end{equation}$$

$$\begin{align} V[\hat{Y}_0] &= V[\bm{x}'\hat{\bm{\beta}}] \\ &= \bm{x}' V[\hat{\bm{\beta}}] \bm{x} \\ &= \bm{x}' (\sigma^2(X'X)^{-1}) \bm{x} \\ &= \sigma^2\bm{x}'(X'X)^{-1}\bm{x} \end{align}$$

$$\begin{align} X'X &= \begin{pmatrix} n & 0 & n(1-p) \\ 0 & n(1-p) & 0 \\ n(1-p) & 0 & n(1-p) \\ \end{pmatrix} \end{align}$$

$$\begin{align} (X'X)^{-1} &= \frac{1}{|X'X|} \begin{pmatrix} (-1)^{1+1}n^2(1-p)^2 & 0 & (-1)^{1+3}\{-n^2(1-p)^2\} \\ 0 & (-1)^{2+2}\{n^2(1-p) - n^2(1-p)^2\} & 0 \\ (-1)^{3+1}\{-n^2(1-p)^2\} & 0 & (-1)^{3+3}n^2(1-p) \\ \end{pmatrix} \\ &= \frac{1}{n^3p(1-p)^2} \begin{pmatrix} n^2(1-p)^2 & 0 & -n^2(1-p)^2 \\ 0 & n^2p(1-p) & 0 \\ -n^2(1-p)^2 & 0 & n^2(1-p) \\ \end{pmatrix} \\ &= \frac{1}{n} \begin{pmatrix} 1/p & 0 & -1/p \\ 0 & 1/(1-p) & 0 \\ -1/p & 0 & 1/\{p(1-p)\} \\ \end{pmatrix} \end{align}$$

$$\begin{align} \sigma^2\bm{x}'(X'X)^{-1}\bm{x} &= \frac{\sigma^2}{n} \begin{pmatrix} 1 & x_0 & {x_0}^2 \end{pmatrix} \begin{pmatrix} 1/p & 0 & -1/p \\ 0 & 1/(1-p) & 0 \\ -1/p & 0 & 1/\{p(1-p)\} \\ \end{pmatrix} \begin{pmatrix} 1 \\ x_0 \\ {x_0}^2 \end{pmatrix} \\ &= \frac{\sigma^2}{n} \begin{pmatrix} 1 & x_0 & {x_0}^2 \end{pmatrix} \begin{pmatrix} 1/p - {x_0}^2/p \\ x_0/(1-p) \\ -1/p + {x_0}^2/\{p(1-p)\} \\ \end{pmatrix} \\ &= \frac{\sigma^2}{n} \left\{ \left(\frac{1}{p} - \frac{{x_0}^2}{p}\right) + \frac{{x_0}^2}{1-p} + {x_0}^2\left(-\frac{1}{p} + \frac{{x_0}^2}{p(1-p)}\right) \right\} \\ &= \frac{\sigma^2}{n} \left\{ \frac{1}{p} - \frac{{x_0}^2}{p} + \frac{{x_0}^2}{1-p} -\frac{{x_0}^2}{p} + \frac{{x_0}^4}{p(1-p)} \right\} \\ &= \frac{\sigma^2}{np(1-p)} \left\{ (1-p) - {x_0}^2(1-p) + {x_0}^2p - {x_0}^2(1-p) + {x_0}^4 \right\} \\ &= \frac{\sigma^2}{np(1-p)} \left\{ {x_0}^4 + (3p -2){x_0}^2 + (1-p) \right\} \\ \end{align}$$

$$\begin{align} E[V[\hat{Y}]] &= \int_{-1}^{1}\frac{\sigma^2}{np(1-p)}\left\{{x_0}^4 + (3p -2){x_0}^2 + (1-p) \right\}\,dx_0 \\ &= \frac{2\sigma^2}{np(1-p)}\int_{0}^{1}\left\{{x_0}^4 + (3p -2){x_0}^2 + (1-p) \right\}\,dx_0 \\ &= \frac{2\sigma^2}{np(1-p)}\left[\frac{1}{5}{x_0}^5 + \frac{3p -2}{3}{x_0}^3 + (1-p)x_0 \right]_{0}^{1} \\ &= \frac{2\sigma^2}{np(1-p)}\left\{\frac{1}{5} + \frac{3p -2}{3} + (1-p) \right\} \\ &= \frac{2\sigma^2}{15np(1-p)}\left\{3 + 5(3p -2) + 15(1-p) \right\} \\ &= \frac{2\sigma^2}{15np(1-p)}\left\{3 + 15p - 10 + 15 - 15p \right\} \\ &= \frac{16\sigma^2}{15np(1-p)} \end{align}$$