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

問題の要約

• 重回帰モデル

  $$\begin{equation} y_i = \beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i} + \epsilon_i~~~~(i = 1, \ldots, n) \end{equation}$$

• $\epsilon_i$ : 誤差

  $$\begin{equation} E[\epsilon_i] = 0 \end{equation}$$
  $$\begin{equation} V[\epsilon_i] = \sigma^2 \end{equation}$$
  $$\begin{equation} Cov[\epsilon_i \epsilon_j] = 0~~~~(i \ne j) \end{equation}$$

• 次のように定義する．

  $$\begin{equation} \bar{x}_1 = \frac{1}{n}\sum_{i=1}^{n}x_{1i} = 0~~\Longleftrightarrow~~\sum_{i=1}^{n}x_{1i} = 0 \end{equation}$$
  $$\begin{equation} \bar{x}_2 = \frac{1}{n}\sum_{i=1}^{n}x_{2i} = 0~~\Longleftrightarrow ~~\sum_{i=1}^{n}x_{2i} = 0 \end{equation}$$
  $$\begin{equation} S_{k\ell} = \frac{1}{n}\sum_{i=1}^{n}x_{k i}x_{\ell i}~~~~(k, \ell = 1, 2) \end{equation}$$
  $$\begin{equation} S_{ky} = \frac{1}{n}\sum_{i=1}^{n}x_{k i}y_{i}~~~~(k, \ell = 1, 2) \end{equation}$$
  $$\begin{equation} S_{yy} = \frac{1}{n}\sum_{i=1}^{n}{y_{i}}^2 \end{equation}$$
  $$\begin{equation} r_{12} = \frac{S_{12}}{\sqrt{S_{11}}\sqrt{S_{22}}} \end{equation}$$
  $$\begin{equation} r_{1y} = \frac{S_{1y}}{\sqrt{S_{11}}\sqrt{S_{yy}}} \end{equation}$$
  $$\begin{equation} r_{2y} = \frac{S_{2y}}{\sqrt{S_{22}}\sqrt{S_{2y}}} \end{equation}$$
  $$\begin{equation} \bm{y} = \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix} \end{equation}$$
  $$\begin{equation} X = \begin{pmatrix} 1 & x_{11} & x_{21} \\ 1 & x_{12} & x_{22} \\ \vdots & \vdots & \vdots \\ 1 & x_{1n} & x_{2n} \\ \end{pmatrix} \end{equation}$$
  $$\begin{equation} \bm{\beta} = \begin{pmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \end{pmatrix} \end{equation}$$
  $$\begin{equation} \bm{\epsilon} = \begin{pmatrix} \epsilon_1 \\ \epsilon_2 \\ \vdots \\ \epsilon_n \end{pmatrix} \end{equation}$$
  $\hat{\bm{\beta}}$ : $\bm{\beta}$ の最小二乗推定量
  $$\begin{equation} \hat{\bm{\beta}} = \begin{pmatrix} \hat{\beta}_0 \\ \hat{\beta}_1 \\ \hat{\beta}_2 \end{pmatrix} \end{equation}$$