第5講回帰分析

回帰係数

回帰係数の表現 (1)

$\hat{\beta}_0,~\hat{\beta}_1$ は， $y_i$ の一次式で表現することができる．

\begin{aligned} {\hat{β}}_{1} & = \frac{S_{x y}}{S_{x x}} = \frac{1}{S_{x x}} {\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})} \\ = \frac{1}{S_{x x}} {\sum_{i = 1}^{n} (x_{i} - \bar{x}) y_{i} - \bar{y} \sum_{i = 1}^{n} (x_{i} - \bar{x})} \\ = \frac{1}{S_{x x}} {\sum_{i = 1}^{n} (x_{i} - \bar{x}) y_{i}} (∵ \sum_{i = 1}^{n} (x_{i} - \bar{x}) = 0) \\ = \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) y_{i} \end{aligned}

$\begin{align} \hat{\beta}_1 &= \frac{S_{xy}}{S_{xx}} = \frac{1}{S_{xx}}\left\{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})\right\} \\ &= \frac{1}{S_{xx}}\left\{\sum_{i=1}^{n}(x_i - \bar{x})y_i - \bar{y}\sum_{i=1}^{n}(x_i - \bar{x})\right\} \\ &= \frac{1}{S_{xx}}\left\{\sum_{i=1}^{n}(x_i - \bar{x})y_i \right\}~~~~\left(\because \sum_{i=1}^{n}(x_i - \bar{x}) = 0\right) \\ &= \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right)y_i \end{align}$

\begin{aligned} {\hat{β}}_{0} & = \bar{y} - {\hat{β}}_{1} \bar{x} \\ = \sum_{i = 1}^{n} \frac{y_{i}}{n} - \bar{x} \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) y_{i} \\ = \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) y_{i} \end{aligned}

$\begin{align} \hat{\beta}_0 &= \bar{y} - \hat{\beta}_1\bar{x} \\ &= \sum_{i=1}^{n}\frac{y_i}{n}- \bar{x}\sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right)y_i \\ &= \sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)y_i \end{align}$

回帰係数の表現 (2)

前節の式に $y_i = \beta_0 + \beta_1 x_i + \epsilon_i$ を代入し， $\epsilon_i$ だけが確率変数であることに注目してみる．ここで，

\begin{aligned} \sum_{i = 1}^{n} (x_{i} - \bar{x}) = 0 \end{aligned}

$\begin{align} \sum_{i=1}^{n}(x_i - \bar{x}) = 0 \end{align}$

\begin{aligned} \sum_{i = 1}^{n} (x_{i} - \bar{x}) x_{i} & = \sum_{i = 1}^{n} (x_{i} - \bar{x}) x_{i} + \sum_{i = 1}^{n} (x_{i} - \bar{x}) \bar{x} \\ = \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2} \\ = S_{x x} \end{aligned}

$\begin{align} \sum_{i=1}^{n}(x_i - \bar{x})x_i &= \sum_{i=1}^{n}(x_i - \bar{x})x_i + \sum_{i=1}^{n}(x_i - \bar{x})\bar{x} \\ &= \sum_{i=1}^{n}(x_i - \bar{x})^2 \\ &= S_{xx} \end{align}$

であることを利用する．

\begin{aligned} {\hat{β}}_{1} & = \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) y_{i} \\ = \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) (β_{0} + β_{1} x_{i} + ϵ_{i}) \\ = β_{0} \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) + β_{1} \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) x_{i} + \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \\ = β_{0} \frac{1}{S_{x x}} \sum_{i = 1}^{n} (x_{i} - \bar{x}) + β_{1} \frac{1}{S_{x x}} \sum_{i = 1}^{n} (x_{i} - \bar{x}) x_{i} + \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \\ = 0 + β_{1} \frac{1}{S_{x x}} S_{x x} + \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \\ = β_{1} + \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \end{aligned}

$\begin{align} \hat{\beta}_1 &= \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right)y_i \\ &= \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right)(\beta_0 + \beta_1 x_i + \epsilon_i) \\ &= \beta_0\sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) + \beta_1\sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) x_i + \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) \epsilon_i \\ &= \beta_0\frac{1}{S_{xx}}\sum_{i=1}^{n}(x_i - \bar{x}) + \beta_1\frac{1}{S_{xx}}\sum_{i=1}^{n}(x_i - \bar{x}) x_i + \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) \epsilon_i \\ &= 0 + \beta_1\frac{1}{S_{xx}}S_{xx} + \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) \epsilon_i \\ &= \beta_1 + \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) \epsilon_i \end{align}$

\begin{aligned} {\hat{β}}_{0} & = \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) y_{i} \\ = \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) (β_{0} + β_{1} x_{i} + ϵ_{i}) \\ = β_{0} \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) + β_{1} \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) x_{i} + \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \\ = β_{0} \sum_{i = 1}^{n} \frac{1}{n} - β_{0} \sum_{i = 1}^{n} \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}} + β_{1} \sum_{i = 1}^{n} \frac{1}{n} x_{i} - β_{1} \sum_{i = 1}^{n} \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}} x_{i} + \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \\ = β_{0} - 0 + β_{1} \bar{x} - β_{1} \bar{x} + \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \\ = β_{0} + \sum_{i = 1}^{n} (\frac{1}{n} - \bar{x} \frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i} \end{aligned}

$\begin{align} \hat{\beta}_0 &= \sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)y_i \\ &= \sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)(\beta_0 + \beta_1 x_i + \epsilon_i) \\ &= \beta_0\sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right) + \beta_1\sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)x_i + \sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)\epsilon_i \\ &= \beta_0\sum_{i=1}^{n}\frac{1}{n} - \beta_0\sum_{i=1}^{n}\bar{x}\frac{x_i - \bar{x}}{S_{xx}} + \beta_1\sum_{i=1}^{n}\frac{1}{n}x_i - \beta_1\sum_{i=1}^{n}\bar{x}\frac{x_i - \bar{x}}{S_{xx}}x_i + \sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)\epsilon_i \\ &= \beta_0 - 0 + \beta_1\bar{x} - \beta_1\bar{x} + \sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)\epsilon_i \\ &= \beta_0 + \sum_{i=1}^{n}\left(\frac{1}{n}- \bar{x}\frac{x_i - \bar{x}}{S_{xx}}\right)\epsilon_i \\ \end{align}$

第5講 回帰分析

回帰係数

回帰係数の表現 (1)

回帰係数の表現 (2)

第5講回帰分析