残差平方和とデータの各平方和

残差平方和とデータの各平方和

$$\begin{eqnarray} S_e&=&\sum_{i=1}^n e_i^2\\ &=&\sum_{i=1}^n \left(y_i - \hat{y_i}\right)^2\;\dots\;\hat{y_i}=\hat{\alpha}+\hat{\beta} x_i\\ &=&\left(\sum_{i=1}^n y_i^2\right) - 2n\bar{y}\hat{\alpha} - 2\left(\sum_{i=1}^n x_i y_i\right)\hat{\beta} + n\hat{\alpha}^2 + 2n\bar{x}\hat{\alpha}\hat{\beta} + \left(\sum_{i=1}^n x_i^2\right)\hat{\beta}^2 \;\dots\;\href{https://shikitenkai.blogspot.com/2020/03/blog-post.html}{残差平方和の展開した式}\\ &=&\left(\sum_{i=1}^n y_i^2\right) - 2n\bar{y}\left(\bar{y}-\frac{S_{xy}}{S_{xx}}\bar{x}\right) - 2\left(\sum_{i=1}^n x_i y_i\right)\frac{S_{xy}}{S_{xx}} + n\left(\bar{y}-\frac{S_{xy}}{S_{xx}}\bar{x}\right)^2 + 2n\bar{x}\left(\bar{y}-\frac{S_{xy}}{S_{xx}}\bar{x}\right)\frac{S_{xy}}{S_{xx}} + \left(\sum_{i=1}^n x_i^2\right)\left(\frac{S_{xy}}{S_{xx}}\right)^2 \;\dots\; \hat{\alpha}=\bar{y}-\frac{S_{xy}}{S_{xx}}\bar{x},\;\hat{\beta}=\frac{S_{xy}}{S_{xx}}\\ &=&\left(\sum_{i=1}^n y_i^2\right) - 2n\bar{y}^2 + 2n\frac{S_{xy}}{S_{xx}}\bar{x}\bar{y} - 2\frac{S_{xy}}{S_{xx}}\left(\sum_{i=1}^n x_i y_i\right) + n\bar{y}^2 - 2n\frac{S_{xy}}{S_{xx}}\bar{x}\bar{y} + n\left(\frac{S_{xy}}{S_{xx}}\right)^2\bar{x}^2 + 2n\frac{S_{xy}}{S_{xx}}\bar{x}\bar{y} - 2n\left(\frac{S_{xy}}{S_{xx}}\right)^2\bar{x}^2 + \left(\sum_{i=1}^n x_i^2\right)\left(\frac{S_{xy}}{S_{xx}}\right)^2\\ &=&\left\{ \left(\sum_{i=1}^n y_i^2\right) - 2n\bar{y}^2 + n\bar{y}^2 \right\} +\left\{ 2n\frac{S_{xy}}{S_{xx}}\bar{x}\bar{y} - 2\frac{S_{xy}}{S_{xx}}\left(\sum_{i=1}^n x_i y_i\right) - 2n\frac{S_{xy}}{S_{xx}}\bar{x}\bar{y} + 2n\frac{S_{xy}}{S_{xx}}\bar{x}\bar{y} \right\} +\left\{ n\left(\frac{S_{xy}}{S_{xx}}\right)^2\bar{x}^2 - 2n\left(\frac{S_{xy}}{S_{xx}}\right)^2\bar{x}^2 + \left(\sum_{i=1}^n x_i^2\right)\left(\frac{S_{xy}}{S_{xx}}\right)^2 \right\}\\ &=&\left\{ \left(\sum_{i=1}^n y_i^2\right) - n\bar{y}^2 \right\} +\left\{ 2n\frac{S_{xy}}{S_{xx}}\bar{x}\bar{y} - 2\frac{S_{xy}}{S_{xx}}\left(\sum_{i=1}^n x_i y_i\right) \right\} +\left\{ - n\left(\frac{S_{xy}}{S_{xx}}\right)^2\bar{x}^2 + \left(\sum_{i=1}^n x_i^2\right)\left(\frac{S_{xy}}{S_{xx}}\right)^2 \right\}\\ &=&\left\{ \left(\sum_{i=1}^n y_i^2\right) - n\bar{y}^2 \right\} -2\frac{S_{xy}}{S_{xx}}\left\{ \left(\sum_{i=1}^n x_i y_i\right) - n\bar{x}\bar{y} \right\} +\left(\frac{S_{xy}}{S_{xx}}\right)^2\left\{ \left(\sum_{i=1}^n x_i^2\right) - n\bar{x}^2 \right\} \;\dots\;\sum_{i=1}^n y_i^2-n\bar{y}^2=S_{yy},\;\sum_{i=1}^n x_i^2-n\bar{x}^2=S_{xx},\;\sum_{i=1}^n x_i y_i-n\bar{x}\bar{y}=S_{xy}\\ &=&S_{yy} -2\frac{S_{xy}}{S_{xx}}S_{xy} +\left(\frac{S_{xy}}{S_{xx}}\right)^2S_{xx}\\ &=&S_{yy} -2\frac{S_{xy}^2}{S_{xx}} +\frac{S_{xy}^2}{S_{xx}}\\ &=&S_{yy} -\frac{S_{xy}^2}{S_{xx}}\\ \end{eqnarray}$$ よって，残差平方和\(S_{e}\)はデータの各平方和(\(S_{xx}, S_{yy}, S_{xy}\))より求めることができる．