平方和分解の途中式を淡々と書く。

$\sum_{}(y_i-\bar{y})^2 = \sum_{}( (y_i - \hat{y_i} ) + ( \hat{y_i} - \bar{y}) )^2\\ = \sum_{}(y_i-\hat{y_i})^2 + 2\sum_{}(y_i-\hat{y_i})(\hat{y_i}-\bar{y}) + \sum_{}(\hat{y_i}-\bar{y})^2 \\$

となるが、ここで上式の第二項を取り出すと

$\sum_{}(y_i-\hat{y_i})(\hat{y_i}-\bar{y})\\ = \sum_{}(y_i - \hat{y_i} )(\hat{\beta}(x_i - \bar{x} ))\\ = \sum_{}(y_i - ( \bar{y} + \hat{\beta}(x_i - \bar{x} ) )) (\hat{\beta}(x_i - \bar{x} ))\\ = \sum_{}( ( y_i - \bar{y} ) - \hat{\beta}(x_i - \bar{x} ) ) (\hat{\beta}(x_i - \bar{x} ))\\ = \sum_{}\hat{\beta}(y_i - \bar{y} )(x_i - \bar{x} ) - \hat{\beta}^2 \sum_{}(x_i - \bar{x} )^2\\ = \hat{\beta}S_{xy} - \hat{\beta}^2S_{xx} = \hat{\beta}(S_{xy} - \hat{\beta}S_{xx} )\\ = \hat{\beta}(S_{xy} - \frac{S_{xy}}{S_{xx}}S_{xx} )\\ = \hat{\beta}(S_{xy} - S_{xy} )\\ = 0\\$

となるので、結局

$\sum_{}(y_i-\bar{y})^2 = \sum_{}(y_i-\hat{y_i})^2 + \sum_{}(\hat{y_i}-\bar{y})^2 \\$

と得る。

なお、
$\hat{y_i} = \hat{\alpha} + \hat{\beta} x_i , \hat{y_i} - \bar{y} = \hat{\beta}(\hat{x_i} - \bar{x})\\ \Leftrightarrow \hat{y_i} = (\bar{y} - \hat{\beta}\bar{x} ) + \hat{\beta} x_i \\ \Leftrightarrow \hat{y_i} - \bar{y} = \hat{\beta}( x_i - \bar{x} )$

および、
$T_{xy} = \sum_{}(x_i - \bar{x} )(y_i -\bar{y}), \\ T_{xx} = \sum_{} (x_i - \bar{x} )^2 , \\ S_{xy} = \frac{1}{n}T_{xy}, \\ S_{xx} = \frac{1}{n}T_{xx}, \\ \hat{\beta} = \frac{ T_{xy} }{ T_{xx} } = \frac{ S_{xy} }{ S_{xx} }$
を用いた。

axjack's blog

### axjack is said to be an abbreviation for An eXistent JApanese Cool Klutz ###

平方和分解の途中式を淡々と書く。