全平方和の分解

全平方和の分解

以下のような和を考える．第一項は回帰直線から推定される値\(\hat{y_i}\)と目的変数の平均値\(\bar{y}\)との差の 平方和であり，第二項は残差平方和である． $$\begin{eqnarray} &&\sum_{i=0}^n\left(\hat{y_i}-\bar{y}\right)^2+\sum_{i=0}^n\left(y_i-\hat{y_i}\right)^2\\ &=&\sum_{i=0}^n\left(\hat{y_i}^2-2\hat{y_i}\bar{y}+\bar{y}^2\right)+\sum_{i=0}^n\left(y_i^2-2y_i\hat{y_i}+\hat{y_i}^2\right)\\ &=&\sum_{i=0}^n\left\{ \left(\alpha+\beta x_i\right)^2-2\left(\alpha+\beta x_i\right)\bar{y}+\bar{y}^2 \right\} +\sum_{i=0}^n\left\{ y_i^2-2y_i\left(\alpha+\beta x_i\right)+\left(\alpha+\beta x_i\right)^2 \right\}\\ &=&\sum_{i=0}^n\left(\alpha^2+2\alpha\beta x_i+\beta^2x_i^2-2\alpha\bar{y}-2\beta\bar{y}x_i+\bar{y}^2\right) +\sum_{i=0}^n\left(y_i^2-2\alpha y_i-2\beta x_i y_i+\alpha^2+2\alpha\beta x_i+\beta^2 x_i^2\right)\\ &=&\sum_{i=0}^n\left\{ \left(\bar{y}-\beta\bar{x}\right)^2 +2\left(\bar{y}-\beta\bar{x}\right)\beta x_i +\beta^2x_i^2 -2\left(\bar{y}-\beta\bar{x}\right)\bar{y} -2\beta\bar{y}x_i+\bar{y}^2 \right\} +\sum_{i=0}^n\left\{ y_i^2-2\left(\bar{y}-\beta\bar{x}\right) y_i-2\beta x_i y_i+\left(\bar{y}-\beta\bar{x}\right)^2+2\left(\bar{y}-\beta\bar{x}\right)\beta x_i+\beta^2 x_i^2 \right\}\\ &=&\sum_{i=0}^n\left( \bar{y}^2-2\beta\bar{x}\bar{y}+\beta^2\bar{x}^2 +2\beta \bar{y}x_i -2\beta^2\bar{x} x_i +\beta^2x_i^2 -2\bar{y}^2 +2\beta\bar{x}\bar{y} -2\beta\bar{y}x_i+\bar{y}^2 \right) +\sum_{i=0}^n\left( y_i^2 -2\bar{y}y_i+2\beta\bar{x}y_i -2\beta x_i y_i +\bar{y}^2-2\beta\bar{x}\bar{y}+\beta^2\bar{x}^2 +2\beta\bar{y} x_i-2\beta^2\bar{x} x_i +\beta^2 x_i^2 \right)\\ &=&\bar{y}^2\left(\sum_{i=0}^n 1\right) -2\beta\bar{x}\bar{y}\left(\sum_{i=0}^n 1\right) +\beta^2\bar{x}^2\left(\sum_{i=0}^n 1\right) +2\beta \bar{y}\left(\sum_{i=0}^n x_i\right) -2\beta^2\bar{x}\left(\sum_{i=0}^n x_i\right) +\beta^2 \left(\sum_{i=0}^n x_i^2\right) -2\bar{y}^2 \left(\sum_{i=0}^n 1\right) +2\beta\bar{x}\bar{y} \left(\sum_{i=0}^n 1\right) -2\beta\bar{y} \left(\sum_{i=0}^n x_i\right) +\bar{y}^2 \left(\sum_{i=0}^n 1\right)\\ &&+\left(\sum_{i=0}^n y_i^2 \right) -2\bar{y}\left(\sum_{i=0}^n y_i\right) +2\beta\bar{x}\left(\sum_{i=0}^n y_i\right) -2\beta \left(\sum_{i=0}^n x_i y_i\right) +\bar{y}^2\left(\sum_{i=0}^n 1\right) -2\beta\bar{x}\bar{y}\left(\sum_{i=0}^n 1\right) +\beta^2\bar{x}^2 \left(\sum_{i=0}^n 1\right) +2\beta\bar{y} \left(\sum_{i=0}^n x_i\right) -2\beta^2\bar{x} \left(\sum_{i=0}^n x_i\right) +\beta^2 \left(\sum_{i=0}^n x_i^2 \right)\\ &=&n\bar{y}^2 -2n\beta\bar{x}\bar{y} +n\beta^2\bar{x}^2 +2n\beta\bar{x}\bar{y} -2n\beta^2\bar{x}^2 +\beta^2 \left(\sum_{i=0}^n x_i^2\right) -2n\bar{y}^2 +2n\beta\bar{x}\bar{y} -2n\beta\bar{x}\bar{y} +n\bar{y}^2\\ &&+\left(\sum_{i=0}^n y_i^2 \right) -2n\bar{y}^2 +2n\beta\bar{x}\bar{y} -2\beta \left(\sum_{i=0}^n x_i y_i\right) +n\bar{y}^2 -2n\beta\bar{x}\bar{y} +n\beta^2\bar{x}^2 +2n\beta\bar{x}\bar{y} -2n\beta^2\bar{x}^2 +\beta^2 \left(\sum_{i=0}^n x_i^2 \right)\\ &=&\left\{ \left(\sum_{i=0}^n y_i^2 \right) +n\bar{y}^2 -2n\bar{y}^2 +n\bar{y}^2 -2n\bar{y}^2 +n\bar{y}^2 \right\} +\left\{ 2\beta^2 \left(\sum_{i=0}^n x_i^2\right) +n\beta^2\bar{x}^2 -2n\beta^2\bar{x}^2 +n\beta^2\bar{x}^2 -2n\beta^2\bar{x}^2 \right\} +\left\{ -2\beta \left(\sum_{i=0}^n x_i y_i\right) -2n\beta\bar{x}\bar{y} +2n\beta\bar{x}\bar{y} +2n\beta\bar{x}\bar{y} -2n\beta\bar{x}\bar{y} +2n\beta\bar{x}\bar{y} -2n\beta\bar{x}\bar{y} +2n\beta\bar{x}\bar{y} \right\}\\ &=&\left\{ \left(\sum_{i=0}^n y_i^2 \right) -n\bar{y}^2 \right\} +2\beta^2\left\{ \left(\sum_{i=0}^n x_i^2\right) -n\bar{x}^2 \right\} -2\beta \left\{ \left(\sum_{i=0}^n x_i y_i\right) -n\bar{x}\bar{y} \right\}\\ &=&S_{yy}+2\beta^2 S_{xx}-2\beta S_{xy}\\ &=&S_{yy}+2\beta \left(\beta S_{xx} - S_{xy}\right)\\ &=&S_{yy}+2\left(\frac{S_{xy}}{S_{xx}}\right) \left\{\left(\frac{S_{xy}}{S_{xx}}\right) S_{xx} - S_{xy}\right\}\\ &=&S_{yy}+2\left(\frac{S_{xy}}{S_{xx}}\right) \left(S_{xy} - S_{xy}\right)\\ &=&S_{yy} = \sum_{i=0}^n \left(y_i - \bar{y}\right)^2\\ &=&S_{T} \end{eqnarray}$$ よって，全平方和\(S_{T}\)である\(S_{yy}\)は回帰直線から推定される値\(\hat{y}\)と目的変数の平均値\(\bar{y}\)との差の 平方和と残差平方和で表せられる．
この回帰直線から推定される値\(\hat{y}\)と目的変数の平均値\(\bar{y}\)との差の 平方和を，回帰による変動の平方和\(S_{R}\)と呼ぶ． $$\begin{eqnarray} \sum_{i=0}^n \left(y_i - \bar{y}\right)^2&=&\sum_{i=0}^n\left(\hat{y_i}-\bar{y}\right)^2+\sum_{i=0}^n\left(y_i-\hat{y_i}\right)^2\\ S_{T}&=&S_{R}+S_{e}\\ S_{yy}&=&S_{R}+S_{e}\\ S_{yy}&=&S_{R}+\left(S_{yy}-\frac{S_{xy}^2}{S_{xx}}\right)\\ S_{R}&=&\frac{S_{xy}^2}{S_{xx}}\\ \end{eqnarray}$$ 回帰による変動の平方和\(S_{R}\)もデータの各平方和(\(S_{xx}, S_{yy}\))より求めることができる．