证明sst=ssr+sse

$$\begin{array}{rl}& y-\overline{y}=(y-\hat{y})+(\hat{y}-\overline{y})\\ & \Rightarrow \sum (y-\overline{y}{)}^2=\sum (y-\hat{y}{)}^2+\sum (\hat{y}-\overline{y}{)}^2+2\sum (y-\hat{y})(\hat{y}-\overline{y})\\ & \end{array}$$

$$\begin{array}{rl}\because \sum (y-\hat{y})(\hat{y}-\overline{y})& =\sum (y-\hat{y})(a+bx-\overline{y})\\ & =\sum (y-\hat{y})\left[(a-\overline{y})+bx\right]\\ & =(a-\overline{y})\sum (y-\hat{y})+b\sum (y-\hat{y})x\\ & =(a-\overline{y})\sum (y-a-bx)+b\sum (y-a-bx)x\end{array}$$
根据最小二乘法的原理，得到
$$\begin{gathered} \sum (y-a-bx)=0,\sum (y-a-bx)x=0 \\ \therefore \sum (y-\hat{y})(\hat{y}-\overline{y})=0 \\ \therefore \sum (y-\overline{y}{)}^2=\sum (y-\hat{y}{)}^2+\sum (\hat{y}-\overline{y}{)}^2 \end{gathered}$$