$$\begin{array}{l}{r}_{我}＝y_{我}-{\stackrel{＾}{y}}_{我}\\ \text{剩余=数据}-\text{适合}\end{array}$$

$$\mathrm{年代}＝{\displaystyle \underset{我＝1}{\overset{n}{\sum }}{r}_{我}^2}＝{\displaystyle \underset{我＝1}{\overset{n}{\sum }}{（y_{我}-{\stackrel{＾}{y}}_{我}）}^2}$$

$$error\sim N（0，{\sigma }^2）$$

$$y＝p_{1}x+p_2$$

$$\mathrm{年代}＝{\displaystyle \underset{我＝1}{\overset{n}{\sum }}{（y_{我}-（p_1{x}_{我}+p_2））}^2}$$

$$\begin{array}{l}\frac{\partial \mathrm{年代}}{\partial p_1}＝-2{\displaystyle \underset{我＝1}{\overset{n}{\sum }}{x}_{我}（y_{我}-（p_1{x}_{我}+p_2））＝0}\\ \frac{\partial \mathrm{年代}}{\partial p_2}＝-2{\displaystyle \underset{我＝1}{\overset{n}{\sum }}（y_{我}-（p_1{x}_{我}+p_2））＝0}\end{array}$$

$$\begin{array}{c}{\displaystyle \sum x_{我}（y_{我}-（b_1{x}_{我}+b_2））＝0}\\ {\displaystyle \sum （y_{我}-（b_1{x}_{我}+b_2））＝0}\end{array}$$

$$\begin{array}{c}{b}_1{\displaystyle \sum x_{我}^2}+b_2{\displaystyle \sum x_{我}}＝{\displaystyle \sum x_{我}}{y}_{我}\\ b_1{\displaystyle \sum x_{我}}+n{b}_2＝{\displaystyle \sum y_{我}}\end{array}$$

$$b_1＝\frac{n{\displaystyle \sum x_{我}{y}_{我}-{\displaystyle \sum x_{我}{\displaystyle \sum y_{我}}}}}{n{\displaystyle \sum x_{我}^2-{（{\displaystyle \sum x_{我}}）}^2}}$$

$$b_2＝\frac{1}{n}（{\displaystyle \sum y_{我}-b_1{\displaystyle \sum x_{我}}}）$$

$$［\begin{array}{c}{y}_1\\ y_2\\ y_{3.}\\ ．\\ ．\\ ．\\ y_n\end{array}］＝［\begin{array}{c}{x}_{1}1\\ x_{2}1\\ x_{3.}1\\ ．\\ ．\\ ．\\ x_{n}1\end{array}］\times ［\begin{array}{l}{p}_1\\ p_2\end{array}］$$

$$\mathrm{年代}＝{\displaystyle \underset{我＝1}{\overset{n}{\sum }}{w}_{我}{（y_{我}-{\stackrel{＾}{y}}_{我}）}^2}$$

$$b＝\stackrel{＾}{\beta }＝{（X^{T}WX）}^{-1}{X}^{T}Wy$$

$$w_{我}＝1/{\sigma }_{我}^2$$