殘差平方和

对于两变量x和y, 它们的数据组的均值分别记为${\displaystyle {\bar {x}},{\bar {y}}}$，则两数据组的皮尔逊相关系数为 ${\displaystyle r={\frac {S_{xy}}{\sqrt {S_{xx}S_{yy}}}}}$，其中， ${\displaystyle S_{xy}=\sum _{i=1}^{n}({\bar {x}}-x_{i})({\bar {y}}-y_{i})}$；${\displaystyle S_{xx}=\sum _{i=1}^{n}({\bar {x}}-x_{i})^{2}}$； ${\displaystyle S_{yy}=\sum _{i=1}^{n}({\bar {y}}-y_{i})^{2}}$.

给定最小二乘回归线方程为 ${\displaystyle {\hat {y}}=ax+b=f(x)}$, 其中 ${\displaystyle b={\bar {y}}-a{\bar {x}}}$ ; ${\displaystyle a={\frac {S_{xy}}{S_{xx}}}}$. 则这时残差平方和可以表示为：

${\displaystyle {\begin{aligned}\operatorname {RSS} &=\sum _{i=1}^{n}(y_{i}-f(x_{i}))^{2}=\sum _{i=1}^{n}(y_{i}-(ax_{i}+b))^{2}=\sum _{i=1}^{n}(y_{i}-ax_{i}-{\bar {y}}+a{\bar {x}})^{2}\\[5pt]&=\sum _{i=1}^{n}(a({\bar {x}}-x_{i})-({\bar {y}}-y_{i}))^{2}=a^{2}S_{xx}-2aS_{xy}+S_{yy}=S_{yy}-aS_{xy}=S_{yy}\left(1-{\frac {S_{xy}^{2}}{S_{xx}S_{yy}}}\right)\end{aligned}}}$

通过皮尔逊相关系数的公式，可以得到 ${\displaystyle \operatorname {RSS} =S_{yy}(1-r^{2})}$.