一元线性回归方程推导

设其为

y=bx+ay=bx+a

记误差为 QQ

Q(a,b)=∑(yi−bxi−ai)2=∑[yi−bxi−(yˉ−bxˉ)+(yˉ−bxˉ)−a]2=∑[yi−bxi−(yˉ−bxˉ)]2+2∑[yi−bxi−(yˉ−bxˉ)][(yˉ−bxˉ)−a]+n[yˉ−bxˉ−a]2\begin{align} Q(a,b)&=\sum (y_{i}-bx_{i}-a_{i})^2 \\ &=\sum[y_{i}-bx_{i}-(\bar{y}-b\bar{x})+(\bar{y}-b\bar{x})-a]^2 \\ &=\sum[y_{i}-bx_{i}-(\bar{y}-b\bar{x})]^{2} \\ &+2\sum[y_{i}-bx_{i}-(\bar{y}-b\bar{x})][(\bar{y}-b\bar{x})-a]+n[\bar{y}-b\bar{x}-a]^{2} \\ \end{align}

观察第二项

2∑[yi−bxi−(yˉ−bxˉ)][(yˉ−bxˉ)−a]=2[(yˉ−bxˉ)−a]∑[yi−yˉ+b(xˉ−xi)]=2[(yˉ−bxˉ)−a](∑[yi−yˉ+b(xˉ−xi)])=0\begin{align} &2\sum[y_{i}-bx_{i}-(\bar{y}-b\bar{x})][(\bar{y}-b\bar{x})-a] \\ &=2[(\bar{y}-b\bar{x})-a]\sum[y_{i}-\bar{y}+b(\bar{x}-x_{i})] \\ &=2[(\bar{y}-b\bar{x})-a](\sum[y_{i}-\bar{y}+b(\bar{x}-x_{i})]) \\ &=0 \end{align}

故

Q(a,b)=n[yˉ−bxˉ−a]2+∑[yi−bxi−(yˉ−bxˉ)]2=\begin{align} Q(a,b)&=n[\bar{y}-b\bar{x}-a]^{2}+\sum[y_{i}-bx_{i}-(\bar{y}-b\bar{x})]^{2} \\ &= \end{align}