PumpkinBook/Chapter3/chapter3.md

3.7

$$ w=\cfrac{\sum_{i=1}^{m}y_i(xi-\bar{x})}{\sum{i=1}^{m}xi^2-\cfrac{1}{m}(\sum{i=1}^{m}x_i)^2} $$

令式（3.5）等于0： $$ 0 = w\sum_{i=1}^{m}xi^2-\sum{i=1}^{m}(y_i-b)xi $$ $$ w\sum{i=1}^{m}xi^2 = \sum{i=1}^{m}y_ixi-\sum{i=1}^{m}bxi $$ 由于令式（3.6）等于0可得$ b=\cfrac{1}{m}\sum{i=1}^{m}(y_i-wxi) $，又$ \cfrac{1}{m}\sum{i=1}^{m}yi=\bar{y} $，$ \cfrac{1}{m}\sum{i=1}^{m}x_i=\bar{x} $，则$ b=\bar{y}-w\bar{x} $，代入上式可得： $$ \begin{aligned}

w\sum_{i=1}^{m}x_i^2 & = \sum_{i=1}^{m}y_ix_i-\sum_{i=1}^{m}(\bar{y}-w\bar{x})x_i \\\\
w\sum_{i=1}^{m}x_i^2 & = \sum_{i=1}^{m}y_ix_i-\bar{y}\sum_{i=1}^{m}x_i+w\bar{x}\sum_{i=1}^{m}x_i \\\\
w(\sum_{i=1}^{m}x_i^2-\bar{x}\sum_{i=1}^{m}x_i) & = \sum_{i=1}^{m}y_ix_i-\bar{y}\sum_{i=1}^{m}x_i \\\\
w & = \cfrac{\sum_{i=1}^{m}y_ix_i-\bar{y}\sum_{i=1}^{m}x_i}{\sum_{i=1}^{m}x_i^2-\bar{x}\sum_{i=1}^{m}x_i}

\end{aligned} $$ 又$ \bar{y}\sum_{i=1}^{m}xi=\cfrac{1}{m}\sum{i=1}^{m}yi\sum{i=1}^{m}xi=\bar{x}\sum{i=1}^{m}yi $，$ \bar{x}\sum{i=1}^{m}xi=\cfrac{1}{m}\sum{i=1}^{m}xi\sum{i=1}^{m}xi=\cfrac{1}{m}(\sum{i=1}^{m}xi)^2 $，代入上式即可得式（3.7）： $$ w=\cfrac{\sum{i=1}^{m}y_i(xi-\bar{x})}{\sum{i=1}^{m}xi^2-\cfrac{1}{m}(\sum{i=1}^{m}x_i)^2} $$

【注】：式（3.7）还可以进一步化简为能用向量表达的形式，将$ \cfrac{1}{m}(\sum_{i=1}^{m}xi)^2=\bar{x}\sum{i=1}^{m}x_i $代入分母可得： $$ \begin{aligned}

 w & = \cfrac{\sum_{i=1}^{m}y_i(x_i-\bar{x})}{\sum_{i=1}^{m}x_i^2-\bar{x}\sum_{i=1}^{m}x_i} \\\\
 & = \cfrac{\sum_{i=1}^{m}(y_ix_i-y_i\bar{x})}{\sum_{i=1}^{m}(x_i^2-x_i\bar{x})}

\end{aligned} $$ 又$ \bar{y}\sum_{i=1}^{m}xi=\bar{x}\sum{i=1}^{m}yi=\sum{i=1}^{m}\bar{y}xi=\sum{i=1}^{m}\bar{x}yi=m\bar{x}\bar{y}=\sum{i=1}^{m}\bar{x}\bar{y} $，则上式可化为： $$ \begin{aligned}

w & = \cfrac{\sum_{i=1}^{m}(y_ix_i-y_i\bar{x}-x_i\bar{y}+\bar{x}\bar{y})}{\sum_{i=1}^{m}(x_i^2-x_i\bar{x}-x_i\bar{x}+\bar{x}^2)} \\\\
& = \cfrac{\sum_{i=1}^{m}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^{m}(x_i-\bar{x})^2} 

\end{aligned} $$ 若令$ \mathbf{X}=(x_1,x_2,...,x_m) $，$\mathbf{X}_{demean}$为去均值后的$ \mathbf{X} $，$ \mathbf{y}=(y_1,y_2,...,y_m) $，$ \mathbf{y}_{demean} $为去均值后的$ \mathbf{y} $，其中$ \mathbf{X} $、$ \mathbf{X}_{demean} $、$ \mathbf{y} $、$ \mathbf{y}_{demean} $均为m行1列的列向量，代入上式可得： $$ w=\cfrac{\mathbf{X}_{demean}\mathbf{y}_{demean}^T}{\mathbf{X}_{demean}\mathbf{X}_{demean}^T}$$