證明

$\widehat{{\beta }_0},\widehat{{\beta }_1}$ 的不同特性

Define

• $S_{XX}={\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x}{)}^2$
• $S_{YY}={\mathrm{\Sigma }}_{i=1}^n(y_i-\bar{y}{)}^2$
• $S_{XY}={\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})$

Given

• $\widehat{{\beta }_1}={\mathrm{\Sigma }}_{i=1}^n\frac{(x_i-\bar{x})}{{\mathrm{\Sigma }}_{j=1}^n(x_j-\bar{x}{)}^2}{y}_i={\mathrm{\Sigma }}_{i=1}^n{w}_i{y}_i$

• $\widehat{{\beta }_0}={\mathrm{\Sigma }}_{i=1}^n(\frac{1}{n}-\frac{(x_i-\bar{x})\cdot \bar{x}}{S_{XX}})y_i={\mathrm{\Sigma }}_{i=1}^n{w}_i{y}_i$

• ${\mathrm{\Sigma }}_{i=1}^{n}Var(y_i)={\sigma }^2$

• ${\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})={\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x})y_i$

Prove

• $E(\widehat{{\beta }_1})=?$
  
  $\begin{array}{rl}E(\widehat{{\beta }_1})& =E({\mathrm{\Sigma }}_{i=1}^n{w}_i{y}_i)={\mathrm{\Sigma }}_{i=1}^n{w}_i({\beta }_0+{\beta }_1{x}_i)\\ & ={\beta }_0\text{color}blue{\mathrm{\Sigma }}_{i=1}^n{w}_i+{\beta }_1\text{color}green{\mathrm{\Sigma }}_{i=1}^n{w}_i{x}_i\\ & ={\beta }_0\cdot \text{color}blue0+{\beta }_1\cdot \text{color}green1\\ & ={\beta }_1\end{array}$
  
  $\therefore E(\widehat{{\beta }_1})={\beta }_1$
  • $\text{color}blue\mathrm{藍}\mathrm{色}\mathrm{部}\mathrm{分}{\mathrm{\Sigma }}_{i=1}^n{w}_i={\mathrm{\Sigma }}_{i=1}^n\frac{(x_i-\bar{x})}{S_{XX}}=\frac{1}{S_{XX}}\underset{=0}{\underbrace{{\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x})}}$
  • $\text{color}green\mathrm{綠}\mathrm{色}\mathrm{部}\mathrm{分}{\mathrm{\Sigma }}_{i=1}^n{w}_i{x}_i={\mathrm{\Sigma }}_{i=1}^n\frac{(x_i-\bar{x})\cdot x_i}{S_{XX}}=\frac{1}{S_{XX}}{\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x})(x_i-\bar{x})=\frac{S_{XX}}{S_{XX}}=1$

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• $Var(\widehat{{\beta }_1})=?$
  
  $\begin{array}{rl}Var(\widehat{{\beta }_1})& =Var({\mathrm{\Sigma }}_{i=1}^n{w}_i{y}_i)={\mathrm{\Sigma }}_{i=1}^n{w}_i^{2}Var(y_i)={\mathrm{\Sigma }}_{i=1}^n{w}_i^2{\sigma }^2\\ & ={\sigma }^2{\mathrm{\Sigma }}_{i=1}^n\frac{(x_i-\bar{x}{)}^2}{(S_{XX}{)}^2}=\frac{{\sigma }^2}{(S_{XX}{)}^2}\underset{=S_{XX}}{\underbrace{{\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x}{)}^2}}\\ & =\frac{{\sigma }^2}{(S_{XX})}\end{array}$
  
  $\therefore Var(\widehat{{\beta }_1})=\frac{{\sigma }^2}{(S_{XX})}$

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• $E(\widehat{{\beta }_0})=?$
  
  $\begin{array}{rl}E(\widehat{{\beta }_0})& =E({\mathrm{\Sigma }}_{i=1}^n{w}_i{y}_i)={\mathrm{\Sigma }}_{i=1}^n{w}_i({\beta }_0+{\beta }_1{x}_i)\\ & ={\beta }_0\text{color}blue{\mathrm{\Sigma }}_{i=1}^n{w}_i+{\beta }_1\text{color}green{\mathrm{\Sigma }}_{i=1}^n{w}_i{x}_i\\ & ={\beta }_0\cdot \text{color}blue1+{\beta }_1\cdot \text{color}green0\\ & ={\beta }_0\end{array}$
  
  $\therefore E(\widehat{{\beta }_0})={\beta }_0$
  • $\text{color}blue\begin{array}{rl}\mathrm{藍}\mathrm{色}\mathrm{部}\mathrm{分}{\mathrm{\Sigma }}_{i=1}^n{w}_i& ={\mathrm{\Sigma }}_{i=1}^n[\frac{1}{n}-\frac{(x_i-\bar{x})\cdot \bar{x}}{S_{XX}}]={\mathrm{\Sigma }}_{i=1}^n\frac{1}{n}-{\mathrm{\Sigma }}_{i=1}^n\frac{(x_i-\bar{x})\cdot \bar{x}}{S_{XX}}\\ & =1-\frac{\bar{x}}{S_{XX}}\underset{=0}{\underbrace{{\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x})}}=1\end{array}$
  • $\text{color}green\begin{array}{rl}\mathrm{綠}\mathrm{色}\mathrm{部}\mathrm{分}{\mathrm{\Sigma }}_{i=1}^n{w}_i{x}_i& ={\mathrm{\Sigma }}_{i=1}^n[(\frac{1}{n}-\frac{(x_i-\bar{x})\cdot \bar{x}}{S_{XX}})\cdot x_i]=\frac{1}{n}{\mathrm{\Sigma }}_{i=1}^n{x}_i-\frac{\bar{x}}{S_{XX}}\underset{=S_{XX}}{\underbrace{{\mathrm{\Sigma }}_{i=1}^n[(x_i-\bar{x})\cdot x_i]}}\\ & =\bar{x}-\bar{x}=0\end{array}$

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• $Var(\widehat{{\beta }_0})=?$
  
  $\begin{array}{rl}Var(\widehat{{\beta }_0})& =Var({\mathrm{\Sigma }}_{i=1}^n{w}_i{y}_i)={\mathrm{\Sigma }}_{i=1}^n{w}_i^2{\sigma }^2={\sigma }^2\cdot {\mathrm{\Sigma }}_{i=1}^n(\frac{1}{n}-\frac{(x_i-\bar{x})\cdot \bar{x}}{S_{XX}}{)}^2\\ & \stackrel{\mathrm{將}{w}_i^2\mathrm{拆}\mathrm{開}}{=}{\sigma }^2[\text{color}blue{\mathrm{\Sigma }}_{i=1}^n\frac{1}{n^2}-\text{color}green2\cdot \frac{1}{n}\cdot {\mathrm{\Sigma }}_{i=1}^n\frac{(x_i-x)\cdot \bar{x}}{S_{XX}}+\text{color}purple{\mathrm{\Sigma }}_{i=1}^n(\frac{(x_i-\bar{x})\cdot \bar{x}}{S_{XX}}{)}^2]\\ & ={\sigma }^2[\text{color}blue\frac{1}{n}-\text{color}green0+\text{color}purple\frac{{\bar{x}}^2}{S_{XX}}]\end{array}$
  
  $\therefore Var(\widehat{{\beta }_0})={\sigma }^2[\frac{1}{n}+\frac{{\bar{x}}^2}{S_{XX}}]$
  • $\text{color}green\mathrm{綠}\mathrm{色}\mathrm{部}\mathrm{分}=\frac{2}{n}\times \frac{\bar{x}}{S_{XX}}\times \underset{=0}{\underbrace{{\mathrm{\Sigma }}_{i=1}^n(x_i-\bar{x})}}=0$
  • $\text{color}purple\mathrm{紫}\mathrm{色}\mathrm{部}\mathrm{分}=\frac{1}{S_{XX}^2}\times {\mathrm{\Sigma }}_{i=1}^n{[(x_i-\bar{x})\cdot \bar{x}]}^2=\frac{{\bar{x}}^2}{S_{XX}^2}\times S_{XX}=\frac{{\bar{x}}^2}{S_{XX}}$