> # 迴歸分析作業02

:::success
**Homework**
證明：
**1.** ${Cov}$($\overline{y}$，$\hat{\beta_1}$)=${0}$
**2.** $\Sigma$($y_i$-$\hat{y_i}$)($\hat{y_i}$-$\overline{y}$)=${0}$
:::
<font size=4>**1.**$Cov$($\overline{y}$，$\hat{\beta_1}$)
=$\cfrac{1}{n‧S_{xx}}$$Cov$($\Sigma$$y_i$，$\Sigma$($x_i$-$\overline{x}$)($y_i$-$\overline{y}$))
#註：${n}$和${S_{xx}}$是常數，故可以提出來。
=$\cfrac{1}{n‧S_{xx}}$$Cov$($\Sigma$$y_i$，($\Sigma$($x_i$-$\overline{x})(y_i)$-($\Sigma$$x_i$-$\overline{x})(\overline{y}$))
#註：($\Sigma$$x_i$-$\overline{x})(\overline{y}$)=$\Sigma$$x_i$-${n}$$*$$\cfrac{\Sigma(x_i)}{n}*(\overline{y}$)=${0}$
=$\cfrac{1}{n‧S_{xx}}$$\Sigma$ $Cov((y_i)，(x_i$-$\overline{x})(y_i)$)
=$\cfrac{1}{n‧S_{xx}}$$\Sigma$ $((x_i$-$\overline{x})Cov(y_i,y_i)$)
#註：$\Sigma$($x_i$-$\overline{x})$=${0}$ & $Cov(y_i,y_i)$=$var(y_i)$=$\sigma^2$
=${0}$

<font size=4>**2.**$\Sigma$($y_i$-$\hat{y_i}$)($\hat{y_i}$-$\overline{y}$)
=$\Sigma$(${e_i}$)($\hat{\beta_0}$+$\hat{\beta_1}$${x_i}$-$\overline{y}$)
=$\Sigma$${e_i}$$\hat{\beta_0}$+$\Sigma$${e_i}$$\hat{\beta_1}$${x_i}$-($\Sigma$${e_i}$)$\overline{y}$
#註：$\Sigma$${e_i}$=$\Sigma$($y_i$-$\hat{y_i}$)=${0}$
=$\hat{\beta_1}$$\Sigma$${e_i}$${x_i}$
=$\hat{\beta_1}$[$\Sigma$($y_i$-$\hat{y_i}$)(${x_i}$)]
=$\hat{\beta_1}$[$\Sigma$($y_i$-($\hat{\beta_0}$+$\hat{\beta_1}$${x_i}))$(${x_i}$)]
=$\hat{\beta_1}$[$\Sigma$$x_i$$y_i$-$\Sigma$$\hat{\beta_0}$${x_i}$-$\Sigma$$\hat{\beta_1}$${x_i}^2$]
=$\hat{\beta_1}$[$\Sigma$$x_i$($\hat{\beta_0}$+$\hat{\beta_1}$${x_i}$)-$\Sigma$$\hat{\beta_0}$${x_i}$-$\Sigma$$\hat{\beta_1}$${x_i}^2$]
=$\hat{\beta_1}$[$\Sigma$$\hat{\beta_0}$$x_i$+$\Sigma$$\hat{\beta_1}$${x_i^2}$-$\Sigma$$\hat{\beta_0}$${x_i}$-$\Sigma$$\hat{\beta_1}$${x_i}^2$]
=${0}$