Least square estimators in the simple linear regression are independent of sum of square of residual

Suppose I have a simple linear regression model: $$y_i = \alpha + \beta x_i + \epsilon_i,$$ and I know the expression for the least square estimators of $$\alpha$$ and $$\beta$$: $$\hat{\alpha} = \bar{y} - \hat{\beta}\bar{x},\\ \hat{\beta} = S^{-1}_{xx}S_{xy} = \left(\sum^n_{i=1}(x_i-\bar{x})^2 \right)^{-1}\sum^n_{i=1}(x_i-\bar{x})(y_i-\bar{y}).$$ And the sum of square of residual is given by: $$RSS = \sum^n_{i=1}\hat{\epsilon}_i^2 = \sum^n_{i=1}(y_i-\hat{y}_i)^2.$$ I was told that $$\hat{\alpha}$$ and $$\hat{\beta}$$ are independent of $$RSS$$, but I can't figure out why.

• Following this answer, we have $Z_1=\sqrt n\,\overline y\,,Z_2=\hat\beta\sqrt{s_{xx}}$ and $RSS=\sum_{i=3}^n Z_i^2$ where the $Z_i$'s are all independently distributed normal random variables. This shows the independence of $\hat\beta$ and $RSS$. Now $\hat\alpha=\overline y-\hat\beta\overline x$ is a function of $Z_1,Z_2$ only which is therefore also independent of $RSS$. A general proof is shown here: stats.stackexchange.com/q/173396/119261. Mar 27 at 19:20

$$(y - \hat{y})^T \beta$$
If you replace $$\beta$$ by the projection matrix and work through the algebra, what do you get?
• I understand that the coefficient vector and the residual are independent. I am just wondering how do I prove the independence for the sum of square of residuals. Besides, what do you mean by replace $\beta$ by the projection matrix? Oct 8 '20 at 4:56