How can I write the sum of squared residuals as a function of the sample mean and variance of $y$, given that the regression equation is:
$y = \beta_0 + \beta_1(x-\bar{x}) + \epsilon$
where $\bar{x}$ is the average of the $x$'s.
I want to calculate the sum of the squared residuals using only the 1st and 2nd sample moments of $x$ and $y$.
Here is my deduction, hope it helps.
Given the regression equation,
\begin{equation} \hat{y} = \beta_0 + \beta_1(x - \overline{x}) + \epsilon \end{equation}
By minimizing the sum of squared errors, we can get
\begin{equation} \hat{\beta_0} = \overline{y}\\ \hat{\beta_1} = \frac{Cov(x, y)}{Var(x)} \end{equation}
Then, assuming there are n samples, we can calculate the sum of squared residuals
\begin{eqnarray*} RSS&=&\sum_{i}(y_i - \hat{y_i})^2\\ &=&\sum_{i}(y_i - \hat{\beta_0} - \hat{\beta_1}(x_i - \overline{x}))^2\\ &=&\sum_{i}(y_i - \overline{y} - \hat{\beta_1}(x_i - \overline{x}))^2\\ &=&\sum_{i}(y_i - \overline{y})^2 + \hat{\beta_1}^2\sum_{i}(x_i - \overline{x})^2 -2\hat{\beta_1}\sum_{i}(x_i - \overline{x})(y_i - \overline{y})\\ &=&n(Var(y) - \frac{Cov(x,y)^2}{Var(x)}) \end{eqnarray*}
