In linear regression, if we know $MSE$, which is $MSE=\frac{1}{n}\sum_{i=1}^{n}(\hat{y_i}-y_i)^2$, can we know $Var(|\hat{y_i}-y_i|)$? The reason I am trying to find the variance of the bias is that I want to know the error level for the individual point. I was trying to find an upper bound for $|\hat{y_i}-y_i|$, which is $|\hat{y_i}-y_i| \leq \sum_{i=1}^{n}(|\hat{y_i}-y_i|)\leq \sqrt{n}\sqrt{\sum_{i=1}^{n}|\hat{y_i}-y_i|^2}=n\sqrt{MSE}$. But I don't like this result because the upper bound for $|\hat{y_i}-y_i|$ depends on $n$, the more points, the higher the upper bound. Instead, if we can get $Var(|\hat{y_i}-y_i|)$, in addition to $E(|\hat{y_i}-y_i|)\leq \sqrt{MSE}$, then we can use $E+\sigma$, $E+2\sigma$, $E+3\sigma$ to show the error level of $|\hat{y_i}-y_i|$. So how to calculate $Var(|\hat{y_i}-y_i|)$ ?
Thank you!
PS: I tried to calculate it using the definition, but didn't seems to reach an solution:
$Var(|\hat{y_i}-y_i|) \\ = E[(\hat{y_i}-y_i)^2] -E^2[(\hat{y_i}-y_i)] \\ =\frac{1}{n}\sum_{i=1}^{n}(\hat{y_i}-y_i)^2-E^2[(\hat{y_i}-y_i)] \\ =MSE-E^2[(\hat{y_i}-y_i)]$