Show that for any linear model, $\sum_{i=1}^{n}\frac{\text{Var}(\widehat{Y_{i}})}{n} = \frac{p\sigma^2}{n}$.
Wasn't too sure where to start here. I know that Bias($\widehat{\sigma^2}$)=$-\frac{p\sigma^2}{n}$, but I am not sure if the two are related?
$\newcommand{\Var}{\operatorname{Var}}$Hint: Remember, $\widehat{Y} = X \widehat{\beta}\newcommand{\trace}{\operatorname{trace}}$, so the covariance matrix of $\widehat{Y}$ is $$\Var\left(\widehat{Y} \right) = X\left(\sigma^2(X^T X)^{-1}\right)X^T,$$ since the covariance matrix of $\widehat{\beta}$ is $\sigma^2(X^T X)^{-1}$. Now use the fact that $$\sum\limits_{i=1}^{n}\Var\left(\widehat{Y}_i\right)=\trace\left(\Var\left(\widehat{Y} \right) \right)$$ and the cyclic property of trace ($\trace(ABC)=\trace(BCA)$).