Proving if $\|H\varepsilon\|^2 \sim \chi^2_{p+1} $ is chi square with $p+1$ degrees of freedom

Question

True or False: $\|H\varepsilon\|^2 \sim \chi^2_{p+1} $ ($H$ is that hat matrix and $\varepsilon\sim\operatorname{N}(0,\sigma^2I_n)$

My current strategy is letting $\varepsilon=(I-H)$ so we have $\|H(I-H)\|^2$

Then, via idempotency of $H$, we get $\|H^2-H\|^2 = \|H-H\|^2 = 0$, which means this is false.

Is this correct? feels to good to be true.

Answer 1

Assume that $\sigma^2=1$, and thus $\hat{\beta}_{p+1} \sim N(\beta, (X'X)^{-1})$, hence \begin{align} \|H\varepsilon \|^2 & = \| H(Y-X\beta) \|^2 = \| X\hat{\beta} - X\beta \|^2 = \|X(\hat{\beta} - \beta) \|^2=(\hat{\beta} - \beta)'X'X(\hat{\beta} - \beta) \\[10pt] & = \sum_{i=1}^{p+1}\frac{(\hat{\beta}_j - \beta_j )^2}{\sigma^2(\hat{\beta}_j)^2} \sim \chi^2(p+1) \,. \end{align}