# Does Cook's distance follow an $F$-distribution?

The question is pretty much contained in the title: in a linear model $$Y=X\beta +\epsilon$$, with $$\epsilon\sim N_n(0,\sigma^2 I_n)$$, does the Cook's distance, defined as $$D_i:=\frac{\frac{1}{p}||X(\hat{\beta}_{(-i)}-\hat{\beta})||^2}{\tilde{\sigma}^2},$$ follow an $$F_{p,n-p}$$-distribution?

Wikipedia seems to suggest it does, but the sentence in the Wikipedia article is cryptic and says "Since Cook's distance is in the metric of an F distribution...".

Can someone point me to a formal proof of the fact that $$D_i\sim F_{p,n-p}$$?

I am aware that $$D_i=\frac{1}{p}\frac{p_i}{1-p_i} \hat{\eta}_i^2$$, but what this seems to indicate is that $$D_i\sim \frac{1}{p}\frac{p_i}{1-p_i} F_{1,n-p}$$, not $$F_{p,n-p}$$.

• I would think that this is a change in variables problem, but I don't have the time to attempt it right now. Nov 23, 2023 at 20:29
• Thank you for answering. Unfortunately I don't see how it could be, because it seems like, even supposing the leverage of $X_i$ is nice, say $p_i \approx 1/2$, we get $D_i \sim 1/p F_{1,n-p}$ which should be approximately $1/p F_{p,n-p}$ and not $F_{p,n-p}$ (using $\chi^2_p/p\approx \chi^2_1$). Nov 23, 2023 at 20:35