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I was reading this answer, where it is explained that the externally studentized residual $$t_i = \frac{e_i}{\hat\sigma_{(i)}\sqrt{1 - h_i}} = \frac{\frac{e_i}{\sigma\sqrt{1 - h_i}}}{\sqrt{\frac{e_{(i)}^Te_{(i)}}{\sigma^2(n-p-1)}}}$$ follows a t-distribution with $(n-1-p)$ degrees of freedom.

I understand why $$\frac{e_i}{\sigma\sqrt{1 - h_i}} \sim \mathcal N(0,1).$$ and $$\frac{{\hat\sigma_{(i)}}}{\sigma^2}=\sqrt{\frac{e_{(i)}^Te_{(i)}}{\sigma^2(n-p-1)}} \sim \sqrt{\chi^2_{n-p-1} / (n-p-1)}$$

but i don't get why $e_i$ and ${\hat\sigma_{(i)}}$ are independent.

I mean, $e_i=((I_n-H)y)_i$ is a function of $y_1,y_2,\cdots,y_{i-1},y_{i},y_{i+1},\cdots ,y_n$
while ${\hat\sigma_{(i)}}$ is a function of $y_1,y_2,\cdots,y_{i-1},y_{i+1},\cdots ,y_n$.
So at first glance they seem pretty dependent.

But, according to the linked answer, since observation $i$ does not appear in ${\hat\sigma_{(i)}}$, we have that $e_i$ and ${\hat\sigma_{(i)}}$ are independent. I do not understand this statement.

So, why are $e_i$ and ${\hat\sigma_{(i)}}$ independent?

abhishek
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  • $\hat{\sigma}_{(i)}$ doesn't depend on $y_i.$ – User1865345 Sep 09 '23 at 16:15
  • But both $e_i$ and $\hat{\sigma}_{(i)}$ depend on $y_j, j\ne i$ – abhishek Sep 09 '23 at 16:17
  • But $e_i$ is a function of $y_i$ and $Y_j$s are independent. – User1865345 Sep 09 '23 at 16:19
  • I dont understand. If $Y_1$ and $Y_2$ are iid standard normals, then $(Y_1+Y_2)$ and $(Y_1)$ would be dependent wouldn't they? – abhishek Sep 09 '23 at 16:57
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    The description you reference is indeed confusing. The numerator is not supposed to be the original residual $e_i:$ it is meant to be the residual based on the model fit with observation $i$ deleted. This guarantees independence and everything is fine. – whuber Sep 09 '23 at 17:26
  • @whuber i dont see how this guarantees independence, as $e_i$ still remains a function of $y_1,y_2,\cdots,y_{i-1},y_{i},y_{i+1},\cdots ,y_n$ and $\hat{\sigma}{(i)}$ still remains a function of $y_1,y_2,\cdots,y{i-1},y_{i+1},\cdots ,y_n$. So they have variables in common which might lead to dependence – abhishek Sep 10 '23 at 00:09
  • nvm got it. thanks – abhishek Sep 10 '23 at 13:08

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