Why leave-one-out residuals (studentized residuals) is t-distribution?

Question

From my text book, there are several types of residuals:

1. residuals (or ordinary residuals) :
   $\hat\varepsilon_i = y_i - \hat y_i = y_i - \boldsymbol{x}_i^T \hat{\beta}$
2. standardized residuals :
   $r_i = \frac{\hat\varepsilon_i}{\hat\sigma\sqrt{1-h_{ii}}}$ ,
   where $h_{ii}$ is the i 'th diagonal element of the hat matrix $H = X(X^TX)^{-1}X^T$.
3. studentized residuals :
   $r_i^* = \frac{ \hat\varepsilon_{i} }{ \hat\sigma_{(i)}\sqrt{1-h_{ii}} }$,
   where $\hat\sigma_{(i)}$ is the estimator of $\sigma$ where i 'th data are not used.

Also, it said that standardized residuals $r_i$ is not t-distribution because $\hat\sigma$ is a function of $\hat\varepsilon_i$, thus the numerator and denominator of $r_i$ are not independent.
On the contrary, studentized residuals $r_i^*$ is t-distribution because "we did not use the i'th observation $y_i$ when calculating $\hat\sigma_{(i)}$".

I am a little confused about this reason. Numerator of $r_i^*$ is using all data ($y_1$, ..., $y_n$), denominator of $r_i^*$ is using all data except $y_i$. So there are still $n-1$ data are shared. Why are they independent?

Answer 1

Independence does not imply that one statistic is not constructed by other. For example, in normal distribution $S^2 = \frac{1}{n}\sum_{i=1}^n(X_i - \bar{X}_n)^2$ is independent of $\bar{X}_n$, but clearly $S^2$ uses $\bar{X}_n$ in its construction. Indepednece simply imply that $$ f_{S^2, \bar{X}_n}(s, x) = f_{S^2}(s)f_{\bar{X}_n}(x). $$
The same idea holds for the studentized residuals.