Extreme value distribution with unknown variance Let $\{X_1,\ldots,X_n\}$ be a sequence of r.v. such that $X_i\sim N(0,\sigma^2)$.
It is usually stated in Extreme Value Theory textbooks that (for suitably chosen $a_n$ and $b_n$)
$$\mathbb{P}\left(\frac{1}{\sigma}\max X_i \leq a_n + b_ny\right)\rightarrow \exp(-e^{-x})$$
My question is if it is also true that
$$\mathbb{P}\left(\frac{1}{\hat\sigma}\max X_i \leq a_n + b_ny\right)\rightarrow \exp(-e^{-x})$$
where $\hat\sigma^2 = \frac{1}{n-1}\sum_{i=1}^n(X_i - \bar{X})^2$.
 A: I couldn't find the right way to use the suggestions in the comments and answers but think about ways of using the Slutsky's theorem (sugested by @Glen_b) I came up with a proof using a nice property of the normal distribution.
Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have
$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$
$$b_n = \frac{1}{\sqrt{2\log n}}$$
$$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$
(a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)
Now
$$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$
and after some manipulation
$$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$
We have
$$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$
$$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$
$$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$
And the result follows from Slutsky's theorem.
A: Yes.  IIRC, you can use the Dominated Convergence Theorem to get that the difference between those decreases and is dominated by another sequence which goes to zero as n goes to infinity.
