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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 $\hat\sigma^2 = \frac{1}{n-1}\sum_{i=1}^n(X_i - \bar{X})^2$.

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    $\begingroup$ In the limit as $n\to\infty$, you should be able to apply Slutsky's theorem, I think. $\endgroup$ – Glen_b Nov 12 '16 at 4:08
  • $\begingroup$ I'm not sure how to apply it. In the question I have $(\max x_i /\sigma - a_n)/b_n \rightarrow G$. Now all I can think of is to multiply and divide $(\max x_i /\hat\sigma - a_n)/b_n$ by $\sigma$ but then I get $\frac{\sigma}{\hat\sigma}(\max x_i /\sigma - (\hat\sigma/\sigma)a_n)/b_n$ And I don't know if $(\max x_i /\sigma - (\hat\sigma/\sigma)a_n)/b_n \rightarrow G$ $\endgroup$ – Mur1lo Nov 12 '16 at 4:25
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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.

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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.

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