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

  • 2
    $\begingroup$ In the limit as $n\to\infty$, you should be able to apply Slutsky's theorem, I think. $\endgroup$
    – Glen_b
    Nov 12, 2016 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, 2016 at 4:25

2 Answers 2


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.


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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