# A reference for a Gaussian inequality ($\mathbb{E} \max_i X_i$)

I am looking for a reference to cite, for the following "folklore" asymptotic behaviour of the maximum of $n$ independent Gaussian real-valued random variables $X_1,\dots, X_n\sim \mathcal{N}(0,\sigma)$ (mean $0$ and variance $\sigma^2$): $$\mathbb{E} \max_i X_i = \sigma\left(\tau\sqrt{\log n}+\Theta(1)\right)$$ (where, if I'm not mistaken, $\tau=\sqrt{2}$). I've been pointed to a reference book of Ledoux and Talagrand, but I can't see the satement "out-of-the-box" there -- only results that help to derive it.

• Actually $E(M_n)\leqslant\sigma\sqrt{2\log n}$ is a one-line computation. – Did Oct 25 '14 at 15:31
• Mmh, I just realized my question was slightly wrong, as phrased. I should have written $\Theta(1)$ instead of $O(1)$ (for both upper and lower bound). – Clement C. Oct 25 '14 at 15:36

## 2 Answers

I eventally found these two references:

• from : the expected value of the maximum of $N$ independent standard Gaussians: Theorem 2.5 and Exercise 2.17, p. 49; for a concentration result, combined with the variance (which is $O(1)$). Exercise 3.24 (or Theorem 5.8 for directly a concentration inequality).
• from , Theorem 3.12

 Concentration Inequalities: A Nonasymptotic Theory of Independence By Stéphane Boucheron, Gábor Lugosi, Pascal Massart (2013)

 Concentration Inequalities and Model Selection, by Pascal Massart (2003)

• @Chill2Macht How so? Look e.g. at the discussion after the proof (specifically equation (3.26)) in ; or as mentioned above Exercise 2.17 in . – Clement C. Feb 18 '18 at 2:10
• The discussion near 3.26  is helpful, but one has to prove first that the max is sub-Gaussian before being able to use it. Thankfully theorem 5.8 from  gives a proof of that, which does help. I have to admit that when I complained I didn't actually bother looking at Theorem 5.8 because I didn't see how it could help with this problem, since the earlier parts of  mentioned did not seem very helpful. – Chill2Macht Feb 18 '18 at 22:35
• @Chill2Macht Starting with 1.: No. Why would you have a $\sqrt{n}$ in the denominator, while the first expression has a $\sqrt{2\log n}$? – Clement C. Sep 21 '18 at 1:44
• @Chilll2Macht i think the best way to proceed is by asking a new question, with a link to this one possibly. – Clement C. Sep 21 '18 at 2:00
• stats.stackexchange.com/questions/367942/… It seems I was wrong on both counts. – Chill2Macht Sep 22 '18 at 23:13

You have an explicit asymptotic result concerning the limit distribution in this post and the associated references.