# Expectation of the maximum of standard Gaussians

I'm strugging to find a neat solution to the following problem.

Let $$X_1,\dots,X_N \sim N(0,1)$$ and let $$X = \max_i X_i$$. Show that

$$\mathbb{E}[X] \geq c \sqrt{\ln N}$$

for some absolute constant $$c$$.

Combining the lower tail bound for standard Gaussians

$$\mathbb{P} (X_i > a) \geq \frac{1}{\sqrt{2 \pi} a + 2} e^{-a^2/2}$$

with the total expecation theorem $$\mathbb{E}[X] = \mathbb{E}[X | X \geq a] \mathbb{P}(X \geq a) + \mathbb{E}[X | X < a]\mathbb{P}(X < a)$$, I managed to prove the inequaility but it requeired a lot of tedious calculations.

Is there a simple and neat way to prove that lower bound?

I have seen a related post Expectation of the maximum of gaussian random variables, where the distribution of $$X$$ is derived, however I was not able to bound from below the expectation integral.

If $$g$$ is a standard Gaussian, then $$\hbox{Prob}\{g>t\}\geq (\frac{1}{t}-\frac{1}{t^3})e^{-t^2/2}$$ (See Feller, Vol. $$1$$, $$3$$rd edition, p. $$175$$) Hence there exists a positive $$\beta>0$$ (smaller than $$1$$) such that $$\hbox{Prob}\{g>\beta(\log n)^{1/2}\}\geq\frac{1}{n}$$ Now if $$g_1,\dots,g_n$$ are independent standard Gaussians, then using the previous inequality, we get $$\hbox{Prob}\{\max g_i \leq\beta(\log n)^{1/2}\}<(1-\frac{1}{n})^n\approx \frac{1}{e}$$ Hence $$\mathbb{E}(\max g_i)>\beta(\log n)^{1/2}\hbox{Prob}\{\max g_i>\beta(\log n)^{1/2}\}>(1-\frac{1}{e})\beta(\log n)^{1/2}$$
• Many thanks, but what I don't like is that the constant $\beta$ (which I'm able to find) depends on $N$, while I would like to come up with a LB where $c$ is an absolute constant. By setting $\beta = \sqrt{2- \frac{2 \log \log N}{\log N}}$ I manged to prove the bound, but has I said there are quite a few calculations involved. Maybe there are simpler choice for $\beta$? May 17, 2021 at 9:39
• Yes, take $\beta = 1/2$. This works for sufficiently large $n$ - how much large can be calculated - and then take care of the remaining finite set of $n$'s by a sufficiently large $c$. By the way, the $\beta$ you pointed out does not work. It has to be smaller than $1$. May 17, 2021 at 9:53
• Thanks, you're right about the choice of $\beta$. Indeed, with the one I initially proposed I get a lower bound of the form $c_0 \sqrt{\log n} - c_0$. May 24, 2021 at 7:00