Let $w_i \stackrel{iid}{\sim} \mathcal N(0, 1)$. Show that:
$$\lim_{n\rightarrow \infty}\mathbb E \max_i^n |w_i| /\sqrt{\log{2n}} = 1$$
my attempt
First, recall the well known upper bound: $\mathbb E \max |w_i| \leq \sqrt{2\log n}$.
Next, we show an asymptotic lower bound, after which the result will follow by sandwiching. By Markov's inequality, for all positive $t$:
$$\mathbb E \max |w_i| \geq t\mathbb P (\max |w_i| \geq t)$$
By independence:
$$\mathbb P (\max |w_i| \geq t) = 1- (1-2\mathbb P(w_i \geq t))^n$$
Taking $t = \sqrt{2\log n}$ all that remains to show is that $(1-2\mathbb P(w_i \geq \sqrt{2 \log n}))^n$ goes to $0$. It is sufficient to show that there is some $N$ such that for all $m \geq N$ we have that $2\mathbb P(w_i \geq \sqrt{2\log m}) < 1$. This is obviously true since $\mathbb P(w_i \geq u)$ is decreasing in $u$ and has minimum $0$.
my question
Is there a mistake in my proof? It is surprisingly simpler than other solutions I have seem, so I expect some issue. See the solution for part (c) here: https://high-dimensional-statistics.github.io/2020/08/27/exercise-2.11.html