tail probability of max of Gaussians

Question

I'm trying to follow an argument in C. Giraud's "High Dimensional Statistics" (2nd Ed, p. 11 / $\S$ 1.2.3). The specific page is accessible via Google Books here but the formatting is awful.

Suppose that $\epsilon_j \sim {\cal N}(0,1), j = 1, \ldots, p$ and these are all independent.

Then $\mathbb{P}[ \max_j |\epsilon_j| \geq x] = 1 - (1 - \mathbb{P}[|\epsilon_1| > x])^p$. I understand the steps to get the expression on the right from the one on the left.

What I don't understand is the next part, where a limit is taken (?) in $p$:

$\mathbb{P}[ \max_j |\epsilon_j| \geq x] = 1 - (1 - \mathbb{P}[|\epsilon_1| \geq x])^p \sim^{p \rightarrow \infty } p \mathbb{P}[ |\epsilon_1| \geq x]$

I guess the result is approximate but I cannot grasp the necessary steps. Any help is appreciated!

Answer 1

The correct version of this formula is $$P(\max_j |\epsilon_j| \ge x) = 1-(1-P(|\epsilon_1| \ge x))^p \underset{x \to\infty}\sim p \,P(|\epsilon_1| \ge x)$$ for each real $p>0$, which follows because $(1-u)^p=1-(p+o(1))u$ as $u\to0$.

(The reproduction quality of the preview of the book is indeed terrible. It is also clear that $1-(1-P(|\epsilon_1| \ge x))^p \underset{p\to\infty}\to1$ for each real $x$.)