Let $X_n = \max (|Z_1|, \dots, |Z_n|) $ where $(Z_i)$ are i.i.d. standard normally distributed random variables.

I'd like to show $\mathbb{E} [X_n] \sim \sqrt{2 \log n}$. I've shown already that $$\lim_{n \rightarrow \infty} \frac{\mathbb{E} [X_n]}{\sqrt{2 \log n}} \leq 1$$

I would appreciate it if someone can explain how to prove 1 is also a lower bound for this limit using an approach that goes like this:

I note that $\mathbb{P} (X_n < x) = (2\Phi(x) -1)^n$ for positive $x$ where $\Phi$ is the cumulative distribution function for a standard normal. Using this, I can write the expectation as $$\mathbb{E}[X_n] = \int_0 ^ \infty 2x n(2\Phi(x)-1)^{n-1} \mathcal{N}(x;0,1) dx$$ where $\mathcal{N}(x;0,1)$ is the density of a standard normal. Now I want to lower bound this somehow to get a lower bound for the limit above. How can I do that?

Thank you.

  • $\begingroup$ Not following your own approach, but please see the very relevant post #1713154, noting that $\max (|Z_1|, \dots, |Z_n|) \geq \max (Z_1, \dots, Z_n)$ $\endgroup$ – Lee David Chung Lin Feb 21 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.