Expected maximum absolute value of $n$ iid standard Gaussians?

I have a problem where my errors are normally distributed and I want to know what the expected maximum error is if I repeat the process $n$ times.

What is the smallest constant $C$ such that the following statement is true for all $n\geq 2$?

Let $X_1, X_2, \cdots, X_n$ be independent standard Gaussian random variables. Then $$\mathbb{E}\left[\max_{i=1}^n \left|X_i\right| \right] \leq C \sqrt{\log_e n}.$$

I can show that the answer is between $1.35$ and $2$.

Solution 1:

The inequality holds with $C=\sqrt{2}$, and this is the optimal constant (the optimality follows from here).

Outline of the proof:

1. Let $\varphi$ and $\Psi$ be the pdf and complementary cdf of standard normal distribution. Using the inequality $\Psi(x)< \varphi(x)/x$ for $x>0$, it is easy to show that $-\log \Psi(x)$ is convex, therefore, its inverse $G(t) = \Psi^{-1}(e^{-t})$ is concave. Note also that $G$ is increasing.

2. Applying the quantile transformation, $X_k = G(Y_k)$, where $Y_k$ are iid $\operatorname{Exp}(1)$. Denoting $X_{(n)} = \max_k X_k$, $Y_{(n)} = \max_k Y_k$ and using the monotonicity and concavity of $G$, we get with the help of Jensen's inequality $$ E[X_{(n)}] = E[G(E_{(n)})]\le G(E[Y_{(n)}]) = G(H_n), $$ where $H_n = 1+\frac12 + \dots + \frac1n$ is the $n$th harmonic number (the distribution of exponential order statistics is well known). Since $H_n\le \log n + 1$ for all $n$, we get $$ E[X_{(n)}]\le G(\log n+1) = \Psi^{-1}\big(\tfrac1{en}\big). $$

3. Using the inequality $\Psi(x)< \varphi(x)/x$ again, $$ \Psi(\sqrt{2 \log n}) \le \frac{1}{2\sqrt{\pi\log n}}e^{-\log n} \le \frac{1}{en}. $$ As $\Psi$ decreases, $$ E[X_{(n)}]\le \sqrt{2\log n}, $$ as required.