# Log concavity of the maximum of dependent Gaussians

Let $$Z_1,\dots,Z_n$$ be dependent Gaussian random variables. Is it true that $$X=\max\{Z_1,\dots,Z_n\}$$ has a log-concave distribution function? This is true for the independent case, but is it true in general?

$$\newcommand{\R}{\mathbb{R}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$$

This is false in general. E.g., let $$Z_1=U$$ and $$Z_2=|V|\,\text{sign}\,U$$, where $$U,V$$ are iid standard normal random variables. Let $$F$$ be the cdf of $$\max(Z_1,Z_2)$$ and $$L:=\ln F$$. Let also $$\Phi$$ denote the standard normal cdf. Then $$Z_1$$ and $$Z_2$$ are each Gaussian and for $$x\ge0$$ \begin{align*} F(x)&=\P(U-x)+\P(0 which yields $$L''(0+)=4/\pi>0$$, so that $$F$$ is not log concave.

However, if $$Z_1,\dots,Z_n$$ are jointly normal, then for $$x\in\R$$ $$$$F(x)=\int_{\R^n}f(z_1,\dots,z_n)\ii{z_1 where $$F$$ is the cdf of $$\max(Z_1,\dots,Z_n)$$, $$f$$ is the pdf of $$(Z_1,\dots,Z_n)$$, and $$\ii\cdot$$ denotes the indicator. So, by the Prékopa–Leindler theorem (see Section Applications in probability and statistics), $$F$$ is log concave -- since $$f$$ is log concave, $$\ii{z_i is log concave in $$(z_i,x)$$ for each $$i$$, and the product of log-concave functions is log concave.

• +1. Can I bring your attention to this seemingly simple inequality math.stackexchange.com/q/843276/64809?
– Hans
Oct 2, 2018 at 6:55
• @Hans : I have added my answer to that question (and a comment to another answer there). Oct 2, 2018 at 21:28
• Thank you, Iosif. I knew you would not disappoint. :-P
– Hans
Oct 3, 2018 at 8:31