Let $Z_1,\dots,Z_n$ be dependent Gaussian random variables. Is it true that $X=\max\{Z_1,\dots,Z_n\}$ has a logconcave distribution function? This is true for the independent case, but is it true in general?
1 Answer
$\newcommand{\R}{\mathbb{R}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$
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,V\,\text{sign}\,U<x) \\ &=\P(U<0)\P(V>x)+\P(0<U<x)\P(V<x) \\ &=\tfrac12+[\Phi(x)\Phi(0)][\Phi(x)\Phi(x)], \end{align*} 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 $x\in\R$ \begin{equation} F(x)=\int_{\R^n}f(z_1,\dots,z_n)\ii{z_1<x}\cdots\ii{z_n<x}\,dz_1\cdots dz_n, \end{equation} 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, $\ii{z_i<x}$ is log concave in $(z_i,x)$ for each $i$, and the product of logconcave functions is log concave.

$\begingroup$ +1. Can I bring your attention to this seemingly simple inequality math.stackexchange.com/q/843276/64809? $\endgroup$– HansOct 2, 2018 at 6:55

$\begingroup$ @Hans : I have added my answer to that question (and a comment to another answer there). $\endgroup$ Oct 2, 2018 at 21:28

$\begingroup$ Thank you, Iosif. I knew you would not disappoint. :P $\endgroup$– HansOct 3, 2018 at 8:31