Log-concavity of the maximum of gaussians

Question

Let $Z_1,\ldots, Z_n$ be independent standard gaussian random variables. Is it true that $X=\max\{Z_1,\ldots,Z_n\}$ has a log-concave distribution function?

Answer 1

From the context, it appears that by "distribution function" the OP meant the pdf (usually, the distribution function is understood as the cdf). Let $F$ and $G$ denote the cdf's of $X$ and $Z_1$, respectively, so that $F=G^n$. (Note that $1-(1-G)^n$ is the cdf of the minimum, not the maximum, of $Z_1,\ldots,Z_n$. Thus, ofer zeitouni actually showed that the pdf of $\min\{Z_1,\ldots,Z_n\}$ is log concave.)

Let us show that the pdf $f=F'$ of $\max\{Z_1,\ldots,Z_n\}$ is log concave. We have $f=F'=nG^{n-1}g$, where $g:=G'$, the standard normal pdf. Next,
\begin{equation} G^2(\ln f)''=-(n-1)gh-G^2,\quad\text{where } h(x):=xG(x)+g(x). \end{equation} Moreover, $h''=g>0$ and $h(-\infty+)=0=h'(-\infty+)$. So, $h>0$ and hence $(\ln f)''<0$, on the entire real line. Thus, the pdf $f$ is indeed log concave.

Remark. For any $j=1,\dots,n$, let $f_j$ denote the pdf of the $j$th order statistic $Z_{n:j}$ for $Z_1,\ldots,Z_n$, so that $Z_{n:1}=\min\{Z_1,\ldots,Z_n\}$ and $Z_{n:n}=\max\{Z_1,\ldots,Z_n\}$. It follows from a known formula for the pdf of an order statistic (see formula (2.1.6) on page 10 in David--Nagaraja) that \begin{equation} f_j=\binom{n-1}{j-1}f_1^{\frac{n-j}{n-1}}f_n^{\frac{j-1}{n-1}}. \end{equation} So, combining the log-concavity of $f_1$ proved by ofer zeitouni with the log-concavity of $f_n$ proved in this answer, we obtain the more general fact that the pdf $f_j$ of any order statistic $Z_{n:j}$ for $Z_1,\ldots,Z_n$ is log concave. Moreover, by shifting and rescaling, we now see that the pdf of any order statistic for any iid normal sample is log concave.

Answer 2

The answer is yes.

Let $F$ be the CDF of $X$, and let $G$ be the CDF of $Z_i$ (that is, $G(z)=P(Z_i\leq z)$). Then $F(x)= 1-(1-G(x))^n$, hence $F'(x)= nG'(x)(1-G(x))^{n-1}$. Thus, $$p_n(x)=\log F(x)'-\log n=\log (G'(x))+(n-1) \log(1-G(x))$$ $$=-x^2/2+(n-1)\log(1-G(x))$$ Hence, with $g(x)=G'(x)$, $$ p_n''(x)=-1-(n-1) \frac{g(x)^2}{(1-G(x))^2}-(n-1)\frac{g'(x)}{1-G(x)}$$ The only issue is really for $x>0$ (when $x\leq 0$, we have $p_n''(x)<0$). But $p_n''(x)<0$ for all $n$ and $x$ iff for all $x>0$, $g(x)^2\geq|g'(x)| (1-G(x))$. Since $g'(x)=-xg(x)$, this is equivalent to requiring that $q(x)=g(x)-x(1-G(x))>0$. Since $q(0)>0$ and $q(\infty)=0$, we differentiate to find $q'(x)= g'(x)+xg(x)- (1-G(x))=-(1-G(x))<0$. Hence $\inf q(x)=q(\infty)=0$, and the claim follows.