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 logconcave distribution function?

1$\begingroup$ @Peter Heinig: Both Maple and Mathematica produce $$PDF(X,t)={\frac { \left( 1/2+1/2\,{\rm erf} \left(1/2\,t\sqrt {2}\right) \right) {{\rm e}^{1/2\,{t}^{2}}}\sqrt {2}}{\sqrt {\pi}}} $$ in the case $n=2$ (executed codes on demand). Is this a Gumbel distribution? Can you give a reference to your statement? $\endgroup$– user64494Feb 17, 2018 at 8:51

$\begingroup$ @PeterHeinig As your reference states, the Gumbel distribution is the limiting case as $n\to\infty$. It doesn't prove that the maximum is logconcave for finite $n$. Also, the OP didn't say that the normals are identical. $\endgroup$– Brendan McKayFeb 17, 2018 at 9:00

$\begingroup$ Please, clarify your question. Do you mean random variables are IID? What do you mean by "distribution function":$PDF(X,t)$ or $CDF(X,t)$? $\endgroup$– user64494Feb 17, 2018 at 9:34

$\begingroup$ Thank you for your observations, I indeed meant that the random variables are IID. $\endgroup$– TOMFeb 17, 2018 at 10:42

1$\begingroup$ @TOM: What do you mean by "distribution function":PDF or CDF? $\endgroup$– user64494Feb 17, 2018 at 11:05
2 Answers
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 $1(1G)^n$ 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 $f=F'=nG^{n1}g$, where $g:=G'$, the standard normal pdf.
Next,
\begin{equation}
G^2(\ln f)''=(n1)ghG^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 DavidNagaraja) that \begin{equation} f_j=\binom{n1}{j1}f_1^{\frac{nj}{n1}}f_n^{\frac{j1}{n1}}. \end{equation} So, combining the logconcavity of $f_1$ proved by ofer zeitouni with the logconcavity 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.

1$\begingroup$ Since the Gaussian distribution is symmetric, the law of the maximum and the minimum are the same, up to a reflection :) $\endgroup$ Feb 18, 2018 at 5:10

1$\begingroup$ @oferzeitouni : That's right! Why didn't I think about that? :) $\endgroup$ Feb 18, 2018 at 5:35

1$\begingroup$ I have now added a remark about the logconcavity of the pdf of any order statistic for $Z_1,\ldots,Z_n$. $\endgroup$ Feb 18, 2018 at 5:36
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 $F(x)= 1(1G(x))^n$, hence $F'(x)= nG'(x)(1G(x))^{n1}$. Thus, $$p_n(x)=\log F(x)'\log n=\log (G'(x))+(n1) \log(1G(x))$$ $$=x^2/2+(n1)\log(1G(x))$$ Hence, with $g(x)=G'(x)$, $$ p_n''(x)=1(n1) \frac{g(x)^2}{(1G(x))^2}(n1)\frac{g'(x)}{1G(x)}$$ 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 $x>0$, $g(x)^2\geqg'(x) (1G(x))$. Since $g'(x)=xg(x)$, this is equivalent to requiring that $q(x)=g(x)x(1G(x))>0$. Since $q(0)>0$ and $q(\infty)=0$, we differentiate to find $q'(x)= g'(x)+xg(x) (1G(x))=(1G(x))<0$. Hence $\inf q(x)=q(\infty)=0$, and the claim follows.