3
$\begingroup$

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?

$\endgroup$
5
  • 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? $\endgroup$
    – user64494
    Feb 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 log-concave for finite $n$. Also, the OP didn't say that the normals are identical. $\endgroup$ Feb 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 $CDF(X,t)$? $\endgroup$
    – user64494
    Feb 17, 2018 at 9:34
  • $\begingroup$ Thank you for your observations, I indeed meant that the random variables are IID. $\endgroup$
    – TOM
    Feb 17, 2018 at 10:42
  • 1
    $\begingroup$ @TOM: What do you mean by "distribution function":PDF or CDF? $\endgroup$
    – user64494
    Feb 17, 2018 at 11:05

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
3
  • 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 log-concavity of the pdf of any order statistic for $Z_1,\ldots,Z_n$. $\endgroup$ Feb 18, 2018 at 5:36
2
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.