For a sample of independent observations $X_1,X_2,...,X_n$ on a continuous distribution $F$, let the ordered sample values be $X_{(1)},X_{(2)},...,X_{(n)}$. From the theory of order statistics, the density function $g(x)$ of the maximum variable $X_{(n)}$ is known to be $$g(x)=n\,f(x)\,[F(x)]^{n-1}, $$ with $f(x)=F'(x)$.

On the other hand, let $q_\alpha=F^{-1}(\alpha)$ be the quantil function with respect to $1/2<\alpha<1$.

Is it possible to prove the following inequality


  • $\begingroup$ For a fixed $n$, if you allow $\alpha$ to be arbitrarily closed to $1$, then you can always have a quantile $q_{\alpha}$ larger than the expected value of the sample maximum. And for a very skewed distribution, even the expected value of sample maximum maybe less than the median. $\endgroup$ – BGM Nov 16 '18 at 15:31
  • $\begingroup$ My guess is that for sufficient large samples the inequality might hold. While, the sample size depends on $\alpha$ such as $n>\alpha/(1-\alpha)$ or so. Do you have a counter example? $\endgroup$ – kaffeeauf Nov 16 '18 at 15:49
  • 1
    $\begingroup$ For sufficiently large $n$, the sample maximum will converge to the supremum of the support, so in other words for a fixed quantile, you can have a sample maximum with expected value larger than the quantile with sufficiently large $n$ $\endgroup$ – BGM Nov 16 '18 at 16:28

if you have F such that $F(-B)=0.3$, $F(0)=0.5$ and $F(1)=1$

You have $q_\alpha>=0$ for $\alpha \ge 0.5$

and $E=\int_{-\infty}^\infty\,x\,g(x)\,dx \le -B(0.3)^n+1$ (as you have a probability $(0.3)^n$ to draw all your sample bellow -B)

you can chose $B\gt\frac{1}{(0.3)^n}$ such that $E \lt 0 \le q_\alpha$

So the inequality is not always true.

  • $\begingroup$ Ok. Could you help me to understand how to obtain the estimate $E\leq 1-B\, 0.3^n$ in your argument? Is $F$ supposed to be a continuous distribution in your example? $\endgroup$ – kaffeeauf Nov 16 '18 at 16:05
  • $\begingroup$ expectation E of the maximum M can be express E(M)=P(A)E(M/A)+(1-P(A))E(M/not A). In my case A is the event : all sample is bellow -B. $\endgroup$ – Arnaud Mégret Nov 16 '18 at 16:11
  • $\begingroup$ Actually, the idea is to let a high probability for very very negative values that have an influence on the average maximum but not on the quantile as it is only influenced by when F reach $\alpha$ and not by how the density is distributed before that point $\endgroup$ – Arnaud Mégret Nov 16 '18 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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