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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

$$q_\alpha\leq\int_{-\infty}^\infty\,x\,g(x)\,dx\quad?$$

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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