Let's say we have 50 independent random variables $X_i$ with the same distribution. Let's also say that $E(X_i) = \mu$ and $Var(X_i) = \sigma^2 > 0$ are known values. Is it possible to determine a (non trivial) lower bound for $E(\max X_i)$?

I've found some questions here concerning uniform distributions, but I was wondering if we could say something about this question when distribution of $X_i$s is unknown (other than ($E(\max X_i)\geq \mu)$).

  • $\begingroup$ I would imagine for a non-trivial result that the R.V.s would have to be on finite support... or are you taking your max over $i$? $\endgroup$ – Emily Oct 29 '12 at 21:42

The best possible universal lower bound on the expectation of $M_n=\max\limits_{1\leqslant i\leqslant n}X_i$ is the trivial one, namely, $\mathbb E(M_n)\geqslant\mu$.

To see this, assume that $\mathbb P(X_i=\mu+x)=\mathbb P(X_i=\mu-x)=a$ and $\mathbb P(X_i=\mu)=1-2a$, where $0\lt a\lt\frac12$ and $x=\sigma/\sqrt{2a}$ ensures that the variance of each $X_i$ is $\sigma^2$.

Then $M_n\in\{\mu-x,\mu,\mu+x\}$ almost surely and $\mathbb P(M_n\ne \mu+x)=(1-a)^n$ hence $$ \mathbb E(M_n)\leqslant\mu+x\cdot (1-(1-a)^n)=\mu+\sigma\cdot u_n(a),\qquad u_n(a)=\frac{1-(1-a)^n}{\sqrt{2a}}. $$ Since $u_n(a)\to0$ when $a\to0^+$, the proof is complete.

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  • $\begingroup$ I think I do understand your example but in that case, if 'a' goes to 0, then isn't the variance going also to 0? Can't we say absolutely nothing when this is not the case? [Sorry to insist here, but it seems a 'shame' that this is the best we can get :)] $\endgroup$ – StatQuestion Oct 30 '12 at 9:55
  • $\begingroup$ The variance of each $X_i$ is fixed and equal to $\sigma^2$, for every $a$. The variance of $M_n$ goes to zero when $a$ goes to zero. $\endgroup$ – Did Oct 30 '12 at 9:59
  • $\begingroup$ I see, so if someone says to me there's a bound like $E(M_n)\geq \mu + \delta$, I can pick a suitable $a > 0$ such that this particular distribution contradicts that bound. $\endgroup$ – StatQuestion Oct 30 '12 at 18:11

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