# Expectation of max of independent (unknown distribution) random variables.

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 ($E(\max X_i)\geq \mu)$).

• 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 $i$? – 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.
• 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. – Did Oct 30 '12 at 9:59
• 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. – StatQuestion Oct 30 '12 at 18:11