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I have some indipendent, discrete variables $X_1$ through $X_n$, which don't necessarily share the same (known) probability distributions or possible values. What is the expected value of $\max\left\{X_1, X_2,\dots, X_n\right\}$?

This question is related but not quite general enough (though I'm sure this could be adapted).

Also, this question comes from a generalization of what I did to answer this code golf question which is about when all variables are dice rolls, for which I found the closed form $6 - \sum_{i=1}^5\left(\frac i 6\right)^n$ (not finding the question I linked above!) and explained it this way which I definitely have no idea how to generalize.

RubenVerg
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  • You must know the individual distributions. – David G. Stork Jul 11 '23 at 18:40
  • It is possible to upper bound the expectation, is that okay? – Andrew Jul 11 '23 at 18:49
  • @AndrewZhang A bound in terms of what? – uniquesolution Jul 11 '23 at 19:03
  • Since the OP says the question is motivated by dice, it would seem OK to assume the $X_i$ are subgaussian. @uniquesolution – Andrew Jul 11 '23 at 19:27
  • @DavidG.Stork they are known, I just wanted to specify that they might not be all the same for every variable (one example I can think of which might be interesting is $X_i$ being a roll of a fair $i$-sided dice) – RubenVerg Jul 11 '23 at 20:35
  • by the way, what's the correct name for what I call a probability table, ie something like a table of probabilities and values for example $\frac 1 4 \to 7$; $\frac 3 8 \to -2$; $\frac 1 4 \to 9$; $\frac 1 8 \to 0$ – RubenVerg Jul 11 '23 at 20:39
  • The correct name is "distribution". btw, a discrete random variable may assume a countable number of values. It may be unbounded. Therefore, even for two discrete random variables, your question is too general, since there is no reason to assume anything. – uniquesolution Jul 11 '23 at 22:53
  • @uniquesolution what do you mean by "unbounded"? I would assume the ability to take the expected value of one of the variables is enough requirements for the ability to take the expected value of the maximum – RubenVerg Jul 12 '23 at 06:21
  • @uniquesolution oh okay I thought distribution was only a continuous term, good to know it's not – RubenVerg Jul 12 '23 at 06:22

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If $X_i>0$ and $F_i(x)=\Pr(X_i\leq x)$ then $\Pr(\max X_i\leq x)=F_1(x)\ldots F_n(x)$ and $E(\max X_i)=\int_0^{\infty}(1-F_1(x)\ldots F_n(x))dx.$

Gérard Letac
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