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A parameter of an object is normally distributed with a mean m and a std. dev. s. If r such objects are randomly selected, what is the maximum expected value M of the parameter in the selection?

Cross-posted from: https://mathoverflow.net/questions/105903/what-is-the-maximum-expected-value-in-a-finite-selection-an-object-with-normal-di

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  • $\begingroup$ The maximum expected value or the expected value of the maximum? (This question is a bit rhetorical; answers can be given for both, but one is more interesting than the other.) Also, I recall a recent duplicate or near-duplicate; it would be worth doing a quick site search. $\endgroup$
    – cardinal
    Commented Aug 31, 2012 at 18:22
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    $\begingroup$ Note that for large $n$, the asymptotic theory result was given here: stats.stackexchange.com/q/34418/5739 $\endgroup$
    – StasK
    Commented Aug 31, 2012 at 18:50
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    $\begingroup$ The expected value is m, so there is no point in asking about its maximum. $\endgroup$
    – StasK
    Commented Aug 31, 2012 at 21:34
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    $\begingroup$ @StasK: (+1) There is an interesting and somewhat subtle feature to note here. Since convergence in distribution does not imply that the means converge to that of the limit distribution, we may wonder whether that holds in this particular case, especially since we are dealing with maxima, where the prospect of mass "escaping to infinity" would seem to be more plausible. In fact, the means do converge, as proven by Pickands (1968). $\endgroup$
    – cardinal
    Commented Sep 1, 2012 at 22:15
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    $\begingroup$ @caardinal, that's an interesting note. $\endgroup$
    – StasK
    Commented Sep 1, 2012 at 23:14

1 Answer 1

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Let $\Phi(x) = P[X\leq x]$ when X is normal with mean m and standard deviation s. Then

$$ \Phi^n(x)= P[\max (X_1,X_2,..., X_n) \leq x]. $$

So $E[\max (X_1,X_2,..., X_n)] = \int_{-\infty}^{\infty} x n \Phi^{n-1}(x) \Phi^{\prime}(x)dx$.

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  • $\begingroup$ What is $\Phi^{\prime}$? $\endgroup$
    – Shashank Sawant
    Commented Aug 31, 2012 at 18:40
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    $\begingroup$ $\Phi'$ is of course the derivative of $\Phi$ $\endgroup$
    – StasK
    Commented Aug 31, 2012 at 18:47

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