[frac=n]n+1[/frac] is of course correct, just as hhsilver has quite cleverly intuited.
No clue why mathdotcom gave him such a hard time.
But just for sh
its and giggles, a rather informal late night mathematical proof might read something like this:
We're looking for the E(sup(Xk)) where Xk is a vector of k realizations from the uniform distribution.
For the degenerate case of k=1, the pdf(sup(X1)) = pdf(X1) = 1. Hence:
[int=0]1||E(sup(X1)) = ||p dp[/int]
= [frac=1]2[/frac] * p[sup]2[/sup] | p=0 [hcc]rarr[/hcc] 1
= [frac=1]2[/frac]
So E(sup(X1)) = E(X1) = [frac=1]2[/frac]
For k=2:
Pr(sup(X2) = p [hcc]isin[/hcc] [0,1] )
= Pr( (x(1) [hcc]le[/hcc] p AND x(2) = p) OR (x(1) = p AND x(2) [hcc]le[/hcc] p) ) (this is the key step)
[hcc]asymp[/hcc] 2*p*[hcc]Delta[/hcc]
Hence:
[int=0]1||E(sup(X2)) = ||2p[sup]2[/sup] dp[/int]
= [frac=2]3[/frac] * p[sup]3[/sup] | p=0 [hcc]rarr[/hcc] 1
= [frac=2]3[/frac]
Continuing with the same logic, it's clear that for k=n:
Pr(sup(Xn) = p ∈ [0,1] )
[hcc]asymp[/hcc] n*p[sup]n-1[/sup]*[hcc]Delta[/hcc]
[int=0]1||E(sup(Xn)) = ||n*p[sup]n[/sup] dp[/int]
= [frac=n]n+1[/frac] * p[sup]n+1[/sup] | p=0 [hcc]rarr[/hcc] 1
= [frac=n]n+1[/frac]
QED
Nicely done, hh. YOu get full credit in my book.