QuoteOriginally posted by: quantystE[m/M]=...=E[X(1)/X(n) | X(1)<X(2)<...<X(n)] =E[E[X(1)/X(n) | X(1)<X(2)<...<X(n), X(n)] | X(1)<X(2)<...<X(n)]=E[(1/X(n)) E[X(1) | X(1)<X(2)<...<X(n), X(n)] | X(1)<X(2)<...<X(n)].This corresponds exactly to my second solution. I'm sorry if my notation was not clear enough in the first solution I sent, but in this case I find my notation clearer :-)The last equation is exactly what I wrote E[m/M] = E[(1/X(n)) E[X(1) | X(1)<X(2)<...<X(n), X(n)] | X(1)<X(2)<...<X(n)] = E[(1/X(n)) E[X(1) | X(n)]] = E[(1/M) E[m|M]]because the X(1)<X(2)<...<X(n) is true by definition: X(1)=m is the minimum and X(n)=M is the maximumQuoteNow, it is easy to compute E[X(1)|X(1)<X(2)<...<X(n), X(n)], where obviously X(1)=m, X(n)=M.Is it easy? You don't need the distribution of the minimum (the beta function I mentioned)? I can't see a simpler way to get it... (of course I don't mean it doesn't exist!).>> I edit the post to add the last step (and a second time to fix a typo) in my solution, in case it was not clear:E[m|M]=M/n because m|M is the minimum of (n-1) variables in [0,M] (the n-th variable is fixed at M)the minimum of (n-1) variables in [0,1] is distributed as Beta[1,n-1] and its expectation is 1/(1+(n-1))=1/nin our case the uniform distribution is on [0,M], not [0,1]: the expectation is M/nPutting everything together: E[m/M]=E[1/M*E[m|M]]=E[1/M*M/n]=E[1/n]=1/n

carlitos wrote:"Is it easy? You don't need the distribution of the minimum (the beta function I mentioned)? I can't see a simpler way to get it... (of course I don't mean it doesn't exist!)."Words like "easy" are notoriously relative! One irresistibly blurts out: "Easy for whom?" Well, easy for the one who has done it already. The probability distribution for m and M, respectively areP{m<r}=1-((1-r)^n) P{M<R}=R^n.The rest follows EASILY!

can we integrate directly?

shinichivn's method is cool. Say M_k is the k-th biggest value of X_1, X_2 ...X_n. One can use shinichivn's method to calculate E(M_k) easily!