Given x1,x2,...xn are independent, and uniformly distributed in the range [a,b], what is the expectation of the maximum of (x1..xn)?
No, this doesn't particularly have anything to do with the Super Bowl, except that it is what I am doing instead of watching.
There's a right way and a wrong way to attack this problem. The wrong way is to guess the answer, and try to prove it by induction. That might even be possible, and it sure did look interesting, but it isn't terribly practical.
The useful approach is to try to work out the probability distribution, then do the integral.
If the distribution is uniform, then the probability of a single trial achieving the value x is dx/(b-a). The probability that a single trial returns a value less than x is (x-a)/(b-a). These probabilities multiply, but there is one last trick, which is to recall that any of the n trials can be the one with the highest value, which pops a factor n in front.
So the probability distribution of the maximum is:
p(x) dx = n (x-a)^(n-1) / (b-a)^n
= d/dx[ (x-a)^n/(b-a)^n ]
which really shouldn't be all that much of a surprise - notice that the integral of this expression is trivially equal to one when evaluated on the range [a,b].
From here, life should be relatively straight forward
= ∫ x p(x) dx
= ∫ x n (x-a)^(n-1) dx / (b-a)^n
= n/((b-a)^n) * ∫ x (x-a)^(n-1) dx
Integration by parts gets us past the next bit
∫ x (x-a)^(n-1) dx
= ∫ d/dx[ 1/n (x-a)^n ((nx+a)/(n+1)) ]
= 1/n (x-a)^n ((nx+a)/(n+1))
At x=a, this cancels trivially. At x=b, the leading terms cancel against the factors we had pulled from the integral above, leaving
The obvious case to check is that in which n=1, in which case we find that the expected value of x is the average of a and b.
February 6, 2005 7:07 PM
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