A friend of mine recently posed an interesting question to me. We can easily see that if we choose a random number between 0 and 1, the expected value should be one-half. Now, in this case, the only number is the same as the largest number, so that the expected value of the largest number is also one-half.
So what occurs if we choose two random numbers? Can we determine the expected value of the largest number in this case? What if there are N numbers? I work through a pretty direct interpretation for the N=2 case, then find a general approach for arbitrary N.
Given the result of N/(N+1), I noticed that this is equal to 1 minus the integral from 0 to 1 of x^N. This would be a really quick solution, but I can't seem to find an interpretation of the situation which lends itself nicely to that expression. Perhaps it's a coincidence, but who knows?
4 comments:
Your pdf does things rather the hard way.
Consider:
[let "_n" stand for "subscript n" and "^n" for "superscript n"]
Let X_1, X_2 ..., X_n be independent standard uniform random variables.
Let Y = max(X_1, X_2 ..., X_n)
P(Y <= y) = P(X_1 <= y) x P(X_2 <= y) x ... x P(X_n <= y)
= y^n (0 < y < 1)
This is the distribution function of the maximum
Hence the density for y is the derivative of this, which is n y^(n-1), (0 < y < 1).
And therefore the expectation is the integral of n.y^n over the same interval.
You may find the wikipedia entry on Order statistics useful (though it could be better written).
[I considered whether the integral for the expectation of the minimum would help (since by symmetry the expectation of the maximum would be 1 minus that). It does yield 1/(N+1), as it should, but I don't see an obvious direct argument that relates the integral for it to the integral for x^N.]
Then I remembered that the expectation of a random variable is equal to the integral of its survivor function. And that seems to yield what you want:
Let S(y) = 1 - F(y) (where F is the distribution function of Y).
That is E(Y) is the integral from 0 to 1 of S(y) dy
= integral from 0 to 1 of 1 - y^n dy
= integral{_0^1} 1 dy - integral{_0^1} y^n dy
= 1 - integral {0^1} y^n dy
which looks like the desired result.
So, no, not exactly a coincidence.
Really nice analysis efrique, the survivor function stuff is something I hadn't run across before. Thanks for the information!
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