Expected Maximum

1. Oct 1, 2006

ghotra

Suppose I have a random variable X with known mean and standard deviation. After n realizations of X, what is the expected maximum of those n realizations?

When n is very large, we know the mean of those realizations will be the mean of the distribution. Is the expected maximum simply the mean plus the standard deviation? If so, how can one show this?

2. Oct 16, 2006

EnumaElish

No reason why E[max] = E[X] + std.

Suppose you are working with the standard uniform distribution F(x) = x. Then E[F] = 1/2, Std[F] = 1/sqrt(12) ===> E[F] + std[F] = 0.79.

Define G = Max{F1, F2} where each of Fi is independently distributed F.

Prob{G < x} = Prob{Max{F1,F2} < x} = Prob{F1 < x and F2 < x} = Prob{F1 < x}Prob{F2 < x} = F(x)^2 = x^2.

E[G] = Integral of (x dx^2) from 0 to 1 = Integral (2x^2 dx) = 2 Integral (x^2 dx) = 2(x^3)/3, which, when calculated from 0 to 1, equals 2/3 < 0.79.

The intuition with the uniform distribution is: the expected values of k independent draws separate the unit interval into k + 1 equal subintervals. So when k = 2, E[min] = 1/3 and E[max] = 2/3. In general, E[min] = 1/(k+1) and E[max] = k/(k+1).

Last edited: Oct 16, 2006
3. Oct 16, 2006

Hurkyl

Staff Emeritus
It depends heavily on the distribution. After all, a random variable that (uniformly) takes the values 0 and 1 will have a much different expected maximum than a random variable that only takes on the value 1/2.

4. Oct 17, 2006

EnumaElish

The relationship E[max] = mean + std does hold in Hurkyl's 2nd example, which is a degenerate (although legitimate) distribution.

If there is a non-degenerate distribution such that the above equality holds, then it has to be a special case.

Although, ghotra, if you replace "standard dev." with the "spread" parameter, defined as the half-length of the interval over which the uniform distribution is defined, then your equality does hold for each and every uniform distribution.

If the uniform dist. is defined over a < x < b, then as the number of draws n ---> infinity, then E[max] ---> b. (In my previous post I had used k for n.)

Define spread = (b - a)/2 = b - (b + a)/2 = b - mean, for any uniform distribution.

It follows that b = mean + spread.

And since Limit[E[max]] = b, the equality "Limit[E[max]] = mean + spread" holds.

Last edited: Oct 17, 2006