Expected Max of n Realizations of X: Mean + SD?

  • Level: Graduate 
  • Thread starter Thread starter ghotra
  • Start date Start date
  • Tags Tags
    Maximum
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
3 replies · 10K views
ghotra
Messages
53
Reaction score
0
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?
 
Physics news on Phys.org

Discussion

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:
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.
 
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: