AndyMC

02-19-2010, 08:49 AM

Let x1, x2, ... xn be independent standard normal variables. (standard normal: normal with 0 mean and variance 1).

Let the range R be defined as R=max(x_i)-min(x_i)

I would like to find the distribution of the range R.

I found a procedure but I end up with an integral I can't solve analytically:

Let fi(x) be the distribution of a standard normal, Fi(x) the cumulative distribution, and g(R) the distribution of the range R , and G(R) the cumulative distribution of the range R.

G(R)=integral (x from -infinity to +infinity) (n*((Fi(x+R)-Fi(x))^(n-1))*fi(x)dx)

It means I choose randomly one of the n variables to be the minimum x, and impose that the other n-1 are between x and x+R, and multiply it by n to consider the n possible choices of minimum. Then I integrate for all possible values of the minimum x.

g(R)=integral (x from -infinity to + infinity) (n*(n-1)*((Fi(x+R)-Fi(x))^(n-2))*fi(x)*fi(x+R)dx)

It means that if I want the probability of the range being exactly R, i choose two variables randomly out of n to be respectively the maximum and the minimum, then I impose the other n-2 are between them, multiply by all the possible choices (n*(n-1)) (because I can switch the roles of the two I've chosen between maximum and minimum), and integrate for all possible values of the minimum x.

You can verify that G(0)=0, lim for R to infinity G(R)=1, g(R) is non negative.

If I can't find analytically the distribution, can I at least find E[R] and var[R], as a function of n?

Thank you!

Let the range R be defined as R=max(x_i)-min(x_i)

I would like to find the distribution of the range R.

I found a procedure but I end up with an integral I can't solve analytically:

Let fi(x) be the distribution of a standard normal, Fi(x) the cumulative distribution, and g(R) the distribution of the range R , and G(R) the cumulative distribution of the range R.

G(R)=integral (x from -infinity to +infinity) (n*((Fi(x+R)-Fi(x))^(n-1))*fi(x)dx)

It means I choose randomly one of the n variables to be the minimum x, and impose that the other n-1 are between x and x+R, and multiply it by n to consider the n possible choices of minimum. Then I integrate for all possible values of the minimum x.

g(R)=integral (x from -infinity to + infinity) (n*(n-1)*((Fi(x+R)-Fi(x))^(n-2))*fi(x)*fi(x+R)dx)

It means that if I want the probability of the range being exactly R, i choose two variables randomly out of n to be respectively the maximum and the minimum, then I impose the other n-2 are between them, multiply by all the possible choices (n*(n-1)) (because I can switch the roles of the two I've chosen between maximum and minimum), and integrate for all possible values of the minimum x.

You can verify that G(0)=0, lim for R to infinity G(R)=1, g(R) is non negative.

If I can't find analytically the distribution, can I at least find E[R] and var[R], as a function of n?

Thank you!