Re: Question relating to order statistics of normal variables

shrishri@xxxxxxxxx wrote:
Hi all,

I am looking for a solution / direction to solve the following problem.

Any pointers will be greatly appreciated!

The problem I am seeking a solution for is as follows.
Let X_1, X_2,... X_N be N i.i.d Gaussian random variables with unit
variance and zero mean.
Suppose we arrange the realizations in increasing values of their
absolute values
(similar to order statistics, except that we are using the absolute
values to sort them), and let the corresponding sequence of
random variables be Y_1, Y_2,...,Y_N.
I am looking for the expected value of sum of squares of the first K
terms of the sequence Y_i, in the asymptotic regime where
N tends to infinity, and K/N is a constant in the open interval (0,1).

If someone has a pointer to the literature, that would be great!
Thanks -Shri.

Don't know the literature in this area, and I can't guarantee I'm getting you any further than you've already gotten (or that it's even right - it's late, and I'm studying for qualifiers), but here's something:

First, note that the Y_i's are half-normal. Then you can get the joint distribution of the order statistics for a half normal using the standard n! prod ( f(y_i) ) 1 [y_1 < ... < y_n ] formula. So now you're left with finding E(Y_1^2 + ... + Y_k^2), where the expectation is taken with respect to the joint distribution of the half-normal order statistics. This, of course, equals E(Y_1^2) + ... + E(Y_k^2). All the terms are of similar form - take E(Y_1^2), for example:

E(Y_1^2) ~= int_0^{y_1} x^2 exp(-x^2 / 2 s^2) * prod_{i=2 to k} int_{y_{i-1}}^k exp(-y_i^2 / 2 s^2)

(I'm leaving out some constants here)

The first part is just the second moment of a half-normal (you can find that), and the second part is a product of F(y_i) - F(y_{i-1}) for i = 2 to k, where F is the half-normal CDF (has to do with the "erf", according to Mathworld).

That's as far as I got; maybe if you put things together, you can get somewhere. In any case, hope it helps a bit.

-J
.

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