# Solving for (or approximating) the value of the $k^{th}$ largest sample in a set of samples from a Gaussian, divided by its standard deviation

### Problem

I have an array with $$r$$ (some relatively small number (say, 5 or 10)) rows and $$c$$ (some very large number) columns. The $$i^{th}$$ row of the array is populated with elements randomly drawn from a Gaussian distribution with mean $$0$$ and standard deviation $$\sigma_i$$. I slide an $$r \times q$$ (with $$q \ll c$$) window across this array. The window starts with its leftmost column aligned with the first column of the array, and move across the array in increments of 1 column. Every time the window moves, I do the following:

1. If at least $$r_{thresh} < r$$ rows within the window contain at least one element exceeding some threshold $$t$$, I record the value of the largest sample of each row, and divide it by the standard deviation of the Gaussian distribution associated with that row in the array. I then record the $$k^{th}$$ largest of these values (call it $$x$$). The process continues.
2. If fewer than $$r_{thresh}$$ rows within the window contain at least one element exceeding $$t$$, the process continues.

My goal is to determine the expectation value of $$\langle x \rangle$$ after some very large number of iterations.

### What I've tried

I began by implementing a slightly simplified version of the algorithm in code. If I assume that all $$i$$ Gaussian distributions are the same (i.e. $$\sigma_1=\sigma_2=\ldots \sigma_i$$), I find that the mean value of $$x$$ after some large number of iterations (independent of the sizes of $$r,q,c$$ and $$r_{thresh}$$) is roughly linear in $$\sigma$$.

I'd like to instead solve for (or approximate) this value analytically, as it is highly computationally expensive to actually run the algorithm. I had originally thought I could simply compute the $$n^{th}$$ Gaussian order statistic, and divide this by the standard deviation, however it seems that all approximations for the Gaussian order statistic are linear in $$\sigma$$, hence dividing by $$\sigma$$ yields a constant value (which is not what I see when I actually do this).