3
$\begingroup$

I would like to test if a sample comes from a standard normal distribution. I want to do that by sorting the sample values, and measuring the Mahalanobis distance to the expected order statistics from this distribution.

To compute the Mahalanobis distance, we need to subtract the ordered sample vector by the expected values, and multiply by the inverse of the covariance matrix. Investigating this problem numerically, I noticed this matrix seems to be symmetric tridiagonal. The second diagonal also seems to be merely the diagonal times -1/2.

My question is basically how can I compute this vector of expected values and inverse of the covariance from the order statistics of a sample from the standard normal distribution with size N? (I'm interested in N = 16)

I hope there is a closed formula for this. If this isn't available, though, can we at least prove that the matrix is tridiagonal, and the coefficients of the secondary diagonals are proportional to the principal diagonal?

$\endgroup$
3
  • 2
    $\begingroup$ Check this related X validated answer: stats.stackexchange.com/a/397749/7224 $\endgroup$
    – Xi'an
    Dec 11, 2021 at 14:15
  • $\begingroup$ Any two order statistics $X_{(i)},X_{(j)}$ except neighbours are independent conditinoal on $\mathbf{X}_{-ij}$. If they were also multivariate normal this would imply that that entry $ij$ of the precision matrix (the inverse of the covariance matrix) would be zero (see e.g. cp. 2 in books.google.no/books?id=RciHDwAAQBAJ). But in this case, things are only approximately multivariate normal so are you sure that the precision matrix is exactly tridiagonal? $\endgroup$ Dec 12, 2021 at 10:17
  • $\begingroup$ Interesting @JarleTufto, thanks. Looking at the numbers it very much looks like so. I believe we might be able to prove it by taking that formula on Wikipedia, and through the discrete sine transform we can arrive at the this (-1, 2, -1) tridiagonal matrix. $\endgroup$ Dec 12, 2021 at 11:10

1 Answer 1

7
$\begingroup$

The exact covariance $\operatorname{Cov}(U_{(k)},U_{(j)})$ between the order statistics of uniform random variables is given wikipedia. From that you should at least be able to obtain a reasonable approximation of $$\operatorname{Cov}(X_{(k)},X_{(j)})=\operatorname{Cov}(\Phi^{-1}(U_{(k)}),\Phi^{-1}(U_{(j)})) $$ via linearizations of $\Phi^{-1}$ around the expected values of $U_{(j)}$ and $U_{(k)}$. This indeed seems to be approach taken in https://stats.stackexchange.com/a/397749/77222 and in references therein.

$\endgroup$
3
  • $\begingroup$ Thanks, that helps a lot! I noticed the covariance matrix from the uniform dist computed with that formula also has a tridiagonal inverse. More specifically, proportional to Toepliz (-1,2,-1). Can that be proven? Seeing this, and the approximation for other distributions, I'm confident now to rely just on numerically-obtained values for the covariance inverse diagonal in my calculations. Using the approximation wasn't great, but I may have made a mistake. I'm interested in this because if we can compute a sophisticated test with a simple algorithm there may be an interesting theory here. $\endgroup$ Dec 11, 2021 at 19:51
  • $\begingroup$ I think you'll end up with something rather similar to the Shapiro-Wilk test. $\endgroup$
    – Glen_b
    Dec 12, 2021 at 2:51
  • $\begingroup$ Yeah, Shapiro-Wilk is where I started this quest. One difference is that I'm testing for the standard normal, and Shapiro-Wilk tests for normality with unknown parameters. I was looking for how to modify it, thinking it was pretty much Mahalanobis applied to the order statistics, but now I'm thinking there's a difference. $\endgroup$ Dec 12, 2021 at 9:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.