4
$\begingroup$

This question is a step in answering this question on the stats.se.

Given a distribution $F(X_1,\ldots,X_n)$ on the nonnegative orthant $\mathbb{R}_+^n$ (i.e. each of the marginals is supported on the nonnegative reals). Where the mean of each marginal is 1 (i.e. $E(X_i)=1$ for all $i$). What are the restrictions on the covariance matrix (assuming that it exists, other than positive semi-definiteness)?

The idea is to be able to recognize a covariance matrix as coming from a nonegative multivariate distribution. For example $\pmatrix{4&-3\\-3& 4}$ is a perfectly fine covariance matrix, it is symmetric and positive definite, but it cannot come from a non-negative multivariate ditribution with mean $\mathbf 1$ because $\text{Cov}(X_1,X_2)=E(X_1X_2)-1\ge-1$ as $E(X_1X_2)$ is positive. I am certain that this is not the only such restriction.

|cite|improve this question
$\endgroup$
  • $\begingroup$ I doubt you can find some relevant restriction. In one dimension, you have the full range ($0\le \sigma^2 \le \infty$) $\endgroup$ – leonbloy Apr 4 '12 at 15:20
  • $\begingroup$ @leonbloy Yes, but there is also the restriction that $\text{Cov}(X_i,X_j)>1$ above and beyond the positive semi definiteness restriction. $\endgroup$ – deinst Apr 4 '12 at 15:41
  • $\begingroup$ You mean, $\text{Cov}(X_1,X_2)>-1$... $\endgroup$ – Xi'an Apr 5 '12 at 11:37
  • $\begingroup$ @Xi'an Yes, of course. It turns out that that is the only needed restriction (as you already know). $\endgroup$ – deinst Apr 5 '12 at 13:04
4
$\begingroup$

I hate answering my own questions, but noone else is doing so.

It turns out that the only restrictions on the covariance matrix are that it is positive definite and that $\text{Cov}(X_i,X_j)>-1$. As demonstrated in the answer to the related question, given a covariance matrix satisfying these restrictions, a lognormal distribution with mean $\mathbf{1}$ can be constructed having the specified covariance.

|cite|improve this answer
$\endgroup$
0
$\begingroup$

Are you assuming that the covariance matrix exists? $E[X_i]=1$ is not enough to ensure this.

Otherwise there are cases where the covariance matrix is not well-defined. For example, if $X_i\sim G$; where $G$ is the CDF of $X/c$, $X$ has a Student's-$t$ distribution with $1.5$ degress of freedom truncated below $0$, $c=E[X]\approx 2.04$. This implies that $E[X_i]=1$ but their variances do not exist and consequently the covariance matrix does not exist.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Yes, We are assuming that the covariance matrix exists $\endgroup$ – deinst Apr 4 '12 at 14:35
  • $\begingroup$ I took the question to mean: Which matrices are covariance matrices of tuples of nonnegative random variables? The fact that some distributions have no covariance matrices has no bearing on that question. $\endgroup$ – Michael Hardy Apr 10 '12 at 0:44

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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