# What are the restrictions on the covariance matrix of a nonnegative multivariate distribution.

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.

• I doubt you can find some relevant restriction. In one dimension, you have the full range ($0\le \sigma^2 \le \infty$) Apr 4, 2012 at 15:20
• @leonbloy Yes, but there is also the restriction that $\text{Cov}(X_i,X_j)>1$ above and beyond the positive semi definiteness restriction. Apr 4, 2012 at 15:41
• You mean, $\text{Cov}(X_1,X_2)>-1$... Apr 5, 2012 at 11:37
• @Xi'an Yes, of course. It turns out that that is the only needed restriction (as you already know). Apr 5, 2012 at 13:04

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.
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.