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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.

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I doubt you can find some relevant restriction. In one dimension, you have the full range ($0\le \sigma^2 \le \infty$) –  leonbloy Apr 4 '12 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. –  deinst Apr 4 '12 at 15:41
    
You mean, $\text{Cov}(X_1,X_2)>-1$... –  Xi'an Apr 5 '12 at 11:37
    
@Xi'an Yes, of course. It turns out that that is the only needed restriction (as you already know). –  deinst Apr 5 '12 at 13:04

2 Answers 2

up vote 4 down vote accepted

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.

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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.

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Yes, We are assuming that the covariance matrix exists –  deinst Apr 4 '12 at 14:35
    
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. –  Michael Hardy Apr 10 '12 at 0:44

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