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Say I define a probability distribution on $P$ dimensional symmetric matrices such that the diagonals are strictly positive but the off diagonals are unrestricted (except for the symmetry constraint). Does this induce a proper probability distribution on the space of symmetric positive definite matrices?

A simple example like what I'm thinking about: $\sigma_{ii}\sim TN(0, 1)$ ($TN$ is the normal distribution truncated to be positive) and $\sigma_{ij}\sim N(0, 1)$ independently with $\sigma_{ji}=\sigma_{ij}$ so that $\pi(\Sigma) \propto [\prod_{p=1}^P N(0, 1) 1(\sigma_{ii}>0)][ \prod_{i<j} N(0, 1)]$.

So my question is if $A$ is the set of SPD matrices, is $\pi_0(\Sigma) \propto \pi(\Sigma)1(\Sigma \in A)$ a probability distribution? Does this hold generally, or does it depend on the distributions for $\sigma_{ii}, \sigma_{ij}$ (assuming that they are all continuous), or on the size of $P$?

Leaving aside for the moment if this is actually useful or a good idea, does it work? To me it seems like...maybe. Intuitively it seems that it should, but I've learned not to trust my intuition. And I'm not sure about a good way to demonstrate that it does or doesn't, either.

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up vote 1 down vote accepted

There is no guarantee that the matrix you suggest will be positive symmetric with full probability. Already in dimension $2$, your matrix is $\left(\begin{matrix}|X| & Z \\ Z& |Y|\end{matrix}\right)$ with $X$, $Y$ and $Z$ independent standard Gaussian, hence its determinant is $|XY|-Z^2$, which may very well be negative. The argument works as soon as the coefficients are independent and the distribution of an off diagonal coefficient is unbounded.

Natural distributions on the space of symmetric random matrices are the so-called generalized inverse gaussian (GIG) measures, see here.

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Sure, but what I'm wondering about is the restriction $\pi_0(\Sigma) \propto \pi(\Sigma)1(\Sigma\in A)$. I suppose I didn't need to take the diagonals to be strictly positive - maybe that's confusing.. –  JMS Feb 10 '11 at 18:55
    
@JMS: Oh, I see. Indeed, for every nonnegative measure $\mu$ and every measurable set $A$ such that $\mu(A)$ is positive and finite, $\mathrm{d}\mu_A=\mathbb{1}_A(\ )\mathrm{d}\mu/\mu(A)$ defines a bona fide probability measure $\mu_A$. –  Did Feb 10 '11 at 19:03
    
So am I understanding correctly that I can conclude $\pi_0$ is a proper probability measure since the SPD cone $A$ is measurable wrt $\pi$? –  JMS Feb 10 '11 at 19:05
    
Yes, all you have to check is that $\pi(A)$ is positive (for instance, $A$ is open and $\pi$ absolutely continuous) and finite (the total mass of $\pi$ is $2^{-P}$). –  Did Feb 10 '11 at 21:13
    
Thought so. Just what I needed, thanks for clearing it up for me! –  JMS Feb 10 '11 at 23:38
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