Marginal distribution of the diagonal of an inverse Wishart distributed matrix

Suppose $X\sim \operatorname{InvWishart}(\nu, \Sigma_0)$. I'm interested in the marginal distribution of the diagonal elements $\operatorname{diag}(X) = (x_{11}, \dots, x_{pp})$. There are a few simple results on the distribution of submatrices of $X$ (at least some listed at Wikipedia). From this I can figure that the marginal distribution of any single element on the diagonal is inverse Gamma. But I've been unable to deduce the joint distribution.

I thought maybe it could be derived by composition, like:

$$p(x_{11} | x_{ii}, i\gt 1)p(x_{22}|x_{ii}, i>2)\dots p(x_{(p-1)(p-1)}|x_{pp})p(x_{pp}),$$

but I never got anywhere with it and further suspect that I'm missing something simple; it seems like this "ought" to be known but I haven't been able to find/show it.

• Proposition 7.9 of Bilodeau and Brenner (the pdf is freely available on the web) gives a promising result for the Wishart (perhaps it carries over for the inverse Wishart). If you partition $X$ in blocks as $X_{11},X_{12};X_{21},X_{22}$, then $X_{22}$ is Wishart, as is $X_{11} - X_{12}X_{22}^{-1}X_{21}$, and they are independent. Oct 23, 2011 at 4:34
• That proposition only applies if you know the whole matrix: if you've only got the diagonal, then you don't know e.g. $X_{12}$, so you can't do the transformation. Sep 18, 2013 at 21:09

$$\Sigma = \text{diag}(\Sigma) \ Q \ \text{diag}(\Sigma)^\top = D\ Q \ D^\top$$ Here $Q$ is the correlation matrix with unit diagonals $q_{ii} = 1$. Thus, the diagonal entries of $\Sigma$ are now a part of a diagonal matrix of variances $D = [D]_{ii} = [\Sigma]_{ii}$. Since the off diagonal entries of the variance matrix are zero $d_{ij} = 0, \ i \ne j$, the joint distribution you are looking for is just the product of the marginal distributions of each diagonal entry.
Now consider the standard inverse-Wishart model for a $d$-dimensional covariance matrix $\Sigma$
$$\Sigma \sim \mathcal{IW}(\nu +d -1, 2\nu \Lambda), \quad \nu > d-1$$
Diagonal elements of $\sigma_{ii} = [\Sigma]_{ii}$ are marginally distributed as $$\sigma_{ii} \sim \text{inv-\chi^2}\left(\nu+d-1,\frac{\lambda_{ii}}{\nu -d + 1}\right)$$