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. 
 A: In general one can decompose a any covariance matrix into a variance-correlation decomposition as
$$ \Sigma =  \text{diag}(\Sigma) \  Q \ \text{diag}(\Sigma)^\top = D\ Q \ D^\top$$ 
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 
$$\sigma_{ii} \sim \text{inv-$\chi^2$}\left(\nu+d-1,\frac{\lambda_{ii}}{\nu -d + 1}\right)$$
A nice reference with a variety of priors for the covariance matrix that decompose into different variance-correlation distributions is given here
