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.

  • 1
    $\begingroup$ 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. $\endgroup$ – shabbychef Oct 23 '11 at 4:34
  • 1
    $\begingroup$ 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. $\endgroup$ – petrelharp Sep 18 '13 at 21:09

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

  • $\begingroup$ Do you know what happens for the off-diagonal elements? $\endgroup$ – An old man in the sea. Apr 21 '20 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.