In the Wikipedia entry for "Inverse-Wishart distribution" (current revision) there is a formula for a Wishart distribution conditioned on a conjugate prior ( $\Psi,\nu$) in which the original Wishart scale matrix $\Sigma$ has been integrated out to yield the conditional distribution

$$ p(X\mid\Psi,\nu) = \frac { |\Psi|^{\nu/2}\Gamma_p( \frac {\nu + n}{2})}{\pi^{np/2} |\Psi+X^TX|^{(\nu+n)/2} \Gamma_p (\frac{\nu}{2})} $$
where $A_{n\times n} $ is the observed data $X^TX $.

This doesn't appear to be a valid Wishart distribution - which I think should be something more like

$$ p(X) = \frac { |\Psi|^{\nu/2}\Gamma_p( \frac {\nu + n}{2})}{\pi^{np/2} |\Psi+A|^{(\nu+n)/2} \Gamma_p (\frac{\nu}{2})} \exp \Big [-\operatorname{tr} \Big ((\Psi +\Sigma)^{-1}A \Big) \Big].$$ So am I misunderstanding this and the formula is a partial result for illustration purposes only?


1 Answer 1


The distribution shown is not a Wishart distribution, it is a multivariate t-distribution. This is the marginal distribution of $X$ after integrating the covariance matrix out of the Gaussian sampling model. The expression may not be correct (I haven't checked carefully, but it doesn't look quite right). A good reference for the derivation is section 1.4.3 of the book Markov Chain Monte Carlo: Stochastic Simulation for Bayesian Inference by Gamerman and Lopes.

  • $\begingroup$ The equation also differs from Wikipedia's own definition of a multivariate t distribution so it needs to be clarified. My problem is in comms and has known zero means everywhere: only the covariance part is of interest. I had referred to Murphy's "Conjugate Bayesian analysis of the Gaussian distribution" previously. $\endgroup$
    – eulogy
    May 9, 2019 at 10:44
  • $\begingroup$ The density given appears to have a typo (not clear if it intends to integrate over the prior or the posterior), but it is closer than it looks after applying the matrix determinant lemma to $|A + \Psi|$, to get $(1 + X^t \Psi^{-1} X)|\Psi|$, which is the more familiar t-distribution form. $\endgroup$
    – R Hahn
    May 9, 2019 at 15:13

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