2
$\begingroup$

How does one prove that the Matrix-Normal-Inverse-Wishart distribution is a conjugate prior for a Linear Model? This prior is a generalization of the Normal-Inverse-Wishart Distribution.

By Matrix-Normal distribution, I mean this distribution.

$\endgroup$

1 Answer 1

1
$\begingroup$

First, I'll assume by Matrix-Normal-Inverse-Wishart distribution, you mean $$ \begin{align} M &\sim \mathcal{MN}\left(M_0, \frac{1}{\lambda_1}U, \frac{1}{\lambda_2}V\right)\\ U &\sim \mathcal{W}^{-1}\left(\Psi,\nu\right)\\ V &\sim \mathcal{W}^{-1}\left(\Phi,\eta\right)\\ \end{align} $$ To prove that a prior is conjugate, you check that the posterior has the same form as the prior: $$ P(M,U,V|X) =P(X|M, \lambda_1, \lambda_2, U, V)P(U|\Psi, \nu)P(V|\Phi,\eta) $$ Typically this is done in log space. I'll skip writing out all the expressions, but it turns out that (via completing the matrix quadratic form) there is an extra term $$-\frac{1}{2}(1+\lambda_1\lambda_2)\text{tr}\left[V^{-1}\left(\frac{\sum_k X_k+\lambda_1\lambda_2M_0}{k + \lambda_1\lambda_2}\right)^{T}U^{-1}\left(\frac{\sum_k X_k+\lambda_1\lambda_2M_0}{k + \lambda_1\lambda_2}\right)\right]$$ which means that this formulation of the Matrix-Normal-Inverse-Wishart distribution is not conjugate. However, other formulations may be conjugate. For example, if we instead fix one of $U$ or $V$ as a hyperparameter, then due to the circular property of the trace, the form of the posterior will match the prior. Alternatively, if $n=p$, we could state that $V$ depends on $U$: $$ \begin{align} M &\sim \mathcal{MN}\left(M_0, \frac{1}{\lambda_1}U, \frac{1}{\lambda_2}V\right)\\ V &\sim \mathcal{W}^{-1}\left(Z^TU^{-1}Z,\eta\right)\\ U &\sim \mathcal{W}^{-1}\left(\Psi,\nu\right)\\ \end{align} $$ for some hyperparameter $Z\in \mathbb{R}^{n \times n}$. Remember, there is nothing special about a conjugate prior except for providing closed-form posteriors. For example, if I want to make inferences about both $U$ and $V$ simultaneously, then the conjugate prior made by fixing one is a poor choice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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