In a Bayesian analysis I want to sample $\sigma \sim \text{inverse-Wishart}(\nu_0,M)$ where $\nu_0$ is the degrees of freedom, equal to dimension+1 and $M$ is a scalar matrix

Then I will sample the $\mu$ from $\mu|\sigma \sim N(\mu_0,\sigma/\kappa_0)$. My question is, what is the value of $\mu_0$ and $\kappa_0$ and how can I determine my dimensions in the inverse-Wishart?


1 Answer 1


$\mu_0$ is the prior mean and $\kappa_0$ is the number of prior measurements, both on the $\sigma$ scale. The dimensions refer to the number of dimensions in the $d \times d$ M matrix which is also the number of dimensions/variables in the multivariate Inverse-Wishart distribution.

A basic noninformative prior distribution for the Inverse-Wishart distribution would be

$M = I = d \times d$ Identity matrix

$\nu_0 = d+1$

This gives noninformative marginal distributions, but the joint distribution is not uniform however. A noninformative $\kappa_0$ can be chosen to create a very large variance. $\mu_0$ would depend on the domain of your problem.

Thoughts on how to create informative priors would require more information about your problem.

More details can be found in section 3.6 of Gelman's Bayesian Data Analysis

  • $\begingroup$ great information thank you . but what about the value of d in the identity matrix how can you determine this value $\endgroup$
    – Sona yu
    May 15, 2015 at 9:27
  • $\begingroup$ @Sonayu d is the number of dimensions/variables in the multivariate Inverse-Wishart distribution $\endgroup$
    – Eric Farng
    May 15, 2015 at 10:20
  • $\begingroup$ I have a system of differential equations have three variables and I want to sample from the posterior but because I have noninformative prior I will sample the sigma then the mu then the posterior. In this case the d will equal to 3 ? $\endgroup$
    – Sona yu
    May 16, 2015 at 4:39
  • $\begingroup$ @Sonayu Unfortunately, I haven't used MCMC to solve differential equations. But if there are 3 variables/parameters that you want posteriors for, then yes $d=3$ $\endgroup$
    – Eric Farng
    May 16, 2015 at 11:20

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.