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When you have a posterior that looks like as one step in Gibber Sampler

$P(\xi | \Sigma_\xi, \theta) ∝ exp\{-1/2 \xi\Sigma_\xi^{-1}\xi\}P(data | \xi, \theta)$

Do you always assume inverse Wishart for $\Sigma_\xi$?

$\Sigma_\xi^{-1} $ ~ $Wishart(\nu_n, \Delta_n)$

where,

$\nu_n = \nu_o + N$, and $\Delta_n^{-1} = \Sigma\xi\xi^{-1} + \Delta_o^{-1}$

When I check the original derivation, it was for the case when the likelihood was in exponential family. Would this be true if $P(data | \xi, \theta)$ was an arbitrary function, not an exponential family?

I saw one author doing exactly this for a very complex likelihood function.

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There is no "default" priors for anything. Wishart distribution is commonly used for variance because it is a conjugate prior for multivariate normal distribution, but that does not make it anyhow "better" then any other choice for prior distribution.

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  • $\begingroup$ I agree with your argument that it is not better than any other. But is it common practice to just default to it since there is nothing else better? $\endgroup$ – bhomass Jun 17 '17 at 20:48
  • $\begingroup$ @biomass better in what sense? $\endgroup$ – Tim Jun 17 '17 at 20:49
  • $\begingroup$ in the sense you meant. I am just wondering whether what this author did is common practice. $\endgroup$ – bhomass Jun 18 '17 at 1:50
  • $\begingroup$ @bhomass but I didn't say it's better. $\endgroup$ – Tim Jun 18 '17 at 6:12
  • $\begingroup$ I didn't say you said it's better. You quoted "better", I am only paraphrasing. To be exact, you said it's not better, I am using it to also mean it's not better, only a common practice. $\endgroup$ – bhomass Jun 18 '17 at 9:24

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