Given the standard linear model $$Y=X\beta+\epsilon$$ a Gaussian likelihood function $Y|b,\sigma^2 \sim N(X\beta,\sigma^2 I)$ and a hierarchical model for the regression coefficient $\beta$ of a form like $$\beta|\sigma^2,\alpha \sim N(\alpha,\sigma^2 \bar{V})\\ \alpha \sim N(\bar{\alpha},\bar{V}_\alpha)\\ \sigma^2\sim G(v_1,v_2) $$ I would like to know whether there is some functional form for the hyper parameters such that the marginal posterior distribution $$\beta|D$$ can be derived in a closed form solution (e.g. I know that this is possible with deterministic $\alpha$)? Any comments, references, ideas are welcome, thank you!

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    $\begingroup$ Integrating out the $\alpha$ means $\beta\sim N(\bar{\bar\alpha},\sigma^2\bar V+\bar{V}_\alpha)$. $\endgroup$ – Xi'an Oct 26 '15 at 14:14
  • $\begingroup$ Thanks @Xi'an! Do you mean that after integrating out the joint posterior distribution of $(\beta,\sigma^2)$ follows a normal-gamma distribution with parameters given by $\bar{\bar{\alpha}},\sigma^2\bar{V}+\bar{V}_\alpha,\bar{v_1}$ and $\bar{v_2}$? And subsequently, integrating out $\sigma^2$ leads to a multivariate $t$ distribution for $\beta$? Or do I misunderstand something? $\endgroup$ – muffin1974 Oct 26 '15 at 14:34
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    $\begingroup$ No you cannot integrate out $\sigma^2$ nicely because of the $\sigma^2\bar{V}+\bar{V}_\alpha$ format... $\endgroup$ – Xi'an Oct 26 '15 at 14:36
  • $\begingroup$ Can you add what the term $\bar{\bar{a}}$ is? When I understand it correctly, I have for the joint posterior distribution $$\beta,\alpha,\sigma^2\propto N(X\beta,\sigma^2I)N(\bar{\alpha},\bar{V}_\alpha)N(\alpha,\sigma^2\bar{V})G(\bar{v}_1,\bar{v}_2)$$. This is a Normal-Gamma distribution with terms (for simplicity) $\beta,\alpha,\sigma^2|D\propto NG(\mu,\Sigma,\bar{v}_1,\bar{v}_2)$. Your point is that integrating over $\alpha$ leads to $\beta,\sigma^2|D \propto NG(\bar{\bar{\alpha}},\sigma^2\bar{V}+\bar{V}_\alpha,\bar{v}_1,\bar{v}_2)$? $\endgroup$ – muffin1974 Oct 26 '15 at 17:07
  • $\begingroup$ Sorry, typo: $\bar a$ not $\bar{\bar a}$. $\endgroup$ – Xi'an Oct 26 '15 at 17:16

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