Having: $$y\sim N_n(X\beta, \sigma^2 I_n)$$ with prior distributions:

$$\beta\sim t_\nu(\beta_0, B_0)$$ and $$\sigma^2 \sim IG(\alpha_0/ 2, \delta_0/2)$$

What would be the conditional posterior of $\beta|\sigma^2, y, x$

I´m trying the following but don´t know if I´m in the right track:

$$\beta|\sigma^2, y, x \propto |B_0|^{N /2} \left(1+\frac{1}{\nu}(\beta-\beta_0)'B_0^{-1}(\beta- \beta_0)\right)^{-(\nu+k)/2}\times exp \left(-\frac{1}{2}\sum \sigma^{-2}(y_i-x_i'\beta)^2 \right)$$

But don't know if this would be a known distribution.

Also I´m thinking in doing: $$\beta\sim N(\beta_0, \lambda^{-1}B_0)$$ $$\lambda\sim G(\nu /2, \nu /2)$$ But I don't know if this would be ok.

  • $\begingroup$ This sounds like an homework question: can you add the self-study tag if so and in any case tell us where you are stuck with this derivation? Note that the prior on $\sigma^2$ is not useful for that question. $\endgroup$
    – Xi'an
    Dec 15, 2015 at 8:39
  • $\begingroup$ (+1) Indeed, this is the right track! Use the Normal-Gamma decomposition, condition on the latent variable to merge both Normal terms together and end up with a compound Normal distribution, expressed as an integral on $\lambda$ that has no closed form expression. $\endgroup$
    – Xi'an
    Dec 15, 2015 at 15:10
  • 1
    $\begingroup$ So $\beta|y,x,\sigma^2,\lambda \sim N(\bar{\beta}, B_1)$ with $\bar{\beta}=B_1\left[\sigma^{-2}X'y +\lambda^{-1} B_0^{-1}\beta_0 \right]$ and $B_1=(\lambda B_0^{-1}+\sigma^{-2} X'X)^{-1}$? $\endgroup$ Dec 15, 2015 at 15:52
  • $\begingroup$ Exactly! You can post it as an answer to your own question! $\endgroup$
    – Xi'an
    Dec 15, 2015 at 22:04

1 Answer 1


The posterior conditional distribution:

$$\beta|y,x,\sigma^2,\lambda \sim N(\hat{\beta}, B_1)$$ with $$\hat{\beta}=B_1[\sigma^{-2}X'y+\lambda^{-1}B_0^{-1}\beta_0 ]$$ and $$B_1=(\lambda B_0^{-1}+\sigma^{-2}X'X)^{-1}$$


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