Could someone please tell me, where this comes from:

$$ p(\beta, \sigma^2 | y, \tau) \propto p(y | \beta, \sigma^2) p(\beta | \tau) p(\sigma^2). $$

Thank you!

  • $\begingroup$ Added some more explanation $\endgroup$
    – Glen_b
    Dec 16, 2012 at 23:39

1 Answer 1


It looks like a Bayesian posterior for the parameters of some model, possibly a regression, which is proportional to the likelihood times the prior for the parameters. The prior for $\beta$ appears to depend on a scale parameter, $\tau$ - at least that's what I'd guess, though there are other possibilities.

$\beta$ may well be a vector of parameters.

That the LHS is proportional to the RHS follows from Bayes' theorem.

Here's an outline of a way to get from LHS to RHS:

Consider: $P(CD) = P(C|D)P(D)$ ... (1) (we will use this more than once)

Hence $P(C|D) \propto P(CD)$ ... (2)

So from (2) $P(\beta, \sigma^2|y) \propto P(y, \beta, \sigma^2)$.

Applying (1) to the RHS:

$P(y, \beta, \sigma^2) = P(y| \beta, \sigma^2) P(\beta, \sigma^2)$

Applying (1) to the last term:

$P(y, \beta, \sigma^2) = P(y| \beta, \sigma^2) P(\beta| \sigma^2) P(\sigma^2)$


$P(\beta, \sigma^2|y) \propto P(y| \beta, \sigma^2) P(\beta| \sigma^2) P(\sigma^2)$.

Now condition everything on both sides on $\tau$, then drop the conditioning on it from anything that's independent of $\tau$:

$P(\beta, \sigma^2|y,\tau) \propto P(y| \beta, \sigma^2,\tau) P(\beta| \sigma^2,\tau) P(\sigma^2|\tau)$

$P(\beta, \sigma^2|y,\tau) \propto P(y| \beta, \sigma^2) P(\beta| \sigma^2,\tau) P(\sigma^2)$.

Now ... presumably because the prior for $\beta$ independent of $\sigma^2$, drop the conditioning on it:

$P(\beta, \sigma^2|y,\tau) \propto P(y| \beta, \sigma^2) P(\beta| \tau) P(\sigma^2)$.

  • $\begingroup$ Thank you very much! This is exactly what I didn't see. $\endgroup$
    – far away
    Dec 17, 2012 at 17:14

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.