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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!

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

1 Answer 1

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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:

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

So

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

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:

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

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

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