In "A First Course in Bayesian Methods", Hoff writes that:

$p(\theta, \sigma^2|y_1,...y_n) = p(\theta|\sigma^2,y_1,...y_n)p(\sigma^2|y_1,...y_n)$

when describing joint inference for the mean and variance for normal data.

How do we arrive at this starting from $p(A|B) = \frac{p(B|A)P(A)}{P(B)}$?

  • $\begingroup$ You don't use Bayes rule. It is just a joint = conditional times marginal statement, so you just use the definition of a conditional probability (while conditioning on y the entire time). $\endgroup$
    – jaradniemi
    Nov 11 '16 at 20:35
  • $\begingroup$ Given that joint=conditional x marginal comes from Bayes rule, I thought there may be a way to apply it here. Put another way, what happens if you do try and apply Bayes rule. Is there a way to get the desired result? Thanks. $\endgroup$
    – julieth
    Nov 11 '16 at 21:19
  • 1
    $\begingroup$ That is incorrect: "joint = conditional x marginal" does not come from Bayes' theorem, it is a rearrangement of the definition of conditional density. Bayes' theorem follows directly from this definition, not the other way around. $\endgroup$
    – Chris Haug
    Nov 12 '16 at 2:03

I think you only need to express the conditional with the marginal, i.e:

$P(A|B) =\frac{P(A,B)}{P(B)} \Rightarrow P(A,B)=P(A|B)P(B) $

$P(A,B)$ is represented by $P(\theta, \sigma^2|y_1,...y_n)$

$P(A|B)$ is represented by $P(\theta|\sigma^2,y_1,...y_n)$

$P(B)$ is represented by $p(\sigma^2|y_1,...y_n)$


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