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We have three events $A$, $B$ and $C$ in question, and given appropriate priors, we derive the posterior $\Pr(A|B)$. Now we want to derive a 'second-order' posterior $\Pr(A|B,C)$ by using the 'first-order' posterior $\Pr(A|B)$ as the prior.

First of all, does this mean that $\Pr(A|B,C)$ is the same as $\Pr((A|B)|C)$? If so, is the following correct:

$$\Pr(A|B,C)=\frac{\Pr(C|(A|B))\cdot\Pr(A|B)}{\Pr(C)}$$

and how do we derive $\Pr(C|(A|B))$?

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    $A|B$ is not an event, and conditioning on $A|B$ makes no sense. What _is_ true is that $$P(A|B,C)=\frac{P(A,B,C)}{P(B,C)}=\frac{P(C|A,B)P(A,B)}{P(C|B)P(B)}=\frac{P(C|A,B)}{P(C|B)}P(A|B)$$2012-02-29
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    @DilipSarwate, and what is $\Pr(C|A,B)$?2012-02-29
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    $\Pr(C|A,B)$ means the probability that the event $C$ occurs, given that the two events $A$ and $B$ both occur.2012-02-29
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    @Henry, I understand the meaning of it, but how do I derive it mathematically? I currently know $\Pr(A|B)$, $\Pr(C|B)$, $\Pr(B|A)$, $\Pr(C|A)$ and $\Pr(A)$.2012-02-29

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