If I am trying to compute the probability $P(Z\mid(A,B))$ using Bayes' Theorem, how would I expand the right-hand side, particularly the evidence $P(A,B)$ in the denominator?


1 Answer 1


I'm guessing that by $(A,B)$ you mean $A\text{ and }B$.

What you do with this might depend on what information you've got.

$$ \Pr(Z\mid A\ \&\ B) = \frac{\Pr(A\ \&\ B\mid Z)\Pr(Z)}{\Pr(A\ \&\ B\mid Z)\Pr(Z) + \Pr(A\ \&\ B\mid\text{not }Z)\Pr(\text{not }Z)}. $$

So there's a question of whether to do something beyond that. If you know, for example $\Pr(B\mid Z)$ and $\Pr(A\mid B\ \&\ Z)$, then you could replace $\Pr(A\ \&\ B\mid Z)$ with the product of those two probabilities. But if you know $\Pr(A\mid Z)$ and $\Pr(B\mid A\ \&\ Z)$, then you could use the product of those instead.

  • $\begingroup$ Thanks for the answer. I was looking for a webpage that showed a right-hand-side with joint probability evidence but couldn't find one. Now, about going beyond the formula, I was reading about Naive Bayes classifiers, and I think you can simplify Pr(A & B | Z) to be Pr(A | Z) x Pr(B | Z) by the chain rule of probability and assuming conditional independence. $\endgroup$ Aug 19, 2013 at 19:27
  • $\begingroup$ That works of $A$ and $B$ are conditionally independent given $Z$, but not otherwise. $\endgroup$ Aug 20, 2013 at 2:03

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