Bayes rule and conditional independence

I have two conditionally independent random variables $A$, $B$ such that $$P(A,B\mid C) = P(A\mid C)P(B\mid C) .$$ I have to find posterior formula $P(C \mid A,B)$.

My result with a straigthforward application of Bayes rule is $$P(C \mid A,B) = \frac{P(B\mid C)P(A\mid C)P(A)}{P(A\cap B)} .$$ with few variants (e.g. get an intersection on numerator).

But I can't get the lecturer's solution that is $$\frac{P(B\mid C)P(C\mid A)}{P(B\mid A)} .$$

• In your formula, what happens if you divide numerator and denominator by $P(A)$ so that in the numerator $P(A)$ disappears while the denominator becomes $P(A\cap B)/P(A)$? – Dilip Sarwate May 14 '14 at 21:53
• Ha! Thanks a lot @DilipSarwate ! Since they're then both correct, is the lecturer one more common somehow? – JTulip May 14 '14 at 22:09
• Well, your lecturer's answer differs from yours in that it has $P(C\mid A)$ while you have $P(A\mid C)$ so I am wondering which one of them is correct, or if one has a typographical error in it. – Dilip Sarwate May 14 '14 at 22:59

$P(C \mid A,B) = \dfrac{P(A,B\mid C) P(C)}{P(A,B)}= \dfrac{P(A\mid C)P(B\mid C) P(C)}{P(A,B)}$
• Perhaps you could also include the simplification that results in the "professor's answer"? I am curious to see how $\dfrac{P(A\mid C)P(B\mid C) P(C)}{P(A,B)}$ can be manipulated into $\dfrac{P(B\mid C)P(C\mid A)}{P(B\mid A)}$, or, assuming that there is a typo in the OP's statement, into $\dfrac{P(B\mid C)P(A\mid C)}{P(B\mid A)}$ – Dilip Sarwate May 15 '14 at 2:19
• @DilipSarwate Perhaps simplification was the wrong word? This was self-study, so I will outline the remaining steps. Combine $P(A \mid C)$ with $P(C)$. Then divide numerator and denominator by $P(A)$ – jsk May 15 '14 at 2:27