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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 $$ \frac{P(B\mid C)P(C\mid A)}{P(B\mid A)} . $$

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  • $\begingroup$ 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 $P(A\cap B)/P(A)$? $\endgroup$ – Dilip Sarwate May 14 '14 at 21:53
  • $\begingroup$ Ha! Thanks a lot @DilipSarwate ! Since they're then both correct, is the lecturer one more common somehow? $\endgroup$ – JTulip May 14 '14 at 22:09
  • $\begingroup$ 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. $\endgroup$ – Dilip Sarwate May 14 '14 at 22:59
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It appears your answer is incorrect. A straight-forward application of Bayes' rule would be that

$P(C \mid A,B) = \dfrac{P(A,B\mid C) P(C)}{P(A,B)}= \dfrac{P(A\mid C)P(B\mid C) P(C)}{P(A,B)}$

After a little bit of simplification, you will get the professor's answer.

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  • $\begingroup$ 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 $\dfrac{P(B\mid C)P(A\mid C)}{P(B\mid A)}$ $\endgroup$ – Dilip Sarwate May 15 '14 at 2:19
  • $\begingroup$ @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 $P(A)$ $\endgroup$ – jsk May 15 '14 at 2:27

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