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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.