I have a problem that I can't work out

I've two conditional independent A,B such as

$P(A,B|C) = P(A|C)P(B|C)$

Now I've to find posterior formula for:

$P(C | A,B)$, now what I got was pretty straigthforward application of bayes:

$\frac{P(B|C)P(A|C)P(A)}{P(A\cap B)}$

With few variants (e.g. get an intersection on numerator)

but I can't get the lecturer solution that is:


Any help?

(note is not homework, but self studying on some pdfs)


1 Answer 1


Note that from the assumption you get: $\mathbb P(B|A,C)=\mathbb P(B|C)$. Therefore: $$ \mathbb P(C|A,B)\mathbb P(B|A)=\mathbb P(B,C|A)=\mathbb P(C|A)\mathbb P(B|C,A)=\mathbb P(C|A)\mathbb P(B|C). $$


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.