Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm going through a revision paper and looking at the solutions and I come across this.

Given a Bayesian Network (sorry I cannot post images):

$A$ and $B$ are parents of $C$. And $C$ is parents of $D$ and $E$.

Solution says:

$\displaystyle P(A | D,B) = P (D | A,B) \frac{P (A | B)}{P (D | B)}$

Can someone explain to be how this happens? I tried joint probabilities and Bayes rule but in the end got something like this:

$P(D,A,B) = P(A,D,B)$ (by solving both sides of the equation).

This does not really make sense to me. As from what I know, for Bayesian Networks, $P(A,B,C) \ne P(C,B,A)$ for example.

Can someone correct me/help me out here?

Thanks.

share|improve this question
    
related question: math.stackexchange.com/questions/142558/… –  amcnabb Aug 16 '13 at 19:12

2 Answers 2

No Bayesian network is needed here, the identity is completely general. To wit, $$ \mathbb P(A|DB)=\frac{\mathbb P(ADB)}{\mathbb P(DB)}=\frac{\mathbb P(D|AB)\mathbb P(AB)}{\mathbb P(D|B)\mathbb P(B)}, $$ and $\mathbb P(AB)=\mathbb P(A|B)\mathbb P(B)$, hence $$ \mathbb P(A|DB)=\frac{\mathbb P(D|AB)\mathbb P(A|B)}{\mathbb P(D|B)}. $$

share|improve this answer

\begin{align} RHS&=\dfrac{P(D|AB)\times P(A|B)}{P(D|B)}\\ &=\dfrac{P(ADB)\times P(AB)\times P(B)}{P(AB)\times P(B)\times P(BD)}\\ &=\dfrac{P(ADB)}{P(BD)}\\ &=P(A|DB) \end{align}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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