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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?


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related question:… – amcnabb Aug 16 '13 at 19:12

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 $$ \mathbb P(A|DB)=\frac{\mathbb P(D|AB)\mathbb P(A|B)}{\mathbb P(D|B)}. $$

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\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}

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