1
$\begingroup$

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):

$$\require{enclose}\def\circle#1{\enclose{circle} #1} \raise{4ex}\circle A\raise{2.5ex}\searrow\hspace{-6.75ex}\lower{4ex}\circle D\lower{2.5ex}\swarrow\circle C\raise{2ex}\swarrow\raise{4ex}\circle B\hspace{-6.5ex}\lower{2ex}\searrow\lower{4ex}\circle E$$

$A$ 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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
0
$\begingroup$

\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}

$\endgroup$

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.