1
$\begingroup$

I have a bayesian belief network with 4 binary variables $A, B, C, D$. I now need to proof that for joint probability distributions factorized according the Bayesian network given below the conditional independency $A⊥⊥D|C$ always holds.

This by using factorization. Now I know that

$$p(A,B,C,D) = p(A)p(B)p(C|A,B)p(D|C)$$

but how can I go from that to proving that $p(A,D|C) = p(A|C)p(D|C)$ (definition conditional independence)

Bayesian belief network

$\endgroup$
1
$\begingroup$

I would do it like this: \begin{align} P(A,D|C) &= P(A|D,C)P(D|C)\\ &=\frac{P(D,C|A)P(A)}{P(D,C)}P(D|C)\\ &=\frac{P(D|C,A)P(C|A)P(A)}{P(D|C)P(C)}P(D|C)\\ &=\frac{P(C|A)P(A)}{P(C)}P(D|C)\\ &=P(A|C)P(D|C) \end{align}

Where passages are just Bayes theorem and axioms of probability. I also used that $P(D|C,A) = P(D|C)$ because of redundancy of the conditioning.

$\endgroup$

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.