CSE 730 Markov Blanket Example

In class today (10/2), there was a question about Markov blankets and conditional independence assumptions. Let's take the following Bayes net:

The claim was that a variable is conditionally independent of all other variables given its Markov blanket (parents, children, and children's parents). This means that B is conditionally independent of F given A, C, and E.

This can only be true if P(B|A,C,E,F) = P(B|A,C,E).

P(B|A,C,E,F) = P(B,A,C,E,F)/P(A,C,E,F)

Now, given the conditional independence assumptions in the network, we can break these joint probabilities down:

P(B,A,C,E,F)=P(A)P(C)P(B|A,C)P(E|B)P(F|E)
P(A,C,E,F)=sum_b P(A)P(C)P(b|A,C)P(E|b)P(F|E)

Given these two equations, and after some cancelling, we can say:
P(B|A,C,E,F) = P(B|A,C)P(E|B) / sum_b P(b|A,C)P(E|b)

Now, let's compute P(B|A,C,E):

P(B|A,C,E) = P(B,A,C,E) / P(A,C,E)
P(B,A,C,E) = P(A)P(C)P(B|A,C)P(E|B)
P(A,C,E,F)=sum_b P(A)P(C)P(b|A,C)P(E|b)
P(B|A,C,E) = P(B|A,C)P(E|B) / sum_b P(b|A,C)P(E|b)

Thus, P(B|A,C,E,F) = P(B|A,C,E). You can also see that P(B|A,C,E) does not equal P(B|A,C).

In the homework, I will ask you to prove something similar about the Markov blanket.

Eric Fosler-Lussier
Last modified: Mon Oct 11 19:41:58 EDT 2004