# Conditional independence in bayesian network

I am referring a Bayesian Net Example from here.

Prob(A=T) = 0.3
Prob(B=T) = 0.6
Prob(C=T|A=T) = 0.8
Prob(C=T|A=F) = 0.4
Prob(D=T|A=T,B=T) = 0.7
Prob(D=T|A=T,B=F) = 0.8
Prob(D=T|A=F,B=T) = 0.1
Prob(D=T|A=F,B=F) = 0.2
Prob(E=T|C=T) = 0.7
Prob(E=T|C=F) = 0.2

In this case how do you solve for P(E=T|A=T) ? Because C is not known E and A are not independent events.

• @gung sorry that was a typo. I meant E and A. – Pdksock Sep 20 '15 at 22:33

A few things.

• The Chain Rule: $$P(E, C | A) = P(E|C,A) \cdot P(C | A)$$
• Conditional Independence: The notation of a Bayes Network implies $$P(E|C,A) = P(E|C)$$ The idea is, if you know that $C$ did or did not happen, it doesn't matter whether or not $A$ happened. So at this point we have: $$P(E,C|A) = P(E|C) \cdot P(C|A)$$
• Marginalization: Removing a variable from a joint distribution (only variables on the left side of the "|") by summing over the possible values they can take. $$P(E|A) = P(E,C=T|A) + P(E,C=F|A).$$

So now you should be able to answer this question. Please let me know if you don't understand any of the steps, or don't see how this applies to your question.

It seems to me that you would need to compute the probability that E = T for each possible value of C, multiply each by the probability of C being that value given A = T, and add everything up. That is

$$\sum_{X}P(E=T|C=X)P(C=X)$$

If every node can be either T or F, that would be

$$P(E=T|C=T)P(C=T|A=T) + P(E=T|C=F)P(C=F|A=T)$$

If I am reading jlimahaverford's answer correctly, I think they end up being the same thing.

Let me know if I am incorrect.

• Yes, my answer is just a justification of this all. You're using the conditional independence implied by the graph structure. – jlimahaverford Sep 29 '15 at 16:33