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I am referring a Bayesian Net Example from here.

Bayesian Net Example

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.

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

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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.

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

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