Variable elimination, Bayesian network

Question

I'm self learning the topic variable elimination and try to solve some questions see if I fully understand the concept. But I'm stuck on this problem's second question for hours. Base on my understanding, if we eliminate a variable then we need to create a new factor which is sum of product of all probabilities that the variable involved. Maybe I didn't describe it precisely but for instance, in question $1$ we have, $$P(A = T, B = T, C = T, E = T, F= T)$$ and since $D$ is missing, we have $$P(A = T, B = T, C = T, E = T, F= T) $$

$$= P(A=T)P(B=T|A=T)P(C=T) \times \sum_D P(D|A=T,B=T,C=T)P(E=T|D)P(F=T|D)$$

If we plug in the corresponding number we will get the solution $0.11088$, this seems straightforward to me. However I'm having tough time to calculate $P(E = T, F = T)$ by eliminating $B$, $A$,and $C$. My approach as, $$P(E =T, F= T) = \sum_D P(E = T| D)P(F = T|D) \sum_C P(C)P(D|A, B ,C) \sum_A P(A)P(D|A,B,C) \sum_B P(B|A) P(D|A,B,C)$$

I don't know if it makes sense, but this looks like a extremely long computation to me, and I couldn't come up a solution as the answer shown. I wonder if anyone could find the same solution and express\explain the full computation would be highly appreciated.

Answer 1

Base on my understanding, if we eliminate a variable then we need to create a new factor which is sum of product of all probabilities that the variable involved.

$\def\={\hspace{0.25ex}{=}\hspace{0.25ex}}\def\P{\operatorname{\sf P}}$Yes, this uses the Law of Total Probability, in your notation: $\P(X\=x,Z\=z)=\sum_{Y}\P(X\=x,Y,Z\=z)$

So $$\small{\P(A\=T,B\=T,C\=T,E\=T,F\=T)={{\P(A\=T)\P(B\=T\mid A\=T)\P(C\=T)\sum_D\P(D\mid A\=T,B\=T,C\=T)\P(E\=T\mid D)\P(F\=T\mid D)}}}$$

And likewise $$\small{\P(E\=T,F\=T)\\=\sum_D\P(E\=T\mid D)\P(F\=T\mid D)\sum_C\P(C)\sum_A\P(A)\sum_B\P(B\mid A)\P(D\mid A,B,C)}$$

Note: You must arrange the factors so all occurrences of the variable being eliminated are to the right of its summation symbol, and each factor only occurs exactly once. Notably you had too many occurrences of $\P(D\mid A,B,C)$ which will surely have thrown off your calculations.

This will still be a tedious computation, as you will be summing sixteen terms, which are each products of five factors. No helping that.