I have 4 random variables A, B, C, D. I know that the joint is $$p(A,B,C,D) = p(A)p(B|C,A)p(C)p(D|B,C) $$ And I want to prove (if true) that $$ A \perp D | B $$
I have tried this: $$p(A,D,B) = \sum_C p(A,B,C,D) = \sum_C p(A)p(B|C,A)p(C)p(D|B,C) = \sum_C p(A)p(B,C|A)p(D,C|B) = p(A)p(B|A)p(D|B)$$ Then $$p(A,D|B) = \frac{p(A,D,B)}{p(B)} = \frac{p(A)p(B|A)}{p(B)}p(D|B) = p(A|B) p(D|B)$$ My question is: is the marginalization correct? I know it works for $p(X)=\sum_Y p(X,Y)$, but does it work in the same way in this case?