Chain rule in conditional probability I'm currently taking a class on computer vision and I have the following problem: I should prove that, if my probability function is given as $p(a,b,c,d)$ and it can be displayed as: $p(a,b,c,d) = p(a) p(d|b,a) p(c|d) p(b).$ We should show that a and b are independent. ($p(a,b)=p(a){}\cdot{}p(b)$) I manged to use the chain rule to get to some other terms, but I end up with: $$p(c|d,b,a){}\cdot{}p(d|a,b){}\cdot{}p(a,b)= p(a) p(d|b,a) p(c|d) p(b),$$ so, how can it be that $$P(b|d,c,a) = P(c|d).$$

In general the statement is not true. It may hold for some models; but in general the only thing you can say is that $$p(a,b,c,d) =p(a)p(b|a)p(c|a,b)p(d|a,b,c)$$ with any other permutation of the variables also valid.

If you set as an assumption that $$p(a,b,c,d) =p(a)p(b)p(c|d)p(d|a,b),$$ Then, we can integrate out (marginalize out) the variables $d$ and $c$ in both expressions and we get that $$p(a)p(b|a) = p(a)p(b),$$ which means that (if the given factorization holds) then $a$ and $b$ are independent.