Let $A,B,C$ be a sequence of exchangeable random variables, i.e. $P(A=a,B=b,C=c) = P(B=b,A=a,C=c)=P(C=c,B=b,A=a)$
and so on for any permutation of $\{A=a,B=b,C=c\}$.
Is $P(A=a,B=b|C=c) = P(A=a|C=c)P(B=b|C=c)$?
I'm working through a proof that assumes this, but I can't seem to prove this simple result (and even suspect it isn't true).
I know we want to show the following equality $\frac{P(A=a,B=b,C=c)}{P(C=c)} = \frac{P(A=a,C=c)}{P(C=c)} \times \frac{P(B=b,C=c)}{P(C=c)}$. But I'm not sure how.
It is not true that $A$ and $B$ are conditionally independent given $C$. For example, shuffle a deck of three cards, two of which are black and one of which is red, and let $A$ (resp. $B,C$) equal $1$ when the first (resp. second, third) card is red, and $0$ otherwise. Then $A,B,C$ are exchangeable, but neither two are independent given the third.