Let $X$ and $Y$ be random variables taking values in the same finite abelian group $G$, so that we can define their sum $X+Y$. Let $(X_1, Y_1)$ and $(X_2, Y_2)$ be conditionally independent copies of $(X,Y)$ relative to $X+Y$; that is, for all $z\in G$, the random variables $(X_1, Y_1 | X+Y = z)$ and $(X_2, Y_2 | X+Y = z)$ are independent and both have the same distribution as $(X,Y | X+Y = z)$. From this can we conclude that $X_1$ and $Y_2$ are conditionally independent relative to $X+Y$ as well? It seems like something that should be true, but perhaps my intuition is flawed because I don't see an obvious proof.
Thanks!
The question translates to whether $$P(X_1 = x_1, Y_2 = y_2 | X+Y = z) = P(X_1 = x_1 | X+Y = z) \cdot P(Y_2 = y_2 | X+Y = z)$$ for all $z \in G$.
For example, in the specific instances where $X$ and $Y$ are to take values in an abelian group $G$ and when conditional on the sum $X+Y$, some kind of algebraic properties would come up. For instance, in an abelian group, the operation (here, addition) is commutative, which might affect the way the sum would interact with the conditioning.
Since $(X_1, Y_1)$ and $(X_2, Y_2)$ are conditionally independent given $X+Y=z$, think of what that means in the distribution of $X_1+Y_1$ and $X_2+Y_2$.
Proving conditional independence without any specific structures for $G$, or the distributions of $X$ and $Y$, might be hard at times. Think of easy examples or counterexamples that would either confirm or disprove the claim. For instance, if $G$ is a finite group, does that simplify the analysis?