Consequences of being both independent and conditionally independent? Suppose $A$, $B$, and $C$ are random variables. If $A$ and $B$ are independent, and they are also conditionally independent given $C$, can we conclude that either $A$ and $C$ are independent or $B$ and $C$ are independent? Or is there a case where given the constraints, $C$ can still be dependent to both $A$ and $B$?
This question was inspired from Bayesian network configurations. I was trying to prove the former with no luck, and I wasn't able to find anything online that helped, so I figured that it might not even be true. Could someone please provide either a proof or a counterexample (or some other reasoning to why it's false)?
 A: What if $C \in \{1, 2, 3, 4\}$ equally likely and
\begin{align}
C=1 \implies A=0, B=0\\
C=2 \implies A=0, B=1\\
C=3 \implies A=1, B=0\\
C=4 \implies A=1, B=1
\end{align}
A: Here is a family of counterexamples:
Let $A$ and $B$ be independent, both almost surely non-zero, both taking positive as well as negative values with non-zero probability.
Let $C$ equal $1$ if $(A,B)$ lies in the first quadrant, $2$ if $(A,B)$ lies in the second quadrant, etc.
$A$ and $C$ are not independent, and neither are $B$ and $C$, but $A$ and $B$ are still independent conditional to $C$.
A: We need the things $A$ teaches us about $C$ have no relevance to $B$ and vice versa. So let $C$ take four values uniformly, coded as the corners of the square. And then, conditional on $C,$ let $A$ and $B$ be constant $0$ or $1$ (thus conditionally dependent, with $A$'s value depending on whether $C$ is on the top or bottom of the square, and $B$'s depending on whether $C$ is on the left or right side of the square. Then $A$ and $B$ will be unconditionally independent as well.
