# For which $C$ is the following true: $P(A \cap B | C)= P(A|C)P(B|C) \iff P(A \cap B) = P(A)P(B)$?

I know that conditional independence does not imply independence, and I know that the converse is also false.

Here I need to show for which events $C$ it is the case that $A$ and $B$ are conditionally independent on $C$ iff $A$ and $B$ are independent $\forall$ events $A, B$.

i.e., for which $C$ is the following true: $P(A \cap B | C)= P(A|C)P(B|C) \iff P(A \cap B) = P(A)P(B)$

I know that the answer is for all $C$ such that $P(C)=1$, but I'm stuck trying to show this.

Informally, equation $P(A \cap B | C) = P(A|C) P(B|C)$ tells you something about what happens when $C$ occurs, but says nothing about what happens when it doesn't.
$P(A) = P(A|C) P(C) + P(A|C^c) P(C^c)$; similarly for $B$ and $A \cap B$. For example, let's suppose $P(A|C^c) = 1/2$ and $P(B|C^c) = 1/2$. $P(A \cap B | C^c)$ could be anything from $0$ to $1/2$. The difference between the two scenarios changes $P(A \cap B)$ by $P(C^c)/2$, leaving $P(A)$ and $P(B)$ the same. If $P(C^c) \ne 0$, and $P(A \cap B) = P(A) P(B)$ in one of the two scenarios, it won't be in the other.