Is Independence of two events generalised by conditional independence of those two events?

Question

If $\mathbb{P}(A\mid C)* \mathbb{P}(B\mid C) = \mathbb{P}(A \cap B\mid C)$ i.e A and B are independent given C has happened, then can we generalise and say that A and B are independent irrespective of C?

My explanation is that as A and B are independent given C that means given C and A happened , our beliefs about B does not change, i.e we gain no new knowledge about B's happening .

Then removing our old knowledge (i.e conditional C is removed), we have no more knowledge of C's happening . With lesser information , we gain no new insight about B , thus with A happening our beliefs or chance of B happening doesn't change implying A and B are independent in generalised way?

Answer 1

If you are an engineer, you may find the following intuition helpful.

Consider $Z_1, Z_2$ to be independent noise random variables. Let $S$ be a random signal.

Let $Y_1 = S + Z_1$ and $Y_2 = S + Z_2$ be two noisy observations of the signal $S$. These observations can be at different times or using different instruments hence we use different, independent random variables $Z_1, Z_2$ to model the noise in each observation.

Conditioned on $S = s_0$, we have $Y_1 = s_0 + Z_1$ and $Y_2 = s_0 + Z_2$. Hence, $Y_1$ and $Y_2$ are conditionally independent given $S = s_0$.

But it is clear that $Y_1$ and $Y_2$ are not independent.

Answer 2

You cannot conclude that $A$ and $B$ are independent. A simple counter-example: Let $A$ be any event with $0 <P(A)<1$ and take $A=B=C$.

$\mathbb{P}(A\mid C)=1, \mathbb{P}(B\mid C) =1$ and $\mathbb{P}(A \cap B\mid C)=1$ in this case.

Answer 3

No, $P(A|C)P(B|C)=P(A\cap B|C)$ does not imply $P(A)P(B)=P(A\cap B)$. Suppose $P(A|C)=P(B|C)=P(A\cap B|C)=1$. But suppose $P(A)=P(B)=\frac 2 3$ and $P(A\cap B)=\frac 1 3$.