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I am trying to prove that conditional independence does not imply independence, ie that $P(A|C)P(B|C)=P(A \cap B|C) \nRightarrow P(A \cap B)=P(A)P(B)$

I guess I need a counter-example but I am struggling to find a way of homing in onto one.

So far I have tried drawing Venn diagrams, and I can see that there is no reason why the sizes of the relevant intersections should multiply as implied by the above, but I am not sure how to proceed from there.

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Let $A,B$ be events that are not independent.

Observe that $P(A\mid A)P(B\mid A)=P(B\mid A)=P(A\cap B\mid A)$ showing that there is conditional independence wrt $A$.

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If $C = A^c,$ then as long as the $P(A) \neq 1,$ then the conditional independence equation will just be $0=0$ while we've learned nothing about whether $A$ and $B$ are independent.

That is, take any two $A, B$ that aren't independent, then $P(A) \neq 1$ (why?) and setting $C=A^c$ implies $A|C, B|C$ are independent.

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  • $\begingroup$ very true !! thank you ! $\endgroup$ – user3203476 Sep 29 '19 at 4:50

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