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I have read a proof of markov property (steps showing how $P(A_3|A_1∩A_2) = P(A_3|A_2)$ ) in other question and this are the steps.

\begin{equation} P(A_3|A_1∩A_2)=\frac{P(A_1∩A_2∩A_3)}{P(A_1∩A_2)}=\frac{P(A_1∩A_3|A_2)P(A_2)}{P(A_1∩A_2)}=\frac{P(A_3|A_2)P(A_1|A_2)P(A_2)}{P(A_1∩A_2)}=\frac{P(A_3|A_2)P(A_1|A_2)P(A_2)}{P(A_1∩A_2)}=\frac{P(A_3|A_2)P(A_1∩A_2)}{P(A_1∩A_2)}=P(A_3|A_2) \end{equation}

I think it's correct but nothing special was said over A1, A2 and A3, so, Could them be any 3 events?

Thanks!

Brian Bohe
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Well, the third equation has employed the fact that $A_1$ and $A_3$ are conditionally independent given $A_2$, since there it is used that $$P(A_1\cap A_3\,|\,A_2)=P(A_3\,|\,A_2)\,P(A_1\,|\,A_2),$$ which is clearly not true in general.

OnoL
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