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I would like to know -- given $P,Q,R$ are three distinct events -- if from $P(R\vert Q,P)= P(R\vert Q)$ follows that $R$ is independent of $P$.

I can see that in order for this to be true it must be that $$P(R\vert Q,P)= \frac{P(R\cap Q\cap P)}{P(Q\cap P)} = \frac{P(R\cap Q) P(P)}{P(Q)P(P)} = P(R\vert Q)$$.

So we must have that $(R\cap Q)\perp P$ and $Q\perp P$ (provided, for this latter conclusion that P is not a subset of Q). But it looks to me like this does not implies that $R\perp P$ in general, is that correct?

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    $\begingroup$ I don't agree with your penultimate sentence - consider the case where $P$, $Q$, and $R$ are all the same event. $\endgroup$
    – fblundun
    Commented Dec 7, 2020 at 13:46
  • $\begingroup$ Ok, got rid of this edge case. $\endgroup$
    – Three Diag
    Commented Dec 7, 2020 at 14:03
  • $\begingroup$ I still don't believe the conclusion $P \perp Q$. For example, suppose we are rolling a die and $P$ is "we roll a 3 or lower", $Q$ is "we roll a 2 or lower", and $R$ is "we roll a 1". Or in general any case where $P$ is always true when $Q$ is true. $\endgroup$
    – fblundun
    Commented Dec 7, 2020 at 14:08
  • $\begingroup$ I see what you mean, I will edit accordingly to rule this case out. Thanks! $\endgroup$
    – Three Diag
    Commented Dec 7, 2020 at 14:18
  • $\begingroup$ That isn't the only problematic case though - for example, suppose $P$ is "we roll a 1", $Q$ is "we roll a 2", and $R$ is "we roll a 1 or a 2". Then $P$ and $Q$ are independent and disjoint. The flaw in your derivation is that if two fractions are equal, it need not follow that their numerators are equal and their denominators are equal. $\endgroup$
    – fblundun
    Commented Dec 7, 2020 at 14:29

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No, it means $R$ is independent of $P$ given $Q$. So, it is conditional independence. Conditional independence doesn't mean that the two events/random variables are independent. Also, if the two events/random variables are independent, it also doesn't mean that they're conditionally independent given some other event/random variable.

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  • $\begingroup$ I get what you mean, but you may want to clarify your wording to make it clear that $R\perp P$ does not hold in general, but there is a specific conditional sense in which it is independent of P (thanks for the name which I had never heard) $\endgroup$
    – Three Diag
    Commented Dec 7, 2020 at 14:06
  • $\begingroup$ @ThreeDiag I've added some more explanation, indicating that neither result can imply the other. $\endgroup$
    – gunes
    Commented Dec 7, 2020 at 14:13

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