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If X and Y two independent events, and Z is a condition that determines the probabilities of X and Y, does $P(X\cap Y\mid Z) = P(X\mid Z) \cdot P(Y\mid Z)$?

For example, X could be the event of flipping a coin twice and the first time landing on heads, while Y could be the event flipping a coin twice and the second time landing on tails. Then Z could be a condition that states the fairness of the given coin. So $P(X\mid Z) = .5$ and $P(Y\mid Z) = .5$, while $P(X\mid Z^{c})=.8$ and $P(Y\mid Z) = .2$

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  • $\begingroup$ PS: Your example has $X,Y$ conditionally independent when given $Z$, but $X,Y$ are not independent. $\endgroup$ – Graham Kemp Sep 21 '17 at 23:35
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Independence does not imply conditional independence. Let $X$ and $Y$ be the event that independent coin flips are heads and $Z$ be the event that either of them is heads (i.e. $Z = X\cup Y$.) Then $P(X|Z) = P(Y|Z) = \frac{2}{3}$ and $P(X\cap Y|Z) = 1/3.$

(Nor does conditional independence imply independence.)

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No, it's not true in general. Independence does neither imply nor is implied by conditional independence. See here for a few examples.

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