If $X$ is independent of $Z$, and $Y$ is independent of $Z$, but $X$ and $Y$ are not independent, is the following true: \begin{align*} P(X|Y, Z) &= \frac{P(X, Y, Z)}{P(Y, Z)}\\ &= \frac{P(X, Y)P(Z)}{P(Y)P(Z)}\\ &= \frac{P(X, Y)}{P(Y)}\\ &= P(X|Y) \end{align*}
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$\begingroup$ Marginally independent does not imply jointly independent: see this earlier answer. $\endgroup$– Xi'anCommented Feb 1, 2021 at 5:37
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$\begingroup$ @Xi'an OP did not say the distributions are marginally independent. I think he/she used "independence" in the common sense. $\endgroup$– stansCommented Feb 1, 2021 at 5:41
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$\begingroup$ This is only true if $X$ is conditionally independent of $Z$ given $Y$. $\endgroup$– mhdadkCommented Feb 1, 2021 at 5:42
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1 Answer
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The following is a counterexample
Say the following events (for binary values of X, Y and Z) have equal probabilities.
X Y Z probability 0 1 1 1/4 1 0 1 1/4 0 0 0 1/4 1 1 0 1/4Then $P(X=1) = P(X=0) = 0.5$ and $P(Y=1) = P(Y=0) = 0.5$ independent from $Z$.
But $P(X| Y,Z) = \text{XOR}(Y,Z)$ and not independent from $Y,Z$ (where $\text{XOR}$ refers to the exclusive or)
answered Feb 1, 2021 at 10:46
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$\begingroup$ Would $P(X|Y, Z) = P(X|Y)$ if $(X, Y)$ is jointly independent of $Z$? In the counterexample above, it is assumed that $X$ and $Y$ are marginally independent of $Z$, correct? $\endgroup$– AdrianCommented Feb 1, 2021 at 15:30