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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*}

Devon
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  • Marginally independent does not imply jointly independent: see this earlier [answer](https://stats.stackexchange.com/a/276348/7224). – Xi'an Feb 01 '21 at 05:37
  • @Xi'an OP did not say the distributions are marginally independent. I think he/she used "independence" in the common sense. – stans Feb 01 '21 at 05:41
  • This is only true if $X$ is conditionally independent of $Z$ given $Y$. – mhdadk Feb 01 '21 at 05:42

1 Answers1

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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   1/4

    Then $P(X=1) = P(X=0) = 0.5$ 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)

Sextus Empiricus
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  • 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? – Devon Feb 01 '21 at 15:30