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One property of the conditional independence relation is

if $X\perp\kern-5pt\perp Y \mid Z$ and $X\perp\kern-5pt\perp Z \mid Y$ then $X\perp\kern-5pt\perp (Y , Z)$.

However the above does not hold universally, but only under additional conditions - essentially that there be no non-trivial logical relationship between Y and Z. A trivial counterexample appears when $X=Y=Z$ with $P(X=1)=P(X=0)=1/2$.

Can anyone explain this counterexample? I don't know how to see this example violates the above relation.

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1 Answer 1

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Assume that $X$ is not a constant. Given $Z$, $X$ and $Y$ are constants and constants are independent. So $X\perp\kern-5pt\perp Y \mid Z$. Similary, $X\perp\kern-5pt\perp Z \mid Y$. But $X$ is not independent of itself (becasue it is not a constant) so $X$ is not independent of $Y$ and $Z$.

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  • $\begingroup$ How to interpret "X is not independent of itself?" $\endgroup$ Dec 12, 2023 at 0:29
  • $\begingroup$ A randon variable is independent of itself if and only if it is a constant. @SofiaDelacruz $\endgroup$ Dec 12, 2023 at 4:44

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