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Suppose we have random variables Y, D and X, where Y is independent of D conditional on X (Y⊥D|X). If there is another variable Z=f(X), where f(.) is a measurable real function, my question is: (1) under what conditions can we have Y⊥D|Z ?; (2) do we need the sigma-algebra σ(Z) belongs to σ(X), so σ(Z) is sub-σ-algebra of σ(X)?

This is crucial to casual inference in econometrics and statistics, where we want to know if the conditional independent assumption (CIA) condition can be relaxed.

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  • $\begingroup$ If $f$ is measurable, $\sigma(Z)\subseteq\sigma(X)$ is automatic. $\endgroup$ Commented Oct 28, 2020 at 7:35
  • $\begingroup$ So, we have Y⊥D|Z at this time as long as σ(Z)⊆σ(X) ? $\endgroup$ Commented Oct 28, 2020 at 7:38
  • $\begingroup$ Suppose $Y$ and $D$ are not independent, but conditionally independent on $X$. Let $f$ be constant. Then $Z$ is constant and being independent conditional on $Z$ is the same as being independent. So the answer is no. $\endgroup$ Commented Oct 28, 2020 at 7:51
  • $\begingroup$ If f is not a constant? $\endgroup$ Commented Oct 28, 2020 at 9:06

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A sufficient condition to have $Y⊥D|Z$ is that $f$ is injective. The sharp condition (if $Y$ and $D$ are not specified) is

$(*)$ $\sigma(X)$ should be contained in the completion of $\sigma(Z)$.

If $(*)$ holds, then conditioning on $X$ is equivalent to conditioning on $Z$. If (*) does not hold, then there is an event $A \in \sigma(X)$ such that $0<P(A|Z)<1$. Take $Y=D=1_A$. Then $Y⊥D|X$ but $Y $ and $D$ are dependent given $Z$.

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