Relaxing conditional independent assumption

Question

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.

Answer 1

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 $Z$.