When does $E(Y\mid X_1,X_2)=E(Y\mid X_2)$ imply $Y\perp X_1\mid X_2$? $Y\perp X_1\mid X_2$, by which I mean that $Y$ is independent of $X_1$ conditional on $X_2$, implies $E(Y\mid X_1,X_2)=E(Y\mid X_2)$. However, the reversed implication does not hold in general since conditional expectations do not in general fully characterize conditional distributions. Nevertheless, under certain assumptions, the reversed implication is true. For example, suppose $Y$ is a dummy variable. Then the distribution of $Y\perp X_1\mid X_2$ is characterized by $E(Y\mid X_1,X_2)=P(Y=1\mid X_1,X_2)$, which gives the reversed implication. But I believe this is not the only example. So, when exactly does $E(Y\mid X_1,X_2)=E(Y\mid X_2)$ imply $Y\perp X_1\mid X_2$? For example, if $E(Y\mid X_1,X_2)=E(Y\mid X_2)$ imply $Y\perp X_1\mid X_2$, does this imply that some property about the random variables is necessarily satisfied?
I found this exercise in a book on causal inference, and I was not sure what kind of answer the author was after.
One idea is that whether the implication is true or not is related to if moments other than the first moment of $Y$ given $(X_1,X_2)$ do not depend on $X_1$.
The question is essentially equivalent (with an added conditional) to the following one:
The implicancy $E[X | Y ] = E[X] \implies X \perp Y$ (where $\perp$ denotes independence) is in general false. Under which additional assumptions it's true?
There is no general answer to this. Practically speaking, there are two important cases:

$(X,Y)$ are jointly gaussian (not the same as &quot;$X,Y$ are both gaussian&quot;!)
$(X,Y)$ are both Bernoulli (takes values in $\{0,1\} \times \{0,1\} $)

This, because in both cases above no correlation implies independence (and no correlation is weaker than $E[X | Y ] = E[X]$).
Assuming the first case is the most typical/important, going back to the original question, we can assert:
If $(Y,X_1)$ conditioned on $X_2$ are jointly gaussian, then $E(Y\mid X_1,X_2)=E(Y\mid X_2) \implies Y\perp X_1\mid X_2$
Or slightly simpler and more in particular:
If $(Y,X_1,X_2)$ are jointly gaussian, then $E(Y\mid X_1,X_2)=E(Y\mid X_2) \implies Y\perp X_1\mid X_2$
The question is essentially equivalent (with an added conditional) to the following one:
The implicancy $E[X | Y ] = E[X] \implies X \perp Y$ (where $\perp$ denotes independence) is in general false. Under which additional assumptions it's true?
There is no general answer to this. Practically speaking, there are two important cases:

$(X,Y)$ are jointly gaussian (not the same as &quot;$X,Y$ are both gaussian&quot;!)
$(X,Y)$ are both Bernoulli (takes values in $\{0,1\} \times \{0,1\} $)

This, because in both cases above no correlation implies independence (and no correlation is weaker than $E[X | Y ] = E[X]$).
Assuming the first case is the most typical/important, going back to the original question, we can assert:
If $(Y,X_1)$ conditioned on $X_2$ are jointly gaussian, then $E(Y\mid X_1,X_2)=E(Y\mid X_2) \implies Y\perp X_1\mid X_2$
Or slightly simpler and more in particular:
If $(Y,X_1,X_2)$ are jointly gaussian, then $E(Y\mid X_1,X_2)=E(Y\mid X_2) \implies Y\perp X_1\mid X_2$