Does $\mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely imply $Y \perp X | Z$

Question

It is clear that $Y \perp X | Z \implies \mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely,but does the reverse implication always hold?

Answer 1

As the other answer mentions, the condition does not suffice. However, if you assume $E[f(Y)|X,Z] = E[f(Y)|Z]$ for all bounded measurable $f$, that does suffice.

Write for any bounded measurable $f,g$ that $$ E[f(Y)g(X) | Z] = E[E[f(Y)g(X)| X, Z]|Z] = E[g(X) E[f(Y)|X,Z]|Z] $$ $$ = E[g(X) E[f(Y)|Z] | Z] = E[g(X) | Z] E[f(Y)|Z]. $$ Thus we have $Y\perp X$ conditional on $Z$.

Answer 2

No.

$\mathbb{E}[Y\mid X] = \mathbb{E}[Y]$ does not imply $Y \perp X $

so conditioning on $Z$ is just as unsatisfactory

Consider the following equally likely sets of values each with probability $\frac18$

X   Y   Z
1   1   3
1   5   3
2   2   3
2   4   3
1   3   5
1   7   5
2   4   5
2   6   5

You find $\mathbb{E}[Y\mid X,Z] = Z = \mathbb{E}[Y\mid Z]$, but $X$ determines the parity of $Y$ so they are not independent in any sense, conditional or otherwise