Implications of mean independence

Question

Let $U$ be a random variable with mean $0$. Take other two random variables $X,Y$. Assume $$ (1)\quad E(U|X,Y)=0. $$ I believe (1) implies $$ E(U\cdot X)=E(U \cdot Y)=0. $$ Does (1) imply $$ E(U \cdot f(X,Y))=0\quad ? $$ where $f$ is any function of $X,Y$? For instance, does (1) imply $$ E(U \cdot X \cdot Y)=0 $$

Answer 1

Yes.

If $\mathcal A$ denotes a $\sigma$-algebra and $Z$ is a random variable that is $\mathcal A$-measurable then:$$\mathbb E[UZ]=\mathbb E[\mathbb E[U|\mathcal A]Z]$$So if $\mathbb E[U|\mathcal A]=0$ then it follows directly that also $\mathbb E[UZ]=0$.

You can apply that here on the $\sigma$-algebra that is generated by the random variables $X$ and $Y$ and is mostly denoted as $\sigma(X,Y)$.

Beware that $\mathbb E[U|X,Y]$ is actually a notation for $\mathbb E[U|\sigma(X,Y)]$.

Then for Borel measurable $f:\mathbb R^2\to\mathbb R$ we find indeed: $$\mathbb E[Uf(X,Y)]=0$$

You state the condition that $U$ has mean zero, but that is a consequence of your statement $(1)$.

This because:$$\mathbb EU=\mathbb E[\mathbb E[U|X,Y]]$$where the RHS takes value $0$ as a consequence of statement $(1)$.