# Implications of mean independence

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 $$E(U \cdot X \cdot Y)=0$$

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 $$(1)$$.