Does conditional symmetry imply mean independence?

Question

suppose I have two random variables $X$ and $Q$. $Q$ is conditionally symmetrically distributed about zero, i.e., its density satisfying satisfying $f(-q|X=x)=f(q|X=x)$ for every $q\in \Omega_{Q|X}$ and $x\in\Omega_X$, where $\Omega_{Q|X}$ is the conditional support for $Q$ and $\Omega_X$ is the support for $X$. Suppose $E(Q|X=x)$ always exist. Does these conditions imply that $E(Q|X)=0$, i.e., $Q$ and $X$ are mean independent?

Answer 1

Yes.

The symmetry implies $$E[Q|X=x]=E[-Q|X=x],$$ and so $E[Q|X=x]=0$, for every fixed $x$ where those expectations exist. In situations where $Q$ actually has well-defined conditional distributions for every $x$ that's basically the definition of $E[Q|X]=0$.