Does conditional symmetry imply mean independence? 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?

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 $E[Q|X]=0$.
