Recall that
$$ \textrm{Cov}(X, Y | Z) = E(XY|Z) - E(X|Z)E(Y|Z) $$
and note that quite trivially we have
$$ X \perp \!\!\! \perp Y | Z \Longrightarrow \textrm{Cov}(X,Y|Z) = 0 \Longrightarrow E \textrm{Cov}(X,Y|Z) $$
I have found examples where $X$ and $Y$ are not conditionally independent given $Z$ but the conditional covariance is zero ($Y = X + \varepsilon_Y$, $Z=X+Y+\varepsilon_Z$ where $X, \varepsilon_Y, \varepsilon_Z \sim \mathcal{N}(0,1)$ independently does the job).
Can anyone think of an example where $X$ and $Y$ are not conditionally independent given $Z$, the conditional covariance is non-zero but the expected conditional covariance is zero? Preferably $X$ and $Y$ should hav densities wrt. Lebesgue but any examples are welcome.
Is it even possible to find an example like this?
EDIT:
I think maybe I've found a solution where $X$ and $Y$ are independent uniforms on $[-1,1]$ and $Z = XY$. Then $E(XY | Z) = XY$ and I think $E(X|Z)=E(Y|Z)=0$ but I have a hard time showing the last bit. Any ideas?
A simple example might be $Z$ and $X$ independently and uniformly distributed on $[-1,1]$, and $Y=XZ$. Then $E(XY|Z)=E(X^2Z|Z)=Z/3$, $E(X|Z)=0$, $E(Y|Z)=E(XZ|Z)=0$, so $\text{Cov}(X,Y|Z)=Z/3$.
But the expected value of the covariance is $E(Z/3)=0$.