Characterizing Identification of a Deconvoluted Density Without Independence Hey, I'm new here and this is my first attempt at a question. My background is in econometrics, so I apologize in advance if I use unfamiliar notation or display ignorance of important work.
Suppose we observe a large amount of data $\{(Z_{i})\}_{i=1}^{n}$
generated i.i.d. by $Z_{i }= X_{i} + Y_{i}$ with $Y$ representing contaminating noise with known distribution.  Interest is in working backwards (deconvolution) to learn (identify) as much as possible about the distribution of $X$.  Let the characteristic functions of these variables be denoted as $\phi_{X}(t) \equiv E(e^{ixt})$ and similar notation for $Y$ and $Z$.
It is well known that the full distribution of $X$ can be learned under the restriction that $X \perp Y$.  This follows because $\phi_{Z}(t)$ can be consistently estimated from the data and $\phi_{X}(t) = \frac{\phi_{Z}(t)}{\phi_{Y}(t)}$.  However, I am interested in relaxing this restriction and replacing it with the weaker requirement that $\forall x\in \text{Supp}(X)$, $E(Y|X=x)=0$.
Under this restriction, I know that (for example), the variance of the unknown distribution $X$ can be learned by:
$Cov(X,Y)=0 \implies Var(X) = Var(Z) - Var(Y).$
However, I am fairly certain that I cannot learn the entire distribution of $X$ with this weaker assumption.  If you can think of a good counter-example, that would be helpful.  More importantly, I would like to make the strongest possible statements about the full distribution of $X$ based on knowledge of the distributions of $Y$ and $Z$.
Edited to add-----------
I have become aware that if I altered my assumption to require that $med(Y|X)=0$ I would pin down a horizontal section of the copula between Y and X and that the bounds in that case have been well studied (for example: tandfonline.com/doi/abs/10.1080/03610920701386976). However, this is a less useful result. Perhaps no aspect of a copula is pinned down by a conditional mean restriction? Would this limit my ability to provide meaningful bounds on quantiles of the $X$ distribution?
 A: Here's a counterexample found through numerical experimentation
Assume $X \in \{ x_1, x_2, x_3\}$ and $Y \in \{ y_1, y_2, y_3 \}$.
Let $P$ be a $3 \times 3$ matrix where $p_{ij} = P(X=x_j \land Y=y_i)$
Take
$$( x_1, x_2, x_3 ) = (1,0,-1) \\\\
( y_1, y_2, y_3 ) = (1,0,-1)$$ and 
$$P = \left\[\begin{array}{ccc}
0.06& 0 & 0.03\\\\
0.06 & 0.37 & 0.39\\\\
0.06 & 0 & 0.03\\\\
\end{array}\right\]$$
The distribution of $Z$ is 
$$\begin{array}{ccc}
p(Z=-2) &=& 0.03\\\\
p(Z=-1) &=& 0.39\\\\
p(Z=0) &=& 0.46\\\\
p(Z=1) &=& 0.06\\\\
p(Z=2) &=& 0.06\end{array}$$
and the marginal distribution of $Y$ is 
$$\begin{array}{ccc}
P(Y=1) &=& 0.09\\\\
P(Y=0) &=& 0.82\\\\
P(Y=-1) &=& 0.09\\\\
\end{array}$$
One can check that $\forall x, E(Y|X=x) = 0$
Suppose now we take
$$( x_1, x_2, x_3 ) = (-1,0,2)$$
and
$$P = \left\[\begin{array}{ccc}
0.03 & 0.06& 0\\\\
0.33& 0.43 & 0.06\\\\
0.03 & 0.06 & 0
\end{array}\right\]$$
This is a different distribution for $X$ yet the distribution for $Z$ and the marginal distribution for $Y$ are unchanged and the same properties hold.
