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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: 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?

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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.

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