Derivation of pmf from convolution Suppose that a discrete random variable (with finite support) $Y$ is given by $Y = X_1 - X_2$, where $X_1$ and $X_2$ are both discrete random variables with finite support and with the same probability mass function. Is it possible to determine the pmf of $X_1$ from the pmf of $Y$?

I will consider the case when $X_1$ and $X_2$ are independent identically distributed and integer-valued.
The following properites hold:


*

*Since $Y$ is discrete with finite support, its pmf is a polynomial in $z$ and $z^{-1}$ with non-negative coefficients.
$$
    Z_Y(z) = \sum_{k=m(Y)}^{n(Y)} \mathbb{P}(Y=k) z^k
$$

*Let $X$ be the purported solution with $Z_X(z) = \sum_{k=m(X)}^{n(X)} \mathbb{P}(X=k) z^k$.

*Since $Y = X_1-X_2$, for $X_1$, $X_2$ iids,  $Z_Y(z) = Z_{X}(z) Z_{X}(z^{-1})$. 
The problem you are posing is whether, given $Z_Y(z)$ one can determine $Z_X(z)$ such that $Z_Y(z) = Z_X(z) Z_X(z^{-1})$. Observe that the solution, if exists, is not unique, since $Z_{X^\prime}(z) = z^k Z_{X}(z)$ would also be a solution for arbitrary $k \in \mathbb{Z}$. 
The necessary condition for existence of a solution is that $Z_Y(z)$ should verify $Z_{Y}(z) = Z_Y(z^{-1})$. Assuming that the given $Z_Y(z)$ satisfies this property, finding $Z_X(z)$ reduces to polynomial factorization. 
