TITLE: Which random variables can be written as the difference of two independent positive random variables?
QUESTION [6 upvotes]: Can we characterize random variables $X$ that satisfy
$$
X\sim Y - Z
$$
for two independent positive random variables $Y$ and $Z$?
Are $Y$ and $Z$ unique in some sense?
Can (one possible choice of) $Y$ and $Z$ be constructed (e.g. formulas for probability density or characteristic function, or sampling algorithms) when they exist?

Possibly simpler question:
Which random variables $X$ satisfy
$$
X\sim Y_1-Y_2
$$
for i.i.d. positive random variables $Y_1\sim Y_2\sim Y$? Since the characteristic function satisfies
$$
\phi_X  = \phi_{Y}\overline{\phi_{Y}}
$$
we must have $\phi_{X}\geq 0$ -- is that sufficient? Is $Y$ unique in some sense? Can it be constructed?
For example, Laplace random variables satisfy $\phi_{X} = (1+x^2)^{-1}=(1+ix)^{-1}\overline{(1+ix)^{-1}}=\phi_{Y}\overline{\phi_{Y}}$ where $Y$ is exponential. Exponentials are positive of course, so we got lucky with this particular decomposition and can write $X=Y_1-Y_2$ as desired. Had we picked $(1+x^2)^{-1}=(1+x^2)^{-1/2}\overline{(1+x^2)^{-1/2}}$ this wouldn't have worked out.
This approach slightly generalizes to $\phi_{X}$ that are rational in $x^2$, but not at all (at least not obviously to me) to only slightly different examples like Linnik random variables, where $\phi_{X} = (1+|x|^{\alpha})^{-1}$, or to limits such as normal random variables, where $\phi_{X}=e^{-\sigma^2x^2}$.

The only result I found that goes remotely in this direction was a theorem by Boas and Kac that positive definite functions with compact support have a convolution square root with half-length compact support. This has a support flavor, but a different one than I'm looking for.

REPLY [3 votes]: As soon as $\phi(x)$ decays too rapidly, you are doomed. For example, if $\phi(x)=e^{-x^2}$, then $\psi(x)=Ee^{itY}$ would have to satisfy $|\psi(x)|=e^{-x^2/2}$, but now the rapid decay will make the distribution of $Y$ absolutely continuous with holomorphic density $f_Y$ (this is well known and easily proved since $f_Y(z)=\int_{-\infty}^{\infty}\psi(t) e^{-itz}\, dt$ converges for all $z\in\mathbb C$), so it's not possible to have $f_Y(y)=0$ for $y<0$.
To state this more formally, this argument shows that if $|\phi_X(x)|\lesssim e^{-c|x|}$ for some $c>0$ and $X= Y_1-Y_2$, with $Y_j$ independent and $Y_1\sim Y_2$, then the distribution of $Y_j$ is of the form $d\mu(y)= f(y)\, dy$ with a real analytic $f$. In particular $P(Y\in A)>0$ for any $A\subseteq\mathbb R$ of positive Lebesgue measure.