Which random variables can be written as the difference of two independent positive random variables?

Question

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?

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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 $\phi_{X}=e^{-\sigma^2x^2}$.

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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.

Answer 1

"we must have $\phi_{X}\geq 0$ -- is that sufficient?"

No. For instance, the function $$\mathbb R\ni t\mapsto(1-t^2+t^4/4)e^{-3t^2/8}$$ is a nonnegative characteristic function -- see the second display after formula (6.3.4) in the book by Lukacs.

However, as stated there in book, this characteristic function is indecomposable, that is, it is not the characteristic function of the sum of any two independent nondegenerate random variables.

Answer 2

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.