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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.

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  • $\begingroup$ $Y$ and $Z$ are not unique. Consider $X$ which has a 1/3 chance of being $1$ and a 2/3 chance of being $2$; you can write it as $Y-Z$ where $Y$ is certain or as $Y'-Z'$ where $Z'$ is certain instead. $\endgroup$
    – Matt F.
    Dec 17 '20 at 17:39
  • $\begingroup$ @MattF. you're right, and I should have figured that out myself. Added the uniqueness question hastily while writing up the question. I'm really more interested in (as explicit as possible) existence. Might still be possible to define better constraints or equivalence relation to get back uniqueness? I'll leave it in and see what people come up with $\endgroup$
    – Bananach
    Dec 17 '20 at 17:43
  • $\begingroup$ @ChristianRemling I thought that's (better) expressed by symmetry of $\phi_{X}$, which is implied by it being real $\endgroup$
    – Bananach
    Dec 17 '20 at 17:48
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"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.

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  • $\begingroup$ Thanks -- I was thinking, but not writing, about this as a question of convolutions and didn't appreciate the difference between a convolution square root and sum of random variables that lies in the positivity requirement on the convolution square root in the latter case. Interested in both $\endgroup$
    – Bananach
    Dec 17 '20 at 20:53
  • $\begingroup$ @Bananach : Sorry, I don't quite understand your comment. $\endgroup$ Dec 17 '20 at 21:29
  • $\begingroup$ I meant that even though the random variable associated to your characteristic function is not the sum of two random variables, its density $p_X$ is still the convolution $f\star f$ for some $f$ that is not a probability density. Similarly, there could be a random variable X that cannot be written as the difference $Y_1-Y_2$ but whose density $p_X$ can be written as $f\star f(-\cdot)$ for some non-density $f$. $\endgroup$
    – Bananach
    Dec 17 '20 at 23:03
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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.

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  • $\begingroup$ Thanks for clarfiying! $\endgroup$
    – Matt F.
    Dec 18 '20 at 15:51
  • $\begingroup$ Glad to see such an easily checked obstruction, which answers in particular the Gaussian case. Although I admit I had secretly hoped for more positive results. $\endgroup$
    – Bananach
    Dec 20 '20 at 14:47

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