Does there exist a (non-trivial) probability distribution on the rational numbers $$\sum_{r\in\mathbb{Q}}p_r=1$$ with $0\leq p_r$, which is stable, meaning that the sum of two i.i.d. random variables with this distribution has a scaling of this distribution (by a rational number, obviously)? Or in other words, does there exist a $p_r$ and $s\in\mathbb{Q}$ such that for each $r\in\mathbb{Q}$ $$\sum_{t\in\mathbb{Q}}p_t p_{r-t}=p_{sr}?$$ I think the answer is no but I can't prove it. The approach I was originally trying to use was to construct a stable distribution by constructing a 'characteristic function' of it on the Pontryagin dual of the rationals, since solving for 'multiplicative scaling' ($f(x)\cdot f(x) \propto f(\alpha x)$) is a lot easier than solving for 'convolutional scaling' ($f(x)\star f(x) \propto f(\alpha x)$), but the Pontryagin dual of the rationals is rather complicated and everything I can think of seems to boil back down to the original problem or seems like a non-starter.

The only approach I can come up with to prove that it's impossible is to think of the distribution as a weird real-valued distribution and to use the central limit theorem (or a generalization) to show that the sum of a sequence of i.i.d. rational-valued distributions always converges in distribution to something that can't correspond to a rational-valued distribution.


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