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If we have two independent random variables each having a Cantor distribution is there an easy way to see that the distribution of their sum is not absolutely continuous?

I am pretty sure that if we let $S_n$ be the set of positive integers having an $n$ digit ternary expansion (leading zeros included) with $n/2$ or more 1's, and let $$T_n = \left\{\frac{2s+1}{3^n}:s\in S_n\right\}$$ Then our random variable has more than a 50-50 chance of being within $3^{-n-1}$ of a member of $T_n$. As the number of intervals grows as $2^n$, and their width shrinks as $3^{-n}$, the measure of the whole thing goes to 0 as $n$ goes to infinity. (It took some handwaving and arithmetic to get here, so don't trust me.)

In the best of all possible worlds, there would be an argument that works for the absolute continuity of the sum of three (or any number) of independent Cantor Random variables.

deinst
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1 Answers1

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The characteristic function of Cantor distribution solves functional equation $$ \varphi_C(t) = \frac12(e^{i2t/3} + 1) \varphi_C(t/3)\tag{1} $$ (there is an explicit formula with cosines, but this is enough for us).

In particular, $\varphi_C(\pi) = \varphi_C(3\pi) = \varphi_C(3^2\pi) = \dots$ If this value were zero, then we would get from (1) that $\varphi(3^{-n}\pi) = 0$, $n\ge 1$, which contradicts the continuity of $\varphi_C$ (otherwise, you can use the formula with cosines to argue that it is not zero).

So for each $n\ge 1$ we have $$0\neq \varphi_C(\pi)^n = \varphi_C(3\pi)^n = \varphi_C(3^2\pi)^n = \dots\tag{2}$$ If the sum of $n$ independent Cantor variables had a density, we would have $\varphi_C(t)^n\to 0$, $t\to\infty$ by the Riemann-Lebesgue lemma, contradicting (2).

I prefer the argument using the recursion (1) to the one with explicit formula, as it (or some clever modification) can be used to prove that some other random series does not have a density.

zhoraster
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  • Exercise :) Why doesn't this argument work for the uniform $[0,1]$ distribution, whose cf solves $\varphi_U(t) = \frac12 (e^{it/2} + 1) \varphi_U(t/2)$? – zhoraster Jan 05 '16 at 17:35
  • Thanks. (The answer to your exercise is that the $\ne$ becomes an $=$ in (2)). This is partly what kept me from looking too closely at the limit $t\to \infty$ of the characteristic function. – deinst Jan 05 '16 at 17:45
  • @deinst, don't forget to accept the answer if you like it. I believe that the exercise will be useful for (at least some of) students. It is always nice to see where an argument fails. – zhoraster Jan 05 '16 at 17:56