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Let $\xi_n$ be iid and uniformly distributed on the three numbers $\{-1,0,1\}$. Set $$X = \sum_{n=1}^\infty \frac{\xi_n}{2^n}.$$ It is clear that the sum converges (surely) and the limit has $-1 \le X \le 1$..

What is the distribution of $X$?

Does it have a name? Can we find an explicit formula? What else can we say about it (for instance, is it absolutely continuous)?

We can immediately see that $X$ is symmetric (i.e. $X \overset{d}{=} -X$). Also, if $\xi$ is uniformly distributed on $\{-1,0,1\}$ and independent of $X$, we have $X \overset{d}{=} \frac{1}{2}(X+\xi)$. It follows that for the cdf $F(x) = \mathbb{P}(X \le x)$, we have $$F(x) = \frac{1}{3}(F(2x+1) + F(2x) + F(2x-1)). \quad (*)$$

The cdf of $\sum_{n=1}^{12} \frac{\xi_n}{2^n}$ looks like this:

cdf of 12th partial sum

It looks something like $\frac{1}{2}(1+\sin(\frac{\pi}{2}x))$ but that doesn't quite work (it doesn't satisfy (*)).

I'd be interested if anything is known about this. Thanks!

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    I think the equation (*) together with the boundary conditions $F(x\leq -1)=0$ and $F(x\geq 1)=1$ determines the cdf $F$ uniquely.2012-04-10
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    Remotely related is the Fabius random variable (see [wiki](http://en.wikipedia.org/wiki/Fabius_function)).2012-04-10
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    It can be shown easily that $F(0)=1/2$, $F(1/2)=5/6$, $F(-x)=1-F(x)$.2012-04-10
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    $F(1/4)=2/3$, $F(1/3)=8/11$, $F(2/3)=10/11$, $F(3/4)=17/18$, ... The values of $F$ for rational arguments can be obtained recursively (but I did not yet figure out an explicit formula).2012-04-10
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    For $x=-1+2^{-k}$ you can get that $F(x)=\frac{1}{2(3^k)}$. The first $k$ have to be negative, and then half of those end up on the left. Thus, for small $\epsilon$, $F(-1+\epsilon)$ is roughly $\frac{1}{2}\epsilon^{\log_2 3}$.2012-04-10
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    +1 for the question. It seems the distribution of $X$ should be purely singular, in the sense that the densitable part and the discrete part are both zero.2012-04-10
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    If you had $3^n$ instead of $2^n$, the question would be really easy to answer.2012-04-10

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