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:
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!