What is the distribution of this random series?

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!

Solution 1:

This is more of a comment, than an answer, yet it's too big, and graphics can't be used in comments.

It is not hard to work out cumulants of $X$: $$ \kappa_X(r) = \sum_{n=1}^\infty \frac{\kappa_\xi(r)}{2^{n r}} = \frac{\kappa_\xi(r)}{2^r-1} = \frac{3^r-1}{2^r-1} \cdot \frac{B_r}{r} \cdot [ r \geqslant 2] $$ Obviously, due to symmetry, odd cumulants and moments vanish. This implies the following low order moments: $$ m_2 = \mathbb{E}(X^2) = \frac{2}{9}, \quad m_4 = \frac{14}{135}, \quad m_6 = \frac{106}{1701}, \quad \ldots $$

The distribution itself appears to not be absolutely continuous, based on simulations:

ClearAll[cdf];
cdf[x_?ExactNumberQ] /; x >= 1 := 1;
cdf[0] = 1/2;
cdf[x_?Negative] := 1 - cdf[-x];
cdf[x_Rational /; EvenQ[Denominator[x]]] /; -1 < x < 1 := 
  cdf[x] = (cdf[2 x] + cdf[2 x - 1] + cdf[2 x + 1])/3;
cdf[x_Rational] := (* set up linear equations and solve them *)
 Block[{f, den = Denominator[x], ru1, ru2, vals, sol, ru3},
  ru1 = {f[z_] :> Divide[f[2 z] + f[2 z + 1] + f[2 z - 1], 3]}; 
  ru2 = {f[z_ /; z <= -1] :> 0, f[z_ /; z >= 1] :> 1, 
    f[z_?Negative] :> 1 - f[-z]};
  ru3 = f[r_Rational /; Denominator[r] < den] :> cdf[r];
  vals = Table[f[k/den], {k, den - 1}] /. ru3;
  sol = Solve[(((vals /. ru1) //. ru2) /. ru3) == vals, 
    Cases[vals, _f]];
  Function[{arg, res}, Set[cdf[arg], res]] @@@ 
   Cases[vals, f[a_] :> {a, f[a] /. First[sol]}];
  cdf[x]
  ]

So, for example,

In[101]:= cdf[1/5]

Out[101]= 31/49

The agreement with simulations is excellent: