# 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 (*)).

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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. – Fabian Apr 10 '12 at 20:20
Remotely related is the Fabius random variable (see wiki). – Sasha Apr 10 '12 at 20:25
It can be shown easily that $F(0)=1/2$, $F(1/2)=5/6$, $F(-x)=1-F(x)$. – Fabian Apr 10 '12 at 20:32
$F(1/4)=2/3$, $F(1/3)=8/11$, $F(2/3)=10/11$, $F(3/4)=17/18$, ... The values of $F$ for rational arguments can be obtained recursively (but I did not yet figure out an explicit formula). – Fabian Apr 10 '12 at 20:48
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 $\frac{1}{2}\epsilon^{\log_2 3}$. – Thomas Andrews Apr 10 '12 at 20:59

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:

Following Fabian's footsteps, it is easy to code computation of CDF at rational points:

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:

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