Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 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: enter image description here


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: enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.