Fix a parameter $\alpha\in(0,1)$ and take an i.i.d. sequence $X_0,X_1,\ldots$ of $\mathbb{R}^n$ valued random variables. Construct the limiting random variable

$X_\infty = (1-\alpha)\sum_{k=0}^\infty \alpha^k X_k.$

Is any general result known about this kind of limit? What if the $X_i$ follow a well known distribution like uniform/Rademacher?

I was motivated by this sum after running into: https://en.wikipedia.org/wiki/Chaos_game. For example if the $X_i$ are uniformly distributed on the 3 vertices of a triangle and $\alpha = 1/2$ the limiting distribution is supported on the associated Sierpinski Triangle. The fact that finite support distributions can give fractal shapes from this construction leads me to believe this is a non-trivial question.

My apologies if this ends up being an exercise in some well known textbook on probability theory. If it is, I'd appreciate a reference for that textbook.

Edit: I was able to locate http://u.math.biu.ac.il/~solomyb/RESEARCH/Bernotes.pdf


1 Answer 1


$\newcommand\al{\alpha}$Let us drop the factor $1-\al$, by considering $$Y:=X_\infty/(1-\al)=\sum_{k=0}^\infty\al^k X_k.$$ By Kolmogorov's three-series theorem, this series will converge almost surely (a.s.) unless at least one of the tails of the distribution of $X_0$ is too heavy.

Assume that the series indeed converges a.s. Then, obviously, $$Y\overset D=X+\al Y,$$ where $\overset D=$ denotes the equality in distribution and $X$ is an independent copy of the $X_k$'s. So, we have the functional equation for $F_Y$: $$F_Y(y)=\int_{-\infty}^\infty F_Y((y-x)/\al)\,dF_X(x)$$ for real $y$, where $F_Z$ denotes the cdf of $Z$. Equivalently, we have the functional equation for $f_Y$: $$f_Y(t)=f_Y(\al t)\,f_X(t)$$ for real $t$, where $f_Z$ denotes the characteristic function of $Z$. Of course, we can also write $$f_Y(t)=\prod_{k=0}^\infty f_X(\alpha^k t)$$ for real $t$.

In the particular case when $X$ is Rademacher, the distribution of $Y$ is the well-studied Bernoulli convolution.

In the particular case when $X$ is $U(0,1)$ and $\alpha=1/2$, $F_Y$ is the well-studied Fabius function.

  • $\begingroup$ Before asking my question I'd not gone much further beyond getting that characteristic function expression. I had not known of the Bernoulli convolution, looks like this has been considered before. For now I'll content myself to generating Python fractals with this. Consider the question answered. Thank you. $\endgroup$ Feb 16, 2020 at 20:18
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    $\begingroup$ I was able to locate u.math.biu.ac.il/~solomyb/RESEARCH/Bernotes.pdf if anyone else is new to the subject and wants further reading. $\endgroup$ Feb 20, 2020 at 7:01

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