Final update on 11/29/2019: I have worked on this a bit more, and wrote an article summarizing all the main findings. You can read it here.

Let us consider $Z = X_1 + X_1 X_2 + X_1 X_2 X_3 +\cdots$ where the $X_i$'s are i.i.d. I proved in my previous answer (see here) that $Z$ has a geometric distribution if and only if $X_i$ has a Bernouilli distribution. Thus the geometric distribution is an attractor for this system. Also, it is an isolated attractor in the sense that there is only one way it can be attained: only if $X_i$ is Bernouilli.


Now let us consider the following shifted Bernouilli distribution: $$P\Big(X_i = \frac{1}{2}\Big) = P\Big(X_i = -\frac{1}{2}\Big) = \frac{1}{2}.$$

Then $Z$ is uniform on $[-1, 1]$. Also $Z$ is uniform on $[-1, 1]$ if and only if $X_i$ has the shifted Bernouilli distribution in question.


The proof of this fact is based on the following relationship for the $k$-th moments (see here): $$E(Z^k) = E[(X_i(1+Z))^k]=E(X_i^k)E[(1+Z)^k].$$ Assume a uniform distribution on $[-1, 1]$ for $Z$. Then we have

  • $E(Z^k) = 1/(k+1)$ if $k$ is odd, and is equal to zero otherwise,
  • $E[(1+Z)^k] = 2^k/(k+1)$,

We also have:

  • $E[X_i^k]= 1/2^k$ if $k$ is odd, and is equal to zero otherwise.

Thus the only way we can have $E(Z^k) = E(X_i^k)E[(1+Z)^k]$ is if all the moments of $Z$ are those of a uniform distribution on $[-1, 1]$. Thus, $Z$ has the prescribed distribution.

To prove the converse, note that in order for $Z$ to be uniform on $[-1, 1]$ then $E[X_i^k]= 1/2^k$ if $k$ is odd, 0 otherwise. The only distribution having these moments is the shifted Bernouilli distribution for $X_i$.

Connection with the binary numeration system

If $Z = X_1 + X_1 X_2 + X_1 X_2 X_3 + \cdots$ and $X_i$ has the shifted Bernouilli distribution mentioned earlier (thus $Z\in [-1, 1]$), and $$Z+1 = \sum_{k=0}^\infty \frac{B_k}{2^k}, \mbox{ with } B_k \in \{0, 1\}$$

then $$X_1 = B_0 -\frac{1}{2} \mbox{ and } X_k = \frac{1}{2}\cdot\frac{2B_{k-1}-1}{2B_{k-2}-1} \mbox{ if } k\geq 2.$$

Conversely, if $k\geq 0$, then

$$B_k = \frac{1}{2}+2^k \cdot \prod_{i=1}^{k+1} X_i.$$

For instance, if $Z = \frac{1}{5}$, then the $k$-th digit ($k = 1, 2, \cdots$) of $Z$ in the new numeration system introduced here is $X_k= (-1)^{k-1}/2$.

Connection with singular distributions

In most cases, if $X_i$ is discrete, then $Z$ is nowhere differentiable. A typical example is the following: if $P(X_i = -\frac{1}{2}) = \frac{1}{4} = 1-P(X_i = \frac{1}{2})$ then see below the percentile distribution for $Z$, it's clearly a singular distribution on $[-1,1]$:

enter image description here

To the contrary, if $P(X_i = -\frac{1}{2}) = \frac{1}{2} = P(X_i = \frac{1}{2})$, then the shape of the percentile distribution is a perfect diagonal, see picture below.

enter image description here

These charts were produced using 20,000 deviates for $Z$, each based on 200 terms $X_1, X_1 X_2,\cdots, X_1 X_2\cdots X_{200}$.

Question (updated)

Can you find other distributions for $X_i$ (a continuous distribution if possible) resulting in $Z$ having a well known distribution?

The key here is that $X_i$ and $1+Z$ act as if they were independent. And the distribution of $Z$ is also that of $X_i(1+Z)$. Characteristic functions (based on the Fourier transform) are of no use. Instead, the Mellin transformation, denoted as $M$, is useful here, as the Mellin transform of a product of independent random variables is the product of the individual Mellin transforms (see Wikipedia article here.) In short, $M(Z)=M(X_i)M(1+Z)$. I tried a log-normal distribution for $X_i$ and $Z$ was very well approximated by a log-normal on $[-3, 3]$, but it is not an exact solution.

Why is Log-normal not working?

I tried the following log-normal distribution: $X_i = \exp(Y_i)/5$ with $Y_i$ being Normal($0, 1$). Thus $X_i$ is log-normal and has the following moments (see here): $E(X_i^k) = 5^{-k}\cdot\exp(k^2/2), k = 0, 1, 2,\cdots$. Since $E(Z^k) = E(X_i^k)E[(1+Z)^k]$, it is possible to compute the exact value of $E(Z^k)$. The approximate values are

  • $E(Z) \approx 0.491967814042304$
  • $E(Z^2) \approx 0.832403517661211$
  • $E(Z^3) \approx 12.796705133476$

These values are obtained using the following formulas:

$$E(Z) = \frac{E(X_i)}{1-E(X_i)}$$

$$E(Z^2) = \frac{E(X_i^2)(1+E(X_i))}{(1-E(X_i))(1-E(X_i^2))}$$

$$E(Z^3) =\frac{E(X_i^3)(1+3E(Z)+3E(Z^2))}{(1-E(X_i^3))}$$

Now let us assume that $Z$ is log-normal. We must have $E(Z^k) = \exp(k\mu + k^2\sigma^2/2)$ for some parameters $\mu$ and $\sigma$. Solving for $\exp(\mu + \sigma^2/2) \approx 0.491967814042304$ and $\exp(2\mu + 4\sigma^2/2)\approx 0.832403517661211$ yields:

  • $\mu \approx -1.32696498721330$
  • $\sigma^2 / 2 \approx 0.61762300397038$

Now $E(Z^3) = \exp(3\mu + 9\sigma^2/2) \approx 4.843860729265$. This is very different from the value computed earlier (namely, $E(Z^3) \approx 12.796705133476$) and thus we must conclude that $Z$ can not be log-normal.

  • $\begingroup$ Each term $X_1 X_2 X_3 ... X_n$ you are either adding or subtracting $\left(\frac{1}{2}\right)^n$, based on a coin flip. So you can see your product also as $\sum_{n=1}^\infty X_i \left(\frac{1}{2}\right)^n$ with $P(X=1) = P(X=-1) = 1/2$. This means that your uniform variable is not uniform on $\mathbb{R}$ $\endgroup$ – Sextus Empiricus Nov 9 '19 at 22:15
  • $\begingroup$ I wonder what numbers form the probability space. It is a discrete distribution with infinitely many possibilities. Does that exist? $\endgroup$ – Sextus Empiricus Nov 9 '19 at 22:22
  • $\begingroup$ Maybe it's a en.wikipedia.org/wiki/Singular_distribution $\endgroup$ – Sextus Empiricus Nov 9 '19 at 22:32
  • $\begingroup$ It looks uniform on $[-1, 1]$. However, most examples involving a discrete distribution for $X_i$ results in a singular distribution for $Z$. The shifted Bernouilli and regular Bernouilli appear to be exceptions, possibly the only ones. $\endgroup$ – Vincent Granville Nov 9 '19 at 23:06
  • $\begingroup$ Uniform on $[-1,1] \in \mathbb{R}$ or $[-1,1] \in \mathbb{Q}$ or something else? For any number $z$ you have $P(Z = z) = 0$. $\endgroup$ – Sextus Empiricus Nov 9 '19 at 23:10

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