Let $(X_n)_{n\ge 0}$ be a process for which $X_0$ is uniform on $(0,1)$ and, for $0<\alpha <1$ and $n\ge 0$, we have:

$X_{n+1} = \alpha X_n + 1 -\alpha$ with probability $X_n$

$X_{n+1} = \alpha X_n$ with probability $1-X_n$.

What are the limiting properties of $X_n$ as $n\to \infty$?

It is easy to see that $0<X_n<1$ and also that $X_n$ is a martingale, and so by the Martingale Convergence Theorem, we must have $\lim_{n\to\infty} X_n =X$ where $X$ is a random variable that is almost surely finite. But I cannot deduce anything else about $X$. Can we even say anything else, and if so, how?

  • $\begingroup$ Assuming that the choices to apply $x\to ax+1-a$ or $x\to ax$ are made independently of the past of the process (which you do not say), the possible limit points are $0$ and $1$ hence the distribution of $X$ is Bernoulli and entirely characterized by $E(X)=P(X=1)=1-P(X=0)$. Surely you can compute $E(X)$... $\endgroup$
    – Did
    Jun 11, 2017 at 15:53

1 Answer 1


Assuming that the choice of values of $X_{n+1}$ is made independently of $\mathcal{F}_n = \sigma(X_0, \cdots, X_n)$, notice that

$$\frac{X_{n+1} - \alpha X_n}{1-\alpha}$$

is a $\{0,1\}$-valued random variable, i.e., a Bernoulli random variable that converges almost surely to $X$. So $X$ is also a Bernoulli random variable. Then the law of $X$ is determined by the parameter

$$p = \Bbb{P}(X = 1) = \Bbb{E}[X]$$

which is easily computed from the martingale nature of $(X_n)$.

(This is essentially what @Did is explaining in the comment.)


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