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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\{B(t), t \in \mathbb{R} \}$ be a two sided brownian motion defined as $$ B(t) = \begin{cases} B_1(t),\quad t >0 \\ 0, \quad t = 0 \\ B_2(-t), \quad t < 0 \end{cases} $$ where $B_1$ and $B_2$ are independent standard Brownian motions on $\mathbb{R}^+$.

Fix $x_0 > 0$ and let $x_k = B(x_{k-1})$ for $k=1,2,\dots$. What can we say about $\lim_{k\rightarrow \infty} x_k$?

If it converges, then I imagine the limit would have to be to $0$ a.s., since the limit $y$ would satisfy $B(y) = y$ a.s.. But I don't know how to show this sequence converges. Any ideas? (this isn't homework, just a problem my friend and I thought up)

share|cite|improve this question
Maybe you can start looking at $\mathbb{E}[x_k | x_{k-1}]$ and use $\mathbb{E}\left[\mathbb{E}[X|Y]\right] = \mathbb{E}[X]$ which would then at least give you a recurrence relation for the expectation values. A similar argument could be tried for the variances. Combining, you might deduce a limit for the actual sequence of random variables. – Raskolnikov Dec 4 '10 at 12:13
I'd like to say I really like this question. There are a lot of things going on here which make things complicated. I hope this link for the wiki article on iterated functions is helpful. – Matt Calhoun Dec 4 '10 at 17:16
up vote 4 down vote accepted

For any fixed $x_0$, there is positive probability that the sequence starting from $x_0$ fails to converge. To see why, take, say, $x_0 = 15$. There is positive probability that $50 < B_t < 60$ for $10 < t < 15$ and $10 < B_t < 20$ for $50 < t < 60$. On this event the sequence oscillates between the intervals $(10,20)$ and $(50,60)$.

On the other hand, almost surely there exist infinitely many $t$ in any interval $(0,\epsilon)$ with $B_t = t$, so even when the sequence does converge the limit need not be 0. For a proof, note that $P(B_t > t) = P(N > \sqrt{t}) \ge 1/4$ for sufficiently small $t$ (here $N$ is a standard normal random variable). So for any sequence $t_n$ decreasing to $0$, we have $P(B_{t_n} > t_n \text{ i.o.}) \ge 1/4$; by the Blumenthal 0-1 law, $P(B_{t_n} > t_n \text{ i.o.}) =1$. However, we also have $P(B_{t_n} < t_n) \ge P(B_{t_n} < 0) = 1/2$ so by a similar argument $P(B_{t_n} < t_n \text{ i.o.}) =1$. The result follows by continuity.

One could ask some other questions:

  1. For a fixed $x_0$, what is the probability that the sequence starting from $x_0$ converges? My guess is $0$ but I don't see a proof offhand.

  2. Consider the (random) set $C$ of $x$ such that the sequence starting from $x$ converges. What is the Lebesgue measure of $C$? My guess is that $m(C) = 0$ a.s. but again no proof.

Edit: Another interesting fact is that almost surely, for every starting point $x_0$, the sequence $x_k$ is bounded, and hence has a convergent subsequence. Let $M_r = \sup_{t \in [-r,r]} |B_t|$. By the strong law of large numbers, $B_t/t \to 0$ a.s. as $t \to \pm \infty$, and it follows that $M_r / r \to 0$ a.s. as $r \to \infty$. In particular, a.s. there exists $r > x_0$ with $M_r < r$, and then $|x_k| \le M_r$ for all $k \ge 1$.

share|cite|improve this answer
very good questions! another way to see $B(t) = t$ i.o. is to notice the set of zero crossings of $X(t) = B(t) - t$ is a perfect set by the strong Markov property, and the last crossing time of $0$ for $X(t)$ has an exponential distribution. – MarkV Dec 4 '10 at 19:56

Your Answer


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.