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)

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  • $\begingroup$ 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. $\endgroup$ – Raskolnikov Dec 4 '10 at 12:13
  • $\begingroup$ 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. $\endgroup$ – Matt Calhoun Dec 4 '10 at 17:16

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 $k \ge 1$.

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  • $\begingroup$ 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. $\endgroup$ – MarkV Dec 4 '10 at 19:56

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