Question about Brownian motion 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)

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:


*

*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.

*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 $k \ge 1$.
