7
$\begingroup$

Suppose $B_t$ is a standard Brownian motion in $\mathbb{R}^2$ and $T = \text{inf}\{t : |B_t| = 1\}$. Let $E$ denote the event that $0$ is contained in the unbounded component of $\mathbb{R}^2 \setminus B[0, T]$. Do there exist $\lambda < \infty$ and $\alpha > 0$ such that for all $|x| < 1$, we have$$\mathbb{P}^x(E) \le \lambda |x|^\alpha?$$

$\endgroup$
  • $\begingroup$ I guess a preliminary question would be, how to see that $E$ is measurable? $\endgroup$ – Nate Eldredge Sep 17 '15 at 21:03
  • $\begingroup$ I changed the title to something a little more self-contained. Feel free to edit further. $\endgroup$ – Nate Eldredge Sep 17 '15 at 21:13
  • $\begingroup$ Related: mathoverflow.net/questions/202944/… $\endgroup$ – Nate Eldredge Sep 17 '15 at 21:14
  • $\begingroup$ In some sense, the question is really asking for the probability that BM does encircle 0 (i.e. what is really needed is a lower bound for the probability that 0 gets encircled). $\endgroup$ – Anthony Quas Sep 17 '15 at 21:56
  • 1
    $\begingroup$ Another related question: "Twisted random walks." $\endgroup$ – Joseph O'Rourke Sep 18 '15 at 1:15
10
$\begingroup$

The probability is asymptotic to $\lambda |x|^{1/4}$. This was proved by Schramm, Werner and myself using the Schramm-Loewner evolution. The exponent is called the disconnection exponent for Brownian motion.

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

Divide the interval $[|x|,1]$ into $-\log_2 |x|$ intervals $[r_i,r_{i+1}]$. Radially, the BM performs a (simple) random walk between the circles of these radii, and there will be at least $-\log_2 |x| $ steps to this random walk before it hits $1$. During each step the tangential component of the BM has probability $>\delta$ to encircle the origin, some $\delta>0$ (the tangential component is a time changed BM on the circle with time change bounded within a constant factor of $r_i^2$). So $$P(E)\leq (1-\delta)^{-\log_2 |x|}$$

This computation does not give the correct exponent $\alpha$.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

Take any $s$ and $r$ such that $0\le s\le r\le1$, and let $p(s,r)$ denote the probability that the Brownian motion in $\mathbb R^2$ starting at a point at distance $s$ from the origin $O$ will encircle $O$ before hitting the circle of radius $r$ centered at $O$. Then for any $u\in[s,r]$ $$(1)\qquad p(s,r)\ge p(s,u)+[1-p(s,u)]p(u,r). $$ The other crucial observation is that $p(s,r)=f(s/r)$ for some function $f\colon[0,1]\to[0,1]$ and all $s$ and $r$ such that $0\le s\le r\le1$ and $r>0$. This observation follows because, for the Brownian motion, space rescaling is equivalent to appropriate time rescaling. So, with $a:=s/u$ and $b:=u/r$, (1) yields $$f(ab)\ge f(a)+[1-f(a)]f(b) $$ and then
$$g(ab)\le g(a)g(b), $$ where $g:=1-f$ and $0<a,b\le1$. Note that $f(1)=0$ and hence $g(1)=1$. Also, $g$ is nonnegative and nondecreasing, since $p(s,r)$ is obviously nondecreasing in $r$. Using ``path corridors'' that are (say) unions of rectangles, one can show that $0<g(1/2)<1$. Take now any $a\in(0,1/2]$ and let $k:=k_a:=\lfloor\log_2\frac1a\rfloor$, so that $a^{1/k}\le1/2$ and $k\ge1$, and hence $k\ge\frac12\,\log_2\frac1a$. Then $$g(a)\le g(a^{1/k})^k\le g(1/2)^k\le g(1/2)^{\frac12\,\log_2\frac1a}=a^\alpha, $$ where $\alpha:=-\frac12\,\log_2 g(1/2)\in(0,\infty)$. That is, $$P^x(E)=1-p(|x|,1)=g(|x|)\le|x|^\alpha$$ for all $x$ with $|x|\le1/2$. If now $|x|\in[1/2,1]$, then $P^x(E)\le1\le(2|x|)^\alpha$. So, $$P^x(E)\le2^\alpha |x|^\alpha$$ for all $x$ with $|x|\le1$, as desired.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't see why (1) is an equality. Let $C_r$ be the circle of radius $r$ centered at $O$. It sounds like you are saying "If the BM is going to encircle $O$ before hitting $C_r$, and it hasn't done so by the time it hits $C_u$ (at some point $y$), then the path started at $y$ must encircle $O$ before hitting $C_r$." But that's not true - what if the path from $x$ to $y$ wound halfway around the origin, and after $y$ it completed the other half? Then the combined path encircles the origin, while neither of the pieces before or after $y$ does so. $\endgroup$ – Nate Eldredge Sep 18 '15 at 14:50
  • $\begingroup$ @NateEldredge However, we do have $g(ab) \leq g(a) g(b)$ and $g(a) \leq 1$, so $g$ is monotonic. This means that for each $\alpha$, either $g(a) \leq \lambda |x|^\alpha$ for some $\lambda$ or $g(a) > |x|^\alpha$. Given that we have upper and lower bounds, this means that there exists a precise asymptotic constant $\alpha$ such that $|x|^\alpha <g(a) < O( |x|^{\alpha+\epsilon})$. $\endgroup$ – Will Sawin Sep 18 '15 at 15:06
  • $\begingroup$ I have corrected the answer, in view of the comments by Nate Eldredge and Will Sawin. $\endgroup$ – Iosif Pinelis Sep 18 '15 at 16:48
3
$\begingroup$

This is an extended comment to observe that $E$ is measurable.

Let $U$ be the set of $\omega \in C([0,1]; \mathbb{R}^2\}$ that have not encircled the origin by time 1; i.e. such that $0$ is in the unbounded component of $\mathbb{R}^2 \setminus \omega([0,1])$. I claim $U$ is open (with respect to the uniform norm $\|\cdot\|_\infty$).

Suppose $\omega \in U$, so that $0$ is in the unbounded component. Choose any $y \in \mathbb{R}^2$ with $|y| > \|\omega\|_{\infty} + 1$, so $y$ is also in the unbounded component. Now the unbounded component is a connected open set, so it is path connected; let $\gamma \in C([0,1]; \mathbb{R}^2)$ be a path joining $0$ to $y$ which does not intersect $\omega$. Set $\epsilon = \inf\{|\omega(t) - \gamma(s)| : s,t \in [0,1]\}$ which is strictly positive by compactness. If $\|\omega - \tilde{\omega}\| < 1 \wedge \epsilon$ then $\gamma$ does not intersect $\tilde{\omega}$ either, so both $0$ and $y$ are in the same component of $\mathbb{R}^2 \setminus \tilde{\omega}([0,1])$. Since $\|\tilde{\omega}\| \le \|\omega\| + \|\omega - \tilde{\omega}\| < \|\omega\|+1$, $y$ is also in the unbounded component of $\mathbb{R}^2 \setminus \tilde{\omega}([0,1])$, hence so is 0. So $\tilde{\omega} \in U$.

By the same argument, for any $t \ge 0$, the set $U_t$, consisting of all those $\omega \in C([0,\infty) ; \mathbb{R}^2)$ which have not encircled the origin by time $t$, is open.

So let $\tau(\omega) = \inf\{t : \omega \notin U_t\}$ be the first time at which $\omega$ has encircled the origin. Since $U_s \supset U_t$ for $s < t$, we can take the infimum over the rationals instead, and see that $\tau : C([0,\infty); \mathbb{R}^\infty) \to [0,\infty]$ is Borel. Indeed, $\tau$ is a stopping time. And $E$ is simply the event $\{\tau > T\}$.

Likewise, the function $p(s,r)$ in Iosif Pinelis's answer does make sense, since we simply have $p(s,r) = P^{x_0}(\tau < T_r)$ where $T_r = \inf\{t : |B_t| = r\}$ and $x_0$ is any point with $|x_0|=s$ (by symmetry it does not matter which such $x_0$ is chosen).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This nicely rounds up the solution of the problem. My only, very minor suggestion here is to remove the assumption $\epsilon<1$ (which may need some justification) and then replace the condition $\|\omega-\tilde{\omega}\|<\epsilon$ by $\|\omega-\tilde{\omega}\|<1\wedge\epsilon$. $\endgroup$ – Iosif Pinelis Sep 18 '15 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.