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$$\mathbb{P}^x(E) \le \lambda x^\alpha?$$

$\begingroup$ I guess a preliminary question would be, how to see that $E$ is measurable? $\endgroup$– Nate EldredgeSep 17, 2015 at 21:03

$\begingroup$ I changed the title to something a little more selfcontained. Feel free to edit further. $\endgroup$– Nate EldredgeSep 17, 2015 at 21:13

$\begingroup$ Related: mathoverflow.net/questions/202944/… $\endgroup$– Nate EldredgeSep 17, 2015 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 QuasSep 17, 2015 at 21:56

1$\begingroup$ Another related question: "Twisted random walks." $\endgroup$– Joseph O'RourkeSep 18, 2015 at 1:15
4 Answers
The probability is asymptotic to $\lambda x^{1/4}$. This was proved by Schramm, Werner and myself using the SchrammLoewner evolution. The exponent is called the disconnection exponent for Brownian motion.
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 $\alpha$.
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 $u\in[s,r]$
$$(1)\qquad p(s,r)\ge p(s,u)+[1p(s,u)]p(u,r).
$$
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
$$f(ab)\ge f(a)+[1f(a)]f(b)
$$
and then
$$g(ab)\le g(a)g(b),
$$
where $g:=1f$ 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,
$$P^x(E)=1p(x,1)=g(x)\lex^\alpha$$
for all $x$ with $x\le1/2$. If now $x\in[1/2,1]$, then $P^x(E)\le1\le(2x)^\alpha$. So,
$$P^x(E)\le2^\alpha x^\alpha$$
for all $x$ with $x\le1$, as desired.

$\begingroup$ 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. $\endgroup$ Sep 18, 2015 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 $x^\alpha <g(a) < O( x^{\alpha+\epsilon})$. $\endgroup$ Sep 18, 2015 at 15:06

$\begingroup$ I have corrected the answer, in view of the comments by Nate Eldredge and Will Sawin. $\endgroup$ Sep 18, 2015 at 16:48
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).

$\begingroup$ 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 $\\omega\tilde{\omega}\<1\wedge\epsilon$. $\endgroup$ Sep 18, 2015 at 22:33