Brownian motion

[Pere Callahan]

This is an exerxise from G. Lawler's book on conformally invariant processes.

Show there exist constants 0<a,c<inf such that the probability p for a planar Brownian Motion B_t starting at x makes a loop around the origin before hitting the unit circle is at least 1-c x^a.

I have not really an idea how to show this. I considered
$$|B_t|=\sqrt{{B_t^{(1)}}^2+{B_t^{(2)}}^2}$$
which is a Bessel-2 process and defined
$$T=\inf{t>0,|B_t|=1}$$

Then for any positive t
$$p\geq \mathbb{P}[\{T\geq t\}\cap \{\exists s,s'\in[0,t]:B_s=B_{s'} \text{ and } \gamma(B_s)-\gamma(B_{s'})\in \mathbb{Z})\}]$$

where $\gamma(z)$ would be some kind of winding number:
$$\gamma(B_t)=\int_0^t{dB_s\frac{1}{B_s}}$$
(I think w.p.1, a planar BM doesn't hit the origin so this should be fine.)

I'd appreciate any input.

Pere

 

[Valkarie]

grin's suggestion is a good one.In this case, we can use the fact that for a Brownian motion, the probability of staying in a circle centered at the origin decays exponentially with the radius of the circle. That is, if B_0 is the starting point of the Brownian motion, then the probability of it not hitting the unit circle before time t is given by P(|B_t| > 1) = e^(-2t/|B_0|^2). Therefore, we have:p \geq 1-cx^a = e^{-2t/|B_0|^2}Taking the natural logarithm of both sides gives:ln(1-cx^a) \leq -2t/|B_0|^2Rearranging terms yields:t \geq -(1/2)|B_0|^2 ln(1-cx^a)This shows that there exist constants 0<a,c<inf such that the probability for a planar Brownian Motion B_t starting at x makes a loop around the origin before hitting the unit circle is at least 1-c x^a.