Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2010 / B / 4 / b / ii

For $z=x+iy$ in the strip,so $\mathrm{Im}f(z)>0$. This is a bijective conformal map onto the upper half-plane. One way to verify bijectivity is to factor it as $w=e^{\pi z/a}$, followed by $(w+w^{-1})/2$. The first map gives $\mathrm{Im}w>0$, $|w|>1$. The two inverse roots of the second map have product one; for any point in the upper half-plane exactly one root has modulus greater than one, and its imaginary part is positive because $\mathrm{Im}[(w+w^{-1})/2]=(1-|w{|}^{-2})\mathrm{Im}w/2$.

The three boundary pieces map as follows: the lower edge goes to $(1,\infty )$, the upper edge to $(-\infty ,-1)$, and the vertical edge to $(-1,1)$. Thus the desired exit event becomes exit from the upper half-plane to the left of $-1$. WriteBy conformal invariance of harmonic measure and the Cauchy exit law from a Brownian half-plane,The strip exit is finite almost surely, since its imaginary coordinate exits $(0,a)$ in finite time; corners have probability zero because individual points are polar points for planar Brownian motion. This is the Top-edge exit probability of a Brownian half-strip.