Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2007 / B / 5 / i

First take the intended annulus configuration $0<\tilde{R}<R<\hat{R}$ with $R\ge 1$. The first stopping time is finite by the allowed fact, and continuity gives $|Z_{H_R}|=R$. After $H_R$, the annulus exit time is bounded by the exit time of the outer disc, which is almost surely finite; use the Strong Markov property at $H_R$. Thus $H_{R,\tilde{R},\hat{R}}<\infty$ almost surely.

The function $\log |z|$ is harmonic on this annulus. Its stopped value is a bounded martingale, so optional stopping theorem and continuity give, if $p$ is the probability of hitting the inner circle first,ConsequentlyFor fixed $\tilde{R}<1$, one may take $R=1$ and let $\hat{R}\to \infty$. The probability of hitting the inner circle before the outer one tends to one. The probability of ever hitting that inner circle is at least each of these probabilities, hence is one. Taking the countable intersection for $\tilde{R}=1/n$, $n\ge 2$, proves visits to balls of arbitrarily small radius almost surely.

The source permits $R<1$ as well. Then $H_R=0$ and the actual starting radius for the next exit is one, rather than $R$. More generally put $s=\max (1,R)$. If $\hat{R}>s$, the answer for the literal definition is $\log (\hat{R}/s)/\log (\hat{R}/\tilde{R})$. If $\hat{R}\le s$, the infimum defining the next exit is already $H_R$ almost surely and the inner-circle probability is zero. Positive $\tilde{R}$ is implicit in an annulus and is necessary for the logarithmic calculation.