= Solution
{c}
Continuity and the regularity of zero for a one-dimensional Brownian coordinate make $S$ the first contact with the imaginary axis. The <Strong Markov property> and symmetry of the subsequent real increments imply that reversing those increments after $S$ preserves the Brownian law. The imaginary coordinate is unchanged, so $\widetilde Z$ is again planar <Brownian motion> started at one.
Reflection preserves distance to the origin, hence $\widetilde T_{D_r}=T_{D_r}$ pathwise. On $T_{D_r}<S$, the path exits through the right semicircle and reflection has not yet acted. On $S<T_{D_r}$, reflection reverses the sign of the exit real part. Thus
$$
\mathbb P(\operatorname{Re}Z_T>0)
=\mathbb P(T<S)+\mathbb P(S<T,\operatorname{Re}Z_T>0),
$$
$$
\mathbb P(\operatorname{Re}\widetilde Z_T<0)
=\mathbb P(S<T,\operatorname{Re}Z_T>0).
$$
The two circular points on the imaginary axis have zero exit probability, so they cause no extra term. Subtracting proves the <reflection identity for Brownian exit from a half-disc> requested in the question.
Both processes have the same exit law. With $p=\pi^{-1}\tan^{-1}((r^2-1)/(2r))$ from part (b),
$$
\boxed{\mathbb P(T_{D_r}<S)=1-2p=\frac4\pi\tan^{-1}(1/r).}
$$
The final trigonometric identity follows from $\tan^{-1}((r^2-1)/(2r))=\pi/2-2\tan^{-1}(1/r)$ for $r>1$.
Back to article page
