The imaginary part
$$
\arg(g_t(z)-U_t)
$$
is a bounded local martingale and hence a martingale. As the simple transient trace passes $z$, this angle converges to $\pi$ if the trace passes to the right of $z$ and to $0$ if it passes to the left. Bounded convergence therefore gives
$$
\arg z
=\pi\,\mathbb P(\gamma\text{ passes to the right of }z),
$$
so the <SLE4 left-passage probability> is
$$
\boxed{\mathbb P(\gamma\text{ passes to the right of }z)
=\frac{\arg z}{\pi}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 3 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-203/3/c/ii/solution