At time zero, $\Upsilon_0(z)=\operatorname{Im}z$. Compactness of $K\subset\mathbb H$ gives
$$
\epsilon_0=\min_{z\in K}\operatorname{Im}z>0.
$$
Also $M_0(z)$ is uniformly bounded above on $K$. On
$$
E_\epsilon=\{\tau_\epsilon<\infty, S_{\tau_\epsilon}\geq1/2\},
$$
one has $M_{\tau_\epsilon}\geq\epsilon^{-p}2^{-q}$. Optional stopping, <Fatou lemma>, and the assumed conditional angular estimate give
$$
\sup_{z\in K}M_0(z)
\geq\mathbb E[M_{\tau_\epsilon};E_\epsilon]
\geq\epsilon^{-p}2^{-q}c_1
\mathbb P(\tau_\epsilon<\infty).
$$
Thus
$$
\boxed{\mathbb P_z(\tau_\epsilon<\infty)
\leq C_K\epsilon^{(8-\kappa)/8}.}
$$
The exponent $(8-\kappa)/\kappa$ requested in the question does not follow and is false as written. The <SLE Green-function estimate> gives probability comparable to $\epsilon^{1-\kappa/8}$, confirming that the denominator in the requested exponent should be $8$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 2 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-203/2/c/ii/solution