 Solution

Let $J_t={\psi }_t^{\prime }(U_t)$, $q_t={\psi }_t^{\prime \prime }(U_t)/{\psi }_t^{\prime }(U_t)$, and $r_t=({\partial }_z^3{\psi }_t)(U_t)/{\psi }_t^{\prime }(U_t)$. Since $d{U}_t=\sqrt{8/3}d{B}_t$, the supplied identity and Itô formula giveFor $M_t=J_t^{\alpha }$, its drift coefficient isThus the nonzero choice isand $M_{t\wedge \tau }$ is a continuous local martingale. The boundary Schwarz lemma for mapping-out maps gives $0\le J_{t\wedge \tau }\le 1$, so $0\le M_{t\wedge \tau }\le 1$. A bounded local martingale is a true martingale. This is the SLE eight-thirds restriction martingale.