Let $J_t=\psi_t'(U_t)$, $q_t=\psi_t''(U_t)/\psi_t'(U_t)$, and $r_t=(\partial_z^3\psi_t)(U_t)/\psi_t'(U_t)$. Since $dU_t=\sqrt{8/3}\,dB_t$, the supplied identity and <Itô formula> give
$$
d\log J_t
=q_t\,dU_t
+\left[-\frac56q_t^2
+\left(-\frac43+\frac12\frac83\right)r_t\right]dt
=q_t\,dU_t-\frac56q_t^2dt.
$$
For $M_t=J_t^\alpha$, its drift coefficient is
$$
-\frac{5\alpha}{6}+\frac12\alpha^2\frac83
=\frac{\alpha(8\alpha-5)}6.
$$
Thus the nonzero choice is
$$
\boxed{\alpha=\frac58,}
$$
and $M_{t\wedge\tau}$ is a continuous local martingale. The boundary Schwarz lemma for mapping-out maps gives $0\leq J_{t\wedge\tau}\leq1$, so $0\leq M_{t\wedge\tau}\leq1$. A bounded local martingale is a true martingale. This is the <SLE eight-thirds restriction martingale>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 4 / c / Solution

{c}
{title2=$\operatorname{SLE}_{8/3}$}

For chordal $\operatorname{SLE}_{8/3}$ and an admissible hull $A$, let $\psi_t$ map out the image of $A$ after removing the SLE hull. Then
$$
M_t=\psi_t'(U_t)^{5/8}
$$
stopped when the trace meets $A$ is a bounded martingale. Its terminal value yields the avoidance probability
$$
\mathbb P(\gamma\cap A=\varnothing)=\psi_A'(0)^{5/8}.
$$


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 Incoming links: SLE eight-thirds restriction martingale