Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 3 / d

Fix $z$ and writeBy assumption, $L(z)$ is a continuous local martingale, so $Z_t(z)=e^{L_t(z)}$ is a semimartingale. The Chordal Loewner equation giveswhich has finite variation. Thereforeis a semimartingale. Thus the Loewner driver $U$ is a continuous semimartingale.

Write the semimartingale decomposition as $U=M+A$, where $M$ is a continuous local martingale and $A$ has finite variation. Applying Itô formula to $\log Z_t(z)$ shows that its finite-variation part isIt vanishes for every $z$. Multiplying by $Z_t(z{)}^2$ givesSubtract this identity for two points with distinct $Z_t$ to obtain $d{A}_t=0$; then $d⟨M{⟩}_t=4dt$. Since the curve starts at zero, $U_0=0$. The Lévy characterization of Brownian motion now gives $U_t=2B_t$. Hence the Loewner chain is