Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 6

A continuous finite-variation path has zero quadratic variation. Hence a continuous finite-variation martingale $M$ satisfies $⟨M⟩=0$. After localizing to make it square-integrable,Letting the localization level tend to infinity shows that $M_t=M_0$ for every $t$ almost surely; continuity makes the equality simultaneous in $t$.

Write $Z_t={\int }_0^t{e}^{-B_s}d{\tilde{B}}_s$, so $X_t=e^{B_t}{Z}_t$. Independence gives $⟨B,\tilde{B}⟩=0$, and Itô formula yieldsThe martingale part has quadratic variation $(1+X_t^2)dt$, so on an enlarged description it equals $\sqrt{1+X_t^2}d{W}_t$. Thus $X$ is a weak solution ofOn the other hand, another application of Itô's formula givesBoth coefficients are Lipschitz continuous, so uniqueness in law gives $X$ and $Y$ the same law.

The variation-of-constants formula gives, when $a\ne 0$,Therefore, with $m=\min (s,t)$,If $a=0$, then $X_t=x+bt+\sigma B_t$ and