Write $Z_t=\int_0^te^{-B_s}d\widetilde B_s$, so $X_t=e^{B_t}Z_t$. Independence gives $\langle B,\widetilde B\rangle=0$, and <Itô formula> yields
$$
dX_t=\frac12X_tdt+X_tdB_t+d\widetilde B_t.
$$
The martingale part has quadratic variation $(1+X_t^2)dt$, so on an enlarged description it equals $\sqrt{1+X_t^2}\,dW_t$. Thus $X$ is a weak solution of
$$
dU_t=\frac12U_tdt+\sqrt{1+U_t^2}\,dW_t,qquad U_0=0.
$$
On the other hand, another application of Itô's formula gives
$$
dY_t=\frac12Y_tdt+\cosh(B_t)dB_t
=\frac12Y_tdt+\sqrt{1+Y_t^2}\,dB_t.
$$
Both coefficients are <Lipschitz continuous>, so <uniqueness in law> gives \b[$X$ and $Y$ the same law].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 6 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-202/6/2/solution