Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 202 / 2 / iv / Solution

On the right-hand side, ${\mathbb{E}}_x$ means that the reference Brownian motion starts at $B_0=x$; on the left it means the stochastic differential equation starts at $X_0=x$. Since $h$ is smooth and compactly supported, $h^{\prime }$ is bounded and globally Lipschitz continuous. Define the stochastic exponentialThe Novikov condition holds on every finite horizon $T$, because $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{h}^{\prime }(B_s{)}^{2}ds)\le \exp (T\Vert h^{\prime }{\Vert }_{\infty }^2/2)$. Thus $D$ is a true martingale. Under $d\mathbb{Q}=D_{T}d\mathbb{P}$, the Girsanov theorem makes ${\hat{B}}_t=B_t-x-{\int }_0^t{h}^{\prime }(B_s)ds$ a standard Brownian motion for $t\le T$. Consequently $B$ under $\mathbb{Q}$ solves the prescribed stochastic differential equation; the existence and pathwise uniqueness theorem for a stochastic differential equation identifies its law with that of $X$.

The Itô formula gives ${\int }_0^t{h}^{\prime }(B_s)d{B}_s=h(B_t)-h(x)-\frac{1}{2}{\int }_0^t{h}^{\prime \prime }(B_s)ds$. Hence, for any bounded measurable $f$,The finite-horizon argument applies separately to every $t\ge 0$.