With $A_t=\int_0^t\mu(s)ds$, the product rule gives
$$
d(e^{-A_t}X_t)=e^{-A_t}X_t\sigma(t)dB_t,
$$
so $X_te^{-A_t}$ is a local martingale under $\mathbb P$.

Set $\theta(t)=\mu(t)/\sigma(t)$. This is bounded and compactly supported, so <Novikov condition> holds and
$$
Z_\infty=\exp\left(-\int_0^\infty\theta(s)dB_s
-\frac12\int_0^\infty\theta(s)^2ds\right)
$$
defines a probability measure $d\mathbb Q=Z_\infty d\mathbb P$. By the <Girsanov theorem>, $W_t=B_t+\int_0^t\theta(s)ds$ is Brownian under $\mathbb Q$, and
$$
dX_t=X_t\sigma(t)dW_t.
$$
Thus \b[$X$ is a local martingale under $\mathbb Q$].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 5 / 3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-202/5/3/solution