The solution is the <geometric Brownian motion>
$$
X_t=x\exp\left((\beta-\tfrac12\sigma^2)t+\sigma B_t\right)>0.
$$
Its <infinitesimal generator> is
$$
Lf(x)=\beta xf'(x)+\frac12\sigma^2x^2f''(x).
$$
For $\gamma=1-2\beta/\sigma^2\ne0$, one has $L(x^\gamma)=0$. Optional stopping of $X_{t\wedge T_r\wedge T_R}^\gamma$ and the boundary values therefore give
$$
\boxed{\mathbb P_x(T_r<T_R)=
\frac{R^\gamma-x^\gamma}{R^\gamma-r^\gamma},
\qquad \gamma=1-\frac{2\beta}{\sigma^2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 4 / 3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-202/4/3/solution