The and part (b) show that, before $\tau=\tau_x\wedge\tau_y$,
$$
F(V_t)=\mathbb P(\tau_x>\tau_y\mid\mathcal F_t).
$$
Thus $F(V_{t\wedge\tau})$ is a bounded <martingale>. From the given stochastic differential equation,
$$
d[V]_t=\frac{(1-V_t)^2}{Y_t^2}dt.
$$
The <Itô formula> says that the drift of $F(V_t)$ is
$$
\frac1{2Y_t^2}\left[(1-v)^2F''(v)+\left(\frac{d-1}{v}+(3-d)v-2\right)F'(v)\right]_{v=V_t}dt.
$$
It must vanish. Dividing by $(1-v)^2/2$ and using the algebraic identity supplied in the question gives
$$
F''(v)+\left(\frac{d-1}{v}+\frac{2d-4}{1-v}\right)F'(v)=0,
\qquad v<0.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-203/3/c/solution