Rearranging part b expresses the last term as
$$
A_t^\epsilon:=\frac12\int_0^t
\frac{\epsilon^2}{(\epsilon^2+B_s^2)^{3/2}}ds
=f_\epsilon(B_t)-\epsilon-
\int_0^t\frac{B_s}{\sqrt{\epsilon^2+B_s^2}}dB_s.
$$
Parts c and d, together with stability of ucp convergence under <addition>, show that
$$
A^\epsilon\xrightarrow{\mathrm{ucp}}
A,
\qquad
A_t=|B_t|-\int_0^t\operatorname{sgn}(B_s)dB_s.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-202/1/e/solution