Put
$$
f_n(x)=\frac1n\log\cosh(nx).
$$
Then $f_n\to|\mathord\cdot|$ uniformly, $f_n'(x)=\tanh(nx)$, and $f_n''(x)=n\operatorname{sech}^2(nx)$. <Itô formula> gives
$$
f_n(X_t)=f_n(X_0)+\int_0^t\tanh(nX_s)\,dX_s
+\frac12\int_0^t n\operatorname{sech}^2(nX_s)\,d[X]_s.
$$
The bounded predictable integrands $\tanh(nX)$ converge pointwise to $\operatorname{sgn}(X)$, with $\operatorname{sgn}(0)=0$. Part (c) therefore makes the stochastic integrals converge u.c.p. The left-hand side converges u.c.p. to $|X|$, so the increasing continuous processes
$$
A_t^{(n)}=\frac12\int_0^t n\operatorname{sech}^2(nX_s)\,d[X]_s
$$
also converge u.c.p. Their limit $A$ has a continuous increasing version: extract almost-sure locally uniform convergence from each compact interval and use a diagonal argument. We obtain
$$
|X_t|=|X_0|+\int_0^t\operatorname{sgn}(X_s)\,dX_s+A_t,
$$
the <Tanaka formula> with $A=L_t^0(X)$. It expresses $|X|$ as a continuous local martingale plus a continuous finite-variation process, so $|X|$ is a continuous semimartingale.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-202/2/d/solution