Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 202 / 2

The bounded continuous local martingale $X$ is a square-integrable martingale. Since $M^{(n)}$ is a discrete predictable transform of $X$, it has mean zero. The identity $A_t^{(n)}=X_t^2-M_t^{(n)}$ therefore givesuniformly in $t$ and $n$.

The Itô isometry for the elementary predictable integrand in $M^{(n)}$ gives

Because $A_t^{(n)}=X_t^2-M_t^{(n)}$, the inequality $(x-y{)}^2\le 2x^2+2y^2$ and part (ii) yield

By the stated Cauchy property and completeness of ${\mathcal{M}}^2$, there is a square-integrable continuous martingale $M$ such thatSet $A_t=X_t^2-M_t$. This process is continuous and adapted, andby the Doob L2 maximal inequality. The process $A$ is the quadratic variation $[X]$.

If $A_t=0$ almost surely, then $X_t^2=M_t$ is a nonnegative martingale starting from zero. A nonnegative random variable of expectation zero vanishes almost surely, so $X_t=0$ almost surely for each $t$. Applying this on the nonnegative rational times and using path continuity shows that $X_t=0$ simultaneously for every $t\ge 0$ almost surely.

Let ${\tau }_m=\inf \{s:|X_s|\ge m\}$. The stopped process $X^{{\tau }_m}$ is bounded, so part (a) gives a continuous adapted quadratic variation $A^{[m]}$. These processes agree before the smaller stopping time, because their dyadic sums agree there and the limits are unique in probability. They therefore paste to a continuous adapted process $A$ with $A_{s\wedge {\tau }_m}=A_s^{[m]}$.

For fixed $t$ and $\epsilon >0$,The second term tends to zero by part (a), while continuity of $X$ on $[0,t]$ makes $\mathbb{P}({\tau }_m\le t)\to 0$. This proves convergence uniformly on compact intervals in probability.

Fix $t$. Uniform continuity of the sample path on $[0,t]$ givesSince $p<2$,The assumed pathwise boundedness of ${\sup }_n{B}_t^{(n)}$ makes the right-hand side tend to zero almost surely. Part (b) also gives $A_t^{(n)}\to A_t$ in probability, so uniqueness of limits in probability yields $A_t=0$ almost surely. The vanishing-quadratic-variation result from part (a), after localization, makes $X$ identically zero. Consequently every $B_t^{(n)}$ is zero and ${\sup }_n{B}_t^{(n)}=0$ almost surely.