Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 202 / 2 / c / Solution

The relevant fact is uniform convergence on compacts in probability under an absolutely continuous measure change. Write $Z=d\tilde{\mathbb{P}}/d\mathbb{P}$. For any event $E$ and $K>0$,The last term tends to zero as $K\to \infty$ because $Z$ is integrable. Consequently, $\mathbb{P}(E_n)\to 0$ implies $\tilde{\mathbb{P}}(E_n)\to 0$. Apply this to $E_n=\{{\sup }_{t\le T}|C_t^n-[M,N{]}_t|>\epsilon \}$ for each finite $T$ and $\epsilon >0$. The given dyadic sums therefore have the same uniform convergence on compacts in probability limit under the new probability measure.

This proves quadratic covariation under an absolutely continuous measure change:Taking $N=M$ proves the corresponding quadratic variation assertion directly; it has not been assumed. The ceiling in the printed sum includes one grid interval extending beyond $t$. This does not affect the limit: on $[0,T]$ the resulting extra product is bounded by the product of the two path oscillations on mesh $2^{-n}$, which tends to zero by uniform continuity on $[0,T+1]$.

The bracket under the new probability measure must be understood as the quadratic covariation of continuous semimartingales. Semimartingale stability under an absolutely continuous measure change ensures this class is preserved; the local martingale property need not be. For example, on a finite horizon, weighting Brownian motion by $\exp (\mu B_T-{\mu }^{2}T/2)$ makes $B_t-\mu t$ a Brownian motion by the Girsanov theorem, so $B$ has nonzero drift under the new measure even though its quadratic variation is still $t$. The PDF additionally assumes that $M,N$ remain local martingales under the new probability measure, so its bracket is also defined directly by the local martingale characterization.