Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 3

Put $t_j=j{2}^{-n}$ and ${\Delta }_j=X_{t_{j+1}}-X_{t_j}$. Telescoping $X_{t_{j+1}}^2-X_{t_j}^2=2X_{t_j}{\Delta }_j+{\Delta }_j^2$ givesThe martingale transform summands are orthogonal in $L^2$: for an earlier summand, conditioning on the sigma-algebra at the start of the later increment makes the cross expectation zero. HenceThe martingale increments themselves are also orthogonal. Since $X_0=0$, their variance sum equals $\mathbb{E}{X}_1^2\le C^2$. ThereforeOnly discrete martingale orthogonality is used here; no pre-existing quadratic variation calculation is needed.

The defining relation gives $A_1^{(n)}=X_1^2-2M_1^{(n)}$. Applying $(r-s{)}^2\le 2r^2+2s^2$ and the bound from part (a),Thusuniformly in the dyadic mesh. This controls the approximations to quadratic variation without assuming that their limits already exist.

For $n>m$, define $r_j=\lfloor j{2}^{m-n}\rfloor 2^{-m}$ and $h_j=X_{j{2}^{-n}}-X_{r_j}$. Each $h_j$ is measurable at time $j{2}^{-n}$, so the summands in the given difference formula are orthogonal martingale transforms. ThereforeLetPath continuity on the compact interval gives ${\omega }_m\to 0$ almost surely, and ${\omega }_m\le 2C$. Since $0\le j{2}^{-n}-r_j<2^{-m}$, $|h_j|\le {\omega }_m$. Thus, by the Cauchy-Schwarz inequality and part (b),The last step is the dominated convergence theorem. The bound is uniform in $n>m$, and the other ordering follows by symmetry. Hence the terminal martingale transforms are Cauchy in $L^2$.

For every $t$, subtraction of the definitions givesThe difference of the two supplied continuous martingales is a square-integrable martingale on $[0,1]$; for each fixed mesh its finite sums are bounded. For this dyadic quadratic variation of a bounded continuous martingale, the Doob L2 maximal inequality therefore givesThus the dyadic approximations to quadratic variation are Cauchy for the expected squared uniform norm, as required. No monotonicity in time of the partially completed squared-increment sums is needed.