Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 35 / 1

Relative to a filtration satisfying the usual conditions, a local martingale is an adapted càdlàg process $X$ for which there exist increasing stopping times $S_n\uparrow \infty$ almost surely such that every stopped process $X_t^{S_n}=X_{t\wedge S_n}$ is an integrable martingale. Thus, for $s\le t$,The sequence is a localizing sequence. A continuous local martingale additionally has continuous sample paths. The usual definition entails integrability of the initial value; local martingales need not be integrable at later times without stopping.

Let $V_t$ be the total variation of $X$ on $[0,t]$, and set ${\tau }_n=\inf \{t:V_t\ge n\}$. Adaptation and continuity of $V$ make these stopping times; finite variation on every compact interval gives ${\tau }_n\uparrow \infty$. The stopped process $U=X^{{\tau }_n}$ satisfies $|U_t|\le n$ and has total variation at most $n$. It is a martingale by the bounded local martingale criterion, and is square-integrable.

Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_m=t$ with mesh tending to zero. The martingale-difference orthogonality of the increments givesPathwise, however,by uniform continuity on $[0,t]$. These sums are bounded by $n^2$, so the dominated convergence theorem gives $\mathbb{E}{U}_t^2=0$. Hence $U_t=0$ almost surely. Take all rational $t$ on a common probability-one event and use continuity to obtain $U\equiv 0$. Then let $n\to \infty$. This proves the continuous finite-variation local martingale is constant result in its zero-starting form:The proof does not assume the uniqueness of quadratic variation that is to be established next.

For two candidates $A,C$, their difference is a continuous adapted finite-variation process. It is also a local martingale, sinceApply part (b) after subtracting its initial value. ConsequentlyThis proves the uniqueness of an increasing square compensator when the initial value is prescribed, in particular under the standard quadratic variation normalization $A_0=C_0=0$.

The printed assertion omits that normalization and is literally false. For standard Brownian motion $Y$, both $A_t=t$ and $C_t=t+1$ are continuous adapted increasing processes; both $Y_t^2-t$ and $Y_t^2-t-1$ are martingales. Thus uniqueness holds after fixing the initial value; otherwise it holds only up to an initial constant.

For a simple predictable process, define its stochastic integral byIt is constant after $t_n$, so the symbol $(H\cdot M{)}_{\infty }$ means its value there. Put ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $k<\ell$, then $Z_k{\Delta }_{k}M{Z}_{\ell }$ is ${\mathcal{F}}_{t_{\ell }}$-measurable, and the conditional expectation of ${\Delta }_{\ell }M$ given that sigma-algebra is zero. Boundedness of the coefficients and square-integrability of $M$ justify the expectations, so all cross terms vanish:For each diagonal term, the martingale increment identity also givesIndeed, the omitted term $2Z_k^2{M}_{t_k}{\Delta }_{k}M$ has expectation zero. Apply the assumed martingale property of $M^2-[M]$ to replace this difference of squares by the corresponding quadratic variation increment. ThusThe intervals $(t_k,t_{k+1}]$ in the Lebesgue–Stieltjes integral exactly match these increments, including possible jumps of $M$.