Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 30 / 1

Let $N$ be the zero-starting continuous local martingale, and let $V_t$ be the total variation of its path on $[0,t]$. Stop when $|N|$ or $V$ reaches $n$, and also at time $n$. The stopped process is bounded, hence a true square-integrable martingale by the bounded local martingale criterion, and its total variation is bounded by $n$.

For a deterministic partition $0=t_0<\cdots <t_m=t$, martingale increments are orthogonal in $L^2$, soThe sum inside the expectation is at most the largest absolute increment times the stopped total variation. As the mesh tends to zero, path continuity makes the largest increment tend to zero. The product is bounded by $2n^2$, so the dominated convergence theorem makes the displayed expectation zero. Thus the stopped process vanishes at every fixed time. Use rational times, continuity and ${\tau }_n\uparrow \infty$ to concludeThis proves that a continuous finite-variation local martingale is constant.

Write a normalized semimartingale decomposition as $X=X_0+N+A$, with both $N_0=A_0=0$, $N$ a continuous local martingale and $A$ of finite variation. The difference of two such martingale parts is also the difference of two finite-variation parts, and hence vanishes by the result just proved. The other parts coincide too. The normalized continuous semimartingale decomposition is unique. Fixing the initial constants is necessary: without normalization a constant can be transferred between the two parts.

Fix a sequence of deterministic partitions with mesh tending to zero and define the completed-increment sumsFor a continuous semimartingale, these sums converge to $[X{]}_t$ in uniform convergence on compacts in probability, the standard path-increment formula for quadratic variation. The sums use only the path of $X$, and do not mention a filtration. If the same process is a semimartingale for two filtrations, the same sums have both proposed quadratic variations as limits. Uniqueness of limits in probability, followed by continuity, makes these limits indistinguishable.

For the change of measure, let $R=d\mathbb{Q}/d\mathbb{P}$. If $\mathbb{P}(A_n)\to 0$, thentruncate $R$ at a large constant, use $\mathbb{E}[R{1}_{\{R>K\}}]\to 0$, and then let $\mathbb{P}(A_n)\to 0$. Apply this argument to the event ${\sup }_{s\le t}|S_s^{(n)}-[X{]}_s^{\mathbb{P}}|>\epsilon$. The same sums converge to the P-version of the bracket in Q-probability, and the Q-semimartingale formula makes them converge to the Q-version. ThereforeThis is quadratic variation under an absolutely continuous measure change. With only $\mathbb{Q}\ll \mathbb{P}$, the equality is asserted under Q: a Q-version is unconstrained on a Q-null set that might have positive P-probability.

For each fixed horizon $t$, the stipulated property is the Class D process condition on $[0,t]$, and imposing it for every finite horizon gives a Class DL process.

First suppose $M$ is a true martingale. For every stopping time $T\le t$, the bounded-time optional sampling theorem givesThe conditional expectations of one integrable variable form a uniformly integrable family. To see the uniformity explicitly, put $V=\mathbb{E}(U\mid \mathcal{H})$. For any $r,K>0$, conditional Jensen and Markov inequalities giveChoose $r$ large and then $K$ large. The bound does not depend on the conditioning sigma-algebra. This proves the required uniform integrability. Bounded-time optional sampling here can also be justified directly: approximate $T$ from above by finite-grid stopping times, use discrete optional sampling with terminal value $M_t$, and use this same conditional-expectation uniform integrability and right continuity to pass to the limit.

Conversely, take a localizing sequence ${\tau }_n$ for the local martingale. For fixed $s\le t$, both families $M_{t\wedge {\tau }_n}$ and $M_{s\wedge {\tau }_n}$ are uniformly integrable, because their times are bounded by $t$. They converge almost surely to $M_t$ and $M_s$, and therefore converge in $L^1$. The stopped martingale identitypasses to the limit by $L^1$ contractivity of conditional expectation. Thus $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$, and all time values are integrable. Hence a local martingale is a true martingale exactly when it is of Class DL. This is a finite-horizon criterion, not a claim of uniform integrability over all times simultaneously.