Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 33 / 1

A continuous adapted process $M$ is a local martingale if there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $M_t^{{\tau }_n}=M_{t\wedge {\tau }_n}$ is a martingale. In this definition $M_0$ is integrable. A continuous adapted process $A$ has finite variation if, almost surely, its paths have finite total variation of a function on each compact time interval:The permitted fact that this total-variation process is continuous and adapted makes its level-hitting times stopping times.

Suppose $M=A$, and put $N=M-M_0$. DefineThese times increase to infinity almost surely. The optional stopping theorem applied to the localized martingale shows that $N^{{\sigma }_n}$ is a martingale; it is also bounded in absolute value by $n$, with total variation at most $n$. Fix $n,t$ and write $L=N^{{\sigma }_n}$. For a deterministic partition of an interval $0=t_0<\cdots <t_r=t$, orthogonality of square-integrable martingale increments givesPathwise the sum on the right is bounded by $n{\max }_j|L_{t_j}-L_{t_{j-1}}|$. For partitions with mesh tending to zero, continuity makes this converge to zero. It is bounded by $2n^2$, so dominated convergence gives $\mathbb{E}{L}_t^2=0$. First use rational $t$, then continuity, to obtain $L\equiv 0$ almost surely. Letting $n\to \infty$ provesThis proves continuous finite-variation local martingale is constant without assuming integrability of the unstopped total variation. Continuity matters: compensated jump martingales can have finite variation without being constant.

The multidimensional Itô formula for continuous semimartingales states that a $C^2$ transformation has first-order stochastic and finite-variation integrals plus one-half its Hessian contracted with the quadratic covariations. A continuous finite-variation process has zero quadratic variation and zero quadratic covariation with a continuous local martingale. Thus in the present caseThe first integrand is continuous adapted, hence predictable, and locally bounded by stopping the two coordinate paths on exiting compact sets. The standard construction of the Itô integral then makes this integral a zero-starting continuous local martingale. The remaining integrals are pathwise Lebesgue-Stieltjes integrals. On each compact time interval their integrands are bounded along the continuous paths, so their total variations are bounded by a constant times $V_t(A)+[M{]}_t$, which is finite. Therefore the normalized terms areThe process $A^f$ is continuous adapted of finite variation and $A_0^f=0$. When $f(M_0,A_0)$ is integrable, $M^f$ is a local martingale in the definition of part (a). Uniqueness follows from part (a): the difference of two normalized decompositions is both a continuous local martingale and a continuous finite-variation process starting at zero, hence vanishes.

An initial-value qualification is necessary under that definition. The printed assumptions do not themselves guarantee $\mathbb{E}|f(M_0,A_0)|<\infty$. Take $M=0$, $A_t=G$ with $G$ a standard normal random variable known at time zero, and $f(x,a)=e^{a^2}$. All path assumptions hold, but $\mathbb{E}{e}^{G^2}=\infty$. Any decomposition with $A_0^f=0$ would force the nonintegrable value $M_0^f=e^{G^2}$, contradicting the local-martingale definition. Without an integrability hypothesis, the always-valid continuous semimartingale decomposition is $X=X_0+N+A^f$, with $N$ the zero-starting integral above. Alternatively one may use the convention that a local martingale permits an arbitrary finite initial value and localizes its centered increments. With deterministic initial data, as in part (c), there is no issue. This is initial-value integrability in normalized semimartingale decompositions.

Here $[M{]}_t=t$ and $d{A}_t=dt$, so the formula in part (b) givesIf $g$ vanishes on $\mathbb{R}\times [0,\infty )$, this integral is zero. Conversely suppose $A_t^f=0$ for all $t$ almost surely. Its integrand is continuous along each path; the fundamental theorem of calculus implies $g(B_t,t)=0$ for all $t$ on that event. For any fixed $t>0$, the normal distribution of $B_t$ has a strictly positive density at every real point. Since $g(\cdot ,t)$ is continuous, a nonzero value at any spatial point would give a nonzero value on an open interval of positive Brownian probability, a contradiction. Thus $g(x,t)=0$ for every $x$ and every positive $t$. Continuity extends this to $t=0$.

Consequently the exact condition isThese are space-time harmonic functions along Brownian motion, satisfying the backward heat equation. The second coordinate never takes negative values, so no equation is imposed for $y<0$, beyond compatibility with the assumed global $C^2$ regularity. The normalized finite-variation term is unique by part (a), so it cannot be removed by choosing a different decomposition.