Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 34 / 1

Use the predictable sigma-algebra generated by $A\times \{0\}$, $A\in {\mathcal{F}}_0$, and $A\times (s,t]$, $A\in {\mathcal{F}}_s$. A previsible process is a measurable function for this sigma-algebra. Set $J_t=1_{\{S<t\le T\}}$, with $J_0=0$. The events $\{S<t\}$ and $\{T<t\}$ are in ${\mathcal{F}}_t$, because they are countable unions of events $\{S\le q\}$ and $\{T\le q\}$ for rational $q<t$. Thus $J$ is an adapted process. Each path is left-continuous, including at the two endpoints, and is zero if $S=T$.

For a direct measurability argument, letEach summand is measurable for the predictable sigma-algebra, since its coefficient is ${\mathcal{F}}_{k{2}^{-n}}$-measurable. At every $t>0$ the sampling time approaches $t$ strictly from below, so left-continuity gives $J_t^{(n)}\to J_t$. ThereforeThe original PDF has a right-closed interval. The open right endpoint in the converted TeX is a transcription error; replacing $(S,T]$ by $(S,T)$ is not valid for arbitrary stopping times.

An elementary predictable process with stopping-time intervals has the form $H={\sum }_{j=0}^{r-1}{h}_j{1}_{({\tau }_j,{\tau }_{j+1}]}$, where the stopping times are ordered and $h_j$ is bounded and ${\mathcal{F}}_{{\tau }_j}$-measurable. Deterministic endpoints give the usual simple predictable process. Define the zero-starting stochastic integral byCommon refinement and telescoping show that this definition is independent of the representation. Use $M_{\infty }$ for an infinite endpoint. The L2-bounded martingale convergence theorem gives convergence of $M_t$ in $L^2$ to $M_{\infty }$ and $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$.

Put $D_j=M_{{\tau }_{j+1}}-M_{{\tau }_j}$. The optional sampling theorem gives $\mathbb{E}[D_j\mid {\mathcal{F}}_{{\tau }_j}]=0$. For $j<k$, the variable $h_j{D}_j{h}_k$ is ${\mathcal{F}}_{{\tau }_k}$-measurable; hence $\mathbb{E}[h_j{D}_j{h}_k{D}_k]=0$. Applying the same theorem to the assumed martingale $M^2-[M]$ givesConsequently the Itô isometry follows:The unbounded stopping times cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ uniformly integrable and makes the family of stopped $M^2$ uniformly integrable, by conditional Jensen applied to $M_{\infty }^2$. Also $\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2<\infty$, from monotone convergence theorem and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating stopping times and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.

The stopping-interval indicator is an elementary predictable process with stopping-time intervals. Its stochastic integral isIt is zero before $S$, equals $M_t-M_S$ between $S$ and $T$, and equals $M_T-M_S$ after $T$. For a martingale with jumps, the jump at $S$ is excluded and the jump at $T$ is included, exactly as required by the half-open interval.

Use the shifted filtration ${\mathcal{G}}_s={\mathcal{F}}_{T+s}$. Each $T+s$ is a stopping time, and the optional sampling theorem implies that $M_s^{(T)}=M_{T+s}-M_T$ is a ${\mathcal{G}}_s$-martingale. It is $L^2$-bounded, for example by $\mathbb{E}|M_{T+s}-M_T{|}^2\le 4\mathbb{E}{M}_{\infty }^2$. Its quadratic variation is $[M^{(T)}{]}_s=[M{]}_{T+s}-[M{]}_T$. The shifted process $H_{T+s}$ is continuous and ${\mathcal{G}}_s$-adapted, hence predictable. Its boundedness and the integrability of the bracket make the right-hand stochastic integral well-defined in $L^2$.

For a deterministic partition $0=s_0<\cdots <s_r=t$, use left-endpoint coefficients $H_{T+s_j}$. Both integrals have the same approximating sumOn the original clock this is the integral of the simple process on the stopping-time intervals $(T+s_j,T+s_{j+1}]$; on the shifted clock it is a deterministic-partition integral. As the mesh tends to zero, continuity of $H$ gives pathwise uniform approximation on the random compact interval $[T,T+t]$. The Itô isometry and domination by a constant times the integrable $[M{]}_{\infty }$ give $L^2$ convergence of both sums. Thus stopping-time shift of a stochastic integral yieldsThe left integral includes increments strictly after $T$, so no jump at the starting time is added.