Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 29 / 2

A càdlàg martingale is an adapted process $X$ with càdlàg paths, integrable $X_t$ at each finite time, andThe càdlàg requirement means that the chosen path version is right-continuous and has a finite left limit at every positive time. With a complete filtration, it suffices that this holds on one common event of probability one, since the process can be redefined on the exceptional null set.

The usual hypotheses for a filtration are completeness and right continuity. Completeness means that ${\mathcal{F}}_0$ contains every subset of every null event of the ambient probability space; right continuity means

Fix $t>0$ and partition $[0,t]$ into $m$ equal intervals, with $r_k=kt/m$. Round $t\wedge T$ upward within this interval. The corresponding sampled value isEvery time in this sum is at most $t$. The stopping time property makes each indicator function ${\mathcal{F}}_t$-measurable, while the adapted process property makes each sampled value ${\mathcal{F}}_t$-measurable. Consequently $Y_m$ is ${\mathcal{F}}_t$-measurable.

The rounded times approach $t\wedge T$ from the right, so right continuity gives $Y_m\to X_{t\wedge T}$ for the chosen pathwise càdlàg version. If path regularity is instead stated only almost surely, the usual complete filtration handles the exceptional null set; on an incomplete filtration one should formulate that case as existence of an adapted version. At $t=0$, the value is simply $X_0$. Thus $X^T$ is an adapted process, by adaptedness of a stopped right-continuous process. The argument actually needs neither the martingale property nor boundedness of $T$.

Recall the stopping-time sigma-algebra:For the proposed random time,The first set belongs to ${\mathcal{F}}_t$. Since $S\le T$, the second can be writtenwhich also belongs to ${\mathcal{F}}_t$. Therefore $U$ is a stopping time, as in pasting ordered stopping times. Moreover $S\le U\le T$, so $U$ is a bounded stopping time.

Fix deterministic $0\le s<t$ and $A\in {\mathcal{F}}_s$. The pasting ordered stopping times argument shows that $U=s{1}_A+t{1}_{A^c}$ is a bounded stopping time. The given stopped-expectation property, applied to $U$ and to the deterministic stopping time $t$, yieldsThis holds for every $A\in {\mathcal{F}}_s$. Since $X_s$ is ${\mathcal{F}}_s$-measurable and both time values are integrable, it is exactly the defining test for conditional expectation:Hence $X$ is a martingale. This proof of the characterization of a martingale by bounded continuous-time stopped expectations uses only deterministic two-time pastings; the càdlàg assumption and the usual conditions for a filtration are stronger than needed for this implication.