Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 30K (Stochastic Financial Models)

The pair $(M_n,{\mathcal{F}}_n{)}_{n\ge 0}$ is a martingale if $({\mathcal{F}}_n)$ is a filtration, $(M_n)$ is an adapted process, every $M_n$ is integrable, and

The process is adapted to its natural filtration. At each fixed time it has finite support and is therefore integrable. Conditional on $X_{n-1}=0$,Conditional on $X_{n-1}\ne 0$,The Markov property now givesso $(X_n)$ is a martingale.

Assume (i), and let $\tau$ be a stopping time. The increments of the stopped process satisfyBecause $\{\tau >n\}\in {\mathcal{F}}_n$, taking conditional expectation given ${\mathcal{F}}_n$ makes the right-hand side zero. Integrability follows froma finite sum of integrable random variables. Hence $M^{\tau }$ is a martingale, proving (ii).

Assume (ii). A martingale has constant expected value by the tower property of conditional expectation, so for every stopping time $\tau$ and every $n$,This is (iii).

Assume (iii). Fix $n\ge 0$ and $A\in {\mathcal{F}}_n$, and defineThis is a stopping time: $\{T\le n\}=A\in {\mathcal{F}}_n$. Applying (iii) at time $n+1$ first to $T$ and then to the deterministic stopping time $n+1$ givesSubtracting yieldsBy the defining characterization of conditional expectation, this says $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Thus $M$ is a martingale, proving (i). Together the three implications establish the characterization of a martingale by stopped expectations.