Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 30K (Stochastic Financial Models)

The pair $(M_n,{\mathcal{F}}_n{)}_{n\ge 0}$ is a martingale when $({\mathcal{F}}_n)$ is a filtration, each $M_n$ is ${\mathcal{F}}_n$-measurable and integrable, and

The variable $M_n$ is ${\mathcal{F}}_n$-measurable. Positivity, independence, and the mean-one assumption giveso it is integrable. Since $Y_{n+1}$ is independent of ${\mathcal{F}}_n$,Thus $(M_n)$ is the product martingale from independent mean-one factors.

Define $A_0=0$ and recursively setThe increment is ${\mathcal{F}}_n$-measurable, so $A_{n+1}$ is ${\mathcal{F}}_n$-measurable by induction; hence $A$ is predictable. Conditional expectation preserves integrability, so $A_n$ andare integrable and adapted. Their increments satisfyso $M$ is a martingale.

For uniqueness, suppose $X=M+A=M^{\prime }+A^{\prime }$ are two such decompositions. Thenis both a martingale and predictable, with $D_0=0$. Since $D_{n+1}$ is ${\mathcal{F}}_n$-measurable,Induction gives $D=0$, and therefore $M=M^{\prime }$ and $A=A^{\prime }$. This proves the Doob decomposition of an adapted integrable process.

Every martingale is a supermartingale, with equality in its defining conditional expectation. Taking expectations and using the tower property of conditional expectation givesso induction yields

Conversely, for a supermartingale defineThe assumed equality of expectations givesA nonnegative random variable of expectation zero vanishes almost surely, sofor every $n$. Hence $X$ is a martingale. Together with part (i), this proves the constant-expectation supermartingale is a martingale criterion.