Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 101 / 3 / a

A discrete-time martingale relative to a filtration $({\mathcal{F}}_n)$ is an adapted process $(M_n)$ with $\mathbb{E}|M_n|<\infty$ andIterating conditional expectation also gives $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$ for $m\ge n$. It is bounded in $L^2$ when each $M_n$ is square integrable andThis is a uniform bound on the second moments, not a pathwise bound on all sample values.

The optional stopping theorem for bounded stopping times says that, if $\sigma \le \tau \le K$ for a deterministic finite $K$, thenHere ${\mathcal{F}}_{\sigma }$ is the stopping-time sigma-algebra. The second assertion follows by taking $\sigma =0$ and then expectations. An $L^2$ bound is not needed: integrability at the finitely many deterministic times suffices.

For a proof, both stopped values are integrable since $|M_{\tau }|,|M_{\sigma }|\le {\sum }_{j=0}^K|M_j|$. ExpandIf $A\in {\mathcal{F}}_{\sigma }$, then $A\cap \{\sigma \le k\}\in {\mathcal{F}}_k$ by the definition of the stopping-time sigma-algebra, and $\{\tau >k\}\in {\mathcal{F}}_k$ because $\tau$ is a stopping time. Thus the indicator of $A\cap \{\sigma \le k<\tau \}$ is ${\mathcal{F}}_k$-measurable. Multiplying the sum by $1_A$, taking expectations, and conditioning each summand on ${\mathcal{F}}_k$ gives zero, by the martingale increment property. Therefore $\mathbb{E}[1_A{M}_{\tau }]=\mathbb{E}[1_A{M}_{\sigma }]$ for every $A\in {\mathcal{F}}_{\sigma }$. Since $M_{\sigma }$ is ${\mathcal{F}}_{\sigma }$-measurable, this is exactly the claimed conditional expectation identity.