Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 30K (Stochastic Financial Models)

The process is adapted because $X_0$ is ${\mathcal{F}}_0$-measurable, $A_k$ is ${\mathcal{F}}_{k-1}$-measurable, and $M_k-M_{k-1}$ is ${\mathcal{F}}_k$-measurable. Using the predictability of $A_n$ and the martingale increment property,The assumed integrability therefore proves that $X$ is the martingale transform of $M$ by $A$ and is a martingale.

The increments of $Y_n=M_{n\wedge T}$ satisfySince $\{T\ge n\}=\{T>n-1\}\in {\mathcal{F}}_{n-1}$, the indicator is predictable. Part (a), applied to this bounded integrand, shows that $Y$ is a martingale. Equivalently, this is the stopped martingale theorem.

Write $\Delta M_n=M_n-M_{n-1}$. Conditional on ${\mathcal{F}}_{n-1}$, the variables $X_{n-1}$, $M_{n-1}$, and $A_n$ are known, while $\mathbb{E}[\Delta M_n\mid {\mathcal{F}}_{n-1}]=0$. HenceThe denominator is positive almost surely, so the recovery of a martingale-transform integrand by conditional covariance gives

Let $D_n=M_n-M_{n-1}$. Given ${\mathcal{F}}_{n-1}$, the new information at time $n$ is only the independent symmetric sign ${\xi }_n$. Thusfor some ${\mathcal{F}}_{n-1}$-measurable $C_n,B_n$. Taking conditional expectation and using the martingale property gives $C_n=0$. More explicitly, one can takewhich is predictable. ThereforeThis is the predictable representation in a Rademacher filtration.

Choose $N$ with $T\le N$. By part (d),Each $B_k{1}_{\{T\ge k\}}$ is ${\mathcal{F}}_{k-1}$-measurable. Distinct summands are orthogonal: if $j<k$, conditioning on ${\mathcal{F}}_{k-1}$ makes every factor except ${\xi }_k$ known and $\mathbb{E}[{\xi }_k\mid {\mathcal{F}}_{k-1}]=0$. The cross terms with $M_0$ vanish in the same way, while ${\xi }_k^2=1$. Expanding the square therefore givesThis is the stopped martingale isometry in a Rademacher filtration.