Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 30L (Stochastic Financial Models)

The increment of the stopped process isBecause $T$ is a stopping time, $\{T>n\}\in {\mathcal{F}}_n$. Taking conditional expectation therefore gives zero. The stopped variables are adapted and integrable, so the stopped martingale in discrete time is a martingale.

The martingale property givesbecause ${\mathcal{F}}_0$ is trivial. Since $T<\infty$ almost surely,
$M_{n\wedge T}\to M_T$ almost surely. The assumed uniform bound permits bounded convergence, so

For the symmetric increment ${\xi }_{n+1}$,The same multiplier holds for $w^{-X_n}$. Thus chooseThen each of $z^n{w}^{X_n}$ and $z^n{w}^{-X_n}$ is a martingale, and every linear combinationis a martingale.

Given $0<z<1$, takeChoose $A,B$ so thatSolving,The stopped discounted martingale is bounded, and $T$ is finite, so part (b) givesTherefore the discounted symmetric random-walk exit transform is