Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 1

Put $q=1-p$. Sincethe martingale condition isThe root in $(0,1)$ is therefore

Let $\tau =T_a\wedge T_b$. The walk exits the finite interval $[a,b]$ almost surely, and the stopped martingale $\varphi (S_{n\wedge \tau })$ is bounded between ${\lambda }^b$ and ${\lambda }^a$. The optional stopping theorem and bounded convergence theorem giveSolving for the first probability gives the biased gambler's ruin probability

As $b\to \infty$, the events $\{T_a<T_b\}$ increase to $\{T_a<\infty \}$: every path that reaches $a$ has a finite maximum before doing so. Since ${\lambda }^b\to 0$,Similarly, as $a\to -\infty$, the events $\{T_b<T_a\}$ increase to $\{T_b<\infty \}$, andHence

The strong law of large numbers givesBecause $0<\lambda <1$, it follows that ${\lambda }^{S_n}\to 0$ almost surely. If this martingale were uniformly integrable, almost-sure convergence would imply convergence in $L^1$, and thereforeBut the martingale has constant expectation $\mathbb{E}[{\lambda }^{S_n}]=1$. This contradiction proves that