The gives
$$
\frac{S_n}{n}\longrightarrow\mathbb E[X_1]=2p-1>0
\qquad\text{almost surely}.
$$
Because $0<\lambda<1$, it follows that $\lambda^{S_n}\to0$ almost surely. If this martingale were <uniformly integrable>, almost-sure convergence would imply convergence in $L^1$, and therefore
$$
\mathbb E[\lambda^{S_n}]\longrightarrow0.
$$
But the martingale has constant expectation $\mathbb E[\lambda^{S_n}]=1$. This contradiction proves that
$$
\boxed{(\phi(S_n))_{n\geq0}\text{ is not uniformly integrable}.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-201/1/d/solution