# Counterexample in optional stopping martingale

Problem: Give an example of submartingale $\{X_n\}$ with $\sup_nE |X_{n-1}-X_n|<\infty$ and stopping time $N$ with $E[N]<\infty$ such that $\{X_{n\wedge N}\}$ is not uniformly integrable.

Attempt: I think this asks us to give a counterexample to the optional stopping theorem with bounded increments. Since we have a finite a.s. stopping time condition, I think the example maybe some kind of low dimensional random walk. But I am not really sure..

Are you sure you have the assumptions correct? This link Here has a Theorem 4.7.5 which states that if $\{X_n\}_{n\geq0}$ is a sub-martingale and $\mathbb{E}\big[|X_{n+1} - X_n|\; \big| \; \mathcal{F}_n\big] < B < \infty$ and $\mathbb{E}[N] < \infty$ then $\{X_{n \wedge N}\}_{n \geq 0}$ is uniformly integrable.
Taking expectations $\mathbb{E}\Big[\mathbb{E}\big[|X_{n+1} - X_n| \; \big| \; \mathcal{F}_n\big]\Big] = \mathbb{E}\big[|X_{n+1} - X_n|\big] < B$. Therefore, it must be that $\{X_n\}_{n\geq1}$ cannot have almost surely uniformly bounded increments for your result to hold.
• Why does $\mathbb{E}(|X_{N \wedge n}|) \leq \infty$ (...which is the essence of the last calculation in your answer) tell you that $\{X_{N \wedge n}\}_n$ is not uniformly integrable? – saz Mar 6 '15 at 6:28
• @john Nevertheless he is right about the fact that under the given assumptions there is no such counterexample. If we drop the assumption $\sup_n \mathbb{E}(|X_n-X_{n-1}|)<\infty$, then it possible to construct a counterexample. – saz Mar 6 '15 at 18:04