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..


1 Answer 1


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.

If you have uniformly bounded increments, then the Theorem suggests that such an example by proof is not possible.

  • $\begingroup$ 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? $\endgroup$
    – saz
    Commented Mar 6, 2015 at 6:28
  • $\begingroup$ i agree. I think it isn't sufficient. $\endgroup$
    – john
    Commented Mar 6, 2015 at 11:59
  • $\begingroup$ @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. $\endgroup$
    – saz
    Commented Mar 6, 2015 at 18:04
  • $\begingroup$ @saz I wrote this out quickly and missed the bound. You are right it is the direction to prove it is uniformly integrable. Since a proof exists showing that under the OPs assumptions there cannot be a counter-example, I am going to delete everything after the citation of that result and proof. $\endgroup$
    – J. Stewart
    Commented Mar 6, 2015 at 19:41

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