Statement

Let $(X_n)$ be a martingale and let $T$ be a stopping time with

$$ \mathbb P(T<\infty)=1. $$

If there exists $B>0$ such that

$$ \mathbb P(|X_n|\le B \text{ for all } n\le T)=1, $$

then

$$ \mathbb E[X_T]=\mathbb E[X_0]. $$

Why it is useful

This is the most common practical way to apply OST in hitting-time problems.

Why it works

The boundedness before stopping gives:

  • $|X_T|\le B$, so $\mathbb E[|X_T|]<\infty$
  • $|X_n|\mathbf 1(T>n)\le B\mathbf 1(T>n)$, and since $\mathbf 1(T>n)\to 0$ a.s.,
$$ \mathbb E[|X_n|\mathbf 1(T>n)]\to 0. $$

So the general OST conditions are satisfied.