Core question

If $(X_n)$ is a martingale and $T$ is a stopping time, when do we still have

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

General statement

Let $(X_n)$ be a martingale and let $T$ be a stopping time. If:

  1. $\mathbb E[|X_T|]<\infty$
$$ \lim_{n\to\infty}\mathbb E\bigl[|X_n|\mathbf 1(T>n)\bigr]=0 $$

then

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

Why extra conditions are needed

Martingale plus stopping time is not enough by itself. A stopping time can be so large, or the process can grow so much before stopping, that the equality fails.

Main idea

The theorem is proved by truncating the stopping time:

$$ T_m=\min(T,m). $$

Each $T_m$ is bounded, so the bounded version of OST applies first, and then one passes to the limit $m\to\infty$.