Let $X=(X_n)_{n\in\mathbb{N}}$ be a stochastic process. The optional stopping theorem (OST) requires $X$ to be a martingale. The OST assures that under certain conditions on the stopping time $\tau$, it holds that $\mathbb{E}[X_\tau]=\mathbb{E}[X_0]$.

But just by being a martingale it would follow that $\mathbb{E}[X_n]=\mathbb{E}[X_0],\ \forall n\in\mathbb{N}$.

This is even used in the proof of the OST. What is so special in the OST?


1 Answer 1


Yes, the equality $\mathbb{E}[X_n]=\mathbb{E}[X_0]$ holds, but $X_T$ is a different random variable, this is not an element of the sequence $X_n$. It depends on the values of $T$ as well. So the expectation of $\mathbb{E}[X_T]$ might be different.

For example, consider the simple random walk. Let $(Y_n)_{n=1}^\infty$ be a sequence of independent random variables such that $\mathbb{P}(Y_n=1)=\mathbb{P}(Y_n=-1)=\frac{1}{2}$. Then we define $S_0=0$ and $S_n=Y_1+...+Y_n$ for all $n\in\mathbb{N}$. Then $(S_n)$ is a martingale with respect to the filtration $\mathcal{F_0}=\{\emptyset, \Omega\}, \mathcal{F_n}=\sigma(Y_1,..,Y_n)$. Obviously $\mathbb{E}[S_n]=0$ for all $n$. But now we can define the stopping time $T=\inf\{n\in\mathbb{N}: S_n=1\}$. It is a not very trivial fact that $\mathbb{P}(T<\infty)=1$, this can be proved after some work. But then from the definition it follows that $\mathbb{E}[S_T]=1$, because $S_T=1$ at every point where $T$ is finite. (which happens almost surely)

So in general $\mathbb{E}[X_T]=\mathbb{E}[X_0]$ might not hold. The optional stopping theorem gives some conditions under which the equality holds.

  • $\begingroup$ This is a good answer! What does $X_\tau$ in terms of measurable mappings mean? Do I understand $X_\tau$ as $X(\omega)_{\tau(\omega)}$? $\endgroup$ Commented Jul 3, 2020 at 16:42
  • 1
    $\begingroup$ It is $X_{\tau(\omega)}(\omega)$. I believe this is what you meant. $\endgroup$
    – Mark
    Commented Jul 3, 2020 at 16:45

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