Optional Stopping Theorem

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?

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