A simple question about the definition of martingales

Question

The definition of Martingale denotes that $E(M_{n+1}\mid\mathcal{F}_n)=M_n$. This implies $E(E(M_{n+1}\mid\mathcal{F}_n))=E(M_n)$. Then does it mean that $E(M_{n+1})=E(M_n)$ using the tower property? If so, what gives a significance to the optional stopping theorem that tells us $E(M_T)=E(M_0)$?

Answer 1

While it is correct that $E(M_n) = E(M_0)$ for all $n$, stating $E(M_T) = E(M_0)$ is pretty different as $T$ is itself a random variable so $M_T$ is not equal to any of the $M_n$'s (as a random variable).

To see that the "almost-surely boundedness" hypothesis is crucial in this theorem, consider a one-dimensional integer random walk $X$.

Then, you can decide to stop the walk at time $T$ whenever you reach a certain value $B \in \mathbb{N}$ fixed in advance. Then, the occurrence of $T \leq t$ only depends on $X_0, \cdots, X_t$, so $T$ is a stopping time.

Thus, you clearly have $E(X_T) = B$, while $E(X_n)=E(X_0)=0$ for all $n$.