# martingale and stopping time

I currently learn martingale and I am confused on martingale with a stopping time.

Dobb's optional stopping says that if $T$ is bounded, $\{X_n\}$ is a martingale, then $E[X_T] = E[X_0]$.

I have two questions:

1. Stopping time $T$ is a random variable and $X_n$ is also a random variable. But how to understand $X_T$?

2. $\{X_n\}$ is a martingale so $\{X_n\}$ have the same expectation already. What's the fancy part of Dobb's optional stopping? I mean why it is important.

• (1) $X_{T}$ is given by $X_{T(\omega)}(\omega)$. (2) Note that the result doesn't hold if $T$ is unbounded. Consider simple symmetric random walk with $T = \inf(n: X_{n} = 2)$ for example.
• Be careful not to confuse $X_T$ with the stopped process $X_{T\wedge n}$, which is sometimes denoted $X^T$. Commented Sep 26, 2016 at 3:15
1. $$X_T : \omega\in\Omega \mapsto X_{T(\omega)}(\omega)\in\mathbb{R}$$ ;
2. $$T$$ is random, but whatever value it takes, on average, $$X_T$$ will have the same value as $$X_0$$ (on average...).