4
$\begingroup$

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.

Thanks in advance.

$\endgroup$
2
  • 1
    $\begingroup$ (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. $\endgroup$
    – aduh
    Commented Sep 25, 2016 at 23:56
  • $\begingroup$ Be careful not to confuse $X_T$ with the stopped process $X_{T\wedge n}$, which is sometimes denoted $X^T$. $\endgroup$
    – Math1000
    Commented Sep 26, 2016 at 3:15

1 Answer 1

0
$\begingroup$
  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...).
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .