I hope I am not radically repeating a question here, but I have some issues understanding the martingale property at stopping times. Say I have a martingale $\{X_n,n\in\mathbb{N}\}$ and a stopping time $\tau$ which is finite:$$\mathbb{P}(\tau<\infty) = 1$$ What do I need to show to be able to say: $$\lim_{n\rightarrow\infty} \mathbb{E}\left[\hat{X}_{\tau\wedge n}\right] = \mathbb{E}\left[\lim_{n\rightarrow\infty}\hat{X}_{\tau\wedge n}\right] = \mathbb{E}\left[\hat{X}_\tau\right] = \mathbb{E}\left[\hat{X}_1\right]$$

I'll summarice the questions that are arising:

  • Do I need to explicitely show dominated convergence to exchange the limit and the expectation?
  • Does a finite stopping time imply that $\tau$ is a bounded stopping time?
  • Does the convergence of the martingale play an important role here?

I very much appreciate your help!

  • 1
    $\begingroup$ The answer is YES for the first question and NO for second. $\endgroup$ Nov 23, 2019 at 11:42
  • $\begingroup$ Thanks @KaboMurphy. How would you normally show the boundness of the stopping time and the dominated convergence? Is there a link to the almost sure or $L^p$ convergence? Ty $\endgroup$ Nov 23, 2019 at 15:01

1 Answer 1

  1. There are more things that enable us to change limit and expectation:

    $\bullet L_1$ convergence (or higher)

    $\bullet$ Dominated convergence ($X_{\tau\wedge n}$ can be bounded from above by an integrable function)

    $\bullet$ Monotone convergence ($X_{\tau\wedge n}$ is strictly positive)

  2. A finite stopping time does not imply $\tau$ to be bounded. A bounded stopping time requires: $$\tau \leq c < \infty$$ Having a finite stopping time yields for example:$$\lim_{n\rightarrow\infty}\tau \wedge n = \tau$$

  3. As previously stated, $L_p$ converges implies that the limit can be exchanged with the expectation.

This answer is not really mathematical but will help noobs like me...


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