Expectation at stopping time = Expectation at the beginning

Question

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!

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...