# Expectation at stopping time = Expectation at the beginning

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!

• The answer is YES for the first question and NO for second. Nov 23, 2019 at 11:42
• 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 Nov 23, 2019 at 15:01

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...