Doob’s Optional Stopping Theorem: $\xi_\tau$ vs $\xi_{\tau\land n}$

I have some troubles in understanding the Optional Stopping Theorem by Doobs.

I have a bit of confusion about the following (Brzezniak, Zastawniak – Basic Stochastic Processes p. 58-59):

Let $ \xi_n$ is a martingale and $ \tau$ is a stopping time wrt a filtration $ \mathcal{F}_{n}$ such that:

1) $ \tau < \infty$ a.s. 2) $ \xi_\tau$ is integrable 3) $ E(\xi_n1_{\tau>n})$ goes to $ 0$ as n go to $ \infty$ .

Then $ E(\xi_\tau) = E(\xi_1)$ .

The proof starts saying

$ \xi_\tau = \xi_{\tau\land n}+(\xi_\tau-\xi_n)1_{\tau>n}$

and then, by applying $ E(.)$

$ E(\xi_\tau) = E(\xi_{\tau\land n})+E(\xi_\tau1_{\tau>n})+E(\xi_n1_{\tau>n})$ .

Now, the crucial point is: I do not understand the difference between $ \xi_\tau$ and $ \xi_{\tau\land n}$ . I mean, the second one is for sure the stopped martingale at the stopping time $ \tau$ while the first one is the martingale evaluated at the instant $ \tau$ .

But if:

$ \xi_{n}=\eta_1+2\eta_2+…+2^{n-1}\eta_n$ (“game martingale”) and $ \tau<n$

then aren’t $ \xi_\tau$ and $ \xi_{\tau\land n}$ the same?

The proof goes on saying $ E(\xi_{\tau\land n})=E(\xi_1)$ thanks to definition of martingale (the stopped process is a martingale itself, so this is clear).

Thank you 🙂