## 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

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 🙂