How can a stopping time be independent of its stochastic process? I was reading about a special case of Wald's equation, which led me to the following question:
If $X_t$ is a sequence of iid RV's, and $\tau$ is a stopping time for this discrete stochastic process, how is it possible for $\tau$ to be independent of $X_i$'s? Wouldn't this violate the definition of a stopping time which says that we should be able to tell if the stopping time is equal to $k$ by looking only at $X_1$,...,$X_k$? Doesn't a stopping time by definition depend on the process?
I'm asking because I saw an example of an independent stopping time in my text: do a random walk on the integers, and at every step flip a coin, when the coin is heads for the first time, stop. But I cannot see why this is a stopping time because if you were given $X_1,...,X_k$ here, you don't have enough information to conclude whether $k$ is the time to stop.
 A: An example of a stopping time which is independent of its stochastic process is a constant stopping time. Say, $\tau=3$ $w.p. 1$
I believe this is also the only such example, since $P(\tau=k|X_1,..X_k) = P(\tau = k) $ since $\tau$, is independent of the process. On the other hand, since $\tau$ is a stopping time, that is the event $\{\tau=k \}$ is completely determined by $\{X_1,...,X_k\}$, $P(\tau=k | X_1,...,X_n) \in \{0,1\}$. So that for all $k \in N$ $P(\tau = k) \in \{0,1\}$ So that $\tau$ is a constant random variable.
A: Indeed $\tau$ need not be independent of the process $\{X_n\}$. In the example where $\mathbb P(X_1=1)=1-\mathbb P(X_1=-1)=p$ and $$\tau=\inf\{n>0:X_n=1\}, $$ we have $$\tau = \sum_{n=1}^\infty p(1-p)^{n-1}\mathsf 1_{\{X_n=1\}\cap\{X_{n-1}=\cdots=X_1=-1\}}, $$ so $\sigma(\tau)\subset\sigma\left(\bigcup_{n=1}^\infty X_n \right)$ as $\tau$ is a measurable function of $\{X_n\}$.
In general, a nonnegative integer-valued random variable $T$ is a stopping time with respect to the filtration $\{\mathcal F_n\}$ if $\{T= n\}\in\mathcal F_n$ for all $n$. This implies that $$\sigma(T)=\sigma\left(\bigcup_{n=0}^\infty \{T=n\}\right)\subset\sigma\left(\bigcup_{n=0}^\infty \mathcal F_n\right),$$ so $T$ is independent of $\{X_n\}$ if and only if $\mathbb P(A)\in\{0,1\}$ for all $A\in\sigma(T)$, i.e. $\mathbb P(T=n)=1$ for some $n$.
