I am trying to get a better understand of filtrations, and I can't seem to find any simple, concrete examples, so I will try to make one here.
Consider a discrete-time Markov chain with states $\{A,B\}$ that we run for 3 time steps. Is the following a correct representation of the corresponding filtration?
$$\mathcal{F}_0 = \{\emptyset, \{AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB\}\}$$ $$\mathcal{F}_1 = \{\emptyset, \{AAA,AAB,ABA,ABB\}, \{BAA,BAB,BBA,BBB\}, \{AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB\}\}$$ $$\mathcal{F}_2 = \{\emptyset, \{AAA,AAB\},\{ABA,ABB\},\{BAA,BAB\},\{BBA,BBB\},\{AAA,AAB,ABA,ABB\}, \{BAA,BAB,BBA,BBB\}, \{AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB\}\}$$ $$\mathcal{F}_3 = \{\emptyset, \{AAA\},\{AAB\},\{ABA\},\{ABB\},\{BAA\},\{BAB\},\{BBA\},\{BBB\}, \{AAA,AAB\},\{ABA,ABB\},\{BAA,BAB\},\{BBA,BBB\},\{AAA,AAB,ABA,ABB\}, \{BAA,BAB,BBA,BBB\}, \{AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB\}\}$$
If this is correct, what are some "stopping times" in this example? Suppose $\tau_A$ is the first time at which the chain is in state $A$. Then for $\tau_A$ to be a stopping time we must have $\{\tau_A=t\} \in \mathcal{F}_t$ for $t=0,1,2,3$? What is the event $\{\tau_A=t\}$ in the above filtration?
This is correct.
The event $\{\tau_A=1\}$ is exactly $\{AAA,AAB,ABA,ABB\}\in\mathcal F_1$. Indeed, this event means that the first step leads to $A$.
The event $\{\tau_A=2\}$ is exactly $\{BAA,BAB\}\in\mathcal F_2$. This event occures when the chain is at $B$ after the first step, and then at $A$ after the second step.
Try to find the event $\{\tau_A=3\}$ in $\mathcal F_3$.