Stopping time and sigma algebra

Question

Let $S, T$ be two stopping times. Then $S\lor T$ and $S\land T$ are salso topping times with respect to $\mathcal{F}_{S\land T}=\mathcal{F}_{S}\cap \mathcal{F}_{T}$. Futhermore, $\{S\le T\}\in \mathcal{F}_{S\land T}$ and $\{S=T\}\in \mathcal{F}_{S\land T}$. For a stopping time $\tau \in \mathcal{F}_t$ and $A\in \mathcal{F}_{\tau}$. How to show that $A\cap \{\tau\le t\}\in \mathcal{F}_{\tau\land t}$?

Answer 1

For any $t\geq0$, $\tau\wedge t$ is also a sopping time. If $A\in\mathcal{F}_\tau$, then by definition of $\mathcal{F}_\tau$, for any $s\geq0$, $A\cap\{\tau\leq s\}\in\mathcal{F}_s$.

To show that $A\cap\{\tau\leq t\}$ belongs to $\mathcal{F}_{\tau\wedge t}$ notice that for any $s\geq0$, $$\begin{align} \Big(A\cap\{\tau\leq t\}\Big)\cap\{\tau\wedge t\leq s\}&=A\cap\Big(\{\tau\leq t\}\cap\{\tau\wedge t\leq s\}\Big)\\ &=A\cap\{\tau\leq s\wedge t\}\in\mathcal{F}_{t\wedge s}\subset\mathcal{F}_s \end{align}$$ This means that $A\cap\{\tau\leq t\}\in\mathcal{F}_{\tau\wedge t}$.

Answer 2

You are not applying the correct definition of $\mathcal F_{\tau \land t}$. You cannot split the proof into to the cases $\tau \leq t$ and $\tau >t$ because $\mathcal F_{\tau \land t}$ is one fixed $\sigma -$ field.

You have to show that $A \cap \{\tau \leq t\} \cap \{\tau \land t \leq s\} \in \mathcal F_s$ for each $s$. You can do this easily by considering the cases $t\leq s$ and $t >s$.