$\sigma$-algebra of the past intersection property

Question

Let $(\Omega,\mathcal{F},\mathcal{F}_{\cdot}=(\mathcal{F}_k)_{k\in\mathbb{N_0}},\mathbb{P})$ be a filtered probability space and $\sigma,\tau:\Omega\to\mathbb{N}\cup\{\infty\}$ two $\mathcal{F}_{\cdot}$-stopping times. Show that $\mathcal{F}_{\tau\wedge\sigma}=\mathcal{F}_{\tau}\cap\mathcal{F}_{\sigma}$, where $\mathcal{F}_{\tau}=\{E\in\mathcal{F}_{\cdot}|E\cap\{\tau\le k\}\in\mathcal{F}_k\,\,\forall k\in\mathbb{N_0} \}$ (the $\sigma$-algebra of the $\tau$-past).

I can prove the inclusion "$\supseteq$". But for the other one I have $E\in\mathcal{F}_{\tau\wedge\sigma}=\mathcal{F}_{\min(\tau,\sigma)}$ so $E\cap\{\tau\wedge\sigma\le k\}\in\mathcal{F}_k$ for all $k\in\mathbb{N_0}$, so $E\cap\{\tau\le k\}\cap\{\sigma\le k\}\in\mathcal{F}_k$ for all $k\in\mathbb{N_0}$ and I have to show that $E\cap\{\tau\le k\},E\cap\{\sigma\le k\}\in\mathcal{F}_k$ for all $k\in\mathbb{N_0}$. I already know that $\{\tau\le k\},\{\sigma\le k\}\in\mathcal{F}_k$ for all $k\in\mathbb{N_0}$ since $\tau$ and $\sigma$ are $\mathcal{F}_{\cdot}$-stopping times. But I am missing some argument to prove it.

Answer 1

For the other inclusion, suppose that $A \cap \{ \tau \wedge \sigma \leq t\}$ is in $\mathcal{F}_t$. You want to prove that $A \cap \{ \tau \leq t \}$ is also in $\mathcal{F}_t$ (and analogously for $\sigma$). This follows since: \begin{equation} A \cap \{ \tau \leq t \} = A \cap \{ \tau \leq t \} \cap\{ \tau \wedge \sigma \leq t\} = \color{red}{A \cap \{\tau \wedge \sigma \leq t \}} \cap \{\tau \leq t\} \in \mathcal{F}_t. \end{equation} The trick is adding that "irrelevant event" in the middle equality: then the red term is in $\mathcal{F}_t$ by assumption, whereas the rightmost term is in $\mathcal{F}_t$ by the definition of a stopping time.

Answer 2

Note that $\tau \wedge \sigma \leq \tau$ and $\tau \wedge \sigma \leq \sigma$ (with probability one). It follows that $$\{\tau \leq k\} \subseteq \{\tau \wedge \sigma \leq k\} \text{ and } \{\sigma \leq k\} \subseteq \{\tau \wedge \sigma \leq k\}$$ Now let $ E \in \mathcal{F}_{\tau \wedge \sigma}$ ,i.e. $$E \cap \{\tau \wedge \sigma \leq k\} \in \mathcal{F}_k \ \forall k$$ Thus, $$E \cap \{\tau \leq k\} = (E \cap \{\tau \wedge \sigma \leq k\}) \cap \{\tau \leq k\} \in \mathcal{F}_k \ \forall k$$ because, since $\tau$ is a stopping time, $\{\tau \leq k\} \in \mathcal{F}_k $ and the intersection of elements in $\mathcal{F}_k$ is also in $\mathcal{F}_k$ by definition of $\sigma$-algebra.

But this means that $E \in \mathcal{F}_{\tau}$.

With a similar reasoning, you get that $E \in \mathcal{F}_{\sigma} $ and you can then conclude that $E \in \mathcal{F}_{\tau} \cap \mathcal{F}_{\sigma}$