first of all i want to ask whether given any two $\{\mathcal{F}_t\}$-stopping times $\sigma, \tau$ is the following properties true:
(i) $\mathcal{F}_{\sigma \wedge \tau} = \mathcal{F}_{\sigma} \cap \mathcal{F}_{\tau}$
(ii) $\mathcal{F}_{\sigma \vee \tau} = \mathcal{F}_{\sigma} \cup \mathcal{F}_{\tau}$
where $\mathcal{F}_{\tau} = \{A \in \mathcal{F}: A \cap\{\tau \leq t \} \in \mathcal{F}_t \forall t \}$
and $a \wedge b = min\{a,b\}, a \vee b = max\{a,b\}$
if yes please provide the simplest easy to understand proof.
if not please correct the statements and provide a proof.
My failed try was:
$\mathcal{F}_{\sigma \wedge \tau}$
$= \{A \in \mathcal{F}: A \cap\{\sigma \wedge \tau \leq t \} \in \mathcal{F}_t \forall t \}$
$= \{A \in \mathcal{F}: A \cap (\{\sigma \leq t \} \cup \{\tau \leq t \}) \in \mathcal{F}_t \forall t \}$
$= \{A \in \mathcal{F}: (A \cap \{\sigma \leq t \}) \cup (A \cap \{\tau \leq t \}) \in \mathcal{F}_t \forall t \}$
$? \ \{A \in \mathcal{F}: A \cap\{\sigma \leq t \} \in \mathcal{F}_t \forall t \}$
$\cup \{A \in \mathcal{F}: A \cap\{\tau \leq t \} \in \mathcal{F}_t \forall t \}$
$=\mathcal{F}_{\sigma} \cup \mathcal{F}_{\tau}$
$\ …$
$\mathcal{F}_{\sigma \vee \tau}$
$= \{A \in \mathcal{F}: A \cap\{\sigma \vee \tau \leq t \} \in \mathcal{F}_t \forall t \}$
$= \{A \in \mathcal{F}: A \cap (\{\sigma \leq t \} \cap \{\tau \leq t \}) \in \mathcal{F}_t \forall t \}$
$= \{A \in \mathcal{F}: (A \cap \{\sigma \leq t \}) \cap (A \cap \{\tau \leq t \}) \in \mathcal{F}_t \forall t \}$
$? \ \{A \in \mathcal{F}: A \cap\{\sigma \leq t \} \in \mathcal{F}_t \forall t \}$
$\cap \{A \in \mathcal{F}: A \cap\{\tau \leq t \} \in \mathcal{F}_t \forall t \}$
$=\mathcal{F}_{\sigma} \cap \mathcal{F}_{\tau}$
Note: these are not any homework but something i am doing for self-study.
Thank You saz. I understand your proof but can you also suggest for the union case:
What i tried was as follows:
I have already proved the following results,
(i) $σ∨τ$ is also $\{\mathcal{F}_t\}$-stopping time
(ii) $\mathcal{F}_{stopping\ time}$ is a σ-algebra
using (i) and (ii),
$\mathcal{F}_\sigma \cup \mathcal{F}_\tau \subseteq \mathcal{F}_{σ∨τ} \implies$
$σ(\mathcal{F}_\sigma \cup \mathcal{F}_\tau) \subseteq \mathcal{F}_{σ∨τ}$
and now i take $E \in \mathcal{F}_σ$ so $E \cap \{σ\leq t\} \in \mathcal{F}_t \ \forall t$
since $E \cap \{σ∨τ \leq t\} = (E\cap \{σ \leq t\}) \cap \{τ \leq t\} $
i have $E \cap \{σ∨τ\leq t\} \in \mathcal{F}_t \ \forall t$
now together with similar case for $E \in \mathcal{F}_\tau$ yields
$\mathcal{F}_\sigma \cup \mathcal{F}_\tau \subseteq \mathcal{F}_{σ∨τ}$
But now how i should prove
$\sigma(\mathcal{F}_\sigma \cup \mathcal{F}_\tau) \supseteq \mathcal{F}_{σ∨τ}$ ?
I am sorry the answer might be very trivial, but i can't see it.
Best Answer
Note that the union of two $\sigma$-algebras is in general not a $\sigma$-algebra; therefore, the second claim has to read
$$\mathcal{F}_{\sigma \vee \tau} = \sigma(\mathcal{F}_{\sigma} \cup \mathcal{F}_{\tau}).$$
Here is the proof of the first claim; I leave the proof of the second one to you (see also this question).
Let $F \in \mathcal{F}_{\sigma \wedge \tau}$. Then, by definition,
$$F \cap \{ \sigma \wedge \tau \leq t\} \in \mathcal{F}_t$$
for any $t \geq 0$. Moreover, $\tau \leq t$ obviously implies $\sigma \wedge \tau \leq t$. Hence,
$$F \cap \{\tau \leq t\} = \underbrace{F \cap \{\sigma \wedge \tau \leq t\}}_{\in \mathcal{F}_t} \cap \underbrace{\{\tau \leq t\}}_{\in \mathcal{F}_t} \in \mathcal{F}_t$$
as $\tau$ is a stopping time. This proves $F \in \mathcal{F}_{\tau}$. Replacing $\tau$ by $\sigma$, the very argumentation shows $F \in \mathcal{F}_{\sigma}$. Consequently, $F \in \mathcal{F}_{\sigma} \cap \mathcal{F}_{\tau}$.
Now let $F \in \mathcal{F}_{\sigma} \cap \mathcal{F}_{\tau}$. As
$$F \cap \{\sigma \wedge \tau \leq t\} = (F \cap \{\sigma \leq t\}) \cup (F \cap \{\tau \leq t\})$$
for any $t \geq 0$ we see that $F \in \mathcal{F}_{\tau \wedge \sigma}$.