A simple proof concerning stopping time - stochastic processes.

Question

Show that $\{\sigma<\tau\}$ belongs to $\mathfrak{F}_\tau$, $\mathfrak{F}_\sigma$ and $\mathfrak{F}_{\tau\wedge\sigma}$.

I am not sure what should be checked here to fulfill the definition of $\sigma$-algebra and how.

Answer 1

Recall (or prove) the following statement:

If $\varrho_1$ and $\varrho_2$ are two stopping times (with respect to a filtration $(\mathcal{F}_t)_{t \geq 0}$) such that $\varrho_1 \leq \varrho_2$ then $\mathcal{F}_{\varrho_1} \subseteq \mathcal{F}_{\varrho_2}$.

Applying this result, we find that it suffices to show that $$\{\sigma<\tau\} \in \mathcal{F}_{\tau \wedge \sigma}.$$ To this end, we have to prove that $$\{\sigma<\tau\} \cap \{\sigma \wedge \tau \leq t\} \in \mathcal{F}_t \qquad \text{for all $t \geq 0$}. \tag{1} $$ Since $$\sigma(\omega)<\tau(\omega) \implies \sigma(\omega) \wedge \tau(\omega) = \sigma(\omega)$$ we have $$\{\sigma<\tau\} \cap \{\sigma \wedge \tau \leq t\} = \{\sigma<\tau\} \cap \{\sigma \leq t\}.$$ Using

$$\{\sigma<\tau\} \cap \{\sigma \leq t\} = \big(\{\sigma<\tau\} \cap \{\sigma < t\} \big) \cup \underbrace{\big(\{\sigma = t\} \cap \{\tau>t\}\big)}_{\in \mathcal{F}_t}$$ and

$$\{\sigma<\tau\} \cap \{\sigma < t\} = \bigcup_{r \in \mathbb{Q} \cap [0,t)} (\underbrace{\{\sigma \leq r\}}_{\in \mathcal{F}_r} \cap \underbrace{\{\tau > r\}}_{\{\tau \leq r\}^c \in \mathcal{F}_r}) \in \mathcal{F}_t$$ we get $(1)$.