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Let us consider the sequence $(\tau_n)_{n\in \mathbb{N}}$ of stopping times that takes values in $\mathbb{N}$ such that $\tau_n \uparrow \tau$, and $\tau < \infty$. Prove the following equality:

$\mathcal{F}_\tau=\sigma(\cup_n \mathcal{F_{\tau_n}})$

I'm having problem with both the inclusions, any suggestions?

saz
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JCF
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  • Do you assume that the underlying filtration is right-continuous...? – saz Oct 18 '17 at 10:58
  • @saz do you mean for the family of filtrations $(\mathcal{F_t})_{t \geq 0}$? In that case no. – JCF Oct 18 '17 at 11:06
  • Suggestion: Because the stopping times are integer valued, the convergence $\tau_n\uparrow\tau$ means that for each $k\in\Bbb N$ you have ${\tau=k}=\cup_n{\tau_n=k}$. – John Dawkins Oct 18 '17 at 16:31
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    @JohnDawkins Why should this be true? $\tau_n(\omega)=k$ for some $n$ does not imply $\tau(\omega)=k$. I guess you mean $${\tau=k} = \bigcup_n \bigcap_{j \geq n} {\tau_j = k}.$$ – saz Oct 19 '17 at 05:34
  • @saz: Yes indeed, ${\tau=k}=\cup_n\cap_{j\ge n}{\tau_j=k}$. And also (for what it is worth) ${\tau=k}=\cap_n\cup_{j\ge n}{\tau_j=k}$, so that even ${\tau=k}=\lim_n{\tau_n=k}$. – John Dawkins Oct 20 '17 at 16:42
  • Related question – saz Jul 29 '19 at 17:45

1 Answers1

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Hints:

  1. Let $S,T$ be two stopping times (with respect to a common filtration). Show: If $S \leq T$, then $\mathcal{F}_S \subseteq \mathcal{F}_T$.
  2. Conclude that $$\sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{F}_{\tau_n} \right) \subseteq \mathcal{F}_{\tau}.$$
  3. Prove that $\{\tau_n = \tau\} \in \mathcal{F}_{\tau}$ for all $n \in \mathbb{N}$.
  4. Show that for any $F \in \mathcal{F}_{\tau}$ and any $n \in \mathbb{N}$ it holds that $F \cap \{\tau_n=\tau\} \in \mathcal{F}_{\tau_n}$.
  5. Conclude that $$F = \bigcup_{n \in \mathbb{N}} (F \cap \{\tau=\tau_n\}) \in \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{F}_{\tau_n} \right) $$ for any $F \in \mathcal{F}_{\tau}$.
saz
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