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Prove that if $S$ is an optional time and $T$ is a positive stopping time such that with $S \le T$ and $S < T$ on $\{S < \infty\}$, then $\mathscr{F}_{S+} \subset \mathscr{F}_T$.

The following is the given solution:

Consider any $A \in \mathscr{F}_{S+}$, which means that $\forall t \ge 0, A \cap \{S \le t\} \in \mathscr{F}_{t+}$. Then:

\begin{align*} A &= \left( \bigcup_{r \in \mathbb{Q}} [A \cap \{S < r < T \}] \right) \cup [A \cap \{S = \infty\}] \\ \end{align*}

For the left clause, $A \cap \{S < r < T\} = A \cap \{S < r \} \cap \{T > r\}$ is in $\mathscr{F}_T$ because $A \cap \{S < r\} \in \mathscr{F}_r$.

For the right clause, $A \cap \{S = \infty\} = [A \cap \{S = \infty\} ] \cap \{T = \infty\}$ is in $\mathscr{F}_T$. Then the conclusion follows.

All of that makes sense except I don't see how we can conclude that $A \cap \{S < r\} \in \mathscr{F}_r$

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  • $\begingroup$ I get that part. I don't get the earlier part how we say that $A \cap \{S < r\} \in \mathscr{F}_t$ $\endgroup$
    – clay
    Commented Mar 22, 2022 at 5:59
  • $\begingroup$ I think there is a typo. In fact no $t$ has been specified. So it should read $A\cap \{S<r\} \in \mathcal F_r$. $\endgroup$ Commented Mar 22, 2022 at 6:06
  • $\begingroup$ yikes, you are right, that should have been $\mathscr{F}_r$. I fixed it. However, I still don't understand how you get that. Of course $\{ S < r \} \in \mathscr{F}_r$, that is the optional time property. But why is $A \cap \{S < r\} \in \mathscr{F}_r$? $\endgroup$
    – clay
    Commented Mar 22, 2022 at 13:17
  • $\begingroup$ Since $A \in \mathscr{F}_{S+}$, that means for any $t \ge 0$, we have $A \cap \{S \le t\} \in \mathscr{F}_{t+}$. We can change notation to get for any $r \ge 0$, we have $A \cap \{S \le r\} \in \mathscr{F}_{r+}$. How can you infer from that that $A \cap \{S < r\} \in \mathscr{F}_r$? Can't we have $A \in \mathscr{F}_{r+}$ but $A \not\in\ \mathscr{F}_r$? $\endgroup$
    – clay
    Commented Mar 22, 2022 at 13:22

1 Answer 1

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This follows from the definition of $\mathcal{F_{S+}}$. Recall that $A \cap \{S \leq t\} \in \mathcal{F_{t+}},\forall t\geq 0$. Now, $A \cap \{S < r\} = A \cap ( \cup_{n=1}^{\infty} \{S\leq r-\frac{1}{n}\}) = \cup_{n=1}^{\infty} (A \cap \{ S \leq r- \frac{1}{n} \})$. But $A \cap \{ S \leq r- \frac{1}{n} \} \in \mathcal{F_{(r-\frac{1}{n})+}} \subset \mathcal{F_{r}} \forall n$. Therefore, the union is in $\mathcal{F_{r}}$.

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  • $\begingroup$ Thank you! That's rather simple in hindsight after you've explained it :) $\endgroup$
    – clay
    Commented Mar 26, 2022 at 18:50

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