I was interested to understand the interpretation of the sigma field associated to a stopping time.
Let $(X_n)_{n \ge 1}$ be a stochastic process and $(F_n)_{n \ge 1}$ its corresponding filtration.
Let $T$ be a stopping time. the associated sigma-algebra is $F_T = \{A \vert A\cap \{T=n\} \in F_n\}$.
I read about this in many many many but I still can't really understand the interpretation. Is it as following? of $F_T$ is as following : if the stopping time has occured at time $n$, i.e. $\{T=n \}$ is realized, then $A \in F_T \iff A \in F_n$. But I can't see step by step why we have this from the definition...
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roi_saumon
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There is no way to make this rigorous, as far as I can see. Note that $F_n$ and $F_T$ are deterministic objects. You say "if the stopping time has occured at time $n$, [...], then $A \in \mathcal{F}_T \iff A \in \mathcal{F}_n$" but this doesn't work since the information on the occurence of the stopping time is not (and cannot be) included in the equivalence $A \in F_T \iff A \in F_n$. We cannot consider the statement "$A \in F_T \iff A \in F_n$" for fixed $\omega$ or something like this. – saz Apr 14 '19 at 06:42
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This is also a related question (one of the many). – saz Apr 14 '19 at 06:55
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@saz, so is it just an implication in one direction? or what is it? – roi_saumon Apr 14 '19 at 09:56
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Neither of them. $A \in F_T$ does not imply $A \in F_n$ and also the other direction is wrong. It does hold that $A \cap {T=n} \in F_n$ for any $A \in F_T$ but that's nothing than the very definition of $F_T$ – saz Apr 14 '19 at 10:45