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In my book, it states that any constant, including, $\infty$, is a stopping time. But why is this? I thought the reason might be that a constant is measurable with regards to the trivial sigma-algebra, i.e. $\{\emptyset, \Omega\}$ and since $\emptyset \in \mathcal{F}_n$ the results follow? Or what is the reason?

Thanks.

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  • $\begingroup$ We're going to need more context here. $\endgroup$
    – eyeballfrog
    Commented Jun 18, 2022 at 15:48

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Let $\tau=k,\,k \in \mathbb{N}\cup \{\infty\}$. Recalling the definition, $\tau$ is a $\mathscr{F}_n$-stopping time if $\{\tau\leq n\}\in \mathscr{F}_n,\,\forall n \in \mathbb{N}$. Now if $k<\infty$ we have $$\{\tau\leq n\}=\begin{cases}\Omega&n\geq k\\ \emptyset &n<k \end{cases}\in \mathscr{F}_n,\,\forall n \in \mathbb{N}$$ while if $k=\infty$, $\{\tau\leq n\}=\emptyset$ for all $n$ so $\tau $ is always a stopping time.

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  • $\begingroup$ Do I understand it correctly: $\{\tau \leqslant n\} = \{\tau(\omega) \leqslant n\}$ is the set of instances where the stopping time $\tau$ is less or equal than $n$? And if $k<\infty$ is $\tau < n$ always true if $n\geqslant k$ and never true if $n<k$? $\endgroup$
    – HelloWorld
    Commented Jan 17, 2023 at 18:03
  • $\begingroup$ @HelloWorld By definition $\{\tau \leq n\}=\{\omega: \tau(\omega) \leq n\}$. Since $\tau(\omega)=k$ for all $\omega$, we have that $\{\omega:\tau(\omega)\leq n\}=\Omega$ for all $n\geq k$ (so, always) and $\{\omega:\tau(\omega)\leq n\}=\emptyset$ for all $n<k$ (so, never). $\endgroup$
    – Snoop
    Commented Jan 17, 2023 at 18:24
  • $\begingroup$ Thanks! Yes I meant the $\{\omega...\}$ not $\{\tau(\omega)\}...$ ;) $\endgroup$
    – HelloWorld
    Commented Jan 17, 2023 at 20:19

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