Let $\tau : \Omega \to \{0,1,2...,\infty\}$ and $\{ X_n :n \geq 0\}$ be an adapted process. Is $\tau $ as topping time?

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Let $\tau : \Omega \to \{0,1,2,\cdots,\infty\}$ and $\{ X_n :n \geq 0\}$ be an adapted process. Which of the following is a better defined stopping time with respect to $\{F_n : n \geq 0 \}$?

  • $\tau = \inf \left\{ n\geq 0 : X_n >X_{n+1}\right\}$
  • $\tau = \inf \left\{ n\geq 0 : \frac{1}{n}\sum_{i=1}^{n} X_{i}<0\right\}$
  • If $\tau$ is stopping time, then $(\tau -1 )\vee 0 $ is a stopping time.

My guess is the last one because the minimum of two stopping times is a stopping time. I am not sure though whether 0 is well defined in this case.

I would appreciate it if you can correct me.

Thanks.

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You are correct that the minimum of two stopping times is a stopping time, and there is not a problem with $0$ being well-defined and a stopping time, but check whether or not $\tau - 1$ is a stopping time.

I think that the second one is is also a stopping time. We would need to show that $\{ \tau \le n \} \in \mathcal F_n$ for all $n$, and $$ \{ \tau \le n \} = \bigcup_{m=1}^n \left\{ \frac 1m \sum_{k=1}^m X_k < 0 \right\}$$

is a finite union of sets in $\mathcal F_n$ so $\{\tau \le n \} \in \mathcal F_n$ and hence $\tau$ is a stopping time.