My task is to describe the following stopping time formally and to prove that it is indeed a stopping time.

Let $A\in \mathcal{B}(\mathbb{R})$ be a fixed set. $(X_t)_{t\ge 0}$ hits A for the first time, leaves A again and hits A for the second time.

I defined the stopping time $\nu = min \{ t\in \mathbb{N} | \exists \tau_1, \tau_2 < t \text{ such that } X_{\tau_1} \in A, X_{\tau_2} \notin A, X_t \in A \}$, where $min\emptyset := \infty$

In order to proof, that it is indeed a stopping time I rewrote $\nu$:

$\{\nu=t\} = \cap_{n=0}^{\tau_1 - 1} \{X_n \notin A\} \cap_{n = \tau_1}^{\tau_2 - 1} \{X_n \in A \} \cap_{n = \tau_2}^{t - 1} \{X_n \notin A \} \cap \{ X_t \in A\}$,

which is "obviously" in $\mathcal{F}^X \subset \mathcal{F}_t$ (the filtration).

My questions are:

  1. Is this a valid definition of a stopping time?
  2. Am I allowed to rewrite the stopping time in terms of the parameters $\tau_1, \tau_2$?

Thank you in advance!

  • $\begingroup$ You seem to have already asked this question. In the future, rather than deleting and reposting, please use the edit button to make improvements to your question. $\endgroup$
    – Xander Henderson
    Commented Mar 22, 2023 at 13:27

1 Answer 1


Don't forget the quantifiers on $\tau_1$ and $\tau_2$ in your definition of $\nu$. Taking them into account you'll have $$ \{\nu=t\}=\cup_{\tau_1=0}^{t-2}\cup_{\tau_2=\tau_1+1}^{t-1}\left[\cap_{n=0}^{\tau_1 - 1} \{X_n \notin A\} \cap_{n = \tau_1}^{\tau_2 - 1} \{X_n \in A \} \cap_{n = \tau_2}^{t - 1} \{X_n \notin A \} \cap \{ X_t \in A\}\right]. $$ Otherwise, it looks okay.

You could also "modularize" things by checking that if $S$ is a stopping time then so is $T(A,S):=\min\{n>S:X_n\in A\}$. Your stopping time is the two-fold iteration of this construction: $$\nu=T(A,T(A^c,T(A))), $$ where $T(A)$ is just the first time to hit $A$.


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