Let $(\mathcal F_t)_{t\in T}$ be a filtration.

Consider $s<t$ in T and $A \in \mathcal F_s$.

I want to show that $\tau=s \mathbf{1}_{A^c}+t\mathbf{1}_A$ is a stopping time.

My thoughts: In both cases, $\omega \in A^c$ and $\omega \in A$, we have that $\tau(\omega)=const.$, so $\tau$ is a stopping time in both cases (since it's constant). Is this a valid 'proof'? Feels wrong somehow.

In case this is wrong, I thought about the following: We have to show that $\{\tau\leq u\}\in \mathcal F_u\ \forall u \in T$.

$\{\tau \leq s\}=A^c\in F_s$

But how can I show that $\{\tau \leq u\}\in F_u$ for $u>s$?


1 Answer 1


The random variable is $$\tau(\omega)=s\mathbf{1}_{A^c}(\omega)+t\mathbf{1}_{A}(\omega),\,\omega \in \Omega$$ Now notice that $$\{\omega \in \Omega: \tau(\omega)\leq u\}=\begin{cases} \Omega&u\geq t\\ A^c& s\leq u<t\\ \emptyset &u<s \end{cases}$$ As $\mathcal{F}_s\subseteq \mathcal{F}_u$, then $A^c \in \mathcal{F}_u,\,\forall u >s$. The empty set is in all $\sigma$-algebras, and the claim follows.

  • $\begingroup$ I am confused, given $\omega \in \Omega$ is $\tau(\omega)$ a constant for every $s$ and $t$? $\endgroup$
    – UBM
    Nov 21, 2021 at 18:37
  • $\begingroup$ If $\omega \in A$, then $\tau(\omega)=t$. If $\omega \in A^c$, then $\tau(\omega)=s$ @UBM $\endgroup$
    – Snoop
    Nov 21, 2021 at 18:44
  • $\begingroup$ Ah yes, thank you. $\endgroup$
    – UBM
    Nov 21, 2021 at 18:46

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