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Let $\sigma_1$ and $\sigma_2$ be stopping times of a filtration $(\mathscr{F}_n)^\infty_{n=0}$, and let $A\in\mathscr{F}_{n+1}$ for some $n\geq 0$.

Question: Is $\sigma_3 = \sigma_1 I_A + \sigma_2 I_{A^c}$ a stopping time? Here $A^c$ is the complement of $A$, and $I_A$ equals $1$ on $A$ and equals $0$ otherwise.

I am reading Peskir and Shiryaev's "Optimal Stopping and Free-Boundary Problems". On page 9 it is claimed that $\sigma_3$ lies in the set $\mathfrak{M}_{n+1}$ of stopping times $T$ satisfying $T\geq n+1$ when $\sigma_1$ and $\sigma_2$ lie in $\mathfrak{M}_{n+1}$ and $A$ is the set on which $\mathbb{E}[G_{\sigma_1}|\mathscr{F}_{n+1}]\geq\mathbb{E}[G_{\sigma_2}|\mathscr{F}_{n+1}]$, where $(G_t)^\infty_{t=0}$ is an adapted process.

What I think matters here is that $A$ lies in $\mathscr{F}_{n+1}$ and that $\sigma_1$ and $\sigma_2$ are stopping times. I know that the sum of two stopping times again is a stopping time. So if I can show that $\sigma_1 I_A$ and $\sigma_2 I_{A^c}$ are stopping times then I know that $\sigma_3$ is a stopping time.

However, if I am not mistaken then for each $t\in\mathbb{N}$ it holds that \begin{equation*} \{\sigma_1 I_A\leq t\} = \{\sigma_1\leq t\}\cup A^c, \end{equation*} which for $t < n+1$ is a union of an element in $\mathscr{F}_t$ and an element in $\mathscr{F}_{n+1}$, hence possibly not an element in $\mathscr{F}_t$. In other words, $\sigma_1 I_A$ may not be a stopping time!

What am I missing here to see that $\sigma_3$ is a stopping time?

Bart
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1 Answers1

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No, it is important that $\sigma_1, \sigma_2 \ge n+1$.

Example. Deterministic $\sigma_1 = 1, \sigma_2 = 2$ are of course stopping times. But if $A \in \mathscr F_{n+1} \setminus \mathscr F_2$, what about $\sigma_3 := \sigma_1 I_A + \sigma_2 I_{A^c} $? Compute

$$ \{\sigma_3 \le 1\} = A \notin \mathscr F_1 . $$

so $\sigma_3$ is not a stopping time.

GEdgar
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