Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We let $\sigma$ and $\tau$ be 2 stopping times, where $\sigma\leq\tau$.

For a set $A\in F_\sigma$, if we define

$\sigma'(\omega)=\sigma(\omega)$ if $\omega\in{A}$, $\tau(\omega)$ if $\omega\in{A^c}$.

Then $\sigma'(\omega)$ is still a stopping time. Can anyone explain the reason? Thanks!

share|improve this question
Although this is not research level, $\lbrace \sigma' \le t \rbrace = \lbrace \sigma \le t \rbrace \cap A \cup \lbrace \tau \le t \rbrace \cap A^c \in F_t$ since $A\in F_\sigma \subseteq F_\tau$. –  Jochen Wengenroth Mar 14 '13 at 10:01
Hi, you meant $\left ( \{\sigma(\omega)\leq{t}\}\bigcap A \right )\bigcup \left ( \{\tau(\omega)\leq{t}\}\bigcap A^c \right )$? –  Alan Mar 14 '13 at 11:59
$\lbrace f\in B\rbrace$ is a shorthand for $\lbrace \omega\in\Omega: f(\omega)\in B\rbrace$. The conventions for $\cap$ and $\cup$ are as for$\cdot$ and $+$. –  Jochen Wengenroth Mar 14 '13 at 13:57
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.