Sum of Stopping Times is Stopping Time
Theorem
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T$ and $S$ be stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.
Then the pointwise sum $T + S$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Proof
Let $t \in \Z_{\ge 0}$.
Then, if for $\omega \in \Omega$ we have $\map T \omega + \map S \omega = t$, we have:
- $\map T \omega \le t$
and:
- $\map S \omega \le t$
If we have:
- $\map S \omega = s \le t$
and:
- $\map S \omega + \map T \omega = t$
we have:
- $\map T \omega = t - s$
So, we have:
- $\ds \set {\omega \in \Omega : \map S \omega + \map T \omega \le t} = \bigcup_{s \in \Z_{\ge 0}, \, 0 \le s \le t} \set {\omega \in \Omega : \map S \omega = s} \cap \set {\omega \in \Omega : \map T \omega = t - s}$
Since $S$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$ we have:
- $\set {\omega \in \Omega : \map S \omega = s} \in \FF_s$
Since $s \le t$ and $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have:
- $\FF_s \subseteq \FF_t$
and so:
- $\set {\omega \in \Omega : \map S \omega = s} \in \FF_t$
Similarly, we have:
- $\set {\omega \in \Omega : \map T \omega = t - s} \in \FF_{t - s}$
and so:
- $\set {\omega \in \Omega : \map T \omega = t - s} \in \FF_t$
Since $\FF_t$ is closed under finite union, we have:
- $\ds \set {\omega \in \Omega : \map S \omega + \map T \omega \le t} \in \FF_t$
So $S + T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
$\blacksquare$