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 $\sequence {\FF_n}_{n \ge 0}$.

$\blacksquare$