Let $\tau$ be a stopping time with respect to some filtration $(\mathcal{F}_t)_{t\geq 0}$. Then we know that $\{ \tau \leq t \} \in \mathcal{F}_t$ for all $t\geq 0$.
But how do I prove that $\tau+s$ for $s \geq 0$ is a stopping time aswell?
Stopping time + constant
measure-theorystopping-times
Best Answer
Suppose that $t \ge s$.
$$ \{ \tau + s \le t \} = \{ \tau \le t -s \} \in \mathcal{F}_{t-s}$$ because $\tau$ is stopping time.
Thus $$ \{ \tau + s \le t \} \in \mathcal{F}_{t-s} \subset \mathcal{F}_{t}$$ by definition of filtration.
Hence, $$ \{ \tau + s \le t \} \in \mathcal{F}_{t}$$ and $\tau + s$ is stopping time.
Now suppose that $t < s$. Thus $\{ \tau + s \le t \} = \{ \tau \le t -s \} = \varnothing \in \mathcal{F}_{t}.$
Is there any questions?