# Stopping time + constant

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?

• $\{\tau+s\leq t\}=\{\tau\leq t-s\}\in \mathcal F_{t-s}\subset \mathcal F_t.$
– Surb
Commented Dec 22, 2020 at 21:49
• But what if $t < s$? Is $\mathcal{F}_{t-s}$ defined then? Commented Dec 22, 2020 at 21:56
• if $t<s$ then $\{\tau<t-s\}=\varnothing \in \mathcal F_t$.
– Surb
Commented Dec 22, 2020 at 21:59

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?

• @Sebastiano, you are welcome! Commented Dec 23, 2020 at 21:23