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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?

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

1 Answer 1

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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?

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    $\begingroup$ @Sebastiano, you are welcome! $\endgroup$ Dec 23, 2020 at 21:23

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