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Let $\sigma$ and $\tau$ be two stopping times in $\mathscr{F}_t$ and let this filtration satisfy all the usual conditions.

Question: Is $\sigma + \tau$ a stopping time?

Attempt at a solution:

I need to demonstrate that $\{ \sigma + \tau \leq t\}\in \mathscr{F}_t$, or that $\{\sigma \leq t - \tau \} \in \mathscr{F}_t$.

Since $\sigma$ is a stopping time we have that $\{\sigma \leq t - \tau\} \in \mathscr{F}_{t - \tau}$, where $t - \tau \in [0,t]$.

Since $t > t - \tau$, we have that $\mathscr{F}_{t-\tau} \subseteq \mathscr{F}_t$ by the definition of $\mathscr{F}$.

This implies that $\{\sigma \leq t - \tau\} \in \mathscr{F}_t$, and that $\sigma + \tau$ is a stopping time.


Is my attempt correct?

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    @StefanHansen Okay then I'm lost in this problem unfortunately.2012-11-08

4 Answers 4

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Following did's comment, we could write this a little more simply.

For each fixed $\omega$, I claim $\sigma(\omega) + \tau(\omega) < t$ iff there are positive rationals $r,s$ with $r+s < t$ and $\sigma(\omega) < r$, $\tau(\omega) < s$. Suppose $\sigma(\omega) + \tau(\omega) < t$; we can find a rational $q$ with $\sigma(\omega) + \tau(\omega) < q < t$. Then $\sigma(\omega) < q - \tau(\omega)$, so we can find $r$ with $\sigma(\omega) < r < q - \tau(\omega)$. Setting $s = q-r$ we see that we have $\tau(\omega) < s$. The reverse implication is obvious.

Thus we have $\{\sigma + \tau < t\} = \bigcup_{r,s \in \mathbb{Q}^+; r+s But $\{\sigma < r\} \in \mathcal{F}_r \subset \mathcal{F}_t$; likewise $\{\tau < s\} \in \mathcal{F}_t$. Thus $\{\sigma + \tau < t\}$ is a countable union of events from $\mathcal{F}_t$, and so it itself in $\mathcal{F}_t$.

I'd like to point out that this is a useless fact; as far as I can see, there's no meaningful interpretation of the sum of two stopping times, since stopping times represent absolute rather than relative times. (If the train to Paris leaves at 5:00, and the train to Berlin leaves at 6:00, what happens at 11:00? Nothing.)

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Revised answer:

I found it easier to look at the complements instead. Then we might as well show that $\{\tau+\sigma>t\}\in\mathscr{F}_t$ for all $t$. For a stopping time $\tau$ we know that $\{\tau and also $\{\tau=t\}\in\mathscr{F}_t$. Now we write our set as

$ \{\tau+\sigma>t\}=\{\tau=0,\,\tau+\sigma>t\}\cup\{0<\taut\}\cup\{\tau\geq t,\, \tau+\sigma>t\}\\ =\{\tau=0,\,\sigma>t\}\cup\{0<\taut\}\cup\{\tau>t,\,\sigma=0\}\cup\{\tau\geq t,\,\sigma>0\}. $

Then $\{\tau=0,\,\sigma>t\}\in\mathscr{F}_t$ and $\{\tau>t,\,\sigma=0\}\in\mathscr{F}_t$, since $\tau$ and $\sigma$ are stopping times. Furthermore $\{\tau\geq t,\,\sigma>0\}\in\mathscr{F}_t$ because $\{\sigma>0\}=\{\sigma=0\}^c\in\mathscr{F}_0$ and $\{\tau\geq t\}=\{\tau. At last we have that $ \{0<\taut\}=\bigcup_{r\,\in\, (0,t)\cap\,\mathbb{Q}}\{r<\taut-r\}\in\mathscr{F}_t. $

I hope that this last equality with the union now holds.

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    @Gibarian: Thanks for the correction.2013-08-08