Sum of two stopping times is a stopping time?

Question

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.

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Is my attempt correct?

Answer 1

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 $$\{\sigma + \tau < t\} = \bigcup_{r,s \in \mathbb{Q}^+; r+s<t} \{\sigma < r\} \cap \{\tau < s\}.$$ 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.)

Answer 2

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 $\{\tau<t\}\in\mathscr{F}_t$ and also $\{\tau=t\}\in\mathscr{F}_t$. Now we write our set as

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

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 $\{\tau\geq t\}=\{\tau<t\}^c\in\mathscr{F}_t$. At last we have that $$ \{0<\tau<t,\tau+\sigma>t\}=\bigcup_{r\,\in\, (0,t)\cap\,\mathbb{Q}}\{r<\tau<t,\,\sigma>t-r\}\in\mathscr{F}_t. $$

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

Answer 3

Nates' answer is on the right track but assumes right continuity. The definition of stopping time is $\{\sigma \le t\}\in \mathcal{F}_t$ (instead of $\{\sigma < t\}\in \mathcal{F}_t$). Stefan's uses of $\{\tau < t\} \in \mathcal{F}_t$ and $\{\tau = t\} \in \mathcal{F}_t$ for stopping time $\tau$ are correct, but the lemmas can use some extra proof. The following is a proof hopefully with the issues avoided.

We claim that $$\{\sigma + \tau > t\} = \left(\bigcup_{r,s \in Q^+, r + s > t, r\le t, s\le t}\{\sigma > r\} \cap \{ \tau > s \}\right)\cup\{\sigma > t\}\cup\{\tau > t\}. $$

It is trivial to see that the RHS $\subset$ LHS, RHS $ \in\mathcal{F}_t$ (note $\{\sigma > r\}\in \mathcal{F}_r \subset \mathcal{F}_t$ for $r \le t$). It remains to show LHS $\subset$ RHS to conclude $\{\sigma + \tau > t\} \in \mathcal{F}_t$.

For any $\omega \in $ LHS, if $\sigma(\omega) = 0$ or $\tau(\omega) = 0$, then $\omega \in \{\sigma > t\} \cup \{\tau > t\}\subset$RHS. Otherwise, we have $\sigma(\omega) + \tau(\omega) >t, \sigma(\omega) > 0, \tau(\omega) > 0.$ We can always find two rational numbers $r, s$ so that $r + s > t, r\le t, s\le t,$ $\sigma(\omega) > r \ge 0,$ $\tau(\omega) > s \ge 0.$ Hence $\omega \in \{\sigma > r\} \cap \{ \tau > s \} \subset$ RHS. This completes the proof that LHS = RHS = $\{\sigma + \tau > t\} \in \mathcal{F}_t$.

It follows that $\{\sigma + \tau \le t\}\in \mathcal{F}_t$, i.e. $\sigma + \tau$ is a stopping time.

Answer 4

There is a brief proof in the book ``S. W. He et al., Semimartingale Theory and Stochastic Calculus, CRC Press Inc, 1992.''(p.84, Th.3.7.(3)).

The proof is based on following fact(Th.3.7.(1)): If $S$ is a stopping time, r.v. $ T\in\mathscr{F}_S $ and $ T\ge S $, then $ T $ is a stopping time.(Proof of this fact: For each $ t\ge 0 $, $ [T\le t]\in\mathscr{F}_S$, and by the definition of $\mathscr{F}_S $ we have $ [T\le t]=[T\le t][S\le t]\in\mathscr{F}_t $, hence $ T $ is a stopping time.)

$ S+T $ is a stopping time: Since $ S+T\ge S \lor T $ and $ S+T\in \mathscr{F}_{S\lor T} $($ S\in \mathscr{F}_S\subset \mathscr{F}_{S\lor T}$), then $ S+T $ is a stopping time.