# Sum of two stopping times is a stopping time?

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?

• It seems like you're working with $\tau$ as if it were a constant. For example: "Since $\sigma$ is a stopping time we have that $\{\sigma\leq t-\tau\}\in\mathscr{F}_{t-\tau}$" - is this clear from the definition of $\sigma$ being a stopping time? Commented Nov 8, 2012 at 8:46
• I thought that that is true because of the following: Since I'm told that $\sigma$ is a stopping time, then $\{\sigma \leq x\} \in \mathscr{F}_x$ for any $x \in [0,\infty)$. Now for $x = t - \tau$, it's true that the image $x(\omega)$ satisfies the above, but I'm not sure if $x$ itself does. I'm pretty bad at maths (unfortunately).
– Jase
Commented Nov 8, 2012 at 9:07
• The problem is that $\{\sigma\leq x\}\in\mathscr{F}_x$ holds for any deterministic (constant) $x\in [0,\infty)$. Now $x=t-\tau$ is random, i.e. $x(\omega)=t-\tau(\omega)$, so we cannot apply the definition on this $x$. Actually $x(\omega)$ may even be negative (if $\tau(\omega)>t$). Commented Nov 8, 2012 at 9:12
• @StefanHansen Okay then I'm lost in this problem unfortunately.
– Jase
Commented Nov 8, 2012 at 9:29

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.)

• r,s are $\mathbb{R}$-valued - why can you restrict to $\mathbb{Q}$? It seems even though $\mathbb{Q}$ is dense in $\mathbb{R}$ you have issue about precise specification Commented Feb 18, 2015 at 17:20
• @JohnFernley: I don't understand your comment. I defined $r,s$ to be rational numbers. Can you explain specifically which step of my argument you think is incorrect or insufficiently proved? Commented Feb 18, 2015 at 17:25
• My issue is that $\{\sigma < \sqrt{2}\} \cap \{\tau < 2-\sqrt{2}\} \subset \{\sigma + \tau < 2\}$ and so I don't see how the union you give is the whole set Commented Feb 23, 2015 at 16:15
• Thank you for the answer! I'd also like to point out that your remark is very witty :) Commented Dec 10, 2015 at 18:28
• I can think of a few reasons we might care about the sum of stopping times. (1) If we just take $\tau = c$ is a constant, then $\sigma + \tau$ represents what is happening at some time after $\sigma$ occurred. E.g. if the train leaves to Paris at 5, where is it 2 hours later? Or e.g. how far are you from a set at a fixed time after hitting it? (2) if e.g. $\sigma$ tells me when I have the opportunity to bet and $\tau$ tells me when I have enough money to bet, then $\tau - \sigma$ tells me how long I have to wait from when an opportunity presents itself to when I can take advantage of it Commented Mar 31, 2019 at 17:21

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.

• There is no such $s$ when $\sigma+\tau=t$ and $t$ is not rational hence this identity does not hold. You might want to use $\lt t+1/n$ instead of $\leqslant t$, and then consider the intersection of these over $n$.
– Did
Commented Nov 8, 2012 at 10:37
• @did, I'm sorry that you, yet again, have to clean up in my answers. I think you should post the answer instead of me, because i'm not entirely sure, and then I will delete this. Commented Nov 8, 2012 at 10:42
• Stefan: PLEASE do not feel sorry for these one or two occurrences (of something I would not describe as you do). Why not revise your answer, I am sure you can do that, and then everybody shall be happy. (Oh, and I forgot to say: your answers are good, from what I have seen... :-))
– Did
Commented Nov 8, 2012 at 10:54
• @did: Thanks for the cheering up! However, I can't seem to express $\{\sigma+\tau<t+1/n\}$ as a countable union of sets belonging to $\mathscr{F}_t$. I think you might as well post it :) Commented Nov 8, 2012 at 11:36
• Use your idea: $\sigma+\tau\lt t$ if and only if there exists $r$ and $s$ rational numbers such that $r+s\lt t$, $\sigma\leqslant r$, $\tau\leqslant s$.
– Did
Commented Nov 8, 2012 at 11:40

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.

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.

• Nate’s answer is complete. Because the filtration satisfies the usual conditions, the filtration is right-continuous. It is therefore sufficient to show that $\{\sigma + \tau < t\} \in \cal F_t$. Commented Oct 16, 2019 at 5:48
• You are correct if "usual conditions" is interpreted as including the filtration being right-continuous. It is still worthwhile to prove/show the conclusion holds without right continuity. Commented Oct 16, 2019 at 22:20
• That is part of the standard interpretation of the phrase “usual conditions”. I agree that it is worthwhile to show this, but perhaps you can remove the statement that Nate’s answer is incomplete, as that is not the case. Commented Oct 17, 2019 at 0:31
• Just modified the statements a bit. Commented Oct 17, 2019 at 1:01