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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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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? – Stefan Hansen Nov 8 '12 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 Nov 8 '12 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 $\tau(\omega)>t$). – Stefan Hansen Nov 8 '12 at 9:12
@StefanHansen Okay then I'm lost in this problem unfortunately. – Jase Nov 8 '12 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.)

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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 – John Fernley Feb 18 '15 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? – Nate Eldredge Feb 18 '15 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 – John Fernley Feb 23 '15 at 16:15
Thank you for the answer! I'd also like to point out that your remark is very witty :) – Sasha Dec 10 '15 at 18:28

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.

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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 Nov 8 '12 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. – Stefan Hansen Nov 8 '12 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 Nov 8 '12 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 :) – Stefan Hansen Nov 8 '12 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 Nov 8 '12 at 11:40

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