Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
2  
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
add comment

2 Answers 2

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.)

share|improve this answer
add comment

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.

share|improve this answer
1  
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
1  
@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
1  
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
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.