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.

Working through the book "Brownian Motion & Stochastic Processes" by Karatzas and Shreve, I found the following problem (page 6, Problem 2.2):

Let $ X $ be a stochastic process and $ T $ a stopping time of $ \{ \mathcal{F}_t^X \}$, where $ \mathcal{F}_t^X := \sigma (X_s, 0 \leq s \leq t) $. Suppose that for any $ \omega, \omega ' \in \Omega $, we have $ X_t (\omega) = X_t ( \omega '), $ for all $ t \in [0, T ( \omega )] \cap [ 0, \infty ) $ .

Show that $ T ( \omega ) = T ( \omega ') $.

Does anybody know how to prove this? Thanks a lot for your efforts! Regards, Si

share|improve this question
add comment

1 Answer 1

up vote 7 down vote accepted

Here is a proof coming from a french Math forum :

http://www.les-mathematiques.net/phorum/read.php?12,376956,377123#msg-377123

I translate the solution for the non french readers.

So here comes the solution (credit goes to egoroff) :

For $T(\omega)<\infty$, fix $\mathcal{H}$ as the collection of sets that do not separate $\omega$ and $\omega'$, i.e. sets $A$ s.t. either $\{\omega,\omega'\}\in A$ or $ \in A^c$. Then it is easy to see that $\mathcal{H}$ is a $\sigma$-field.

This was the first step. Next for every $(n+1)$-tuple $t_0<...<t_n\le T(\omega)$ and every Borel sets $A_{t_i}$, the set $(X_{t_i})_{i=0,...,n}\in \Pi_{i=0}^n A_{t_i}$ is in $\mathcal{H}$, by hypothesis over $\omega$ and $\omega'$, so $\mathcal{F}_t\subset \mathcal{H}$ for every $t\le T(\omega)$ as those set generate $\mathcal{F}_t$ .

Now $T(\omega)$ is known and finite we have :

-$S=T\vee T(\omega)$ is a stopping time and moreover $S\in \mathcal{F}_{T(\omega)}\subset \mathcal{H}$ we have $S(\omega)=S(\omega')$, and so $T(\omega)\le T(\omega')$.

-On the other and the event $\{T<T(\omega)\}$ is in $\mathcal{F}_{T(\omega)}$, as $T$ is a stopping time so it is in $\mathcal{H}$, and $\omega\in \{T\le T(\omega)\}$ and so $\omega'$ too, and $T(\omega')\le T(\omega)$.

Finally we have shown that $T(\omega)=T(\omega')$ over $T(\omega)<\infty$ which was the claim to be proved.

Best regards

PS : I also have a solution of mine based on a variant of Doob's lemma but as it is longer, more technical and far less elegant than this one, I do not post it here.

share|improve this answer
    
Nice! Thanks a lot (and thanks also for the translation, my french is really not the best ;-))! –  Mad Si Nov 22 '11 at 14:29
    
@Mad Si : Tout le plaisir est pour moi. ;-) –  TheBridge Nov 22 '11 at 15:03
add comment

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.