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
Best Answer
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 :
-$S=T\wedge T(\omega)$ 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 hand the event $\{T\le T(\omega)\}$ 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.