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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

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up vote 7 down vote accepted

Here is a proof coming from a french Math forum :,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.

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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

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