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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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3 Answers 3

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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.

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  • $\begingroup$ Nice! Thanks a lot (and thanks also for the translation, my french is really not the best ;-))! $\endgroup$
    – Mad Si
    Commented Nov 22, 2011 at 14:29
  • $\begingroup$ By the way this property is known as Galmarino's test. $\endgroup$
    – TheBridge
    Commented Jan 26, 2022 at 12:43
  • $\begingroup$ @TheBridge I looked up Galmarino's test which looks similar to this question, but is actually very different. Galmarino's test imposes conditions on sample paths that happen to agree. This question quite literally says that all sample paths agree up until some time $T_0$. $\endgroup$
    – Mark
    Commented Dec 21, 2023 at 1:43
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I'm studying the same problem. I found the accepted solution hard to follow and I believe I have a much simpler solution:

Seeking a contradiction, suppose that $T(\omega) \neq T(\omega')$. Then one must be greater than the other, it doesn't matter which. We will suppose $T(\omega) < T(\omega')$ although the logic for the other case is the same. Then label $T(\omega) = t_0$ such that $T(\omega) = t_0 < T(\omega')$. Then $\omega \in \{T \le t_0\}$ and $\omega' \not\in \{T \le t_0\}$.

$\{T \le t_0\} \in \mathscr{F}_{t_0}^X$. Every set in a $\sigma$-algebra $\mathscr{F}_{t_0}^X$ is of the form $\cap_i \{X_{t_i} \in B_i\}$ for countable indices $0 \le t_1 < t_2 < \cdots \le t_0$ and Borel sets $B_i$. So $\{T \le t_0\}$ is one such $\cap_i \{X_{t_i} \in B_i\}$. Since all $t_i \le t_0$, then $X_{t_i}(\omega) = X_{t_i}(\omega')$. Therefore, it is a contradiction that $\omega \in \cap_i \{X_{t_i} \in B_i\}$ and $\omega' \not\in \cap_i \{X_{t_i} \in B_i\}$.

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    $\begingroup$ Nice! For those of us who studied Exercise 1.8 (or its solution), this is the same kind of logic that was used in that problem. $\endgroup$
    – Alex Ortiz
    Commented Jan 15, 2023 at 18:38
  • $\begingroup$ You said every set in $\mathcal{F}_{t_0}^X$ can be written as an intersection $\cap_i X_{t_i}^{-1}(B_i)$. This isn't true is it? You need also countable unions, and intersections of unions, ad finitim. $\endgroup$
    – Mark
    Commented Dec 20, 2023 at 22:22
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Fix an arbitrary $\omega_0 \in \Omega$. Define $T_0 = T(\omega_0)$ and $X_{0,t} = X_t(\omega_0)$.

Note that $X_{0,t}$ is a deterministic function on a deterministic interval $[0, T_0]$. It is a fixed sample path. And by assumption, all sample paths of $X$ agree at least until time $T_0$. So $X$ is deterministic at least until time $T_0$.

For any deterministic "random variable" $D$, we trivially have $\sigma(D) = \{ \emptyset, \Omega \}$. It follows that $\mathcal{F}_{T_0}^X = \{ \emptyset, \Omega \}$.

Since $\{T \le T_0 \} \in \mathcal{F}_{T_0}^X = \{ \emptyset, \Omega \}$, we then have $\{T \le T_0 \} = \emptyset$ or $\{T \le T_0 \} = \Omega$ . However, we can rule out the first option since $\omega_0 \in \{T \le T_0 \}$ so then we must have $\{T \le T_0 \} = \Omega$.

So for all $\omega \in \Omega$, we have $T(\omega) \le T_0$.

Now $\omega_0$ was arbitrary, so for every $\omega_0, \omega \in \Omega$, we have $T(\omega) \le T(\omega_0)$. Swapping $\omega_0$ and $\omega$, we also have $T(\omega_0) \le T(\omega)$. So $T(\omega_0) = T(\omega)$.

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