Assume that you ahve a stochastic process $\{X_t\}$ on the probaiblity space $(\Omega, \mathcal{F},P)$, equip this space with a filtration($\{\mathcal{F}_t\}$). And you have a random variable $\Omega\rightarrow [0, \infty]$, where the set $T^{-1}(-\infty,t]\in \mathcal{F}_t$, Then T is called a stopping time. I have two questions regarding this subject.

It is stated that it can be shown that if $X_t$ is cadlag, then it can be shown that $X_T$ is $\mathcal{F}_T$-measurable. But what does this mean? That $X_T^{-1}(B)\in \mathcal{F}_T$? The problem is that $T$ is a function of $\omega$. If I look at $X_T^{-1}(B)$, I get those $\omega$'s, such that $X_{T(\omega)}(\omega) \in B$. Now, all these $\omega$'s together is a subset of $\Omega$, but what does it mean that they are contained in $\mathcal{F}_T$? I mean $\mathcal{F}_T$ depends on $\omega$ aswell?

My second question is this: If we forget cadlag etc. Can it be shown that $X_T$ is just a random variable? The problem is showing measurability, that is, that $X_T^{-1}(B) \in \mathcal{F}$? Does this require a lot of work? Or maybe it can not be done? I've tried, but got nowhere.

  • 1
    $\begingroup$ No, the $\sigma$-field $\mathcal{F}_T$ doesn't depend on $\omega$. Reread its definition. $\endgroup$ – Nate Eldredge Mar 4 '16 at 3:25

Given a stopping time $\tau$, we define $$\mathcal F_\tau = \sigma\left(A\in\mathcal F_\infty: A\cap\{\tau\leqslant t\}\in \mathcal F_t,\ t\geqslant 0 \right) $$ where $$\mathcal F_\infty = \sigma\left(\bigcup_{t\geqslant 0} \mathcal F_t\right).$$ If $X_t$ is càdlàg and adapted to $\{\mathcal F_t\}$ (i.e. when $\{\mathcal F_t\}$ is the natural filtration), then $X_t$ is progressively measurable, that is, for each $t$, the map $(t,\omega)\mapsto X_t(\omega)$ is $\mathcal B([0,t])\otimes \mathcal F_t$-measurable. For a fixed $t$, $$X_\tau = X_{t\wedge\tau} = X_t\mathsf 1_{t\leqslant\tau} + X_\tau\mathsf1_{t>\tau} $$ and $t\wedge\tau$ is $\mathcal F_\tau$-measurable, so $X_\tau$ is $\mathcal B([0,\infty)\otimes\mathcal F_t$-measurable and therefore $X_\tau$ is progressively measurable (and hence adapted).


Refer to Math1000's answer for the definition of $\mathscr{F}_T$, when $T$ is a stopping time. The following theorem answers your first question.

Theorem (Protter). Let $T$ be a finite stopping time. Then $\mathscr{F}_T$ is the $\sigma$-algebra generated by random variables of the form $X_T$ where $X$ is an adapted cadlag process.

The theorem says that $\mathscr{F}_T$ is generated by sampling cadlag processes at the stopping time $T$. In particular, if $X$ is adapted and cadlag, then $X_T$ is $\mathscr{F}_T$-measurable.

Proof of theorem. If $A \in \mathscr{F}_T$ then $X_t := 1_A 1_{t \geq T}$ is an adapted cadlag process with $X_T = 1_A$, showing $\mathscr{F}_T \subseteq \sigma\{X_T:X \text{ adapted cadlag}\}$. Next let $X$ adapted cadlag be given. Fix a borel $B$ and $t \geq 0$. We need to show $\{X_T \in B\} \cap \{T \leq t\} \in \mathscr{F}_t$. We can view $X_T$ as a composition $X\circ \phi$ where $\phi(\omega) := (T(\omega),\omega)$. Since $X$ is adapted and cadlag, $X$ is progressively measurable, so that $X|_{[0,t]\times\Omega}$ is $(\mathscr{B}([0,t])\otimes \mathscr{F}_t)/\mathscr{B}(\mathbb{R})$ measurable. Since $\{T \leq t\} \in \mathscr{F}_t$ we have $\phi|_{\{T \leq t\}}$ is $(\mathscr{F}_t \cap \{T \leq t\})/(\mathscr{B}([0,t])\otimes \mathscr{F}_t)$ measurable (check each component separately). Thus $X|_{[0,t]\times \Omega} \circ \phi|_{\{T \leq t\}}$ is $(\mathscr{F}_t \cap \{T \leq t\})/\mathscr{B}(\mathbb{R})$ measurable. Thus $\{X_T \in B\} \cap \{T \leq t\} = \{X|_{[0,t]\times \Omega} \circ \phi|_{\{T \leq t\}} \in B\} \in \mathscr{F}_t \cap \{T \leq t\} \subseteq \mathscr{F}_t$. Completing the reverse inclusion.

As for your second question, there is no guarantee that $X_T$ will be a random variable unless some assumptions are made. The weakest assumption one usually encounters is that $X$ is measurable, i.e. $(\mathscr{B}([0,\infty)) \otimes\mathscr{F}) / \mathscr{B}(\mathbb{R})$ measurable and $T\geq 0$ is a random time, i.e. $\mathscr{F}/\mathscr{B}([0,\infty))$ measurable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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