Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to figure this out, and I feel I'm pretty close to why it's the case. I just can't quite get the details to work.

Let $X$ be an adapted process on $(\Omega,\mathcal{F},\mathbb{P})$ and $T$ a finite stopping time. Show that $X_T$ is $\mathcal{F}$-measurable.

As I understand it, the process $X_T = \{X^T_n\}_{n\geq1}$ is defined as $X^T_n(\omega) := X_{T(\omega) \wedge n}(\omega)$. Since $X$ is an adapted process we have that $X_n$ is $\mathcal{F}_n$-measurable for all $n$, and since $\mathcal{F}_n \subset \mathcal{F}$ this intuitively should carry over to $X_T$. The $\mathcal{F}_n's$ are from the filtration of $X$. My problem is with $T(\omega)$ being dependant on $\omega$, and that $T$ isn't necessarily bounded which means you can't bound $T(\omega)\wedge n$.

Any tips on how I should approach this?

EDIT: some clarification

share|cite|improve this question
You should explain the notations: what is $\mathcal F$? I also think the process is discrete (by your attempt), but you have to tell it. Furthermore, I think a $n$ is missing in the definition of $X_T^n(\omega)$. – Davide Giraudo Oct 21 '12 at 20:00
Your definitions are ambiguous: usually, $X_T$ is not a process but the random variable $X_T:\omega\mapsto X_{T(\omega)}(\omega)$. The process $(X_{T\wedge n})_n$ is often abbreviated as $X^T$. – Did Oct 21 '12 at 21:05
Ah, I see what you mean. I'm still getting to terms with all the notation, it can be a little confusing. – BallzofFury Oct 21 '12 at 21:08
up vote 5 down vote accepted

Hint Show that everything in the RHS of the formula below is measurable with respect to $\mathcal F$: $$ X_T=\sum_n\mathbf 1_{T=n}\cdot X_n $$ Note If really the object of interest is the process $X^T=(X_{T\wedge n})_n$, use the formula $$ X^T_n=X_{T\wedge n}=\mathbf 1_{T\geqslant n}\cdot X_n+\sum_{k\lt n}\mathbf 1_{T=k}\cdot X_k $$ and show that everything in the RHS is measurable with respect to $\mathcal F_n$.

share|cite|improve this answer
I finally had some time to look at your reply properly. It seems I was looking at the wrong definition of $X_T$. Since $\{T=n\} \in \mathcal{F}_n$ by the definition of $T$, we have that $1_{\{T=n\}}$ is measurable. And since $X_n$ is also per definition $\mathcal{F}_n$-measurable, the product is also $\mathcal{F}_n$-measurable. Since $\mathcal{F}_n \subset \mathcal{F}$ for all $n \in \mathbb{N}$, we have that $X_T$ is $\mathcal{F}$-measurable. – BallzofFury Oct 22 '12 at 18:01

The idea is to decompose over the sets on which we know $T(\omega)$. Write $$\small (X^T_n)^{-1}(B)=\bigcup_{k\geq 0}(X_{k\wedge n}^{-1}(B)\cap \{\omega\mid T(\omega)=k\})=\bigcup_{k=0}^{n-1}(X_k^{-1}(B)\cap\{\omega\mid T(\omega)=k\}) \cup (X_n^{-1}(B)\cap \{\omega\mid T(\omega)\leq n\}^c)\in\mathcal F_n.$$

share|cite|improve this answer

Your Answer


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.