1
$\begingroup$

Let

  • $(\Omega,\mathcal A)$ be a measurable space
  • $I$ be an at most countable set
  • $\mathbb F=(\mathcal F_t)_{t\in I}$ be a filtration in $(\Omega,\mathcal A)$
  • $X=(X_t)_{t\in I}$ be an $\mathbb F$-adapted stochastic process with values in a measurable space $(E,\mathcal E)$
  • $\tau$ be a $\mathbb F$-stopping time
  • $X_\tau(\omega):=X_{\tau(\omega)}(\omega)$ for $\omega\in\Omega$
  • $\mathcal F_\tau:=\left\{A\in\mathcal A:A\cap\left\{\tau\le t\right\}\in\mathcal F_t\;\text{for all }t\in I\right\}$

I'm curious whether or not we need to force $\tau$ to be finite, if we want to show, that $X_\tau$ is $\mathcal F_\tau$-measurable.


Let's take a look at a possible proof. I will use the following elementary result:

Lemma 1$\;\;\;$Since $I$ is almost countable and $\tau$ is a $\mathbb F$-stopping time, $$\left\{\tau=t\right\}\in\mathcal F_t\;\;\;\text{for all }t\in I\;.$$

Now, let $A\in\mathcal E$ and $t\in I$. If $s\in I$ with $s\le t$, then $$\left\{\tau=s\right\}\in\mathcal F_t$$ by (Lemma 1) and, since $X$ is $\mathbb F$-adapted, $$X_s^{-1}(A)\in\mathcal F_t\;,$$ i.e. $$X_s^{-1}\cap\left\{\tau=s\right\}\in\mathcal F_t\;.$$

Since $I$ is almost countable, $$X_\tau^{-1}(A)\cap\left\{\tau\le t\right\}=\bigcup_{s\in I:s\le t}\left(X_s^{-1}\left(A\right)\cap\left\{\tau=s\right\}\right)\in\mathcal F_t$$

Is there any part of the proof which doesn't work until $\tau<\infty$ is enforced?

$\endgroup$
  • $\begingroup$ How do you define $X_{\tau}(\omega)$ for $\omega \in \{\tau=\infty\}$? $\endgroup$ – saz Sep 28 '15 at 17:39
  • $\begingroup$ @saz You're right, that's an obvious problem. Maybe we can define $X_\tau(\omega)$ as a suitable limit, but honestly, I don't know exactly how we could do that. $\endgroup$ – 0xbadf00d Sep 28 '15 at 18:03
  • $\begingroup$ @saz Maybe you should delete your comment and create an answer. $\endgroup$ – 0xbadf00d Sep 28 '15 at 18:23
2
$\begingroup$

$X_{\tau}(\omega)$ is not well-defined for $\omega \in \{\tau=\infty\}$. Since the limit

$$X_{\infty}(\omega) := \lim_{i} X_i(\omega)$$

does, in general, not exist, we cannot simply set $X_{\tau}(\omega) := X_{\infty}(\omega)$ for $\omega \in \{\tau=\infty\}$. One possibility to fix this is defining

$$X_{\tau}(\omega) = 0$$

for $\omega \in \{\tau=\infty\}$, i.e.

$$X_{\tau}(\omega) := X_{\tau(\omega)}(\omega) 1_{\{\tau<\infty\}}(\omega).$$

Then, using your argumentation, it is not difficult to see that this random variable is $\mathcal{F}_{\tau}$ measurable.

$\endgroup$

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.