• $(\Omega,\mathcal A)$ 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?

  • $\begingroup$ 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

$X_{\tau}(\omega)$ 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.


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