3
$\begingroup$

Let

  • $(\Omega,\mathcal A)$ be a measurable space
  • $I\subseteq\mathbb R$
  • $(\mathcal F_t)_{t\in I}$ be a filtration of $\mathcal A$
  • $\tau$ be an $\mathcal F$-stopping time, i.e. $\tau:\Omega\to I\cup\sup I$ is $\mathcal A$-$\mathcal B(I\cup\sup I)$-measurable ($\mathcal B(E)$ denotes the Borel $\sigma$-algebra on $E\subseteq[-\infty,\infty])$ and $$\left\{\tau\le t\right\}\in\mathcal F_t\;\;\;\text{for all }t\in I\tag1$$
  • $t\in I$

How can we show that $\tau\wedge t$ is $\mathcal F_t$-measurable?

Let $\iota$ denote the inclusion of $I\cup\sup I$ into $\overline{\mathbb R}$. Note that $$\mathcal B(A)=\left.\mathcal B(\overline{\mathbb R})\right|_A=\left\{A\cap B:B\in\mathcal B(\overline{\mathbb R})\right\}\;\;\;\text{for all }A\subseteq\overline{\mathbb R}\tag2$$ and hence $\tau$ is $\mathcal A$-$\mathcal B(I\cup\sup I)$-measurable iff $\iota\tau$ is $\mathcal A$-$\mathcal B(\overline{\mathbb R})$-measurable.

Let $t\in I$. By the former argument, $\tau\wedge t$ is $\mathcal F_t$-measurable iff $$\left\{\tau\wedge t\le b\right\}\in\mathcal F_t\tag3\;\;\;\text{for all }b\in\mathbb R\;.$$ If $b\ge t$, then $$\left\{\tau\wedge t\le b\right\}=\left\{\tau\le b\right\}\cup\underbrace{\left\{t\le b\right\}}_{=\:\Omega}=\Omega\in\mathcal F_t\;.\tag4$$ If $b\in(-\infty,t)$, then $\left\{t\le b\right\}=\emptyset$. In that case, if $(-\infty,b]\cap I=\emptyset$, then $$\left\{\tau\wedge t\le b\right\}=\left\{\tau\le b\right\}=\emptyset\in\mathcal F_t\tag5\;.$$

Now, we arrived at the crucial part: If $(-\infty,b]\cap I$ is closed and nonempty, then $$s:=\max\left((-\infty,b]\cap I\right)<t$$ is well-defined and hence $$\left\{\tau\wedge t\le b\right\}=\left\{\tau\le b\right\}=\left\{\tau\le s\right\}\in\mathcal F_s\subseteq\mathcal F_t\tag6\;.$$ But what can we do otherwise?

$\endgroup$
  • $\begingroup$ $\tau$ is a stopping time just means that it is a random variable in your setup? Not the usual $(\tau \leq t)\in \mathcal{F}_t$ ? $\endgroup$ – Conformal Jun 12 '17 at 20:58
2
$\begingroup$

Edit answer to clarify your comment:

Continuing from where you stopped:

if $b\geq \sup I$ then $(\tau \leq b)=\Omega\in \mathcal{F}_t$. If $b<\sup I$, define $s=\sup I\cap(-\infty,b]$. If $s\in I$ we are done. Otherwise, put $s_n \uparrow s,s_n\in I $.

It holds that $$(\tau\leq b) = \bigcup_{n}(\tau\leq s_n)$$

The reason for the equality is that $\tau$ takes value in $I$ and therefore cant hit the interval $[s,b]$.

The union consists of $\mathcal{F}_{s_n}\subset\mathcal{F}_{t}$ sets.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I was unsure why $\left\{\tau\le b\right\}=\bigcup_{n\in\mathbb N}\left\{\tau\le s_n\right\}$. In particular, if $(t_n)_{n\in\mathbb N}\subseteq I$ is nondecreasing with $$t_n\xrightarrow{n\to\infty}t:=\sup I\;,$$ then clearly $$\bigcup_{n\in\mathbb N}\left\{\tau\le t_n\right\}\subseteq\left\{\tau\le t\right\}\;.$$ However, the other inclusions doesn't seem to hold and I think we can only show $$\left\{\tau<t\right\}\subseteq\bigcup_{n\in\mathbb N}\left\{\tau\le t_n\right\}\;.$$ $\endgroup$ – 0xbadf00d Jun 15 '17 at 16:19
  • $\begingroup$ We should be able to fix argument, since $\tau$ is $\mathcal F_t$-measurable iff $\left\{\tau\color{red}{<}b\right\}\in\mathcal F_t$ for all $b\in\overline{\mathbb R}$. $\endgroup$ – 0xbadf00d Jun 15 '17 at 18:39
  • $\begingroup$ I don't understand your first comment. I guess your t is my s? In that case your comment is only relevant for $b\geq sup I$ because otherwise the other inclusion does hold , as s\not\in I. But in this case $(tau\leq b)=\Omega$ (written from mobile) $\endgroup$ – Conformal Jun 15 '17 at 20:05
  • $\begingroup$ I saw your link. I just jumped into your proof, and thought the case $b < \sup I$ case was assumed. I edited my answer to include the $b\geq \sup I$ case. $\endgroup$ – Conformal Jun 15 '17 at 22:36
  • $\begingroup$ We have $\left\{\tau\le b\right\}=\left\{\tau\le s\right\}$. That's clear. However, $$\left\{\tau\le s\right\}=\bigcup_{n\in\mathbb N}\left\{\tau\le s_n\right\}$$ is wrong. We only can show that $$\left\{\tau<s\right\}=\bigcup_{n\in\mathbb N}\left\{\tau<s_n\right\}\;.$$ The latter union is in $\mathcal F_t$. Now, we can assume that $s\not\in I$ (as you've mentioned) and this even yields $$\left\{\tau\le t\right\}=\left\{\tau<s\right\}\;.$$ Hence, we're done. $\endgroup$ – 0xbadf00d Jun 16 '17 at 14:06

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.