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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?

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  • $\begingroup$ $\tau$ 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
    Commented Jun 12, 2017 at 20:58

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented Jun 15, 2017 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
    Commented Jun 15, 2017 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
    Commented Jun 15, 2017 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
    Commented Jun 15, 2017 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
    Commented Jun 16, 2017 at 14:06

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