proof of hitting times are stopping times for right-continuous processes

Let $(\mathcal{F})_{t\in\mathbb{T}}$ be a filtration (http://planetmath.org/FiltrationOfSigmaAlgebras) on the measurable space $(\Omega,\mathcal{F})$ , It is assumed that $\mathbb{T}$ is a closed subset of $\mathbb{R}$ and that $\mathcal{F}_{t}$ is universally complete for each $t\in\mathbb{T}$ .

Let $X$ be a right-continuous and adapted process taking values in a metric space $E$ and $S\subseteq E$ closed. We show that

\tau=\inf\left\{t\in\mathbb{T}:X_{t}\in S\right\}

is a stopping time. Assuming $S$ is nonempty and defining the continuous function $d_{S}(x)\equiv\inf\{d(x,y)\colon y\in S\}$ , then $\tau$ is the first time at which the right-continuous process $Y_{t}=d_{S}(X_{t})$ hits $0$ .

Let us start by supposing that $\mathbb{T}$ has a minimum element $t_{0}$ .

If $\mathbb{P}$ is a probability measure on $(\Omega,\mathcal{F})$ and $\mathcal{F}^{\mathbb{P}}_{t}$ represents the completion (http://planetmath.org/CompleteMeasure) of the $\sigma$ -algebra $\mathcal{F}_{t}$ with respect to $\mathbb{P}$ , then it is enough to show that $\tau$ is an $(\mathcal{F}^{\mathbb{P}}_{t})$ -stopping time. By the universal completeness of $\mathcal{F}_{t}$ it would then follow that

\left\{\tau\leq t\right\}\in\bigcap_{\mathbb{P}}\mathcal{F}^{\mathbb{P}}_{t}=% \mathcal{F}_{t}

for every $t\in\mathbb{T}$ and, therefore, that $\tau$ is a stopping time. So, by replacing $\mathcal{F}_{t}$ by $\mathcal{F}^{\mathbb{P}}_{t}$ if necessary, we may assume without loss of generality that $\mathcal{F}_{t}$ is complete with respect to the probability measure $\mathbb{P}$ for each $t$ .

Let $\mathcal{T}$ consist of the set of measurable times $\sigma\colon\Omega\rightarrow\mathbb{T}\cup\{\infty\}$ such that $\{\sigma<t\}\in\mathcal{F}_{t}$ for every $t$ and that $\sigma\leq\tau$ . Then let $\sigma^{*}$ be the essential supremum of $\mathcal{T}$ . That is, $\sigma^{*}$ is the smallest (up to sets of zero probability) random variable taking values in $\mathbb{R}\cup\{\pm\infty\}$ such that $\sigma^{*}\geq\sigma$ (almost surely) for all $\sigma\in\mathcal{T}$ .

Then, by the properties of the essential supremum, there is a countable sequence $\sigma_{n}\in\mathcal{T}$ such that $\sigma^{*}=\sup_{n}\sigma_{n}$ . It follows that $\sigma^{*}\in\mathcal{T}$ .

For any $n=1,2,\ldots$ set

\sigma_{1}=\inf\left\{t\in\mathbb{T}:t\geq\sigma^{*},Y_{t}<1/n\right\}.

Clearly, $\sigma_{1}\leq\tau$ and, choosing any countable dense subset $A$ of $\mathbb{T}$ , the right-continuity of $Y$ gives

\{\sigma_{1}<t\}=\bigcup_{\begin{subarray}{c}s<t,\\ s\in A\end{subarray}}\{\sigma^{*}<s,Y_{s}<1/n\}\in\mathcal{F}_{t}.

So, $\sigma_{1}\in\mathcal{T}$ , which implies that $\sigma_{1}\leq\sigma^{*}$ with probability one. However, by the right-continuity of $Y$ , $\sigma_{1}>\sigma^{*}$ whenever $\sigma^{*}$ is finite and $Y_{\sigma^{*}}>1/n$ , so

\mathbb{P}(\sigma^{*}<\infty,Y_{\sigma^{*}}>0)\leq\sum_{n}\mathbb{P}(\sigma^{*% }<\infty,Y_{\sigma^{*}}>1/n)=0.

This shows that $Y_{\sigma^{*}}=0$ and therefore $\sigma^{*}\geq\tau$ whenever $\sigma^{*}<\infty$ . So, $\sigma^{*}=\tau$ almost surely and $\tau\in\mathcal{T}$ giving,

\left\{\tau\leq t\right\}=\{\tau<t\}\cup\{Y_{t}=0\}\in\mathcal{F}_{t}.

So, $\tau$ is a stopping time.

Finally, suppose that $\mathbb{T}$ does not have a minimum element. Choosing a sequence $t_{n}\rightarrow-\infty$ in $\mathbb{T}$ then the above argument shows that

\tau_{n}=\inf\left\{t\in\mathbb{T}:t\geq t_{n},Y_{t}=0\right\}

are stopping times so,

\left\{\tau\leq t\right\}=\bigcup_{n}\left\{\tau_{n}\leq t\right\}\in\mathcal{% F}_{t}

as required.

Title	proof of hitting times are stopping times for right-continuous processes
Canonical name	ProofOfHittingTimesAreStoppingTimesForRightcontinuousProcesses
Date of creation	2013-03-22 18:39:12
Last modified on	2013-03-22 18:39:12
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 60G05
Classification	msc 60G40