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

Let ()t𝕋 be a filtrationPlanetmathPlanetmath ( on the measurable spaceMathworldPlanetmathPlanetmath (Ω,), It is assumed that 𝕋 is a closed subset of and that t is universally complete for each t𝕋.

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


is a stopping time. Assuming S is nonempty and defining the continuous functionMathworldPlanetmathPlanetmath dS(x)inf{d(x,y):yS}, then τ is the first time at which the right-continuous process Yt=dS(Xt) hits 0.

Let us start by supposing that 𝕋 has a minimum element t0.

If is a probability measureMathworldPlanetmath on (Ω,) and t represents the completionPlanetmathPlanetmath ( of the σ-algebra t with respect to , then it is enough to show that τ is an (t)-stopping time. By the universalPlanetmathPlanetmath completeness of t it would then follow that


for every t𝕋 and, therefore, that τ is a stopping time. So, by replacing t by t if necessary, we may assume without loss of generality that t is completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath with respect to the probability measure for each t.

Let 𝒯 consist of the set of measurable times σ:Ω𝕋{} such that {σ<t}t for every t and that στ. Then let σ* be the essential supremumMathworldPlanetmath of 𝒯. That is, σ* is the smallest (up to sets of zero probability) random variableMathworldPlanetmath taking values in {±} such that σ*σ (almost surely) for all σ𝒯.

Then, by the properties of the essential supremum, there is a countableMathworldPlanetmath sequence σn𝒯 such that σ*=supnσn. It follows that σ*𝒯.

For any n=1,2, set


Clearly, σ1τ and, choosing any countable dense subset A of 𝕋, the right-continuity of Y gives


So, σ1𝒯, which implies that σ1σ* with probability one. However, by the right-continuity of Y, σ1>σ* whenever σ* is finite and Yσ*>1/n, so


This shows that Yσ*=0 and therefore σ*τ whenever σ*<. So, σ*=τ almost surely and τ𝒯 giving,


So, τ is a stopping time.

Finally, suppose that 𝕋 does not have a minimum element. Choosing a sequence tn- in 𝕋 then the above argumentMathworldPlanetmath shows that


are stopping times so,


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