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Suppose that we are working with a Filtration which is right continuous. I know then, that the first hitting time of a right continuous process into an open set is a stopping time. Is the same true, if we replace right continuity with left continuity?

hulik

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    @DavideGiraudo No, sorry if that wasn't clear. The filtration is again right continuous, just the process should be left continuous instead of right continuous.2012-08-04

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For a right continuous filtration $(\mathcal F_t)_{t\geq 0}$, a left continuous process and $O$ an open set, the map $T_O\colon\omega\mapsto \inf\{t>0, X_t(\omega)\in O\}$ is a stopping time.

Since the filtration is right continuous, it's enough to show that $\{T_O for all $t>0$. We have $\{T_O Indeed, let $\omega$ such that $T_O(\omega). We can find $t' such that $X_{t'}(\omega)\in O$. By left continuity and using openness of $O$, we can find a rational $q such that $X_q(\omega)\in O$.

Conversely, if $X_q(\omega)\in O$ for some rational $q, we have $T_O(\omega)\leq q.

Since $\{X_q\in O\}\in\mathcal F_q\subset\mathcal F_t$ for each $q, we are done.