1
$\begingroup$

Let $(\Omega, \mathcal{A}, P)$ be a probability space, $(E, \mathcal{E})$ a measurable space and $(X_t)_{t \in [0, \infty)}$ a continuous-time process in $E$. For a subset $A \subseteq E$ denote by $\tau_A(\omega) := \inf \{ t > 0 \mid X_t(\omega) \in A \}$ the hitting time of $A$.

If we are given a filtration $\mathcal{F}_t$ on $\Omega$ then there are several theorems that provide sufficient conditions for $\tau_A$ to be a stopping time.

I am rather searching for examples such that $\tau_A$ is not even measurable. These example should be not too constructive meaning that (if possible) (i) $(\Omega, \mathcal{A}, P)$ should be complete, (ii) $E$ is a standard space (e.g. $E = \mathbb{R}$ and $\mathcal{E}$ the Borel $\sigma$-algebra), (iii) $A \in \mathcal{E}$ is measurable (or even open or closed if $E$ is a topological space) and (iv) $X_t$ has nice sample path properties which are nevertheless not enough for measurability of $\tau_A$.

$\endgroup$
  • $\begingroup$ To clarify: you want stopping times that are not even measurable, rather than stopping times that are just not adapted to the natural filtration? $\endgroup$ – Aaron Montgomery Jul 10 '18 at 16:55
  • $\begingroup$ No, I am searching for examples of hitting times that are not $\mathcal{A}$-measurable. A hitting time $\tau_A$ is a stopping time if moreover $\{ \tau_A \leq t \} \in \mathcal{F}_t$ for each $t$ which implies $\mathcal{A}$-measurability. $\endgroup$ – yadaddy Jul 10 '18 at 16:57
  • $\begingroup$ Err, right -- that's what I intended to say (but did not) in my comment. $\endgroup$ – Aaron Montgomery Jul 10 '18 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.