Taking the definition from wikipedia,

With $X = (X_t : t \geq 0) $ as a stochastic process on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with natural filtration $\{ \mathcal{F}(t) \}_{t \geq 0}$.

$$ \mathcal{F}^+(t) = \bigcap_{s>t} \mathcal{F}(s) = \bigcap_{\varepsilon>0} \mathcal{F}(s+\varepsilon) $$

$$ \mathcal{F}^+(\tau)=\{A \in \mathcal{F}: \{\tau=t\} \cap A \in \mathcal{F}^+(t) ,\, \forall t \geq 0\ \} $$

$X$ is said to have the strong Markov property if, for each stopping time $\tau$, conditioned on the event $\{ \tau < \infty \}$, we have that for each $ t \geq 0$, $X_{\tau + t}$ is independent of $\mathcal{F}^+(\tau)$ given $X_\tau$.

What is the importance of being independent of $\mathcal{F}^+(\tau)$ instead of $\mathcal{F}(\tau)$? I know that $\mathcal{F}^+(t)$ is right continuous, i.e.

$$ \bigcap_{\varepsilon > 0} \mathcal{F}^+(t+\varepsilon) = \mathcal{F}^+(t) $$ I understand this as there being no additional information for each infinitesimally small step.

Am I interpretting $\mathcal{F}^+(t)$ correctly as a $\sigma$-algebra of events $X \subseteq \mathcal{F}^+(t)$ s.t. $\{ \tau \leq t \}$ is observable?

It would be most helpful if this was explained with Brownian motion as an example.


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