What is $\mathcal F_{t^+}=\bigcap_{u>t}\mathcal F_u$ ? Is $\mathcal F_{t^+}$ anticipative?

Question

Let $(\mathcal F_t)_{t\geq 0}$. I know that $$\mathcal F_{t^+}=\bigcap_{u>t}\mathcal F_u,$$ is a filtration. I know that if the filtratino is right continuous, then $\mathcal F_{t^+}=\mathcal F_t$. But suppose it's not, what is exactely $\mathcal F_{t^+}$ ? is this filtration anticipative ? I mean, does it see a bit in the futur ? Because I don't understand in what is it interesting. For example, what would mean $X_s$ being $\mathcal F_{t^+}$ measurable ? (if $X$ is $\mathcal F_t$ adapted).

Answer 1

Yes, it's a bit like an infinitesimal glimpse into the future. A typical example of an $\mathcal{F}_{t+}$-measurable random variable is the first hitting time

$$\tau = \inf\{t \geq 0; X_t \in U\}$$

of an open set $U$, where $(X_t)_{t \geq 0}$ is some nice process (e.g. with continuous sample paths).

Consider, for instance, some continuous process starting at $X_0=0$ and the open set $U=(a,\infty)$ for $a>0$. If we have the information about the sample paths $X_t(\omega)$, say, up to time $T>0$, then we cannot decide, in general, whether $\tau(\omega) \leq T$ or $\tau(\omega)>T$. Why? If $X_t(\omega)<a$ for all $t \leq T$, then certainly $\tau(\omega)>T$ because we have continuous sample paths - this is the "good" case. However, if $X_t(\omega)<a$ for $t<T$ and $X_T(\omega)=a$, then we cannot tell (in general) whether the process will next move upwards (hence, hitting $U$) or downwards (hence, avoiding $U$), i.e. we do not know whether $\tau(\omega) \leq T$ or $\tau(\omega)>T$. This means that $\tau$ is not an $\mathcal{F}_t$-stopping time. Having a small glimpse into the future, allows us to get rid of the problem, i.e. $\tau$ is an $\mathcal{F}_{t+}$-stopping time. See this answer for small illustration.