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Given a set $\Omega$ and a filtration $(\mathcal{F_t}, t\in T)$ on $\Omega$, where $T\subseteq\mathbb{R}$, we say that such a filtration is right-continuous if for every $t\in T$ it holds that $\mathcal{F_t}=\bigcap\limits_{\varepsilon>0}\mathcal{F_{t+\varepsilon}}$.

I was wondering: what is the actual meaning of this property? What is the main intuition behind this definition?

Did
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Jack London
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    Roughly speaking, the identity $\mathcal F_{t+}=\mathcal F_t$ means that there is no information "just after" time $t$ which is not already given at time $t$ or before. Example: Consider the canonical filtration $(\mathcal F^X_t)$ of the process $(X_t)$ defined by $X_t=\max(N-t,0)$, where $N$ is some given integer valued random variable. Then $[N=n]\in\mathcal F^X_{n+}\setminus\mathcal F^X_n$ for every $n$ such that $[N=n]$ and $[N\geqslant n+1]$ are both nonempty, hence this is a case when $\mathcal F_{t+}\ne\mathcal F_t$, at least for some times $t$. – Did Mar 08 '16 at 20:39
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    @ Jack London : concerning the "why" we need this "bizarre" notion, we can say that the right continuity of filtration (which is refered to the usual hypothesis together with the integration of negligible sets in the $\sigma$-algebra $\mathcal{F}_0$) is important to show that there exists a càdlàg (right continuous with left limit) modification of any local martingales. This permits to evacuate all weird bad behaviors of measure theoretic nature on path of the process and to concentrate on the fact that a local martingale is essentially a with process càdlàg paths. Best regards – TheBridge Mar 09 '16 at 13:22
  • @Did: referring to your example, I would say that since $X_{n-\varepsilon}$ is measurable w.r.t. $\mathcal{F}^X_n$ for some fixed $\varepsilon>0$ then $[N=n]=[X_{n-\varepsilon} = \varepsilon]\in\mathcal{F}^X_n$. So it should be false that $[N=n] \in \mathcal{F}_{n+}^X \setminus \mathcal{F}_n^X$. – Jack London Mar 09 '16 at 15:25
  • Sorry I mistyped, the counterexample is $X_t=\max(t-N,0)$. – Did Mar 09 '16 at 20:13
  • @Did: ok, perfect, now I can see your point! To my understanding the right-continuity of the filtration is a key issue when stopping times come into play. Referring to your example one may be interested in the quantity "exit time from the zero state", say $\mathcal{T}$; in this case the event $[n⩾\mathcal{T}]$ does not belong to $\mathcal{F}_n^X$ and $\mathcal{T}$ is not a stopping time. – Jack London Mar 10 '16 at 08:40
  • @TheBridge: thank you as well. – Jack London Mar 10 '16 at 09:24

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