# Filters for Stochastic Process

Let $$\Omega$$ be a set and $$\{F_t\}_{t \geqslant 0}$$ be a filtration. That is, it is a family of $$\sigma$$-algebras on $$\Omega$$ such that $$F_{s} \subseteq F_t$$ whenever $$s \leqslant t$$. Fix $$t > 0$$. Define :

$$\begin{equation*} F_{t+} = \bigcap_{\epsilon > 0} F_{t + \epsilon} \end{equation*}$$ $$\begin{equation*} F_{t-} = \sigma (\bigcup_{s < t} F_{s}) \end{equation*}$$

And:

$$\begin{equation*} F_{t++} = \bigcap_{\epsilon > 0} F_{(t + \epsilon)+}= \bigcap_{\epsilon > 0} \bigcap_{\delta > 0} F_{t + \epsilon + \delta} = \bigcap_{\epsilon > 0} F_{t + \epsilon} = F_{t+} \end{equation*}$$ $$\begin{equation*} F_{t+-} = \bigcap_{\epsilon > 0} F_{(t + \epsilon)-} = \bigcap_{\epsilon > 0} \sigma (\bigcup_{s < t + \epsilon} F_{s}) \end{equation*}$$ $$\begin{equation*} F_{t-+} = \sigma (\bigcup_{s < t} F_{s+}) = \sigma (\bigcup_{s < t} \bigcap_{\epsilon > 0} F_{s+\epsilon}) \end{equation*}$$ $$\begin{equation*} F_{t--} = \sigma (\bigcup_{s < t} F_{s-}) = \sigma (\bigcup_{s < t} \sigma(\bigcup_{u < s} F_{u})) \end{equation*}$$

The $$F_{t++}$$ simplifies really well. The other three not so much. Is there any simpler form one can obtain for the other three? In particular, do we get examples where (at least one of the following proper inclusion) :

$$\begin{equation*} F_{t--} \subset F_{t-} \subset F_{t-+} \subset F_t \subset F_{t+-} \subset F_{t+} \end{equation*}$$

• The question is well posed but where does that come from? In 35 years of working with stochastic processes I have never seen ${\cal F}_{t+-}$ and ${\cal F}_{t-+}\,.$ Even about ${\cal F}_{t-}$ I don't remember much. Commented Apr 28 at 4:54
• It does not come from a source. It is just something I wonder. If $\{F_t\}_{t \geqslant 0}$ is a filter, and if we use $\{F_{t+}\}_{t \geqslant 0}$ or $\{F_{t-}\}_{t \geqslant 0}$ as new filters (for fun), what change will it bring and what does the structure look like.
– 温泽海
Commented Apr 28 at 14:29

1. $$F_{t+-} = F_{t+}$$ because $$F_u\subset F_{s-}$$ whenever $$u. (Incidentally, I'd have labelled this one $$F_{t-+}$$, but that's just me.)
2. Likewise $$F_{t-+} = F_{t-}$$ because $$F_{u+}\subset F_{s}$$ if $$u.
3. $$F_{t--} = F_{t-}$$.