To this day I still ignore the reason why ,to define ,for instance, a filtration that is continuous to the right (based on an existing one),We write:

$D_{t}=D_{t+}^{0}=\bigcap_{\epsilon>0} D_{t+\epsilon}^{0}$ (isn't $\bigcap_{\epsilon \geq \alpha} D_{t+\epsilon}^{0}= D_{\alpha}^{0}?$)

instead of just writing

$D_{t}=lim_{\epsilon \rightarrow 0^+ } D_{t+\epsilon}^{0}$

(there is something similar with unions too)

Thank you in advance for your answers and insights

  • $\begingroup$ How do you define $\lim_{\epsilon \to 0} \mathcal{F}_{\epsilon}$ for some family of $\sigma$-algebras $\mathcal{F}_{\epsilon}$? (That's what you need to define in order to write $\lim_{\epsilon \to 0+} D_{t+\epsilon}^0$.) $\endgroup$
    – saz
    Feb 27 '15 at 13:50
  • $\begingroup$ Thanks saz. I see what you mean although I can present it in rigorous terms. What I have found confusing was the use of intersection the while we know that in the case of filtrations for instance we have the inclusion $D_r \subseteq D_s$ for $r \leq s$. besides the definitions found in textbooks of limsup and liminf to define the limits (when they exist) are not natural/intuitive and well explained in books. I bet first mathematicians haven't invented those notions out of nowhere. $\endgroup$
    – Averroes
    Feb 27 '15 at 14:51
  • $\begingroup$ Well, $\bigcap_{\epsilon>0} D_{t+\epsilon}^0$ is the smallest $\sigma$-algebra which contains all $D_{t+\epsilon}^0$, $\epsilon>0$. Obviously, we can introduce a new notation and write $$D_{t+}^0 = \lim_{\epsilon \to 0} D_{t+\epsilon}^0,$$ but what's the advantage of this? (Mind that $\limsup$ and $\liminf$ are defined for sequences of sets and not sequences of $\sigma$-algebras; and both notions have a natural meaning.) $\endgroup$
    – saz
    Feb 27 '15 at 15:10
  • $\begingroup$ Again thank you for pointing out at me confusing sets and filtrations :). Could you please give me an example of a trivial filtration that doesn't respect continuity? $\endgroup$
    – Averroes
    Feb 27 '15 at 15:57

By definition,

$$D_{t+}^0 = \bigcap_{\epsilon>0} D_{t+\epsilon}^0$$

is the smallest $\sigma$-algebra which contains $D_{t+\epsilon}^0$ for all $\epsilon>0$. We can introduce a new notation and write

$$D_{t+}^0 = \lim_{\epsilon \to 0} D_{t+\epsilon}.$$

(Mind that we have to give $\lim_{\epsilon \to 0}$ a meaning; in general, it is not clear how to define the limit of a sequence of $\sigma$-algebras.)

An easy example for a filtration which is not right-continuous is the following: Take

$$D_t := \{\emptyset,\Omega\} \qquad \text{for all} \, \, t \leq 1$$


$$D_t := \mathcal{A} \qquad \text{for all} \, \, t>1$$

for some (non-trivial) $\sigma$-algebra $\mathcal{A}$ on $\Omega$. Then

$$D_{1+} = \mathcal{A} \neq D_1 = \{\emptyset,\Omega\}.$$

More sophisticated examples are $\sigma$-algebras generated by stochastic processes, i.e.

$$D_t := \sigma(X_s; s \leq t), \qquad t \geq 0;$$

they are (in general) not right-continuous.

Two remarks:

  • One may ask: Why do we need right-filtrations? Well, one typical application is the following: Suppose that $X \in L^1$ is a random variable and $(D_t)_{t \geq 0}$ a right-continuous filtration. Then $$\lim_{s \downarrow t} \mathbb{E}(X \mid D_s) = \mathbb{E}(X \mid D_t).$$
  • Even if a given filtration is not right-continuous, it is often possible to enlarge the filtration in such a way that it becomes right-continuous; therefore many authors assume ("without loss of generality") that the given filtration is right-continuous (the so-called usual conditions).

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.