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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

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  • $\begingroup$ 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, 2015 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, 2015 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, 2015 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, 2015 at 15:57

1 Answer 1

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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

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

and

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

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).
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