Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(B)_{t \geq 0}$ be a standard Brownian motion. Then $B$ is adapted to its natural filtration $(\mathcal{F}^B_t)_{t\geq 0}$. Often, we want to consider a slightly bigger filtration, ones satisfying the right-continuity condition. In that case, we define for every $t \geq 0$: \begin{align*} \mathcal{F}_t^+ = \bigcap_{s > t} \mathcal{F}^B_s. \end{align*} My questions is that is it true that, according to the definition above, \begin{align*} \mathcal{F}_t^+ = \bigcap_{n \in \mathbb{N}} \mathcal{F}^B_{t+1/n}. \end{align*}

My reasoning is as follow:

Clearly, the set on the left is contained by the set on the right. But for every $s > t$, we can find $n \in \mathbb{N}$ such that $t + 1/n < s$, so I think this implies that the set on the left is also contained by the set on the right. Is my assertion true? Is this how we actually prove that $\mathcal{F}_t^+$ is actually a $\sigma$-algebra (because countable intersection of $\sigma$-algebras is also a $\sigma$-algebra)?

share|improve this question
2  
Your reasoning is correct. But: Any intersection of $\sigma$-algebras (not necessarily only countably many) is a $\sigma$-algebra. –  martini May 16 '12 at 11:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.