Filtration of a probability space

  • Context: MHB 
  • Thread starter Thread starter gnob
  • Start date Start date
  • Tags Tags
    Probability Space
Click For Summary
SUMMARY

The discussion centers on the relationship between a filtration of a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and the generated $\sigma$-algebra $\mathcal{F}_{\infty}$. It is established that $\mathcal{F}_{\infty} = \sigma\left(\bigcup_{t} \mathcal{F}_t \right)$ does not equal $\mathcal{F}$ unless strict inclusion is assumed. A counterexample is provided where $\mathcal{F}_t = \mathcal{G}$, indicating that $\mathcal{G}$ is a strictly contained $\sigma$-algebra, thus demonstrating the necessity of strict inclusion in the definition of filtration.

PREREQUISITES
  • Understanding of probability spaces and their components: $(\Omega,\mathcal{F},\mathbb{P})$
  • Knowledge of $\sigma$-algebras and their properties
  • Familiarity with the concept of filtrations in probability theory
  • Basic grasp of counterexamples in mathematical proofs
NEXT STEPS
  • Study the properties of $\sigma$-algebras in detail
  • Learn about the implications of strict inclusion in filtrations
  • Explore counterexamples in probability theory to strengthen understanding
  • Investigate the role of $\mathcal{F}_{\infty}$ in stochastic processes
USEFUL FOR

Mathematicians, statisticians, and students of probability theory who are exploring advanced concepts in filtrations and $\sigma$-algebras.

gnob
Messages
11
Reaction score
0
Hi again, I hope you can clarify me on this:

Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and let $\{\mathcal{F}_t\}_{0\leq t<\infty}$ be a filtration on it. Define
$\mathcal{F}_{\infty} = \sigma\left(\bigcup_{t} \mathcal{F}_t \right)$ where $t \in [0,\infty).$

My question: Is $\mathcal{F}_{\infty} = \mathcal{F}$?

I came across with an answer that it is not, but I forgot the source nor I remember if there is a counterexample. Can anyone please help me? Thanks. Please also give me some reference on it.

Thanks, thanks, thanks...:o
 
Physics news on Phys.org
If in the definition of filtration we don't assume strict inclusion we get a counter example taking $\mathcal F_t=\mathcal G$, where the latest is a strictly contained $\sigma$-algebra.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K