- Thread starter
- #1
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...
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...