$\mathcal{F}_{T}$, when $T$ is constant.

Question

let $T$ be a stopping time, assume it's constant, find $\mathcal{F}_{T}$ :

$\mathcal{F}_{T}=\left\{A \in \mathcal{F}_{\infty}, A \cap\{T=n\} \in \mathcal{F}_{n}, \forall n \geq 1\right\}$

since $\{T=n\}$ is either the empty set or the universal set and the empty set is always an element of a sigma algebra then :

$\mathcal{F}_{T}=\left\{A \in \mathcal{F}_{\infty}, A \in \mathcal{F}_{T}\right\}$

so $\mathcal{F}_{T} =\mathcal{F}_{T}$, ?

I'm not sure of the point of the question, is there something I'm missing or is it that simple?

Answer 1

The answer is yes, I think you got lost in notation. Let $j$ be such that $T=j$ in the whole probability space. As you said, $A\cap[T=n]\in \{\emptyset, A\}$. Moreover, to define $\mathcal{F}_T$, we assume that $\{\mathcal{F}_t\}_t$ is a filtration and take $\mathcal{F}_{\infty}= \vee_{t} \mathcal{F}_t$ (the $\sigma$-algebra generated by the union of the finite time filtrations), hence $ \mathcal{F}_\infty \cap \mathcal{F}_n= \mathcal{F}_n$. Therefore, $$\mathcal{F}_{T}=\left\{A \in \mathcal{F}_{\infty}: A \cap \Omega \in \mathcal{F}_j, A\cap \emptyset \in \mathcal{F}_n, \forall n \neq j \right\} = \mathcal{F}_j.$$