A problem about the supremum of countable stopping times

Asked Jun 30 '16 at 15:28

Active Jun 30 '16 at 16:42

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Let $\left(\Omega\ ,\mathcal{F}\ ,\mathbb{P} \right)$ be a probability space with a countable filtration $F=\left\{ \mathcal{F}_0,\mathcal{F}_1,\cdots,\mathcal{F}_n,\cdots \right\}$.
$\left\{ T_n \right\}$ is a sequence (countable infinite) of stopping times.

Let $S_1=\wedge_n{T_n}$ and $S_2=\vee_n{T_n}$ , it is easy to show that $S_1$ and $S_2$ are also stopping times and that $\mathcal{F}_{S_1} = \bigcap_{n}{\mathcal{F}_{T_n}}$ .

It is also easy to show that $\mathcal{F}_{T_1\vee T_2}=\mathcal{F}_{T_1}\vee\mathcal{F}_{T_2}=\sigma\left(\mathcal{F}_{T_1}\cup\mathcal{F}_{T_2}\right)$ , but I wonder whether $$\mathcal{F}_{S_2} = \vee_{n}{\mathcal{F}_{T_n}}$$ still holds.

edited Jun 30 '16 at 16:42

Davide Giraudo

181,608

asked Jun 30 '16 at 15:28

Wang Ke

See this question: https://math.stackexchange.com/q/2478216/36150 – saz Oct 18 '17 at 17:30
... and this question – saz Jul 29 '19 at 17:09
... and also this question – saz Feb 06 '20 at 16:19
Hi @saz, what does $\mathcal{F}{T_1}\vee\mathcal{F}{T_2}$ mean? – Akira Feb 18 '20 at 14:43
@Navier_Stokes $\mathcal{F}{T_1} \vee \mathcal{F}{T_2}$ is the smallest $\sigma$-algebra which contains $\mathcal{F}{T_1}$ and $\mathcal{F}{T_2}$, i.e. $\mathcal{F}{T_1} \vee \mathcal{F}{T_2} = \sigma(\mathcal{F}{T_1},\mathcal{F}{T_2})$. – saz Feb 18 '20 at 15:18
@saz you meant $\mathcal{F}{T_1}\vee\mathcal{F}{T_2}:=\sigma\left(\mathcal{F}{T_1}\cup\mathcal{F}{T_2}\right)$? – Akira Feb 18 '20 at 15:20
1

@Navier_Stokes Well, it's just a matter of notation... if you prefer to write it like this, then go ahead (for me both things are (by definition) the same). – saz Feb 18 '20 at 15:22

A problem about the supremum of countable stopping times

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