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Let

  • $(\Omega,\mathcal A)$ be a measurable space;
  • $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$ and $$\mathcal F_\infty:=\sigma(\mathcal F_t,t\ge0);$$
  • $\tau:\Omega\to[0,\infty]$.

I'm used to the definition $$\mathcal F_\tau:=\left\{A\in\mathcal A:A\cap\left\{\tau\le t\right\}\in\mathcal F_t\text{ for all }t\ge0\right\}.$$ However, I've also seen a different definition: $$\tilde{\mathcal F}_\tau:=\left\{A\in\mathcal F_\infty:A\cap\left\{\tau\le t\right\}\in\mathcal F_t\text{ for all }t\ge0\right\}.$$ Are they really different or do they coincide?

If $A\in\mathcal F_\tau$, we can clearly write $$A=\underbrace{\bigcup_{n\in\mathbb N}\left(A\cap\{\tau\le n\}\right)}_{\in\:\mathcal F_\infty}\cup(A\cap\{\tau=\infty\}).$$ Since $$\{\tau<\infty\}=\bigcup_{n\in\mathbb N}\{\tau\le n\},$$ we know that $$\{\tau=\infty\}=\{\tau<\infty\}^c\in\mathcal F_\infty\tag2;$$ at least if we assume that $\tau$ is an $\mathcal F$-stopping time; which I'm clearly willing to do.

However, it seems like we are only able to conclude $A\cap\{\tau<\infty\}\in\mathcal F_\infty$.

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1 Answers1

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They do not coincide. Consider $T\equiv \infty$ and $\mathcal{F}_t=\{\emptyset,\Omega\}$. Then $\mathcal{F}_{\infty}$ is trivial as well, and since $\{T\le t\}=\emptyset$ for all $t\in\mathbb{R}_{+}$, $$ \mathcal{F}_T^1:=\{A\in \mathcal{A}:A\cap\{T\le t\}\in\mathcal{F}_t \text{ for all }t\ge 0\}=\mathcal{A}, $$ but $$ \mathcal{F}_T^2:=\{A\in \mathcal{F}_{\infty}:A\cap\{T\le t\}\in\mathcal{F}_t\text{ for all }t\ge 0\}=\mathcal{F}_{\infty}, $$ Also, as you noticed if $T<\infty$ everywhere $\mathcal{F}_T^1\subset \mathcal{F}_{\infty}$, i.e., $\mathcal{F}_T^1=\mathcal{F}_T^2$.

After looking at different textbooks, it seems that there is no convention about the definition of $\mathcal{F}_T$. For example, Dellacherie & Meyer use $\mathcal{F}_T^2$, while Karatzas & Shreve use $\mathcal{F}_T^1$. Yet another construction is given in Çinlar's book, where $$ \mathcal{F}_T=\{A\in \mathcal{A}:A\cap \{T\le t\}\in \mathcal{F}_t \text{ for each } t\in \bar{\mathbb{R}}_{+}\} $$ such that $\mathcal{F}_T\subset \mathcal{F}_{\infty}$ automatically.