Definition: Let $\mathcal{F}_t$ be a filtration. A mapping $T:\Omega\rightarrow\mathbb{R}_{\geq 0}$ is called a stopping time, if $\{T\leq t\}\in\mathcal{F}_t$ for all $t\geq 0$. A stopping time $T:\Omega\rightarrow\mathbb{R}_{\geq 0}$ is called predictable, if the set $\{(\omega,t)\in\Omega\times\mathbb{R}_{\geq 0}|T(\omega)>t\}$ is an element of the predictable $\sigma$-algebra.

There is the following theorem:

If $(T_n)_{n\in\mathbb{N}}$ is a sequence of predictable stopping times and $\Omega=\bigcup_{m\in\mathbb{N}}\{\bigwedge_{n\in\mathbb{N}}T_n=T_m\}$, then $\bigwedge_{n\in\mathbb{N}}T_n$ is a predictable stopping time.

There is the following example, why $\bigwedge_{n\in\mathbb{N}}T_n$ is not necessarily a predictable stopping time, when $\Omega=\bigcup_{m\in\mathbb{N}}\{\bigwedge_{n\in\mathbb{N}}T_n=T_m\}$ is not fulfilled: Let $S$ be a stopping time, which is not predictable and define $T_n=S+\frac{1}{n}$ for $n\in\mathbb{N}_{> 0}$. Then $T_n$ is a predictable stopping time for each $n\in\mathbb{N}$ but $S=\bigwedge_{n\in\mathbb{N}} T_n$ is not predictable by assumption.

My problem is to see why $\bigcap_{n\in\mathbb{N}}\{(\omega,t)\in\Omega\times\mathbb{R}_{\geq 0}|T_n(\omega)>t\}=\{(\omega,t)\in\Omega\times\mathbb{R}_{\geq 0}|\bigwedge_{n\in\mathbb{N}}T_n(\omega)>t\}$ is true when $\Omega=\bigcup_{m\in\mathbb{N}}\{\bigwedge_{n\in\mathbb{N}}T_n=T_m\}$ holds and not true otherwise.


1 Answer 1


If $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}$ is a sequence of real numbers, then the implication $$\forall n \in \mathbb{N}: a_n > t \implies \inf_{n \in \mathbb{N}} a_n>t \tag{1}$$ does, in general, not hold true (just consider e.g. $a_n := t+1/n$). If, however, there exists $m \in \mathbb{N}$ such that $\inf_{n \in \mathbb{N}} a_n = a_m$, then $(1)$ obviously holds true.

For $a_n := T_n(\omega)$ (with $\omega \in \Omega$ fixed), this shows that the implication $$\forall n \in \mathbb{N}: T_n(\omega)>t \implies \inf_{n \in \mathbb{N}} T_n(\omega)>t$$ does, in general, fail to hold, and therefore the set $$\bigcap_{n \geq 1} \{(\omega,t); T_n(\omega)>t\} = \{(\omega,t); \forall n \in \mathbb{N}: T_n(\omega)>t\}$$ does, in general, not equal $$\{(\omega,t); \inf_{n \geq 1} T_n(\omega)>t\}. $$However, if we know that for any $\omega \in \Omega$ there exists $m \in \mathbb{N}$ such that $$\inf_{n \geq 1} a_n = \inf_{n \geq 1} T_n(\omega) = T_m(\omega) = a_m,$$ then both sets are equal.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .