2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.