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Let $\tau$ be the Markov moment with respect to the stream $(\mathcal{F}_{t}, t \in T)$. Prove that $$ \mathcal{F}_{\tau}=\{A \in \mathcal{F}: A \cap \{ \tau \leq t \} \in \mathcal{F}_t, \quad \forall t \in T \setminus \{ \infty\} \} $$

is a $\sigma$-algebra.

Seems I have to check all axioms for $\sigma$-algebra? But how to use the Markov moment?

Cettt
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Leox
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  • What is a Markov moment....? Is this a synonym for "stopping time"? – saz Dec 19 '16 at 17:48
  • yes, it is "stopping time" – Leox Dec 19 '16 at 18:05
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    I see. Well... yes, you have to check all axioms. Why not simply give it a try? You will see where you need that $\tau$ is a stopping time as soon as you start proving the assertion. – saz Dec 19 '16 at 20:17

1 Answers1

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The definition of a Markov moment or stopping time is a random variable $\tau: \Omega \rightarrow T \subseteq [0,\infty]$ such that $$ \{\tau \leq t\} \in \mathcal{F}_t, \quad t \in T. $$ In order to show that $\mathcal F_\tau $ is a $\sigma$-algebra, we have to show three properties.

  • First, we have to show that $\Omega \in \mathcal F_\tau$. This is simple since for all $t \in T$: $$\Omega\cap \{\tau \leq t\} = \{\tau \leq t\} \in \mathcal F_t,$$ hence $\Omega \in \mathcal F_\tau$.
  • Second we have to show that $A \in \mathcal F_\tau $ implies that $A^c$ is also in $\mathcal F_\tau$. So let $A \in \mathcal F_\tau$. We have to show that $A^c \cap \{\tau \leq t\} \in \mathcal F_t$. Observe that $$ A^c \cap \{\tau \leq t\}= \underbrace{\{\tau \leq t\}}_{\in \mathcal F_t} \setminus \underbrace{\Bigl(A \cap \{\tau \leq t\} \Bigr)}_{\in \mathcal F_t} \in \mathcal{F}_t, $$ hence $A^c \in \mathcal{F}_\tau$.
  • Finally, we have to show that if a countable family $(A_i)_{i \in I}$ lies in $\mathcal F_\tau$ then also have that $\bigcup_{i \in I} A_i \in \mathcal F_\tau$. So assume $(A_i)_{i \in I}$ is in $\mathcal{F}_\tau$. Then for all $ t \in T$: $$ \Bigl(\bigcup_{i \in I} A_i \Bigr) \cap \{\tau \leq t\} = \bigcup_{i \in I}\Bigl(\underbrace{A_i \cap \{\tau \leq t\}}_{\in \mathcal{F}_t} \Bigr) \in \mathcal{F}_t, $$ since $\mathcal F_t$ is a $\sigma$-algebra..
Cettt
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