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Let $(\Omega, \mathcal F,P)$ be a probability space, $I$ a countable subset of $\mathbb R$ and $(\mathcal F_i)_{i\in I}$ a filtration.

Let $\tau$ be a random variable with values in $I\cup \{\infty\}$. Prove that $\tau$ is a stopping time if and only if $\forall t\in I, (\tau=t)\in \mathcal F_t$

I've found the implication $\tau$ stopping time $\implies \forall t\in I, (\tau=t)\in \mathcal F_t$ surprisingly tedious to prove. Give some $t\in I$, I considered two cases, whether $t$ can be approached from below by elements of $I$ or not.

  • If it can, $$\displaystyle (\tau = t)= (\tau \leq t) \cap \bigcap_{\substack{a\in I\\a<t}} (\tau \leq a)^c \in \mathcal F_t$$ (note that countability of $I$ comes into play).

  • It it cannot, there is some $\epsilon >0$ such that $(t-\epsilon,t)\cap I =\emptyset$. If $(-\infty,t)\cap I =\emptyset$, we're done. Otherwise, I consider $u=\sup \{u<t, u\in I\}$. Then I checked two cases again:

    1. If $u\in I$, $(\tau = t) = (\tau \leq t) \cap (\tau > u) \in \mathcal F_t$

    2. If $u\notin I$, $$\displaystyle (\tau = t)= (\tau \leq t) \cap \bigcap_{\substack{a\in I\\a<u}} (\tau \leq a)^c \in \mathcal F_t$$

This proof is quite dull and lengthy, do you know a more concise one ?

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2 Answers 2

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The following proof is shorter, but the idea of the proof is actually the same:

Since $I$ is countable there exists an enumeration $(s_n)_{n \in J}$, $J \subseteq \mathbb{N}$ , of the set

$$\{s \in I; s < t\}.$$

Thus,

$$\{\tau=t\} = \{\tau \leq t\} \cap \left( \bigcup_{n \in J}\{\tau \leq s_n\} \right)^c \in \mathcal{F}_t,$$

and this proves the assertion.

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In the second case $\{\tau=t\}=\{\tau\le t\}\setminus\{\tau\le t-\epsilon\}$.

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