Defining stopping time without making use of filtrations

Asked Jun 01 '21 at 16:35

Active Jun 01 '21 at 16:35

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I am currently writing a document in which I want to make use of the concept of a "stopping time" but avoiding a digression into the concept of a "filtration." Is the following definition correct?

Let $\{ X_t \}_{t \ge 0}$ be a stochastic process. A stopping time with respect to $\{ X_t \}_{t \ge 0}$ is a random variable $\tau \ge 0$ such that if one conditions on the collection of random variables $\{ X_t \}_{0 \leq t \leq T}$ for some $T \geq 0$ then $\mathbb{1}(\tau \leq T)$ is independent of the random variables $\{ X_t \}_{t > T}$ (where $\mathbb{1}$ is an indicator function).

asked Jun 01 '21 at 16:35

UtilityMaximiser

I don't think so. Consider $\tau$ to be independent of ${X_t}_{t\geq 0}$. – Leander Tilsted Kristensen Jun 01 '21 at 16:46
Would that not be a stopping time? – UtilityMaximiser Jun 01 '21 at 16:50
1

If $\tau$ is a stopping time with respect to ${X_t}{t\geq 0}$ then for any $T$ we have $${\tau \leq T} \in \sigma (X_t : | : t \leq T) \subseteq \sigma(X_t : | : t\geq 0)$$ But if $\tau$ is also independent of ${X_t}{t\geq 0}$ that would imply that the event ${\tau \leq T}$ is independent of itself implying that $\mathbb{P}(\tau \leq T) \in {0,1}$. So $\tau$ would be constant. – Leander Tilsted Kristensen Jun 01 '21 at 16:57
Is it the case that if I condition on ${ X_t }{0 \leq t \leq T}$ then $\mathbb{1}{\tau \leq T}$ is deterministic? – UtilityMaximiser Jun 01 '21 at 17:06
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Yes, that would be the case. And also a decent way to avoid speaking of filtrations. – Leander Tilsted Kristensen Jun 01 '21 at 17:20

Defining stopping time without making use of filtrations

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