0
$\begingroup$

I'm trying to find an example where the last hitting time $\theta = \sup\{k \ge 0 : X_k \in B\}$ of a set $B$ by the stochastic process $(X_k)_{k\ge 0}$ is not a stopping time.

$\endgroup$
  • $\begingroup$ Usually, you cannot tell at time $\theta$ whether you'll come back to $B$ in the future, so you cannot expect $\theta$ to be a stopping time. What kind of specific example are you looking for and what do you expect to learn from it? $\endgroup$ – Mars Plastic Feb 28 at 12:38
  • $\begingroup$ That makes sense to me, for it would not be measurable in 𝑋𝑘. However, I'm failing to see an example where this would happen. $\endgroup$ – Felipe Bpm Feb 28 at 12:40
1
$\begingroup$

Define a stochastic process $(X_k)_{k \geq 0}$ by

$$X_0 := 0 \qquad X_1 := Y, \quad X_k := 1, \quad k \geq 2,$$

for a random variable $Y$ such that $\mathbb{P}(Y=1)=\mathbb{P}(Y=0)=1/2$. The associated canonical $\sigma$-algebra is given by $$\mathcal{F}_0 = \{\emptyset,\Omega\}, \qquad \qquad \mathcal{F}_k = \sigma(Y), \quad k \geq 1. \tag{2}$$ If we consider $$\tau := \sup\{k \geq 0; X_k = 0\}$$ then $\tau$ is not a stopping time. Indeed: Since $$\{\tau=0\} = \{Y=1\}$$ we have $\{\tau=0\} \notin \mathcal{F}_0$ because of $(2)$.

$\endgroup$

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.