# Show that $L=\sup\{t>0; X_t \geq 2\}$ is not a stopping time

Let $X_t$ be a stochastic process such that $\lim\limits_{t\to \infty} X_t = -3$ a.s. Let $\mathcal{F}_t := \sigma(X_s, s\leq t)$. Finally, let $L := \sup\{t > 0 : X_t \geq 2\}$. We are also given that $L < \infty$ a.s. Show that $L$ is not a stopping time.

1) Is the assumption that $L < \infty$ a.s. necessary in the problem if we know that $X_t \to -3$ a.s.? For almost all $\omega$ s.d. $X_t(\omega) \to -3$, choose $\epsilon = 5$ and there exists $K_\epsilon(\omega)$ s.d. $\forall t \geq K_\epsilon(\omega)$, $|X_t(\omega) + 3| < \epsilon \implies X_t(\omega) < 2$ in particular.

2) My intuition is that $L$ obviously cannot be determined without looking into the future. If $\{L \leq t\} \in \mathcal{F}_t$, then this event can be observed only with the information of how the process could behave up to time t. Meaning, the fact that $X_s > 2$ for all $s > t$ is determined by $\mathcal{F}_t$. From this, I tried the following: assume $L$ is a stopping time then,

$$\{L \leq t\} = \bigcap\limits_{s > t} \{X_s > 2\} \in \mathcal{F}_t$$ I want to discretize the problem to make the intersection at least countable, but I am stuck.

1) Yes, you are right about that (with the convention that $\sup (\emptyset)=0$.)
2) As it is stated right now, the assertion is wrong. Consider for instance the process $(X_t)_{t \geq 0}$ defined by $$X_t(\omega) := \begin{cases} 2, & t \in [0,1], \\ -3, & t>1, \end{cases}, \qquad \omega \in \Omega. \tag{1}$$ The process satisfies the assumptions of your problem and we have $$L = \sup\{t>0; X_t \geq 2\} = 1.$$ Since $L$ is constant it is trivially a random variable with respect to any $\sigma$-algebra, and in particular it is a stopping time.
Even under the additional assumption that $(X_t)_t$ has continuous sample paths, the assertion is still wrong; we can construct similar counterexamples as in $(1)$, instead of jumping directly to $-3$ we can use, for instance, a linear interpolation.
• 1) I agree, but we take the convention that $\sup(\emptyset)) = 0$ Commented Feb 11, 2018 at 15:39