Shift operator of a stopping time, what does it mean exactly?

Question

I'm trying to figure out this question:

Let $X$ be a canonical, right-continuous Markov process with values in a Polish state space $E$, equipped with Borel $\sigma$-algebra $\mathcal{E}$. Assume $t \mapsto \mathbb{E}_{X_t}f(X_s)$ right-continuous everywhere for each bounded continuous function $f : E \mapsto R$. For $x \in E$ consider the stopping time $\sigma_x = \inf\{t > 0 | X_t \neq x\}$.

I've shown that there exists an $a \in [0,\infty]$ such that $\mathbb{P}_x(\sigma_x > t)= e^{-at}$. Now suppose that $x \in E$ such that the above $a \in (0,\infty)$. I want to show that $$ \{X_{\sigma_x} = x,\, \sigma_x < \infty\} \subseteq \{\sigma_x \circ \theta_{\sigma_x} = 0,\, \sigma_x < \infty\}.$$

The problem is I'm only farmiliar with the definition of $\theta_\tau$, where $\tau$ is a stopping time, in the context of a stochastic process. In other words, the expression $$ (X_t \circ \theta_\tau)(\omega) = X_{\tau(\omega) + t}(\omega).$$

Since the stopping time $\sigma_x$ isn't time dependent, I don't know how to interpret the expression $\sigma_x \circ \theta_{\sigma_x}(\omega)$.

Any ideas?

Answer 1

Well, you better think of $\omega$ being an element of a canonical space. I don't know whether left-limits matter here, but let's assume that $X_t$ is cadlag and that $\Omega = D_E([0,\infty))$ is the space of all cadlag functions $\omega:[0,\infty)\to E$. That is, $X_t(\omega) = \omega(t)$ is a coordinate map on $\Omega$.

Whenever you have an expression of the form $\xi\circ \eta(\omega)$ where $\eta:\Omega\to\Omega$ and $\xi:\Omega \to \dots$ you shall think of the following: first you change the original trajectory $\omega$ according to the transformation $\eta$, and then apply your functional $\xi$ to the transformed trajectory.

For example, forgetting about measurability issues, given a trajectory $\omega$ you know for sure a real value $s:=\sigma_x(\omega)$. Now, such value is fixed for any $\omega$ and thus $\omega' := \theta_{\sigma_x}(\omega)$ is just $\theta_s(\omega)$ which is just obtained from the original $\omega$ by truncating its first $[0,s)$ interval. Now, if you apply $\sigma_x$ to $\omega'$, then you are looking for the first appearance of the satisfactory condition for $\sigma_x$ (that is $X_t\neq x$) over the truncated/shifted trajectory. Which actually means looking for the second appearance of the satisfactory condition for the original trajectory. In general, you shall think of $\sigma\circ\theta_\tau$ as a first appearance of $\sigma$ one the trajectory that starts from the first appearance of $\tau$. However, things are actually more tricky in the case of continuous time and space than the intuition described here - which in particular addresses you problem: you want to show that if $X_{\sigma_x} = x$ then second (third, forth etc.) appearance of $\sigma_x$ happens at the same time.

I am not sure whether it completely resolves your confusion, just let me know if not.

Answer 2

For those interested:

Let $\omega \in \Omega$ such that $X_{\sigma_x(\omega)}(\omega) = \omega_{\sigma_x(\omega)} = x$ and $\sigma_x(\omega) < \infty$. Now since $\sigma_x$ is defined as $\inf\{t > 0: X_t \neq x\}$, there exists a sequence $t_n \downarrow \sigma_x(\omega)$ such that $\omega_{t_n(\omega)} \neq x$. Now define $t_n' = t_n - \sigma_x(\omega)$. We now have a sequence $t_n' \downarrow 0$ such that $\omega_{t_n' + \sigma_x(\omega)} \neq x$. In other words $\sigma_x \circ \theta_{\sigma_x}(\omega) = \inf\{ t > 0: \omega_{t + \sigma_x(\omega)} \neq x\} = 0$, since $\omega_{\sigma_x(\omega)} = x$. This finishes the inclusion $\{X_{\sigma_x} = x,\, \sigma_x < \infty\} \subseteq \{\sigma_x \circ \theta_{\sigma_x}=0,\, \sigma_x < \infty\}$.