# Is jump time in a continuous-time Markov chain a stopping time?

Assume $$V$$ is a countable state space and $$L:V^2 \to \mathbb R$$ the infinitesimal generator of a continuous-time Markov chain $$(X_t)_{t \ge 0}$$ on the probability space $$(\Omega, \mathcal{G}, \mathbb{P})$$. Then we can define such random variables as first passage time, holding times, and jump times.

Given $$\omega \in \Omega$$, we define a sequence of random jump times $$(\sigma_n)$$ recursively as follows:

• First, let $$\sigma_0 := 0$$.

• Second, let $$i := X_{\sigma_n} (\omega) \in V$$ and $$L(i) := - L(i,i)$$ where $$X_{\sigma_n} (\omega) := X_{\sigma_n (\omega)} (\omega)$$. Then the time until transition out of state $$i$$ is $$\sigma_{n+1} -\sigma_{n} \sim \operatorname{Exp}(L(i))$$.

It's clear that the first passage time is a stopping time. I would like to ask if jump time is a stopping time too.

• Do you mean the time when the jump occurs, or the increment in time between two jumps? In the case of the latter, it is definitely not a stopping time except when $n=0$, but the reason is kind of obvious (you're comparing apples to oranges, since the definition wants you to restrict to times measured relative to the start of the process but you are really measuring times relative to the arrival at the current state).
– Ian
May 4 '20 at 15:49
• That being said, the actual time of the jump itself is a stopping time provided that you've defined your conventions consistently (i.e. all the inequalities involved are nonstrict).
– Ian
May 4 '20 at 15:49

Do you mean the time when the jump occurs, or the increment in time between two jumps? In the case of the latter, it is definitely not a stopping time except when $$n=0$$, but the reason is kind of obvious (you're comparing apples to oranges, since the definition wants you to restrict to times measured relative to the start of the process but you are really measuring times relative to the arrival at the current state).