Let $\mathbb{S}$ be countable and $\Omega$ be the set of all right continuous functions $\omega:[0,\infty)\rightarrow \mathbb{S}$. Let $X_t(\omega) = \omega(t)$ denote a continuous Markov chain.

Liggett states in proposition 2.29 that

Suppose $X(t)$ is a Markov chain, and let $\tau = \inf\{t\geq 0:X_t\ne X_0\}$ be the time of the first jump. Then $$P^x(\tau>t) = e^{-c(x)t}$$ for some $c(x)\in [0,\infty]$.

So the problem is that if $c(x) = \infty$, then $P^x(\tau > t)=0$ for all $t>0$. Hence, if we take $t\downarrow 0$, then $P^x(\tau > 0)=0$. But, by right continuity, we have $P^x(\tau > 0)=1$ and we reach a contradiction. Is there something wrong with my logic?

  • $\begingroup$ For the case $c(x)=\infty$, if $\tau$ is the random time to stay in state $x$, it looks like $\tau=0$ with probability 1. So $P[\tau=0]=1$. I do not follow your sentence "by right continuity $P[\tau>0]=1$" so I do not know how you are obtaining a contradiction. I also do not follow your first sentence with notation "$X_t(\omega) = \omega(t)$", that notation looks designed to confuse. $\endgroup$ – Michael Jun 4 '18 at 3:45
  • $\begingroup$ @Michael I'm assuming that $\mathbb{S}$ is countable and thus equipped with the discrete metric (maybe I should've mentioned this). Since $\omega$ is right continuous, there exist $\delta>0$ such that if $0<t<\delta$, then $d(\omega(t),\omega(0))<1/2$. Since the discrete metric $d$ can only be 0 or 1, we see that $\omega(t)=\omega(0)$ for all $0<t<\delta$. Hence, $\tau (\omega) > \delta>0$. SInce this is true for all $\omega \in \Omega$, we see that $P^x(\tau >0)=1$. $\endgroup$ – Andrew Yuan Jun 5 '18 at 4:08
  • $\begingroup$ @Michael For the notation, I'm just using the analogy form of the discrete Markov chain $X_n (\omega) = \omega_n$. You can think of it as $X(t,\omega) = \omega(t)$ which is used by Liggett. $\endgroup$ – Andrew Yuan Jun 5 '18 at 4:10
  • $\begingroup$ I thought you meant right-continuity of CDF functions $P[X\leq x]$. You mean of the sample path...Yes I agree that if sample paths are required to be right-continuous then there is a contradiction here with $P[\tau=0]=1$. The case of a transition rate being "infinity" is so unusual though, one would expect some type of singularity or exception in this case. $\endgroup$ – Michael Jun 5 '18 at 4:14
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    $\begingroup$ Sounds like "Markov chain theory made hard." I suppose the shift operator is supposed to simplify some ergodic results or something. Anyway, I would just consider $c(x) \in [0, \infty)$ rather than $c(x) \in [0, \infty]$. I don't know what it would mean to stay in a discrete state for time 0 anyway, though it might make sense for uncountably infinite state with continuous sample paths. $\endgroup$ – Michael Jun 5 '18 at 8:51

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