My textbook, Modeling and Analysis of Stochastic Systems, third edition, by Kulkarni, introduces continuous-time Markov chains (CTMCs) as follows:

In Chapters 2, 3, and 4 we studied DTMCs. They arose as stochastic models of systems with countable state-space that change their state at time $n = 1, 2, \dots$, and have Markov property at those times. Thus, the probabilistic nature of the future behavior of these systems after time $n$ depends on the past only through their state at time $n$.

In this chapter we study a system with a countable state-space that can change its state at any point in time. Let $S_n, n \ge 1$, be time of the $n$th change of state or transition. $Y_n = S_n - S_{n - 1}$, (with $S_0 = 0$), be the $n$th sojourn time, and $X_n$ be the state of the system immediately after the $n$th transition. Define

$$N(t) = \sup\{ n \ge 0 : S_n \le t \}, \ \ \ t \ge 0.$$

Thus $N(t)$ is the number of transitions the system undergoes over $(0, t]$, and $\{ N(t), t \ge 0 \}$ is a counting process generated by $\{ Y_n, n \ge 1 \}$. it has piecewise constant sample paths that start with $N(0) = 0$ and jump up by $+1$ at time $S_n, n \ge 1$.

Under the typical mathematical definition of "continuous", I do not understand how CTMCs are "continuous" -- it seems to me that the time is still discrete, just as for DTMCs. I would greatly appreciate it if people would please take the time to explain the difference here.

  • $\begingroup$ @Xi'an "The Markov chain can thus move to another location on a countable state space at any time in the future." I don't understand exactly what this means. Why is this notable? How is it relevant to continuity? DTMCs can also move to other locations on a countable state space at any time in the future, assuming the DTMC is irreducible (that is, all classes communicate); or, maybe they can't, depending on what you mean by your statement, since DTMCs can't just jump from any state to any other, but, rather, must take some path. Like I said, I'm not sure exactly what your statement means. $\endgroup$ Commented May 25, 2020 at 5:58

1 Answer 1


Time is not discrete. $N(t)$ is a continuous-time process. Its sample paths $N(\omega, t)$ are cadlag functions $$ N(\omega, t): [0, \infty) \rightarrow \mathbb{Z}_+\,\, $$ $\omega$-almost surely, and $$ N(t) - N(t_-) \leq 1. $$ At each given $t \in [0, \infty)$, $N(\omega, t)$ is the number of times the process has switched states up to time $t$. Each jump of $N(t)$, i.e. when $N(t) - N(t_-) = 1$, the CTMC undergoes a state transition.

The sample paths of the corresponding CTMC $M(\omega, t)$ are piecewise constant cadlag functions $$ M(\omega, t): [0, \infty) \rightarrow \mathcal{S} $$ where $\mathcal{S}$ is the state space. By definition, $N(t)$ has the same discontinuities as $M(t)$.

In contrast, the sample path of a DTMC $X(\omega, t)$ is a sequence $$ X(\omega, t): \mathbb{N} \rightarrow \mathcal{S}. $$


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