Continuous-time Markov chain

Remarks

The condition is the Markov property in continuous time, and it implies the Chapman–Kolmogorov equation $P(s + t) = P(s) P(t)$. For a finite $S$ there is a rate matrix $Q = P'(0)$, the generator, with $q_{ij} \ge 0$ for $i \ne j$ and zero row sums, and $P(t) = e^{tQ}$ is a matrix exponential. The chain stays in state $i$ for an exponential time with rate $q_i = -q_{ii}$, then jumps to $j \ne i$ with probability $q_{ij}/q_i$, when $q_i > 0$. Example: the Poisson process of rate $\lambda > 0$ is a chain on $\N$ with $q_{i, i+1} = \lambda$ and $q_{ii} = -\lambda$, and $p_{ij}(t) = e^{-\lambda t} (\lambda t)^{j-i} / (j-i)!$ for $j \ge i$. Non-example: a process on $\Z$ that moves up by $1$ at the times $1, 2, 3, \ldots$ is not time-homogeneous. Indeed $P(X_{t+1/2} = X_t + 1)$ is $1$ for $t = 3/4$ and $0$ for $t = 1/4$, so no single $P(1/2)$ works.