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For a finite-state continuous-time Markov chain with Q-matrix $Q$, the backward and forward equations are $P^{\prime }(t)=QP(t)$ and $P^{\prime }(t)=P(t)Q$, respectively.

A time-homogeneous continuous-time Markov chain on a countable state space is a stochastic process with the Markov property: conditional on the current state, the law of its future depends on neither its past nor the current time. In the usual conservative, nonexplosive setting its sample paths are right-continuous step functions. Its Q-matrix $Q=(q_{ij})$ has $q_{ij}\ge 0$ for $j\ne i$, $q_i=-q_{ii}={\sum }_{j\ne i}{q}_{ij}<\infty$, and infinitesimal transition probabilities ${\mathbb{P}}_i(X(h)=j)=q_{ij}h+o(h)$ for $j\ne i$.

On entering $i$, the holding time has distribution $\mathrm{Exp}(q_i)$; independently, the next state is $j$ with probability $q_{ij}/q_i$. The successive states form the jump chain, whose transition probabilities areIf $q_i=0$, the state is absorbing, its holding time is infinite, and the jump-chain convention is $p_{ii}=1$. Nonexplosion ensures this construction defines the process for all finite times rather than accumulating infinitely many jumps in finite time.

For the finite state space $S$, the generator or Q-matrix $G=(g_{ij})$ is defined byIts off-diagonal entries are nonnegative jump rates and $g_{ii}=-{\sum }_{j\ne i}{g}_{ij}$. The transition semigroup of a continuous-time Markov chain is $P(t)=e^{tG}$.

An invariant distribution of a continuous-time Markov chain is a probability row vector $\pi$ satisfying $\pi P(t)=\pi$ for every $t\ge 0$. Differentiation at zero gives $\pi G=0$; conversely, $\pi G=0$ implies $\pi e^{tG}=\pi$. ThusInvariant distributions need not be unique: each closed communicating class supports one, and their convex combinations are invariant. A finite irreducible chain has a unique invariant distribution.

Let $0=T_0<T_1<\cdots$ be the arrival times of the independent rate-$\lambda$ Poisson process. The increments $T_{n+1}-T_n$ are independent exponential variables of rate $\lambda$. Conditional on $Y_n=X_{T_n}=i$, the Markov property, time homogeneity, and independence of the next increment giveHence $(Y_n)$ is a discrete-time Markov chain. Its transition matrix is the Poisson-sampled continuous-time Markov chain resolvent

If $\pi G=0$, then $\pi e^{tG}=\pi$ and integration gives $\pi P_{\lambda }=\pi$. Conversely, if a probability row vector $\nu$ satisfies $\nu P_{\lambda }=\nu$, multiplyingon the right by $\lambda I-G$ yields $\nu G=0$. Therefore the sampled chain and the original chain have exactly the same invariant distributions.

Conditional on $Y_m=i$, the holding time $V_m=U_m/g(i)$ has an exponential distribution of rate $g(i)$, and after that holding time the next state is $j$ with transition probability $p_{ij}$. The memoryless property of exponential random variables and the Markov property of $(Y_m)$ therefore make $(X_t)$ a continuous-time Markov chain.

Moreover $g(i)<1/\epsilon$, so $V_m\ge \epsilon U_m$. The strong law of large numbers gives ${\sum }_m{U}_m=\infty$ almost surely, hence $J_m\to \infty$ and the process is nonexplosive. Its Q-matrix isEquivalently, if $D_g$ is the diagonal matrix with entries $g(i)$, then

A Q-matrix satisfies $q_{ij}\ge 0$ for $i\ne j$ and $q_{ii}=-{\sum }_{j\ne i}{q}_{ij}$; on a countable state space the exit rate $-q_{ii}$ is required to be finite. For a finite-state continuous-time Markov chain with transition semigroup $P(t)$,The backward and forward Kolmogorov equations are respectively

Use a three-state continuous-time multi-state model with transient infected state $I$ and absorbing recovered and dead states $R$ and $D$. If recovery and death have constant transition intensities $\gamma$ and $\mu$, its Q-matrix iswith state order $(I,R,D)$. This is also a competing risks model: recovery and death are the two mutually exclusive first events.

Add a hospital state $H$ and estimate the enlarged Q-matrix from the cohort. If infection starts in state $I$, the expected hospital occupancy generated by one infection is the Markov reward model integralEquivalently, if $Q_T$ is the subgenerator on all transient states, $m_H$ is the $(I,H)$ entry of the fundamental matrix of an absorbing continuous-time Markov chain $(-Q_T{)}^{-1}$. The number infected in the population has expectation $Np$, so linearity of expectation gives total expected hospital use $Np,m_H$ days. If only occupancy within a finite period counts, replace the upper integration limit by the remaining follow-up time for each infection and average over infection times.

For a continuous-time Markov chain, $P_{ij}(t)={\mathbb{P}}_i(X_t=j)$ and the matrices satisfy $P(s+t)=P(s)P(t)$. On a finite state space with Q-matrix $Q$, $P(t)=e^{tQ}$.