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Consider a birth-death process on $\{0,1,2,\dots \}$ with constant birth rate $\lambda >0$, constant death rate $\delta >0$ at every positive state, and no deaths from zero. The reflecting boundary is essential: the result changes if zero is an absorbing state. Detailed balance would giveWhen $\lambda <\delta$, these weights normalize to a Geometric distribution, so the irreducible chain is positive recurrent. When $\lambda =\delta$, the embedded jump chain is the symmetric reflected random walk: it returns to zero almost surely, but its constant equilibrium weights have infinite sum, so the chain is null recurrent. When $\lambda >\delta$, its upward-biased embedded random walk has probability $\delta /\lambda <1$ of hitting zero from state one, making the chain transient. The holding rates are bounded, so nonexplosion creates no additional qualification here.

A Continuous-time Markov chain on a finite or countable state space is specified by transition rates $q_{ij}\ge 0$ for $i\ne j$, with finite total exit rate $q_i={\sum }_{j\ne i}{q}_{ij}$. At a state with $q_i>0$, it waits an exponential time of rate $q_i$ and then moves to $j$ with probability $q_{ij}/q_i$. A state with $q_i=0$ is absorbing: it has infinite holding time rather than a zero-rate exponential waiting law.

Its transition-rate matrix is $Q$, with off-diagonal entries $q_{ij}$ and diagonal entries $q_{ii}=-q_i$, so each row sums to zero. The rates describe first-order transition probabilities: $\mathbb{P}(X_{t+h}=j\mid X_t=i)=q_{ij}h+o(h)$ for $j\ne i$. On a finite state space, the transition semigroup is $P_t=e^{tQ}$ and explosion is impossible. On a countably infinite state space, finite individual exit rates do not alone exclude infinitely many jumps in finite time; nonexplosion must be checked separately.

For two states with rates $q_{01}=a>0$ and $q_{10}=b>0$, the mean holding times are $1/a$ and $1/b$. Detailed balance gives Stationary distribution $(b/(a+b),a/(a+b))$, andThis starts at one and relaxes to the stationary mass of state zero. Changing holding rates can preserve an embedded jump chain while changing the proportions of elapsed time spent in its states.

An Equilibrium distribution of a continuous-time Markov chain is a probability row vector $\pi$ invariant under its transition semigroup:It satisfies $\pi Q=0$ when the generator sums are defined. An invariant law is a property of the entire evolution, so formal generator equations must be interpreted with the process's nonexplosion and domain conditions.

For a finite irreducible chain, the equilibrium distribution is unique and every transition row converges to it. For a general finite chain, the limiting row from state $i$ is a mixture of the stationary distributions ${\pi }^C$ of its closed irreducible classes:The chain eventually enters one of these classes and then mixes within it. If every closed class is a single absorbing state, this reduces to absorption probabilities at those states. Merely having some absorbing states does not exclude other closed classes.

Holding times explain the link to an embedded jump chain. For a finite irreducible chain with strictly positive exit rates $q_i$ and jump-chain Stationary distribution $\alpha$, the time-stationary distribution isVisit frequency is multiplied by mean time $1/q_i$ per visit. In the countable setting, these time weights need a finite positive normalizer and a nonexplosive process before they define an equilibrium probability. For an alternating two-state jump chain, $\alpha =(1/2,1/2)$ and rates $a,b$ give $\pi =(b/(a+b),a/(a+b))$: slower holding rates increase time occupancy.

Suppose a transient nearest-neighbour jump chain on $\mathbb{N}$ has uniformly bounded expected visits to each state and the holding rate at $i$ grows geometrically. Thenso the continuous-time chain explodes almost surely.

Fix a finite embedded jump path $i_0,\dots ,i_n$ whose holding rates $q_{i_k}$ are positive and finite. Conditional on those successive states, the complete holding times are independent exponential random variables with the corresponding rates. To observe exactly these $n$ jumps by a fixed time $t$, the final holding time is censored rather than completed.

Write $H_{i_0\cdots i_n}(t)$ for this conditional holding-time probability. With $u_n=t-{\sum }_{k<n}{u}_k$,The completed holding times supply their density factors; the last portion supplies only an exponential survival factor. The fixed observation time imposes the simplex constraint.

Multiplying by the terminal rate completes the symmetric product of all $n+1$ rates. Reversing the time portions and the state sequence preserves the simplex integral, with unit absolute Jacobian. HenceThe right-hand holding kernel can be defined by this formula even if the reverse embedded path is not feasible; its conditional path interpretation requires that feasibility. For actual unconditioned path probabilities, the holding kernel must also be multiplied by the product of jump-chain transition probabilities. Those products need not agree under reversal, so the holding identity alone is not a reversibility theorem for the chain.

For an explosive chain, a normalized solution of the formal balance equations $\pi Q=0$ need not be an invariant probability for the minimal transition semigroup. Explosion sends Probability mass to a cemetery state, and the generator equations on the original states do not include that boundary loss.

A concrete example uses the reflected jump chain on $\{0,1,2,\dots \}$ which moves from zero to one, and from $i\ge 1$ moves up with probability $2/3$ and down with probability $1/3$. Give state $i$ holding rate $q_i=4^i$. The embedded chain has invariant weights ${\mu }_0=1$, ${\mu }_i=32^{i-1}$ for $i\ge 1$. Dividing by the holding rates and normalizing yieldsThese weights even satisfy formal Detailed balance: ${\pi }_0{q}_{01}={\pi }_1{q}_{10}=2/5$, and for $i\ge 1$ both opposing edge flows equal $(2/5)2^i$. Thus $\pi Q=0$ holds at each state.

Nevertheless the process explodes. Starting from zero, the embedded walk has expected visit count two at zero and three at every positive level. The zero-state return probability is $1/2$; at a positive state it is $2/3$, giving those geometric visit means. Therefore the expected total elapsed holding time isThe lifetime is finite almost surely. The same bounded expected-visit argument works from every initial state, so the killed minimal semigroup loses all its mass as time tends to infinity. The displayed $\pi$ is consequently a formal generator-balance solution, not a stationary probability of that killed evolution. A conservative continuation after explosion would require an extra boundary rule.

For strictly positive finite holding rates $q_i$, a measure $\pi$ satisfies $\pi Q=0$ exactly when ${\mu }_i=q_i{\pi }_i$ is invariant for the jump chain. Thus ${\pi }_i={\mu }_i/q_i$ when normalization is possible.

For a continuous-time Markov chain with strictly positive finite holding rates whose jump times tend to infinity, the chain and its jump chain have the same directed paths of positive-probability jumps. They are therefore irreducible simultaneously, and a state is recurrent for one exactly when it is recurrent for the other: the event of returning after departure depends only on the sequence of visited states.

A kernel specifying the jump rate from state $x$ for each displacement $z$. With the displacement convention the infinitesimal generator is $L\phi (x)=\int (\phi (x+z)-\phi (x))K(x,dz)$. The destination convention instead records a rate measure on the new state, translating each displacement by $x$. For finite positive total rate, that total is the exponential holding rate and the normalized measure is the next-jump distribution.

If the jump chain visits a state of finite positive holding rate infinitely often, the sum of its independent positive holding times at that state diverges almost surely, preventing explosion.

If every holding rate of a continuous-time Markov chain is at most $0<C<\infty$, its holding times can be coupled as $E_n/q_{Y_n}\ge E_n/C$, where the $E_n$ are independent rate-one exponential random variables. The strong law of large numbers gives ${\sum }_n{E}_n=\infty$ almost surely, so the jump times tend to infinity and the chain cannot explode.

A continuous-time Markov chain with finite holding rates is nonexplosive when its jump times have no finite accumulation almost surely. It then remains in its specified state space at every finite time, rather than being lost to a cemetery state after infinitely many jumps. Bounded rates are sufficient for nonexplosion, but finite rates at individual states alone are not.

For $h>0$, the fixed-time skeleton has matrix $P(h)$. If $q_i$ is the holding rate, then for $nh\le t\le (n+1)h$,Thus ${\int }_0^{\infty }{p}_{ii}(t)dt$ diverges exactly when ${\sum }_n{p}_{ii}(nh)$ does, proving recurrence equivalence for an irreducible chain.

Even when the jump chain has invariant probability $\mu$, the continuous-time invariant measure ${\mu }_i/q_i$ may have infinite mass if holding rates become too small, producing infinite mean return time.