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For a continuous-time Markov chain with transition rates $q_{ij}$, detailed balance isfor every pair of distinct states. A probability distribution satisfying these equations is an invariant distribution of a continuous-time Markov chain, because summing over incoming rates gives $\pi Q=0$. Under the standard nonexplosion assumptions, the equilibrium chain is reversible.

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.

Let $\nu$ be an invariant distribution for the continuous-time Markov chain. Using $Q=D_g(P-I)$, its stationarity equation $\nu Q=0$ becomesThus the measure with weights ${\nu }_{i}g(i)$ is invariant for the irreducible positive-recurrent jump chain. Its invariant distribution $\pi$ is unique, so ${\nu }_{i}g(i)=c{\pi }_i$. Normalization gives the invariant-measure transfer between a jump chain and a CTMC formulaThe denominator is finite and nonzero because $\epsilon <g(j)<1/\epsilon$. Thus $\nu$ is a probability distribution and $(X_t)$ is positive recurrent.

A continuous-time Markov chain is a reversible Markov chain in equilibrium if, when started in a stationary distribution $\pi$, the process $(X_t:0\le t\le T)$ has the same finite-dimensional distributions as $(X_{T-t}:0\le t\le T)$ for every $T>0$.

For transition rates $q_{ij}$, the detailed balance for a continuous-time Markov chain equations areThey imply invariance directly. For each state $j$,because $q_{jj}=-{\sum }_{k\ne j}{q}_{jk}$. Thus $\pi Q=0$, so $\pi$ is an invariant distribution of a continuous-time Markov chain.

Here the total holding rate isand therefore the jump chain has state-independent transitionsIt is an upward biased random walk on $\mathbb{Z}$. By the strong law of large numbers, $Y_n/n⟶1/3$ almost surely, so $Y_n\to +\infty$. The jump chain, and hence $X$, is transient.

Counting measure is invariant for this jump chain: for every $j$, its incoming mass is $1/3+2/3=1$. The invariant-measure transfer between a jump chain and a CTMC therefore gives an invariant measure proportional to $q_i^{-1}=3^{-|i|-1}$. Sincethe normalized invariant distribution of a continuous-time Markov chain isThis does not contradict transience because an invariant distribution of an explosive chain need not imply positive recurrence.

It remains to verify explosion. Starting from zero, the probability that the biased walk ever reaches $-k$ is $2^{-k}$. Once it reaches any state, its probability of returning after departure isso its expected total number of visits to that state, conditional on reaching it, is $1/(1-2/3)=3$. If $V_i$ is the number of visits to $i$, thenConsequently the expected sum of all holding times isThe nonnegative random variable $T_{\infty }$ is therefore finite almost surely. By explosion by summable holding times, $X$ makes infinitely many jumps by time $T_{\infty }$, so it is explosive.

Writing $\pi =({\pi }_1,{\pi }_2)$, the equations $\pi Q=0$ and ${\pi }_1+{\pi }_2=1$ giveThis is the invariant distribution. Since $e^{-st}\to 0$, both rows of $P(t)$ converge to $\pi$, directly verifying convergence to equilibrium.