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Poisson arrivals of rate $\lambda$ and $s$ independent exponential servers of rate $\mu$ give a birth-death chain with departure rate $\mu \min (n,s)$. The process is positive recurrent for $\lambda <s\mu$, null recurrent at equality, and transient above capacity.

A birth-death chain on ${\mathbb{N}}_0$ has generator entries $q_{n,n+1}={\lambda }_n$, $q_{n,n-1}={\mu }_n$ for $n\ge 1$, ${\mu }_0=0$, $q_{nn}=-({\lambda }_n+{\mu }_n)$, and zero elsewhere. Assume the usual conservative nonexplosive chain. The detailed balance for a birth-death process are ${\pi }_n{\lambda }_n={\pi }_{n+1}{\mu }_{n+1}$.

If they hold, each incoming term cancels an outgoing term, giving $\pi Q=0$ and thus invariance. Conversely the invariance equation at zero gives ${\pi }_0{\lambda }_0={\pi }_1{\mu }_1$. At state $n$, write the stationarity equation as equality of the neighbouring probability currentsInduction from the zero boundary current gives detailed balance for a birth-death process for every edge. Thus invariant measures are exactly the detailed-balance measures in this standard birth–death setting. The argument is the generator characterization; a stationary probability law additionally needs finite normalization. For potentially explosive processes, formal generator balance alone would require additional boundary qualifications.

Write $a=\mu /\lambda <1$. The embedded birth-death chain moves upward with probability $\lambda /(\lambda +\mu )$ and downward with probability $\mu /(\lambda +\mu )$ away from zero. The gambler's ruin formula, obtained by solving its two-point difference equation on $\{0,\dots ,N\}$ and then letting $N\to \infty$, gives ${\mathbb{P}}_k(\text{hit }0)=a^k$. Translation gives return probability $a$ from $r+1$ to $r$. From $r-1$ the chain hits $r$ almost surely: before that hit it remains in a finite reflecting interval, from every state of which a uniformly positive finite-step chance of hitting $r$ is available. This also proves that starting at zero it hits each $r\ge 1$ almost surely.

After departure from $r\ge 1$ the return probability is thereforeThe holding rate is $q_r=\lambda +\mu$, so the exponential occupation-time parameter is $q_r(1-r_r)=\lambda -\mu$. Applying the Strong Markov property at the almost-sure first hit givesAt zero the holding rate is $\lambda$ and return probability after departure is $a$, again giving parameter $\lambda -\mu$. From initial state $k$, the expected occupation time at zero is $a^k/(\lambda -\mu )$, since the time is zero unless zero is reached. Averaging over the initial law with probability generating function $G$ yieldsFor $\mu =0$, the interpretation is $a^0=1$, $a^k=0$ for $k>0$, and $G(0)=\mathbb{P}(X(0)=0)$; the same formulas hold.

An M-M-1 queue has a Poisson arrival process of rate $\lambda$, independent exponential service times of rate $\mu$, and one server, with an unlimited waiting room and the usual first-come-first-served discipline. Its number-in-system process is a continuous-time birth-death chain with $q_{j,j+1}=\lambda$ and $q_{j,j-1}=\mu$ for $j\ge 1$. For $\rho =\lambda /\mu <1$, detailed balance givesThis sums to one and satisfies ${\pi }_j\lambda ={\pi }_{j+1}\mu$. The nonexplosive irreducible chain therefore has an invariant probability and is positive recurrent. Equivalently its negative drift above zero gives finite mean return cycles. Here queue length includes the customer in service.

In equilibrium, PASTA gives an arriving customer's number of predecessors the distribution $\pi$. Conditional on $j$ predecessors, memoryless property makes its total waiting-plus-service time a sum of $j+1$ independent $\mathrm{Exp}(\mu )$ variables. Its Laplace transform isThereforeThis is the sojourn time including service, not just the time until service begins.

The chain is a birth-death chain. Detailed balance between $0$ and $1$ givesFor $i\ge 1$, detailed balance between $i$ and $i+1$ givesConsequentlyThe expected occupation time of the positive even states during a return cycle to state $1$ is therefore