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For the M-M-1 queue with Bernoulli feedback, assume $0<\delta \le 1$. The number of service visits before departure is geometric on $1,2,\dots$, with probabilities $\delta (1-\delta {)}^{j-1}$, independent of the independent exponential visit lengths. The transform of the total time actually being served isThusThis excludes the waiting between visits. A feedback completion changes the customer's position but not the number in the system. By exponential memoryless property, the latter process has births of rate $\lambda$ and true deaths of rate $\delta \mu$ while nonempty; unsuccessful completions are self-events. It is therefore an M/M/1 number process with effective service rate $\delta \mu$, givingFor positive $\lambda$ equality is null recurrent, larger arrival rate is unstable, and $\delta =0$ admits no departure-based equilibrium.

To prove the output claim, state the relevant time reversal of a continuous-time Markov chain theorem: an equilibrium continuous-time chain satisfying detailed balance has the same reversed path law. In this reversible birth-death process, final departures are precisely downward jumps; time reversal turns them into upward jumps. Upward jumps are supplied by a state-independent Poisson clock of rate $\lambda$, so their process is Poisson. Hence the stationary final-departure process is Poisson of rate $\lambda$. This is the output conclusion of Burke theorem for the effective number process; it does not say the process of all service completions has that rate.

The Poisson output theorem for a feedback queue, equivalently the Burke theorem applied to the effective queue-length process with service rate $\delta \mu$, shows that successful departures in equilibrium form a Poisson process of rate $\lambda$.

The process of all entries into the queue is different. Its long-run rate isImmediately after any queue-entry event the server is busy, so the conditional probability of another entry in the next interval of length $h$ isthe next entry can be external or can be a service completion followed by feedback. A Poisson process of the displayed long-run rate would instead give $(\lambda /\delta )h+o(h)$. These coefficients are unequal when $\lambda <\delta \mu$, sinceHence the combined arrival process is not Poissonian.

In the stable regime, $L$ is exactly the queue-length process of an $M/M/1$ queue with arrival rate $\lambda$ and effective service rate $\mu (1-p)$. By Burke theorem, when this queue is in equilibrium its external departure process is a Poisson process of rate