 Solution

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.