= 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 is
$$
\lambda+(1-\delta)\mu\,\mathbb P(Q>0)
=\lambda+(1-\delta)\mu\frac{\lambda}{\delta\mu}
=\frac\lambda\delta.
$$
Immediately after any queue-entry event the server is busy, so the conditional probability of another entry in the next interval of length $h$ is
$$
[\lambda+(1-\delta)\mu]h+o(h):
$$
the 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 $\lambda<\delta\mu$, since
$$
\lambda+(1-\delta)\mu-\frac\lambda\delta
=(1-\delta)\left(\mu-\frac\lambda\delta\right)>0.
$$
\b[Hence the combined arrival process is not Poissonian.]
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