Poisson arrivals to Jackson network I'm really confused as to what can be considered Poisson arrivals or departures in an open Jackson network. Say we have a Jackson network of with $K$ servers, with exogenous Poisson arrivals but with no feedback loops between any of the servers. Are the arrivals to any of the $K$ servers Poisson? What about the departures, are they Poisson too?
Now what if there is a feedback loop within the open Jackson network? Does this make every arrival rate to every one of the $K$ servers not Poisson? Or is it only the servers that are directly connected to the server that has a feedback loop that don't have Poisson arrivals?
Thanks for your help!

Recall from Burke's theorem that in equilibrium, the departure process of a $M/M/c$ queue is a Poisson process with the same rate as the arrival process. Since Poisson processes are closed under merging and splitting, it's clear that an acyclic Jackson network has Poisson arrival and departure processes at each queue.
When there is feedback, this is no longer the case. A simple example is a network with a single node. Let the exogenous arrival process be Poisson with rate $\alpha$, the service rate $\mu$, and the probability that a customer finishing service returns to the queue $p$. Then the net arrival rate to the queue $\lambda$ satisfies in equilibrium $$\lambda =  \alpha + p\lambda,$$ so that $\lambda=\frac\alpha{1-p}.$ (Assume that $\frac\alpha{(1-p)\mu}<1$ so the system is stable.) However, the net arrival process depends on the queue length. Conditioned on $N(t)=0$, the time until the next arrival is $\mathsf{Exp}(\lambda)$-distributed. Conditioned on $N(t)=n$, let $T$ be the time until the next exogenous arrival, $S_i$ the service time of the $i^{\mathsf{th}}$ customer in the queue, $B_i\sim\mathsf{Ber}(p)$ equal to one if customer $i$ returns to the queue and zero otherwise, and $\tau=\inf\{i: B_i=1 \}$. Then the time until the next arrival is
$$\hat T=\min\left\{T, \sum_{i=1}^\tau S_i  \right\} $$ (with the convention $\sum_{i=1}^\tau S_i=\infty$ if $\tau=\infty$.) For $n=1$, we have \begin{align}\mathbb E[\hat T] &= (1-p)\mathbb E[T] + p\mathbb E[T\wedge S_1]\\
&= \frac{1-p}\alpha + \frac p{\alpha+\mu}\\
&= \frac{(1-p)(\alpha+\mu) + p\alpha}{\alpha+\mu}\\
&= \frac{\alpha+(1-p)\mu}{\alpha+\mu}.
\end{align}
