Arrival distribution of an M/M/1 queue 
Show that the arrivals $A_n$ of an M/M/1 queue $X$ with initial distribution $\eta_i := \rho^{i-1}(1-\rho)$ ($i \ge 1$), where $\rho$ is the traffic intensity, satisfy $X_{A_n} \sim \ \eta$.

I understand that the stationary distribution of the queue is $\rho^i(1-\rho)$ for $i \ge 0$, and at an arrival the queue cannot be empty, so the queue size should intuitively have the same geometric property but excluding the state 0. How do I show this?
 A: Recall that $\rho=\lambda/\mu$, where $\lambda$ is the arrival rate and $\mu$ the service rate.
For $j \ge 2$, $$\mathbb{P}(X_{A_1}=j) = \sum_{i=j-1}^\infty\mathbb{P}(\text{there are exactly} \ n\ \text{departures before the first arrival}) \\ = \sum_{i=j-1}^\infty\rho^{i-1}(1-\rho)\left(\frac{\mu}{\lambda+\mu}\right)^{i-j+1}\left(\frac{\lambda}{\lambda+\mu}\right)\\ =\sum_{k=j-2}^\infty\rho^k\left(\frac{\mu}{\lambda+\mu}\right)^k(1-\rho)\left(\frac{\lambda}{\lambda+\mu}\right)\left(\frac{\mu}{\lambda+\mu}\right)^{-j+2}\\=\sum_{k=j-2}^\infty\left(\frac{\lambda}{\lambda+\mu}\right)^k(1-\rho)\left(\frac{\lambda}{\lambda+\mu}\right)\left(\frac{\lambda+\mu}{\mu}\right)^{j-2}\\=\frac{\left(\frac{\lambda}{\lambda+\mu}\right)^{j-2}}{\frac{\mu}{\lambda+\mu}}(1-\rho)\left(\frac{\lambda}{\lambda+\mu}\right)\left(\frac{\lambda+\mu}{\mu}\right)^{j-2}\\=\left(\frac{\lambda}{\mu}\right)^{j-2}(1-\rho)\left(\frac{\lambda}{\mu}\right)\\=\rho^{j-1}(1-\rho)$$ as required. For $j=0$, note that the queue is non-empty when a customer has just arrived so $\mathbb{P}(X_{A_1}=0)=0$. For $j=1$, the distribution must sum to 1, yielding $\mathbb{P}(X_{A_1}=1)=1-\rho$.
Hence $X_{A_1} \sim \eta$. By the strong Markov property (for the M/M/1 queue), $(X_{A_n})_{n\ge0}$ is a discrete-time Markov chain which starts in $\eta$ and has its first arrival given by $\eta$, so the Markov property (for the chain), $\forall n\ge0: X_{A_n} \sim \eta$.
$\square$
