Example of a continuous Markov renewal process Clients arrive at 1 server according to a Poisson process with intensity $a$ and leave according to a Poisson process with intensity $b$. However if there are already $N$ clients waiting, a new client will also wait with probability $1/(N+1)$ or leave with probability $N/(N+1)$.

  
*
  
*How do I show that the stationary distribution of the number of clients present is Poisson distributed with expectation $a/b$?
  
*What is the average number of clients that arrive per unit time?
  

What I know:
Our state space is $\mathbb{N}_0$, the number of clients present.
I thought that a state $i$ can only transition to $i-1,i,i+1$ with intensity $b,-(a+b),a$ resp. for $i<N$. For $i\geq N$ the intensity to $i+1$ is $a/(N=1)$, to $i$ is $-a/(N+1)-b$ and to $i-1$ remains $b$. These are the entries of the transition rate matrix $G$.
The transition rate matrix G is given by
$   g_{ij} = \left\{
     \begin{array}{lr}
       v_ip_{ij} & : i \neq j\\
       -v_i & : i=j
     \end{array}
   \right.$
where the residence time in state $i$ is distributed exponentially with parameter $v_i$ and $p_{ij}$ are the transition probabilities.
Further the stationary distribution is $P=(P_j)_j$ where $P_j=\lim_{t\rightarrow\infty}p_{ij}(t)$. And $P$ is the solution of $v_jx_j=\sum_{k\neq j}g_{kj}x_k$.
How can I use these to find 1. and 2.?
 A: I suppose that $N$ is not a fixed number, but for all $N\in\mathbb{N}=\left\{0,1,2,\cdots\right\}$. In other words, whenever there are $N$ clients at the server, a new arriving client would choose to wait with probability $1/\left(N+1\right)$ and to leave with probability $N/\left(N+1\right)$.



*

*Let $N_t$ be the total number of clients at the server at the moment $t$.

*Let $A_t$ be the total number of arriving clients by the moment $t$, with
$$
A_t\sim\text{Poisson}(at).
$$

*Let $\xi_t$ be the wait-or-leave choice of an arriving client at the moment $t$, with
$$
\xi_t|N_t\sim\text{Bernoulli}\biggl(\frac{1}{N_t+1}\biggr).
$$

*Let $L_t$ be the total number of leaving clients by the moment $t$, with
$$
L_t\sim\text{Poisson}(bt).
$$

*Let $A_t$, $L_t$ and $\xi_t|N_t$ be independent processes.


With these settings, the model reads
$$
{\rm d}N_t=\xi_t{\rm d}A_t-{\rm d}L_t,
$$
or in the infinitesimal increment form,
$$
N_{t+{\rm d}t}-N_t=\xi_t\left(A_{t+{\rm d}t}-A_t\right)-\left(L_{t+{\rm d}t}-L_t\right).
$$

Denote
$$
p_n(t)=\mathbb{P}\left(N_t=n\right).
$$
We have
\begin{align}
p_n(t+{\rm d}t)&=\mathbb{P}\left(N_{t+{\rm d}t}=n\right)\\
&=\mathbb{P}\left(N_{t+{\rm d}t}=n|N_t=n-1\right)\mathbb{P}\left(N_t=n-1\right)\\
&\quad+\mathbb{P}\left(N_{t+{\rm d}t}=n|N_t=n\right)\mathbb{P}\left(N_t=n\right)\\
&\quad+\mathbb{P}\left(N_{t+{\rm d}t}=n|N_t=n+1\right)\mathbb{P}\left(N_t=n+1\right).
\end{align}
This relation eventually lead us to
\begin{align}
p_0'(t)&=-ap_0(t)+bp_1(t),&&n=0,\\
p_n'(t)&=\frac{a}{n}p_{n-1}(t)-\left(\frac{a}{n+1}+b\right)p_n(t)+bp_{n+1}(t),&&n\ge 1,
\end{align}
where
$$
p_n'(t)=\lim_{{\rm d}t\to 0^+}\frac{p_n(t+{\rm d}t)-p_n(t)}{{\rm d}t}.
$$
As for stationary distribution, we have $p_n'(t)=0$. Thus
\begin{align}
-ap_0+bp_1&=0,&&n=0,\\
\frac{a}{n}p_{n-1}-\left(\frac{a}{n+1}+b\right)p_n+bp_{n+1}&=0,&&n\ge 1.
\end{align}
In addition, the conservation of probability requires
$$
\sum_{n=0}^{\infty}p_n=1.
$$
The above iterative scheme can be solved inductively. Note that the first three terms read
\begin{align}
p_1&=\frac{a}{b}p_0,\\
p_2&=\frac{1}{2}\left(\frac{a}{b}\right)^2p_0,\\
p_3&=\frac{1}{6}\left(\frac{a}{b}\right)^3p_0.
\end{align}
These inspire to show inductively that
$$
p_n=\frac{1}{n!}\left(\frac{a}{b}\right)^np_0
$$
for all $n\in\mathbb{N}$. Finally, the conservation of probability yields
$$
1=\sum_{n=0}^{\infty}p_n=p_0\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{a}{b}\right)^n=p_0\exp\left(\frac{a}{b}\right)\iff p_0=\exp\left(-\frac{a}{b}\right).
$$
Consequently,
$$
p_n=\frac{1}{n!}\left(\frac{a}{b}\right)^n\exp\left(-\frac{a}{b}\right)
$$
for all $n\in\mathbb{N}$.
Obviously, this is a Poisson distribution with parameter $a/b$.

As per the average number of arriving clients per unit time, it should be
$$
\mathbb{E}\left(A_1\right)=a.
$$
