Let $X(t)$ denote the number of customers in a system at time $t$.
The process $\{ X(t), t \geq 0 \}$ at a $M/M/s$ queue is a reversible Markov process.
Is this statement true or false for:
(a) $s=1$?
(b) General $s$?
My attempt:
A reversible Markov process means that
$P(X_{t-1} = j | X_t = i, X_{t+1} = i_{t+1}, X_{t-2} = i_{t+2}, \dots ) = P(X_{t-1} = j | X_t = i)$.
So is it possible to find the probability $X_{t-1} = j$ given that $X_t = i$? Intuitively I would say no, but here is the point that I get stuck.
Any irreducible birth--and--death process that has positive recurrent states is reversible.
First, an irreducible continuous-time Markov process with positive recurrent states has a probability distribution $\mathbf{p}$ that is the unique solution to the equilibrium equations $\mathbf{p} Q = 0$ and the normalization condition $\mathbf{p} \mathbf{1} = 1$, where $Q$ is the transition rate matrix and $\mathbf{1}$ a vector of ones of appropriate dimension.
We employ Theorem 1.3 of Kelly, Reversibility and Stochastic Networks (1979). If $\mathbf{p}, ~ \mathbf{p} \mathbf{1} = 1$ satisfies
\begin{equation} p(i) q(i,j) = p(j) q(j,i) \tag{1} \end{equation}
for all states $i$ and $j$, then the Markov process is reversible and $\mathbf{p}$ is the equilibrium distribution.
The state space of a birth--and--death process is $\{0,1,2,\ldots\}$. The process is called a birth--and--death process because the process can only transition from state $i$ to states $i - 1$ or $i + 1$. This means that $(1)$ is trivially satisfied for $|i - j| > 1$.
Next, draw a barrier between state $i$ and $i + 1$, splitting the state space in two sets of states: a finite set $A$ and a countably infinite set $A^c$. Since the Markov process is in equilibrium (stationary), the rate at which the process enters the set of states $A$ is equal to the rate at which the process leaves the set of states $A$. Given that the process is currently in state $i$, it leaves the set $A$ with rate $q(i,i+1)$. The probability that the Markov process is in state $i$ in equilibrium is $p(i)$. So, the rate at which the Markov process leaves the set $A$ is $p(i) q(i,i + 1)$. Similarly, the rate at which the Markov process enters the set $A$ can be determined. Ultimately, we get
\begin{equation} p(i) q(i,i + 1) = p(i + 1)q(i + 1,i), \end{equation}
which has a solution since the Markov process is irreducible and has positive recurrent states (express all $p(i)$ in terms of $p(0)$ and use the normalization condition $\mathbf{p} \mathbf{1} = 1$). This proves the claim.
As a special case, an irreducible Markov process that models the $M/M/s$ queue is a birth--and--death process and if is it positive recurrent, it is reversible. Say the arrival rate is $\lambda$ and the service rate is $\mu$, then the Markov process associated with the $M/M/s$ queue is positive recurrent if $\lambda/(s\mu) < 1$.