I "believe" that generator, $\bf Q$, of the waiting time distribution for the $M/M/1$ queue is given by the following (I'm not 100% sure if this is even correct):
$\bf Q$ = $\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & 0\\ \mu & -\mu & 0 & 0 & 0\\ 0 & \mu & -\mu & 0 & 0 \\ 0 & 0 & \mu & -\mu & \dots \end{array} \right) $
But the question I have is that I am unclear how to solve this Markov chain. That is, I'm looking for an analytic solution to
$\bf{p} Q = 0$
I think $\bf p$ should look something like
$\bf p$ = [$1-\rho, \dots]$,
but again, I am unclear how to solve these problems.
Thanks for help in these matters.
The transition rates of the CTMC are given by \begin{align} q_{n,n+1} &= \lambda\\ q_{n+1,n} &= \mu \end{align} for all $n$. It follows that the generator matrix $Q$ has entries $$q_{ij} = \begin{cases}-\lambda,& i=j=0\\ \lambda,& j=i+1\\ \mu,& j=i-1\\ -(\lambda+\mu),& i=j>0.\end{cases} $$ Hence $\pi Q=0$ implies \begin{align} -\lambda\pi_0 +\mu\pi_1 &= 0\\ \lambda\pi_{n-1} -(\lambda+\mu)\pi_n + \mu\pi_{n+1} &= 0, \; n\geqslant 1. \end{align} Let $\rho=\frac\lambda\mu$ It follows that $\pi_1=\rho\pi_0$ and $$\pi_{n+1} = (1+\rho) \pi_n - \rho\pi_{n-1}, n\geqslant 1. $$ Let $P(s)=\mathbb E[s^\pi]$ be the generating function of $\pi$, then multiplying the recurrence by $s^n$ and summing over $n$ yields $$\sum_{n=1}^\infty \pi_{n+1}s^n = \sum_{n=1}^\infty (1+\rho) \pi_n s^n - \sum_{n=1}^\infty \rho \pi_{n-1} s^n, $$ or $$s^{-1}(P(s)-\pi_0-\rho-\rho\pi s) = (1+\rho)(P(s)-\pi_0) - \rho sP(s). $$ Solving for $P(s)$, we have $$P(s) = \frac{\pi_0}{1-\rho s} = \sum_{n=0}^\infty \pi_0\rho^ns^n. $$ From $P(1)=1$ we obtain $$\pi_0 = \left(\sum_{n=0}^\infty \rho^n\right)^{-1} = 1-\rho,$$ and hence $$\pi_n = (1-\rho)\rho^n,\; n\geqslant0. $$