Solving Probabilities for M/M/1 Queue Waiting Time Generating Function 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.
 A: 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
$$\pi_n = (1-\rho)\rho^n,\; n\geqslant0.  $$
