 Solution

Let $(m,n)$ denote the clock position and the queue population, including any customer in service. By the memoryless property of the exponential distribution, the nonzero transition rates areSet $\rho =\nu /\mu <1$. We claim thatwhere $p_m$ is the clock's stationary distribution from part (a). To check global balance for a continuous-time Markov chain, divide the incoming probability flux by $\pi (m,n)$. The clockwise contribution is $\nu p_{m-1}/p_m={\lambda }_m$. The arrival contribution, present only for $n>0$, is ${\lambda }_{m+1}(p_{m+1}/p_m){\rho }^{-1}=\mu$. The service contribution is $\mu \rho =\nu$. Their sum is the outgoing rate $\nu +{\lambda }_m+\mu 1_{\{n>0\}}$.

The proposed probability distribution is normalized, so it is the equilibrium law. In particular, the clock position and queue population are independent random variables at any fixed equilibrium time, and the population has the geometric distribution of a stable M/M/1 queue. Independence of these two random variables does not assert independence of the two entire processes.