 Solution

Write $a=\mu /\lambda <1$. The embedded birth-death chain moves upward with probability $\lambda /(\lambda +\mu )$ and downward with probability $\mu /(\lambda +\mu )$ away from zero. The gambler's ruin formula, obtained by solving its two-point difference equation on $\{0,\dots ,N\}$ and then letting $N\to \infty$, gives ${\mathbb{P}}_k(\text{hit }0)=a^k$. Translation gives return probability $a$ from $r+1$ to $r$. From $r-1$ the chain hits $r$ almost surely: before that hit it remains in a finite reflecting interval, from every state of which a uniformly positive finite-step chance of hitting $r$ is available. This also proves that starting at zero it hits each $r\ge 1$ almost surely.

After departure from $r\ge 1$ the return probability is thereforeThe holding rate is $q_r=\lambda +\mu$, so the exponential occupation-time parameter is $q_r(1-r_r)=\lambda -\mu$. Applying the Strong Markov property at the almost-sure first hit givesAt zero the holding rate is $\lambda$ and return probability after departure is $a$, again giving parameter $\lambda -\mu$. From initial state $k$, the expected occupation time at zero is $a^k/(\lambda -\mu )$, since the time is zero unless zero is reached. Averaging over the initial law with probability generating function $G$ yieldsFor $\mu =0$, the interpretation is $a^0=1$, $a^k=0$ for $k>0$, and $G(0)=\mathbb{P}(X(0)=0)$; the same formulas hold.