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Add independent rate-$\mu$ resets to zero to a linear birth process with immigration. The resulting irreducible continuous-time Markov chain is nonexplosive by domination of births and finiteness of Poisson reset counts. When $\mu =\lambda$, an excursion from one reaches $n$ with probability $2/(n+1)$, so its jump chain is recurrent with infinite expected jump count. The elapsed return-cycle time instead has meanThus the continuous-time chain is positive recurrent while its jump chain is null recurrent. Its stationary probabilities are ${\pi }_0=1/2$ and ${\pi }_n=1/[(n+1)(n+2)]$ for $n\ge 1$.

For this linear birth process with catastrophes, the off-diagonal entries of the Q-matrix arewith all other off-diagonal entries zero. The diagonal entries are $q_{00}=-\lambda$ and $q_{nn}=-[\lambda (n+1)+\mu ]$ for $n\ge 1$. At zero there is no separate catastrophe transition, because it would not change the state.

The chain is irreducible: every positive state can jump to zero, and from zero any finite sequence of births has positive probability. To prove nonexplosion, couple it with the linear birth process with immigration having birth rate $\lambda (n+1)$ and no catastrophes. Its mean population is finite at each finite time, obtained from $m^{\prime }=\lambda (m+1)$, or its exponential holding-time sum diverges since ${\sum }_{n}1/[\lambda (n+1)]=\infty$. Couple births monotonically and implement catastrophes with an independent rate-$\mu$ Poisson clock, resetting the smaller population. On a finite time interval the dominating pure-birth process has finitely many births, and the catastrophe clock finitely many rings. Thus the original chain is nonexplosive.

For its jump chain, the transitions are

For ${\lambda }_n=\alpha n+\beta$,so part (b) ensures nonexplosion. The Markov jump-process generator acts on the identity function $f(n)=n$ asConsequently $m(t)=\mathbb{E}N(t)$ satisfiesSolving this linear ordinary differential equation gives the mean of the linear birth process with immigration: