 Solution

With population $n$, the transition diagram has the two outgoing arrowsThus this is a batch-birth linear-death process. If $P_n(t)=\mathbb{P}(N_t=n)$ and $P_j=0$ for $j<0$, probability enters state $n$ from $n-3$ by a birth and from $n+1$ by a death. Hence the master equation is

The Markov jump-process generator isFor $f(n)=n$, this gives $Ln=3\lambda -\beta n$. Therefore the expected value $m(t)=\mathbb{E}[N_t]$ satisfiesso

For $f(n)=n^2$,Writing $s=\mathbb{E}[N_t^2]$ and using the generator identity givesThe variance $v=s-m^2$ consequently obeysIf $v(0)=v_0<\infty$, substitution of the formula for $m$ yieldsAll transient terms vanish, and the moments of a batch-birth linear-death process therefore give