= Solution
The pair-birth rate here is the constant immigration rate $\lambda$, while deaths occur at total rate $\beta n$. This interpretation is fixed by the requested mean formula. With $P_n(t)=\Pr(N_t=n)$ and $P_j=0$ for $j<0$, the <master equation> is
$$
\boxed{P_n'=\lambda P_{n-2}+\beta(n+1)P_{n+1}-(\lambda+\beta n)P_n.}
$$
The <Markov jump-process generator> acts as $Lf(n)=\lambda[f(n+2)-f(n)]+\beta n[f(n-1)-f(n)]$. Taking $f(n)=n$ gives $m'=2\lambda-\beta m$, hence
$$
\boxed{m(t)=\frac{2\lambda}{\beta}(1-e^{-\beta t})+n_0e^{-\beta t}.}
$$
For the <second moment> $s=\mathbb E N_t^2$, the two jumps give $s'=4\lambda m+4\lambda-2\beta s+\beta m$. Differentiating $V=s-m^2$ therefore yields $V'=4\lambda+\beta m-2\beta V$. Substitution of the mean and solution of this scalar linear equation give
$$
V(t)=\frac{3\lambda}{\beta}+(V(0)-3\lambda/\beta)e^{-2\beta t}
+(n_0-2\lambda/\beta)(e^{-\beta t}-e^{-2\beta t}).
$$
Thus \b[$V(t)\to3\lambda/\beta$], assuming $\beta>0$ and finite initial <second moment>. This is <pair immigration with linear deaths>; births at rate $\lambda n$ would be a different model and would not give the printed mean.
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