 Solution

With two offspring always, the backward equation becomes ${\varphi }^{\prime }=\mu ({\varphi }^2-\varphi )$. Separation and the initial value giveThus $M_t$ is geometric on the positive integers, andThis is a Yule process shifted down by one. It is not an inhomogeneous Poisson process. Given $N_t=j$, its instantaneous jump rate is $\mu (j+1)$, a state-dependent random rate rather than a deterministic function of time. Also $E[N_t]=e^{\mu t}-1$ whereas $\mathrm{Var}(N_t)=e^{\mu t}(e^{\mu t}-1)$, unequal to the mean for $t>0$. Its mean arrival intensity $\mu e^{\mu t}$ does not turn its dependent increments and geometric marginals into Poisson ones.