Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 26I (Applied Probability)

Let $m(t)=E[M_t]$, with finite offspring mean $\rho$. Conditioning on the original cell's first exponential lifetime givesEach of its $k$ offspring starts an independent copy of the process, which explains the factor $\rho$. Changing to $u=t-s$ and differentiating yields $m^{\prime }=\mu (\rho -1)m$, $m(0)=1$. HenceFinite-mean truncation and this integral bound justify finite expectations and exclude explosion in finite time; it is not necessary to assume a bounded offspring count.

For the probability generating function, let $P(s)={\sum }_{k\ge 0}{p}_k{s}^k$. Independence of offspring populations gives the conditional generating function $P({\varphi }_{t-s}(s))$ at a first division time. Thusand differentiation of the changed-variable integral provesFor $0\le s\le 1$, the integrand is bounded, so this generating-function argument has the needed convergence even without higher offspring moments.

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.