Section 5.5

Age-dependent branching processes

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Problem 5.5.1

Let Z(t)Z(t) be the population-size at time tt in an age-dependent branching process, the lifetime distribution of which is exponential with parameter λ\lambda. If Z(0)=1Z(0) = 1, show that the probability generating function Gt(s)G_{t}(s) of Z(t)Z(t) satisfies

∂∂tGt(s)=λ{G(Gt(s))−Gt(s)} \frac{\partial }{\partial t} G_{t}(s) = \lambda \left\{ G\left(G_{t}(s)\right)-G_{t}(s)\right\}

where GG is the probability generating function of a typical family-size. Show in the case of 'exponential binary fission', when G(s)=s2G(s) = s^{2}, that

Gt(s)=se−λt1−s(1−e−λt) G_{t}(s) = \frac{s e^{-\lambda t}}{1-s\left(1-e^{-\lambda t}\right)}

and hence derive the probability mass function of the population size Z(t)Z(t) at time tt.

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Problem 5.5.2

Solve the differential equation of Exercise (5.5.1) when λ=1\lambda = 1 and G(s)=12(1+s2)G(s) = \frac{1}{2}\left(1+s^{2}\right), to obtain

Gt(s)=2s+t(1−s)2+t(1−s) G_{t}(s) = \frac{2 s+t(1-s)}{2+t(1-s)}

Hence find P(Z(t)≥k)\mathbb {P}\left(Z(t) \geq k\right), and deduce that

P(Z(t)/t≥x∣Z(t)>0)→e−2x as t→∞ \mathbb {P}\left(Z(t) / t \geq x \mid Z(t) > 0\right) \rightarrow e^{-2 x} \quad \text{ as } t \rightarrow \infty
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