Continuous-time branching processes

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This exercise explores a continuous time version of a simple branching process.

Consider a population of primitive organisms which do nothing but procreate and die. In particular, the population starts at time 00 with one organism. This organism has an exponentially distributed lifespan T0T_{0} with rate μ\mu (i.e., P{T0≥τ}=e−μτ\mathbb {P}\left\{ T_{0} \geq \tau \right\} = e^{-\mu \tau }). While this organism is alive, it gives birth to new organisms according to a Poisson process of rate λ\lambda. Each of these new organisms, while alive, gives birth to yet other organisms. The lifespan and birthrate for each of these new organisms are IID to those of the first organism. All these and subsequent organisms give birth and die in the same way, again independently of all other organisms.

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(a)

Let X(t)X(t) be the number of (live) organisms in the population at time tt. Show that {X(t);t≥0}\left\{ X(t); t \geq 0\right\} is a Markov process and specify the transition rates between the states.

(b)

Find the embedded Markov chain {Xn;n≥0}\left\{ X_{n}; n \geq 0\right\} corresponding to the Markov process in (a). Find the transition probabilities for this Markov chain.

(c)

Explain why the Markov process and Markov chain above are not irreducible.

(d)

For purposes of analysis, add an additional transition of rate λ\lambda from state 00 to state 11. Show that the Markov process and the embedded chain are irreducible. Find the values of λ\lambda and μ\mu for which the modified chain is positive recurrent, null recurrent, and transient.

(e)

Assume that λ<μ\lambda < \mu. Find the steady-state process probabilities for the modified Markov process.

(f)

Find the mean recurrence time between visits to state 00 for the modified Markov process.

(g)

Find the mean time T‾\overline{T} for the population in the original Markov process to die out.

Remark.
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Note for (c): The majority of results you have seen for Markov processes assume the process is irreducible, so be careful not to use those results in this exercise.

Note for (g): We have seen before that removing transitions from a Markov chain or process to create a trapping state can make it easier to find mean recurrence times. This is an example of the opposite, where adding an exit from a trapping state makes it easy to find the recurrence time.

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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Find the generating function of the number of particles at time tt for a branching process with continuous time and generating function of infinitesimal parameters:

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(a)

f(z)=a2z2+a1z−(a1+a2),a2>0f(z) = a_{2} z^{2}+a_{1} z-\left(a_{1}+a_{2}\right), a_{2} > 0;

(b)

f(z)=zk−z,k⩾2f(z) = z^{k}-z, k \geqslant 2;

(c)

f(z)=1−z−1−zf(z) = 1-z-\sqrt{1-z};

(d)

f(z)=λ(1−z)[1+ln⁡(1−z)],λ>0f(z) = \lambda (1-z)[1+\ln (1-z)], \lambda > 0.

Find the extinction probabilities of the branching processes from problem 10.57.

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(a)
(b)
(c)
(d)

Let XtX_{t} be a branching process with generating function of infinitesimal parameters f(z)=a2z2+a1z−(a1+a2)f(z) = a_{2} z^{2}+a_{1} z-\left(a_{1}+a_{2}\right), where a2>0,a1+2a2<0a_{2} > 0, a_{1}+2 a_{2} < 0. Prove that

lim⁡t→∞e−(a1+2a2)tP(Xt>0)=1+a2a1+a2. \lim _{t \rightarrow \infty } e^{-\left(a_{1}+2 a_{2}\right) t} \mathbb {P}\left(X_{t} > 0\right) = 1+\frac{a_{2}}{a_{1}+a_{2}} .
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Prove that for the branching process XtX_{t} with generating function f(z)=3−5z+z2+z3f(z) = 3-5 z+z^{2}+z^{3}

lim⁡t→∞t−1P(Xt>0)=14. \lim _{t \rightarrow \infty } t^{-1} \mathbb {P}\left(X_{t} > 0\right) = \frac{1}{4} .
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References

[Gal13] Gallager R.G. Stochastic Processes: Theory for Applications. Cambridge University Press, 2013.
[GrSt20] Grimmett G.R., Stirzaker D.R. Probability and Random Processes. 4th ed., Oxford University Press, 2020.
[ПУУ86] Прохоров А.В., Ушаков В.Г., Ушаков Н. Задачи по теории вероятностей. Наука. Главная редакция физико-математической литературы, 1986.