Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 28K (Applied Probability)

A pure birth process $N=(N(t):t\ge 0)$ is a continuous-time Markov chain on ${\mathbb{N}}_0$ whose only possible transition from state $n$ isEquivalently, after entering state $n$ it waits an independent exponential time $X_n\sim \mathrm{Exp}({\lambda }_n)$ and then moves to $n+1$.

The process is nonexplosive when it makes only finitely many jumps during every bounded time interval. If it starts from zero, this is equivalent to its explosion timebeing infinite almost surely.

IfthenA nonnegative random variable with finite expectation is finite almost surely, so the process explodes.

Conversely, suppose the series diverges. For $s>0$, independence and the Laplace transform of an exponential variable giveThe sum in the exponent diverges. If infinitely many ${\lambda }_n\le s$, infinitely many terms are at least $\log 2$; otherwise, eventually $s/{\lambda }_n\le 1$ andThus $\mathbb{E}{e}^{-s{T}_{\infty }}=0$. Since $e^{-s{T}_{\infty }}$ is strictly positive on $\{T_{\infty }<\infty \}$, this forces $T_{\infty }=\infty$ almost surely. Hence the nonexplosion criterion for a pure birth process is

For ${\lambda }_n=\alpha n+\beta$,so part (b) ensures nonexplosion. The Markov jump-process generator acts on the identity function $f(n)=n$ asConsequently $m(t)=\mathbb{E}N(t)$ satisfiesSolving this linear ordinary differential equation gives the mean of the linear birth process with immigration: