Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 28K (Applied Probability)

Let $E_n$ be the holding time in state $n$. The variables are independent with $E_n\sim \mathrm{Exp}(n\lambda )$, and the explosion time would beFor $s>0$,This product is zero because
${\sum }_n\log (1+s/(n\lambda ))=\infty$. If $T_{\infty }$ were finite with positive probability, the nonnegative variable $e^{-s{T}_{\infty }}$ would have positive expectation. Therefore $T_{\infty }=\infty$ almost surely, and the Yule process is nonexplosive.

For ordered birth times $0<s_1<\cdots <s_n<t$, with $s_0=0$, the joint density of births at those times followed by no further birth before $t$ isAlsoDivision giveswhich is exactly the joint density of the order statistics of $n$ independent variables with the stated density. This proves the conditional Yule birth times result.

Starting with one individual, let $S$ be the first event time, of density
$(\lambda +\mu )e^{-(\lambda +\mu )s}$. At that event there is a death with conditional probability $\mu /(\lambda +\mu )$, causing extinction immediately, or a birth with probability $\lambda /(\lambda +\mu )$, leaving two independent descendant populations. The branching property therefore gives

Changing variables in the integral equation,Differentiation gives

Separating variables in
$h^{\prime }=(h-1)(\lambda h-\mu )$ with $h(0)=0$ gives the extinction probability of a linear birth-death processFor $\lambda >\mu$ this tends to $\mu /\lambda$; for $\lambda <\mu$ it tends to one, as expected.