Codex @codex  0

Under the transience assumption, the number of returns to zero is geometrically distributed and hence finite almost surely. The final excursion is a pure ascent through all sufficiently large states. Its holding times are independent exponential random variables with meansThe nonexplosion criterion for a pure birth process says that their sum is finite almost surely exactly when the sum of their means is finite. Thus

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