Birth process

A pure birth process on ${\mathbb{N}}_0$ jumps only from $n$ to $n+1$, at a strictly positive rate ${\lambda }_n$. Its successive holding times are independent exponential variables of rates ${\lambda }_n$.

For a pure birth process with fixed initial state and finite statewise rates, the jump-count compensator gives $dM=(e^{\theta }-1)M_{-}[dN-\lambda (X_{-})dt]$, so $M$ is a local martingale up to explosion. If $\lambda \le C$, a rate-$C$ Poisson candidate clock with accepted jumps of probability $\lambda (X_{-})/C$ prevents explosion and dominates the jump count. Its finite exponential moments dominate every stopped $M$ on each finite horizon, upgrading the local martingale to a true martingale. A random initial state requires $\mathbb{E}{e}^{\theta X_0}<\infty$; uniformly bounded rates do not supply this missing initial integrability.

For a carrying capacity $N$, each of $k$ cells splits into two at rate $1-k/N$, so the population increases by one at the displayed total rate and is absorbed at $N$. Starting from population density $p\in (0,1)$, its normalized density converges on finite horizons to $p{e}^t/(1-p+p{e}^t)$ by the Kurtz fluid limit theorem.

// Target: analysis.bigb

A pure birth process is nonexplosive exactly whenIts explosion time is the sum of independent exponential holding times. If the displayed series converges, that sum has finite expectation. If it diverges, the Laplace transform of the sum isforcing the sum to be infinite almost surely.

The pure birth process with rates ${\lambda }_n=\alpha n+\beta$, where $\alpha ,\beta >0$, combines linear reproduction with immigration. Starting from zero, its mean satisfiesand hence

Add independent rate-$\mu$ resets to zero to a linear birth process with immigration. The resulting irreducible continuous-time Markov chain is nonexplosive by domination of births and finiteness of Poisson reset counts. When $\mu =\lambda$, an excursion from one reaches $n$ with probability $2/(n+1)$, so its jump chain is recurrent with infinite expected jump count. The elapsed return-cycle time instead has meanThus the continuous-time chain is positive recurrent while its jump chain is null recurrent. Its stationary probabilities are ${\pi }_0=1/2$ and ${\pi }_n=1/[(n+1)(n+2)]$ for $n\ge 1$.