Linear Birth-Death Branching Processes

The Linear Birth-Death Model

In a continuous-time branching process, individuals act independently, and each individual’s births and deaths change the total population. For the linear birth-death process, every individual gives birth to one new individual at rate $\lambda$ and dies at rate $\mu$. Thus a birth changes the population by $+1$, while a death changes it by $-1$. The word linear refers to the fact that when the population is $n$, the total birth and death rates are $n\lambda$ and $n\mu$.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $(\mathcal{F}_t)_{t\ge0}$ record the process history through time $t$. The population process $(Z_t)_{t\ge0}$ has state space $S=\{0,1,2,\ldots\}$, continuous time, and starts with $Z_0=1$ unless another initial population is specified. State $0$ is absorbing. Each individual follows independent birth and death clocks, with exponential waiting times of rates $\lambda$ and $\mu$, respectively.

Assume $\lambda,\mu\ge0$. If both rates are zero, the population stays at its initial size forever. Otherwise, the process does not have infinitely many events in a finite time interval. One way to see why is to couple its births with a pure-birth process whose per-individual birth rate is $\lambda$: deaths can only reduce the population, so the birth-death population is bounded by that pure-birth population. For the pure-birth process stopped upon reaching $m$, the expected stopped size by time $t$ is at most $e^{\lambda t}$, by its rate equation and Gronwall’s inequality. Consequently, the probability it reaches $m$ by time $t$ is at most $e^{\lambda t}/m$, which tends to zero as $m$ grows. The population therefore remains bounded on bounded time intervals almost surely; while bounded, its event rate is bounded as well. This rules out infinitely many events in finite time.

The Generating Function and Its Solution

For one initial individual, define the probability generating function $F(s,t)=\mathbb{E}[s^{Z_t}]$, for $0\le s\le1$, using the convention $s^0=1$, including at $s=0$. Independence of descendants implies the branching property: if there are $k$ initial individuals, their descendant populations are independent copies, so the generating function at time $t$ is $F(s,t)^k$.

Result (finite-time generating function): If $\lambda\ne\mu$, put $r=\lambda-\mu$ and $D_t=\lambda e^{rt}-\mu$. For $t\ge0$, $$ F(s,t)=q_t+(1-q_t)\frac{(1-\alpha_t)s}{1-\alpha_t s}, \qquad q_t=\frac{\mu(e^{rt}-1)}{D_t}, \qquad \alpha_t=\frac{\lambda(e^{rt}-1)}{D_t}. $$ If $\lambda=\mu=a>0$, then $$ F(s,t)=\frac{at}{1+at} +\frac{1}{1+at}\frac{\frac{s}{1+at}}{1-\frac{at}{1+at}s}. $$ In both formulas, $q_t=\mathbb{P}(Z_t=0)$. The special case $\lambda=\mu=0$ has $F(s,t)=s$.

We derive these formulas from the first event. Starting with one individual, during a short interval of length $h$, a birth occurs with probability $\lambda h+o(h)$, a death with probability $\mu h+o(h)$, and no event with probability $1-(\lambda+\mu)h+o(h)$. After a birth there are two independent descendant populations for the remaining time; after a death there are none. Conditioning on this initial interval and taking a difference quotient gives the differential equation and initial condition below.

\frac{\partial F}{\partial t} =\lambda F^2-(\lambda+\mu)F+\mu =(F-1)(\lambda F-\mu), \qquad F(s,0)=s.

Proof. For fixed $s\in[0,1]$, the branching property and the initial-interval probabilities give $F(s,t+h)=[1-(\lambda+\mu)h]F(s,t)+\lambda hF(s,t)^2+\mu h+o(h)$. Subtract $F(s,t)$, divide by $h$, and let $h$ decrease to zero. This gives the displayed differential equation; at time zero the population is exactly one, so $F(s,0)=s$. To solve when $\lambda\ne\mu$ and $\lambda>0$, the two constant solutions of the differential equation are $1$ and $c=\mu/\lambda$. Away from those roots, differentiation gives $\frac{d}{dt}\log\big((F-1)/(F-c)\big)=\lambda-\mu=r$. Solving this relation with $F(s,0)=s$ gives the rational expression in the result after collecting terms. At a root, or at a value obtained by continuity from nearby $s$, the same expression follows by continuity and uniqueness for the differential equation, whose right-hand side is a polynomial and hence locally Lipschitz. If $\lambda=0\ne\mu$, the equation is $F'=\mu(1-F)$, whose solution with initial value $s$ is $1-(1-s)e^{-\mu t}$; direct substitution into the stated formula gives that solution. When $\lambda=\mu=a>0$, the equation becomes $F'=a(F-1)^2$. Separating variables and imposing $F(s,0)=s$ gives $F=1-(1-s)/(1+at(1-s))$, which is algebraically the critical-case expression displayed above. Uniqueness of the differential equation identifies these solutions with the generating function. $\square$

For $\lambda\ne\mu$, the parameters $q_t$ and $\alpha_t$ lie in $[0,1)$ for finite $t$. Expanding the geometric series in the generating function therefore gives the full population law. The critical formula has the same geometric form, with $q_t=\alpha_t=at/(1+at)$.

Finite-time population probabilities: For $\lambda\ne\mu$, with $q_t,\alpha_t$ as above, $$ \mathbb{P}(Z_t=0)=q_t,\qquad \mathbb{P}(Z_t=n)=(1-q_t)(1-\alpha_t)\alpha_t^{\,n-1},\quad n\ge1. $$ For $\lambda=\mu=a>0$, $$ \mathbb{P}(Z_t=0)=\frac{at}{1+at},\qquad \mathbb{P}(Z_t=n)=\frac{1}{(1+at)^2}\left(\frac{at}{1+at}\right)^{n-1},\quad n\ge1. $$

Why the probabilities sum to one. For $0\le\alpha_t<1$, the positive-state probabilities sum to $(1-q_t)(1-\alpha_t)\sum_{n=1}^{\infty}\alpha_t^{n-1}=1-q_t$, since the geometric series is $1/(1-\alpha_t)$. Adding the probability $q_t$ of state zero gives total probability one. This also applies to the critical formula by using its stated $q_t=\alpha_t$. At $t=0$, $\alpha_0=0$, so the law assigns probability one to $Z_0=1$, as required.

Worked Example: A Subcritical Process at a Fixed Time

Take $\lambda=1$, $\mu=2$, and $t=\log 2$, with one initial individual. Here $r=1-2=-1$, so $e^{rt}=e^{-\log 2}=1/2$. The denominator is $D_t=1\cdot(1/2)-2=-3/2$. Thus $q_t=2(1/2-1)/(-3/2)=2/3$, and $\alpha_t=1(1/2-1)/(-3/2)=1/3$.

For $n\ge1$, the probability is $(1-q_t)(1-\alpha_t)\alpha_t^{n-1} =(1/3)(2/3)(1/3)^{n-1} =(2/9)(1/3)^{n-1}$. In particular, the probability of one individual is $2/9$, and that of two individuals is $2/27$. Summing the positive-state probabilities gives $(1/3)(2/3)\sum_{j=0}^{\infty}(1/3)^j =(2/9)(3/2)=1/3$. Adding the zero-state probability $2/3$ gives one.

Check: The mean from the law is $(1-q_t)/(1-\alpha_t)=(1/3)/(2/3)=1/2$. The mean equation for this model gives $e^{(\lambda-\mu)t}=e^{-\log 2}=1/2$, in agreement.

Worked Example: The Critical Process

Set $\lambda=\mu=1$ and $t=1$. The critical parameters are $q_t=1/(1+1)=1/2$ and $\alpha_t=1/(1+1)=1/2$. Hence $\mathbb{P}(Z_1=0)=1/2$, and, for $n\ge1$, $\mathbb{P}(Z_1=n)=(1/2)(1/2)(1/2)^{n-1}=(1/4)(1/2)^{n-1}$.

The positive probabilities sum to $(1/4)\sum_{j=0}^{\infty}(1/2)^j=(1/4)\cdot2=1/2$; with the zero-state probability, the total is one. Their contribution to the mean is $(1/4)\sum_{n=1}^{\infty}n(1/2)^{n-1}=1$, using $\sum_{n=1}^{\infty}nx^{n-1}=1/(1-x)^2$ at $x=1/2$. Thus $\mathbb{E}[Z_1]=1$, also equal to $e^{(1-1)1}=1$.

Eventual Extinction

Define the extinction time $T_0=\inf\{t\ge0:Z_t=0\}$, with $\inf\varnothing=\infty$. Because zero is absorbing, $\{T_0\le t\}=\{Z_t=0\}$. Consequently, the eventual extinction probability for one initial individual is the increasing limit of $q_t=\mathbb{P}(Z_t=0)$ as $t$ tends to infinity.

Theorem (eventual extinction): For one initial individual, the probability $q=\mathbb{P}(T_0<\infty)$ is $$ q= \begin{cases} 1, & 0\le\lambda\le\mu,\quad \lambda+\mu>0,\\[2pt] \mu/\lambda, & \lambda>\mu,\\[2pt] 0, & \lambda=\mu=0. \end{cases} $$ For $k$ initial individuals, the extinction probability is $q^k$, with $q^0=1$.

Proof. If $\lambda\ne\mu$, use $q_t=\mu(e^{rt}-1)/(\lambda e^{rt}-\mu)$, where $r=\lambda-\mu$. When $\lambda>\mu$, $e^{rt}$ tends to infinity; dividing numerator and denominator by $e^{rt}$ shows that $q_t$ tends to $\mu/\lambda$. When $0\le\lambda<\mu$, $e^{rt}$ tends to zero, and the ratio tends to $(-\mu)/(-\mu)=1$. This includes $\lambda=0<\mu$. When $\lambda=\mu=a>0$, the finite-time formula $q_t=at/(1+at)$ tends to one. Since zero is absorbing, the union of the events $\{Z_n=0\}$ over integer $n\ge0$ is exactly $\{T_0<\infty\}$; continuity of probability for increasing events identifies the limit with the eventual extinction probability. If $\lambda=\mu=0$, a single individual never changes state, so extinction probability is zero. Finally, for $k$ initial individuals, extinction occurs exactly when every one of the $k$ independent descendant populations becomes extinct. The probability is therefore the product $q^k$; with no initial individuals extinction has already occurred, consistent with $q^0=1$. $\square$

Worked Example: A Supercritical Process

Take $\lambda=2$, $\mu=1$, and $t=\log 2$. Then $r=1$, $e^{rt}=2$, and $D_t=2\cdot2-1=3$. The finite-time parameters are $q_t=(2-1)/3=1/3$ and $\alpha_t=2(2-1)/3=2/3$. Thus, for $n\ge1$, $\mathbb{P}(Z_t=n)=(2/3)(1/3)(2/3)^{n-1}=(2/9)(2/3)^{n-1}$.

The total probability of a positive population is $(2/9)\sum_{j=0}^{\infty}(2/3)^j=(2/9)\cdot3=2/3$. Adding $q_t=1/3$ gives one. The mean is $(1-q_t)/(1-\alpha_t)=(2/3)/(1/3)=2$, agreeing with $e^{(2-1)\log 2}=2$. The eventual extinction probability is $\mu/\lambda=1/2$, which is larger than the finite-time extinction probability $1/3$, as it must be.

Interpreting the Three Regimes

Rate comparison	Mean population at time $t$, starting from one	Eventual extinction probability
Subcritical: $\lambda<\mu$	$e^{(\lambda-\mu)t}$, decreasing in $t$	$1$
Critical: $\lambda=\mu>0$	$1$	$1$
Supercritical: $\lambda>\mu$	$e^{(\lambda-\mu)t}$, increasing in $t$	$\mu/\lambda<1$

A mean alone does not determine the chance of extinction. In the critical case the mean remains one at every finite time, yet eventual extinction has probability one. In the supercritical case the population can grow in expectation while still becoming extinct with positive probability. For $k$ initial individuals, both the finite-time generating function and eventual extinction probability follow from independence: the generating function is $F(s,t)^k$, and the eventual extinction probability is $q^k$.

Common mistake. The positive population probabilities are not $(1-q_t)\alpha_t^{n-1}$; a geometric probability mass also contains $1-\alpha_t$. In the subcritical example, omitting that factor would give the purported positive-population total $(1-q_t)\sum_{j\ge0}\alpha_t^j=(1/3)/(1-1/3)=1/2$. That is incorrect. The correct total is $(1-q_t)(1-\alpha_t)\sum_{j\ge0}\alpha_t^j=(1/3)(2/3)/(1-1/3)=1/3$, which equals $1-q_t$. The missing factor is essential both for normalization and for every individual positive-state probability.

Check Your Understanding

Use the generating-function form and verify each probability calculation by summing the relevant geometric series.

For $\lambda=1$, $\mu=3$, and $t=\log 2$, calculate $q_t$ and $\alpha_t$, then find $\mathbb{P}(Z_t=1)$.
When $\lambda=\mu=a>0$, find the probability of exactly three individuals at time $t$ in terms of $a$ and $t$.
Explain why the event $\{T_0<\infty\}$ equals the union of $\{Z_n=0\}$ over nonnegative integer $n$.
For a supercritical process with one initial individual, why can the eventual extinction probability be positive even though the expected population grows?
Starting from $k$ individuals, write the finite-time generating function and the eventual extinction probability.

Rate comparison	Mean population at time \(t\), starting from one	Eventual extinction probability
Subcritical: \(\lambda<\mu\)	\(e^{(\lambda-\mu)t}\), decreasing in \(t\)	\(1\)
Critical: \(\lambda=\mu>0\)	\(1\)	\(1\)
Supercritical: \(\lambda>\mu\)	\(e^{(\lambda-\mu)t}\), increasing in \(t\)	\(\mu/\lambda<1\)

Linear Birth-Death Branching Processes

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