Extinction Probability in a Linear Birth-Death Population

Extinction as a Hitting Probability

A population can have a growing expected size and still have a positive chance of disappearing: expectation averages across outcomes, while extinction asks whether the population ever reaches zero. In a linear birth-death model, each individual gives birth at rate $b$ and dies at rate $d$, independently of the others. The key question is whether the balance between these rates makes extinction certain or leaves a positive probability of survival.

Let $(Z_t)_{t\ge0}$ be a continuous-time Markov chain on the state space $S=\{0,1,2,\ldots\}$, defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, with its natural filtration $(\mathcal{F}_t)_{t\ge0}$. In state $k\ge1$, the total birth rate is $bk$ and the total death rate is $dk$, where $b,d\ge0$. Thus the generator $Q=(q_{ij})$ has $q_{k,k+1}=bk$, $q_{k,k-1}=dk$, and $q_{kk}=-(b+d)k$; all other entries are zero. Its rows sum to zero. State $0$ is absorbing. We call extinction eventual if $T_0=\inf\{t\ge0:Z_t=0\}$ is finite. If the initial population is $n$, write $\mathbb{P}_n$ for the corresponding probability.

Theorem (Extinction probability): Suppose $b,d\ge0$, $b+d>0$, and the initial population is $n$. If $n=0$, extinction has probability $1$. If $n\ge1$, then $$ \mathbb{P}_n(T_0<\infty)= \begin{cases} 1, & b\le d,\\ (d/b)^n, & b>d. \end{cases} $$ In particular, when $b>d$, the probability of never becoming extinct is $1-(d/b)^n>0$.

The rates depend on the current population, but the direction of the next jump does not: conditional on a jump from a positive state, it is a birth with probability $p=b/(b+d)$ and a death with probability $q=d/(b+d)$. We first calculate the probability of reaching zero before a finite upper population level. Taking that upper level to infinity will give the eventual extinction probability.

Why the Jump-Chain Calculation Applies

At a positive state $k$, the holding time until the next jump is exponential with rate $(b+d)k$. If $b+d>0$, this holding time is finite almost surely, and the next jump is upward with probability $p$ and downward with probability $q$. The successive states at jump times therefore form a nearest-neighbour random walk on the nonnegative integers, absorbed at zero.

We also need to rule out infinitely many jumps in a finite amount of time. Here is a bound that does so. For an integer $K>n$, let $T_K=\inf\{t\ge0:Z_t=K\}$. The process stopped on reaching $K$ has bounded rates. Its expected population satisfies $$ \mathbb{E}_n[Z_{t\wedge T_K}] =n+\mathbb{E}_n\!\left[\int_0^{t\wedge T_K}(b-d)Z_s\,ds\right] \le n+b\int_0^t\mathbb{E}_n[Z_{s\wedge T_K}]\,ds. $$ The first equality follows by summing the birth and death changes, whose conditional mean rate in state $k$ is $bk-dk$; the inequality uses $(b-d)k\le bk$. Gronwall's inequality gives $\mathbb{E}_n[Z_{t\wedge T_K}]\le ne^{bt}$. Since $Z_{T_K}=K$ on $\{T_K\le t\}$, $$ \mathbb{P}_n(T_K\le t)\le \frac{ne^{bt}}{K}. $$ This tends to zero as $K$ tends to infinity. An infinite number of jumps while the population remains bounded is also impossible in finite time: below any fixed bound, all jump rates are bounded, so the holding times cannot accumulate in finite time. Consequently the process is non-explosive, and hitting zero in the jump chain is equivalent to eventual extinction in continuous time.

Finite Barriers and the Extinction Formula

Fix an upper level $M>n$. Let $h_i^{(M)}$ be the probability that the jump chain, started at $i$, hits zero before it hits $M$, for $0\le i\le M$. The boundary values are $h_0^{(M)}=1$ and $h_M^{(M)}=0$. For $1\le i\le M-1$, conditioning on the first jump gives $$ h_i^{(M)}=p h_{i+1}^{(M)}+q h_{i-1}^{(M)}. $$ When $p,q>0$, solving this recurrence gives $$ h_i^{(M)}= \begin{cases} \dfrac{M-i}{M}, & p=q,\\[6pt] \dfrac{r^i-r^M}{1-r^M}, & p\ne q, \end{cases} \qquad r=\frac{q}{p}=\frac{d}{b}. $$ To verify the second expression, the recurrence has characteristic equation $p x^2-x+q=0$, whose roots are $1$ and $q/p=r$. Thus its solutions have the form $A+B r^i$. The boundary values determine the displayed expression. When $p=q$, the repeated root is $1$, so solutions are linear in $i$, and the boundary values give $(M-i)/M$. These solutions are unique: the recurrence together with the two boundary values determines all intermediate values, or, equivalently, a difference of two solutions has zero boundary values and must be identically zero by the same recurrence.

How the barrier limit works: As $M$ increases, the events “hit zero before hitting $M$” increase. Their union is the event of ever hitting zero: a path that reaches zero in finitely many jumps has a finite maximum population before that hit, so it hits zero before every sufficiently large $M$. Therefore the eventual extinction probability is $\lim_{M\to\infty}h_n^{(M)}$.

Subcritical case, $b<d$.
Here $r=d/b>1$. Rewriting the formula as $(r^M-r^n)/(r^M-1)$ shows that its limit is $1$, since $r^n/r^M\to0$ and $1/r^M\to0$.

Critical case, $b=d>0$.
Here $p=q$, so $h_n^{(M)}=(M-n)/M=1-n/M$, which tends to $1$.

Supercritical case, $b>d$.
Here $0\le r<1$, so $(r^n-r^M)/(1-r^M)\to r^n=(d/b)^n$. If $d=0$, births are the only jumps and a positive population never reaches zero; the same answer is $0^n=0$.

Boundary cases.
If $b=0$ and $d>0$, every jump from a positive state is downward, so zero is reached after at most $n$ jumps and extinction is certain. If $n=0$, extinction has already occurred at time zero, including when both rates are zero. These cases complete the result under the stated assumption $b+d>0$. $\square$

Worked Calculations

Worked Example: A Supercritical Population

Suppose each individual gives birth at rate $b=4$ and dies at rate $d=1$, and the initial population is $n=2$. In state $k$, the generator entries are $q_{k,k+1}=4k$, $q_{k,k-1}=k$, and $q_{kk}=-5k$, whose sum is $4k+k-5k=0$. The jump probabilities are $p=4/5$ and $q=1/5$, which sum to $1$. Since $b>d$, the extinction probability is

\mathbb{P}_2(T_0<\infty)=\left(\frac{1}{4}\right)^2=\frac{1}{16}.

For a finite-barrier check, take $M=5$. Then $r=1/4$, $r^2=1/16$, and $r^5=1/1024$, so $$ h_2^{(5)} =\frac{1/16-1/1024}{1-1/1024} =\frac{63/1024}{1023/1024} =\frac{63}{1023} =\frac{21}{341}. $$ This is below $1/16$, as expected: reaching zero before level $5$ is more restrictive than eventually reaching zero. As $M$ increases, the barrier probability tends to $1/16$. The formula also checks in the jump recurrence: $h_i=(1/4)^i$ gives $\tfrac45h_{i+1}+\tfrac15h_{i-1}=(1/4)^i$ for every $i\ge1$.

Worked Example: Extinction Is Certain Below the Threshold

Let $b=1$, $d=2$, and $n=2$. In state $k$, the generator entries are $k$, $2k$, and $-3k$, which sum to zero. The jump probabilities are $p=1/3$ and $q=2/3$, with sum $1$, and $r=q/p=2$. For the finite barrier $M=5$, $$ h_2^{(5)}=\frac{2^2-2^5}{1-2^5} =\frac{-28}{-31} =\frac{28}{31}. $$ The value is less than $1$, but as the barrier grows, $$ h_2^{(M)}=\frac{2^M-2^2}{2^M-1}\longrightarrow 1. $$ Thus extinction is certain, even though the population can reach arbitrarily high finite levels before it becomes extinct. The jump recurrence verifies the finite-barrier expression: it has boundary values $h_0^{(5)}=1$, $h_5^{(5)}=0$, and for $1\le i\le4$, $$ \frac13h_{i+1}^{(5)}+\frac23h_{i-1}^{(5)}=h_i^{(5)}, $$ as follows by substituting the form $(2^i-2^5)/(1-2^5)$.

Worked Example: The Critical Case

Let $b=d=1$, with initial population $n=3$. In state $k$, the generator entries are $k$, $k$, and $-2k$, summing to zero. The jump probabilities are $p=q=1/2$, which sum to $1$. For any integer barrier $M>3$, $$ h_3^{(M)}=\frac{M-3}{M}=1-\frac{3}{M}. $$ For example, at $M=6$, this is $3/6=1/2$; at $M=12$, it is $9/12=3/4$. Both are finite-barrier probabilities, not the probability of eventual extinction. Their limit as $M\to\infty$ is $1$, so extinction is certain at the critical balance $b=d$.

Check: The linear expression $h_i^{(M)}=(M-i)/M$ satisfies the recurrence because $$ \frac12\frac{M-(i+1)}{M}+\frac12\frac{M-(i-1)}{M} =\frac{2M-2i}{2M} =\frac{M-i}{M}. $$ It also gives $h_0^{(M)}=1$ and $h_M^{(M)}=0$, as required.

Interpretation and a Common Mistake

The threshold compares per-individual rates, not the total rates at the initial population: in state $k$, both total rates are multiplied by the same $k$, so the jump probabilities remain $b/(b+d)$ and $d/(b+d)$. The initial population affects extinction through the exponent $n$. In the supercritical case, $(d/b)^n$ decreases as $n$ increases, but it remains positive for every finite $n$ when $d>0$. When $d=0$, a positive starting population has zero extinction probability.

Common mistake: treating the critical case as if it had a positive survival chance. Substituting $b=d$ into the supercritical formula and declaring extinction probability $1$ without checking the boundary limit can obscure the reasoning. The finite-barrier calculation settles the case directly: for each $M>n$, $h_n^{(M)}=(M-n)/M$, and this tends to $1$. Another error is to use $(d/b)^n$ when $b<d$; then the expression exceeds $1$, so it cannot be a probability. The correct subcritical limit is $1$, obtained from the finite-barrier formula.

Extinction probability is not the same quantity as expected population size. A statement about the expected size at a fixed time does not, by itself, determine whether the process ever hits zero. The hitting calculation uses the jump probabilities and the boundary at zero. It also distinguishes extinction from survival: in the supercritical case with $d>0$, extinction has probability $(d/b)^n$, and survival has the complementary probability $1-(d/b)^n$.

Check Your Understanding

Use the finite-barrier recurrence and the theorem to answer the following.

For $b=3$, $d=1$, and $n=2$, find the eventual extinction probability and the survival probability.
With $b=d=2$ and $n=1$, write $h_1^{(M)}$ and take its limit as $M\to\infty$.
For $b=1$, $d=3$, and $n=1$, calculate $h_1^{(4)}$ and explain why it is not the eventual extinction probability.
What happens when $b=0$, $d>0$, and $n\ge1$? Justify your answer using the jump directions.
Why is the probability of hitting zero before $M$ no greater than the probability of ever hitting zero?