Linear Birth-Death Models and Expected Population Growth

From Per-Capita Rates to Mean Population

In a linear birth-death model, a larger population has more opportunities both to produce births and to experience deaths. If each individual gives birth at rate \(b\) and dies at rate \(d\), then a population of size \(n\) has total birth rate \(bn\) and total death rate \(dn\). These event rates determine how the population changes on average, but the mean is not obtained by following one smooth, deterministic population path: the population remains a random jump process.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a continuous-time Markov chain with state space \(S=\{0,1,2,\ldots\}\), natural filtration \(\mathcal{F}_t=\sigma(N_s:0\le s\le t)\), and fixed initial population \(N_0=n_0\), where \(n_0\) is a nonnegative integer. For \(n\ge1\), its only possible transitions are from \(n\) to \(n+1\) at rate \(bn\) and from \(n\) to \(n-1\) at rate \(dn\), where \(b,d\ge0\). State \(0\) is absorbing. The rates are per unit time, so \(b\) and \(d\) have units of inverse time.

The Mean-Population Equation

Proof. We first justify the expectation calculation on the unbounded state space. For an integer \(K>n_0\), let \(\tau_K=\inf\{t\ge0:N_t\ge K\}\) and stop the chain on first reaching \(K\). The stopped chain has finite state space \(\{0,\ldots,K\}\), with \(K\) absorbing. For a finite-state continuous-time Markov chain with generator \(Q_K\), the identity $$ \mathbb{E}[f(N_{t\wedge\tau_K})] =f(n_0)+\int_0^t\mathbb{E}\!\left[(Q_Kf)(N_{s\wedge\tau_K})\right]\,ds $$ holds for any function \(f\) on that finite state space. Indeed, its transition matrix is \(e^{tQ_K}\); differentiating the finite matrix expression for the expectation gives the integrand \(Q_K f\), and integrating that derivative gives the identity. Before \(\tau_K\), the generator applied to \(f(n)=n\) is \((b-d)n\), and at the stopped state its value is zero. Thus $$ \mathbb{E}[N_{t\wedge\tau_K}] =n_0+(b-d)\int_0^t\mathbb{E}[N_s\mathbf{1}_{\{s<\tau_K\}}]\,ds \le n_0+b\int_0^t\mathbb{E}[N_{s\wedge\tau_K}]\,ds. $$ The inequality uses \(b-d\le b\), nonnegativity of the integrand, and \(N_s\mathbf{1}_{\{s<\tau_K\}}\le N_{s\wedge\tau_K}\). Gronwall's inequality gives \(\mathbb{E}[N_{t\wedge\tau_K}]\le n_0e^{bt}\). Consequently, \(\mathbb{P}(\tau_K\le t)\le n_0e^{bt}/K\), since \(N_{t\wedge\tau_K}=K\) on \(\{\tau_K\le t\}\). This bound tends to zero as \(K\) tends to infinity. An infinite number of jumps while the population remains bounded is also impossible: on any bounded set of states the total jump rate is bounded by \((b+d)K\). Hence the chain is nonexplosive.

We also need enough control to pass expectations through the limit. Apply the same finite-state identity to \(f(n)=n^2\). Before stopping, $$ (Q_Kf)(n)=bn(2n+1)+dn(-2n+1) =2(b-d)n^2+(b+d)n \le 2b n^2+(b+d)n. $$ At the stopped state the generator contribution is zero. Write \(Z_K(t)=N_{t\wedge\tau_K}\). The first-moment bound above therefore gives $$ \mathbb{E}[Z_K(t)^2] \le n_0^2+\int_0^t\left(2b\,\mathbb{E}[Z_K(s)^2]+(b+d)n_0e^{bs}\right)\,ds. $$ Gronwall's inequality bounds \(\mathbb{E}[Z_K(t)^2]\) uniformly in \(K\) on every bounded time interval. Thus \(Z_K(t)\) is uniformly integrable there. Nonexplosion implies \(Z_K(t)\to N_t\) almost surely as \(K\to\infty\), so \(\mathbb{E}[Z_K(t)]\to\mathbb{E}[N_t]\). Also, the first-moment bounds and monotone convergence give convergence of the time integrals in the stopped first-moment identity. We obtain $$ m(t)=n_0+(b-d)\int_0^t m(s)\,ds,\qquad m(t)=\mathbb{E}[N_t]. $$ The right side makes \(m\) continuous and then differentiable, with \(m'(t)=(b-d)m(t)\) and \(m(0)=n_0\). Multiplying by \(e^{-(b-d)t}\) shows that \(e^{-(b-d)t}m(t)\) has derivative zero, so \(m(t)=n_0e^{(b-d)t}\). If \(n_0=0\), the absorbing state gives \(m(t)=0\), in agreement with the formula. If \(b=d\), the same argument gives the constant solution \(m(t)=n_0\). \(\square\)

The exponent \(r=b-d\) is the net per-capita growth rate of the mean. When \(r>0\), the expected population grows exponentially; when \(r<0\), it declines exponentially; and when \(r=0\), it stays constant. In the last case, constant expectation does not mean that the population size is constant: births and deaths may continue, and their average effects cancel. Nor does exponential growth of the mean say that every realized population grows. The equation describes an expectation over possible paths.

Worked Regimes

Why the Equation Is Useful—and a Common Error

The mean equation gives a concise summary of average population growth and lets us compare rates without simulating individual paths. Its exponential form also makes clear that the sign of \(b-d\), not the birth rate alone, determines whether the mean increases or decreases. The equation depends on linear rates: at size \(n\), each total rate must be proportional to \(n\). If rates change with population size, a constant exponential rate need not describe the mean; such models require a different analysis.

Common mistake: forgetting that total event rates scale with population size. For \(N_0=3\), \(b=2\), and \(d=1\), an incorrect calculation might use the per-capita difference \(b-d=1\) as a constant population drift and write \(m'(t)=1\), leading to \(m(t)=3+t\). The correction is to multiply the per-capita net rate by the current population: births contribute \(2m(t)\) and deaths contribute \(-m(t)\), so \(m'(t)=2m(t)-m(t)=m(t)\). The correct solution is \(m(t)=3e^t\). At \(t=\ln 2\), it gives \(3e^{\ln 2}=6\); the incorrect expression gives \(3+\ln 2\), which does not satisfy the mean equation \(m'=m\).

Keep three interpretations separate: \(b\) and \(d\) are per-individual event rates; \(b-d\) is the net per-capita rate of change in the mean; and \(n_0e^{(b-d)t}\) is the expected population at time \(t\). In particular, the critical case means zero net mean drift, not an absence of events or a guarantee of a fixed population size.