Density Dependence in Finite-Population Birth-Death Models

Density Dependence in a Finite Population

When resources become scarce, the per-individual birth rate may fall as population size grows. A finite-population birth-death model represents this mechanism by allowing births to become less frequent near a carrying-capacity level, while deaths continue. Its deterministic approximation can have a stable positive equilibrium, but that equilibrium does not make the stochastic population stay there: random upward and downward jumps remain possible, and in this model zero is absorbing.

Fix an integer \(K\ge1\), and let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on the probability space \((\Omega,\mathcal F,\mathbb P)\), with state space \(S=\{0,1,\ldots,K\}\) and filtration \(\mathcal F_t=\sigma(X_s:0\le s\le t)\). The state \(n\) is the number of individuals. For parameters \(\beta>0\) and \(\delta>0\), set the birth and death rates to

From state \(n\), the chain moves to \(n+1\) at rate \(\lambda_n\) and to \(n-1\) at rate \(\mu_n\), when those states are in \(S\). A rate is the parameter of the corresponding exponential waiting clock. At \(n=0\), both rates are zero. At \(n=K\), the birth rate is zero, but the death rate is \(\delta K>0\).

The parameter \(K\) is the level at which density-dependent births cease. It is not generally the positive equilibrium of the deterministic population equation. The distinction matters because deaths reduce the population even below the birth-saturation level.

Drift and the Deterministic Equilibrium

For a differentiable population-size approximation \(x(t)\), using the birth and death rates as the expected upward and downward changes gives the ordinary differential equation

If \(\beta>\delta\), the equilibria are \(x=0\) and \(x^*=K(1-\delta/\beta)\). The positive equilibrium lies strictly between \(0\) and \(K\). If \(\beta\le\delta\), there is no positive equilibrium: for every \(x>0\), \(f(x)<0\). These statements concern the deterministic equation, not a claim that the stochastic chain settles at \(x^*\).

Proof. Factor the drift as \(f(x)=\frac{\beta}{K}x(x^*-x)\). For \(0<x<x^*\), \(f(x)>0\), so a solution increases while it is below \(x^*\). For \(x>x^*\), \(f(x)<0\), so a solution decreases while it is above \(x^*\). Uniqueness for this polynomial differential equation prevents a solution from crossing the equilibrium solution \(x(t)=x^*\). Therefore a solution starting below \(x^*\) is increasing and bounded above by \(x^*\); one starting above is decreasing and bounded below by \(x^*\). In either case it has a limit \(L>0\). If \(f(L)\ne0\), continuity of \(f\) makes the derivative eventually bounded away from zero with a fixed sign, contradicting convergence to \(L\). Hence \(f(L)=0\). The only positive zero of \(f\) is \(x^*\), so \(L=x^*\). Starting exactly at zero gives \(x(t)=0\). The sign of the drift also shows that the distance to \(x^*\) cannot increase along a solution starting near \(x^*\). \(\square\)

The stochastic chain has a corresponding instantaneous drift. For a function \(g:S\to\mathbb R\), its generator acts by \((Qg)(n)=\sum_jq_{nj}(g(j)-g(n))\). Taking \(g(n)=n\) gives \((Qg)(n)=\lambda_n-\mu_n\), the expected instantaneous change in population size. The rate of squared jumps is \(\lambda_n+\mu_n\), because each birth or death changes the population by one. These quantities distinguish the deterministic drift from stochastic variation: zero drift does not mean zero jumps.

Random Movement Near Equilibrium

In continuous time, a state with equal birth and death rates is not a state in which the chain stops. If both rates are positive, the next jump is still either upward or downward. Given state \(n\), the holding time before the next jump is exponential with rate \(\lambda_n+\mu_n\). Conditional on a jump, the probability it is a birth is \(\lambda_n/(\lambda_n+\mu_n)\), and the probability it is a death is \(\mu_n/(\lambda_n+\mu_n)\), provided the total rate is positive.

The deterministic equilibrium describes where the drift changes sign; it is not a stationary population size for each realization. In particular, with a finite population and no immigration, the chain can eventually reach zero even when the deterministic equation has a positive stable equilibrium.

Why Zero Is Absorbing, and Why Extinction Occurs

At zero, the birth rate \(\beta n(1-n/K)\) and death rate \(\delta n\) both vanish. The process therefore has no transition out of zero. For this model, positive death rates at every nonzero state also imply eventual extinction, not merely the possibility of extinction.

Proof. If the chain starts at zero, the claim holds immediately. From a positive state \(n\), the total rate \(\lambda_n+\mu_n\) is positive, and the probability that the next jump is downward is \(d_n=\mu_n/(\lambda_n+\mu_n)>0\). Define \(\eta=\prod_{n=1}^K d_n\), which is positive because this is a finite product of positive numbers. Starting at any \(i>0\), the probability of taking consecutive downward jumps until reaching zero is \(\prod_{n=1}^i d_n\ge\eta\); the inequality holds because the omitted factors lie in \((0,1]\).

Consider successive blocks of \(K\) jumps, as long as absorption has not occurred. From any positive state, the conditional probability of absorption within the next \(K\) jumps is at least \(\eta\). Therefore the probability of remaining unabsorbed after \(m\) such blocks is at most \((1-\eta)^m\), which tends to zero. The number of jumps before absorption is thus finite almost surely. Each holding time before one of these finitely many jumps is finite almost surely, since its exponential rate is positive. Hence the absorption time is finite almost surely. Finally, if \(T_0=\inf\{t\ge0:X_t=0\}\), then \(\mathbb P(X_t\ne0)=\mathbb P(T_0>t)\to0\) as \(t\to\infty\), since \(T_0<\infty\) almost surely. \(\square\)

Finiteness of the state space and positive death rates are important conditions. If the death-rate condition is removed, a positive state need not lead to extinction. For example, on \(\{0,1,2\}\), take \(\lambda_1=1\), \(\lambda_2=0\), and \(\mu_1=\mu_2=0\). Starting at \(1\), the chain moves to \(2\) and then stays there; it does not reach zero. This is why the extinction conclusion cannot be inferred from the existence of an absorbing zero state alone.

A Common Misreading

A common error is to call \(K\) the deterministic equilibrium without accounting for deaths. With \(K=10\), \(\beta=1\), and \(\delta=\tfrac12\), this incorrect calculation would substitute \(x=10\) and claim zero net growth because the birth rate is zero there. In fact, the deterministic drift at \(x=10\) is \(0-\tfrac12(10)=-5\), not zero. The positive equilibrium solves \(\beta(1-x/K)=\delta\), giving \(x=10(1-\tfrac12)=5\). The level \(K\) is where density-dependent births vanish; the positive equilibrium is where births and deaths balance.

The same distinction prevents another mistake: zero drift at the positive equilibrium does not imply that the stochastic population is fixed there. At that state, births and deaths can have equal, positive rates. The chain continues to jump, and because zero is absorbing and there is no immigration, its long-run behavior differs from the deterministic equilibrium.