Linear Birth-Death Rates And Population Models

Population Size and Per-Individual Rates

In a population model, it is useful to separate an individual's event rate from the rate of events in the whole population. If every individual gives rise to a new individual at rate \(\lambda\), then a population of size \(n\) has total birth rate \(n\lambda\). If each individual dies at rate \(\mu\), its total death rate is \(n\mu\). The rates grow with the population because more individuals can experience an event.

Let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on \(S=\{0,1,2,\ldots\}\), defined on a probability space \((\Omega,\mathcal F,\mathbb P)\). Its natural filtration is \(\mathcal F_t=\sigma(X_s:0\le s\le t)\). Suppose \(\lambda,\mu\ge0\) are fixed. From a positive population size \(n\), the chain jumps up by one at rate \(n\lambda\) and down by one at rate \(n\mu\). A jump changes the population by exactly one; in particular, these rates do not describe immigration or simultaneous events. State \(0\) is absorbing because both rates there are zero.

The word “linear” refers to the dependence of each total rate on the current population size. The parameters \(\lambda\) and \(\mu\) are per-individual rates, while \(n\lambda\) and \(n\mu\) are population-level rates. We focus here on the behavior over a short time interval from a specified state, not on a stationary distribution or a long-time limit.

Holding Times and Short-Time Probabilities

At a fixed state \(n\ge1\), write \(c_n=n(\lambda+\mu)\) for the total rate. If \(c_n>0\), the time until the next jump is exponential with rate \(c_n\). Conditional on a jump, it is a birth with probability \(\lambda/(\lambda+\mu)\) and a death with probability \(\mu/(\lambda+\mu)\). These probabilities are meaningful when \(\lambda+\mu>0\); if both parameters are zero, no jump occurs. At state zero there is likewise no jump.

Proof. Fix \(n\ge1\), and put \(c=\lambda+\mu>0\), so the exit rate at \(n\) is \(nc\). The probability of no jump by time \(h\) is \(e^{-nch}=1-nch+O(h^2)\). The probability that the first jump by time \(h\) is a birth is \(\int_0^h n\lambda e^{-ncs}\,ds=\frac{\lambda}{c}(1-e^{-nch})=n\lambda h+O(h^2)\); replacing \(\lambda\) by \(\mu\) gives \(n\mu h+O(h^2)\) for a first death.

It remains to control paths with at least two jumps. The probability of a first jump by time \(h\) is at most \(nch\), using \(1-e^{-x}\le x\) for \(x\ge0\). After that jump, the state is \(n-1\) or \(n+1\), and its exit rate is at most \((n+1)c\). Conditional on the first jump and its time, the probability of another jump in the remaining interval is at most \((n+1)ch\). Consequently the probability of two or more jumps is at most \(nc(n+1)ch^2\), which is \(O(h^2)\). A path ending at a state other than \(n-1,n,n+1\) must have at least two jumps. Also, the probability of ending at \(n+1\) differs from the probability that the first jump is a birth by at most the probability of two or more jumps; the same argument applies to \(n-1\). The probability of ending at \(n\) is the no-jump probability plus a contribution from paths with at least two jumps. These bounds prove all three expansions. At zero both rates vanish, so the chain stays there. \(\square\)

The \(O(h^2)\) terms account for paths with multiple jumps. They are not a claim that such paths are impossible: for a small positive \(h\), a birth followed by a death can occur, but its probability is of smaller order than \(h\).

Generator, Drift, and Local Fluctuation

For a bounded function \(f:S\to\mathbb R\), the generator at state \(n\) is the short-time expected change per unit time: \(\lim_{h\downarrow0}\big(\mathbb E_n[f(X_h)]-f(n)\big)/h\), when this limit exists. The short-time probabilities give the following formula.

Proof. At \(n\ge1\), separate the expectation of \(f(X_h)-f(n)\) according to the endpoint. The probabilities of endpoints \(n+1\) and \(n-1\), divided by \(h\), converge to \(n\lambda\) and \(n\mu\). The no-jump contribution is zero because \(f(X_h)-f(n)=0\) when \(X_h=n\). Since \(f\) is bounded, the absolute contribution from endpoints other than \(n-1,n,n+1\), after division by \(h\), is at most \(2\sup_k|f(k)|\,O(h^2)/h\), which tends to zero. The limit is therefore the displayed expression. At zero the chain stays put, so the limit is zero. \(\square\)

Applying the generator to a function that agrees with the population size around the current state gives the local drift. A bounded function can be chosen to agree with \(k\mapsto k\) at \(n-1,n,n+1\), so this calculation does not require applying the bounded-function result to an unbounded function. The upward jump changes the population by \(+1\), and the downward jump by \(-1\); the rate-weighted mean change is \(n\lambda-n\mu\).

One can also see the variance-rate formula through the local square function. For \(k\mapsto k^2\), the changes at \(n+1\) and \(n-1\) are \(2n+1\) and \(-2n+1\). Thus the local generator expression is \(n\lambda(2n+1)+n\mu(-2n+1)\). Subtracting \(2n\) times the local drift \(n(\lambda-\mu)\) leaves \(n(\lambda+\mu)\). Equivalently, this is the rate-weighted squared jump size. This describes instantaneous fluctuation even when the average local change is zero.

Worked Rate Calculations

Interpretation and a Common Pitfall

The sign of \(\lambda-\mu\) determines the sign of the local drift at every positive population size: it is positive when births are faster per individual, negative when deaths are faster, and zero when the per-individual rates match. The magnitude of the local drift scales with \(n\). The fluctuation rate also scales with \(n\), and it depends on the sum \(\lambda+\mu\), not the difference. These statements describe local behavior from the current state; they do not alone determine the population's distribution at a later time.

Common mistake: using per-individual rates as total rates. If \(n=4\), \(\lambda=\tfrac12\), and \(\mu=\tfrac14\), it is wrong to say that the total birth rate is \(\tfrac12\) and the total death rate is \(\tfrac14\). Those are per-individual rates. The correct totals are \(4\cdot\tfrac12=2\) births per unit time and \(4\cdot\tfrac14=1\) death per unit time. The local drift is \(2-1=1\), and the variance rate is \(2+1=3\). The diagonal generator entry is \(-3\), and the row sum \(2+1-3=0\) confirms the calculation.