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Stochastic Models in Biology, Finance and Epidemics · Tutorial 959 of 994

From Population Mechanisms to Birth And Death Rates

Translate individual-level event mechanisms into state-dependent population transition rates, and check the resulting generator at each state.

Intermediate 10 min read

What You'll Learn

  • Distinguish an individual’s per-capita hazard from the total rate for a population.
  • Add rates when separate mechanisms produce the same population transition.
  • Derive first-event probabilities from competing exponential clocks.
  • Translate pairwise interactions into population-wide density-dependent rates.
  • Construct and check a finite-state birth-and-death generator.

From Events to Population Rates

A population model starts with mechanisms: an individual may give birth at a certain hazard, die at a certain hazard, or take part in an interaction that can change the population. A continuous-time population process does not record which person acted; it records the resulting population size. The modeling step is to add the rates of all mechanisms that produce the same change in that size. In particular, an individual-level hazard is not automatically the rate for the whole population.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a continuous-time Markov chain with state space \(S\subseteq\{0,1,2,\ldots\}\), where \(N_t=n\) means there are \(n\) individuals at time \(t\). Use the natural filtration \(\mathcal{F}_t=\sigma(N_s:0\le s\le t)\). A rate \(r\) means an event occurs at instantaneous hazard \(r\), conditional on the current state and the specified event mechanism. We assume rates are finite at each state considered. For a birth-and-death model, each event changes the count by one: a birth takes \(n\) to \(n+1\), and a death takes \(n\) to \(n-1\).

Definition: For distinct states \(i,j\), the transition rate from \(i\) to \(j\) is \(q_{ij}=\lim_{h\downarrow0}\mathbb{P}_i(N_h=j)/h\), when this limit exists. The generator is \(Q=(q_{ij})\), with \(q_{ii}=-\sum_{j\ne i}q_{ij}\) when the sum is finite. Thus each generator row sums to zero. In a birth-and-death model, write \(\lambda_n=q_{n,n+1}\) for the total birth rate and \(\mu_n=q_{n,n-1}\) for the total death rate; at \(n=0\), set \(\mu_0=0\).

These are population-level rates. If each of \(n\) individuals has a birth hazard \(b_n\), then the total birth rate is \(n b_n\), provided each birth adds one individual and the individual hazards are the stated conditional hazards. If each individual has a death hazard \(d_n\), the total death rate is \(n d_n\). The per-capita rates \(b_n,d_n\) may depend on the current population size. Distinct mechanisms leading to the same change must also be added: for example, ordinary deaths and interaction-caused deaths both contribute to \(q_{n,n-1}\).

Why Rates Add

The elementary calculation behind rate aggregation is the competing-clock calculation. Suppose that, at a fixed state, there are finitely many possible event mechanisms. Mechanism \(\ell\) has an exponential waiting time with rate \(r_\ell\), and the clocks are conditionally independent given the current state. Let \(r=\sum_\ell r_\ell\). Then the probability that no event occurs during an interval of length \(h\) is the product of the probabilities that none of the clocks rings. The probability that a particular mechanism rings first is proportional to its rate.

Theorem (Competing event rates): Under the finite, conditionally independent exponential-clock assumptions above, the time to the first event is exponential with rate \(r=\sum_\ell r_\ell\). If \(r>0\), the probability that mechanism \(\ell\) is first is \(r_\ell/r\). If several mechanisms cause the same transition, their rates add to give that transition’s rate.

Proof. Write \(T_\ell\) for the waiting time of mechanism \(\ell\), and \(T=\min_\ell T_\ell\). For \(t\ge0\), independence and the exponential survival formula give \(\mathbb{P}(T>t)=\prod_\ell e^{-r_\ell t}=e^{-rt}\). This is the survival function of an exponential waiting time of rate \(r\); if \(r=0\), every \(r_\ell=0\) and no event occurs. When \(r>0\), the probability density that mechanism \(\ell\) rings at time \(t\) and every other clock is still silent is \(r_\ell e^{-r_\ell t}\prod_{k\ne\ell}e^{-r_k t}=r_\ell e^{-rt}\). Integrating over \(t\ge0\) gives \(\int_0^\infty r_\ell e^{-rt}\,dt=r_\ell/r\). If several mechanisms lead to the same state \(j\), the events that each is first are disjoint, so their probabilities add. Equivalently, their first-event densities add to \((\sum_{\ell:\,\ell\text{ leads to }j}r_\ell)e^{-rt}\). The rate for that transition is therefore the sum of those mechanism rates. \(\square\)

For the nearest-neighbor population models used below, the small-time interpretation of these rates can also be checked directly. From a fixed state \(n\), the only possible first destinations are \(n-1\) and \(n+1\), and the rates at those states are finite. Let \(r_n=\lambda_n+\mu_n\), and let \(M\) be the maximum total rate at the possible first destinations. The chance of a first jump by time \(h\) is at most \(r_n h\), since \(1-e^{-r_n h}\le r_n h\). Conditional on a first jump, the chance of another jump in the remaining time, which is at most \(h\), is at most \(Mh\). Thus the chance of two or more jumps by time \(h\) is at most \(r_n M h^2\). This local bound justifies ignoring multiple jumps in the first-order rate calculation for these nearest-neighbor models. It uses their finite set of possible first destinations; it is not an assertion about arbitrary countable-state chains.

Worked Examples

Worked Example: Constant Per-Capita Hazards

Suppose the population currently has \(n=3\) individuals. Each individual gives birth at rate \(b=\tfrac14\) and dies at rate \(d=\tfrac16\), independently conditional on the current state. Compute the total birth and death rates, the holding rate, and the probability the next event is a birth.

There are three birth clocks, each of rate \(\tfrac14\), so \(\lambda_3=3(\tfrac14)=\tfrac34\). There are three death clocks, each of rate \(\tfrac16\), so \(\mu_3=3(\tfrac16)=\tfrac12\). The total rate out of state \(3\) is \(r_3=\tfrac34+\tfrac12=\tfrac54\). The mean holding time is \(1/r_3=\tfrac45\). By the competing-rate result, the probability that the next event is a birth is \(\lambda_3/r_3=(\tfrac34)/(\tfrac54)=\tfrac35\); the probability it is a death is \(\mu_3/r_3=(\tfrac12)/(\tfrac54)=\tfrac25\).

Check: The event-type probabilities sum to \(\tfrac35+\tfrac25=1\), and the generator row at state \(3\) has entries \(q_{3,2}=\tfrac12\), \(q_{3,3}=-\tfrac54\), and \(q_{3,4}=\tfrac34\). Their sum is \(\tfrac12-\tfrac54+\tfrac34=0\), as required.

Worked Example: Pairwise Competition Adds a Density-Dependent Death Rate

Suppose there are four individuals. Each has a background death hazard \(\tfrac15\). In addition, each unordered pair has an interaction clock of rate \(\tfrac1{10}\); when it rings, exactly one member of that pair dies, chosen with equal probability. Find the total death rate, and separate the background and interaction contributions.

There are \(\binom42=4\cdot3/2=6\) unordered pairs. The rate of interaction events is therefore \(6(\tfrac1{10})=\tfrac35\). Every such event removes one individual, so all interaction events contribute to the transition from \(4\) to \(3\). The background death rate is \(4(\tfrac15)=\tfrac45\), giving a total death rate of \(\tfrac35+\tfrac45=\tfrac75\).

To check the per-individual interpretation of the interaction part, fix one individual. It belongs to three pairs. In each of those pairs, the interaction clock rings at rate \(\tfrac1{10}\), and this individual is selected with probability \(\tfrac12\). Its interaction-caused death hazard is \(3(\tfrac1{10})(\tfrac12)=\tfrac3{20}\). Summing over four individuals gives \(4(\tfrac3{20})=\tfrac35\), the same population-wide interaction rate.

Check: The total death rate from the pair count is \(\binom42(\tfrac1{10})+\tfrac45=\tfrac35+\tfrac45=\tfrac75\). From the individual calculation it is \(4(\tfrac3{20}+\tfrac15)=4(\tfrac3{20}+\tfrac4{20})=\tfrac{28}{20}=\tfrac75\). The two calculations agree.

Worked Example: A Finite-Capacity Population

Let the population size be restricted to \(S=\{0,1,2,3,4\}\). At size \(n\), each individual gives birth at per-capita rate \(b_n=\tfrac12(1-\tfrac n4)\), and each dies at per-capita rate \(d_n=\tfrac14\). A birth adds one individual; at \(n=4\), the birth hazard is zero. Compute all total rates and write the generator in the state order \(0,1,2,3,4\).

The total rates are \(\lambda_n=n b_n\) and \(\mu_n=n d_n\), with \(\mu_0=0\). At \(n=0\), both rates are \(0\). At \(n=1\), \(\lambda_1=1(\tfrac12)(\tfrac34)=\tfrac38\), \(\mu_1=\tfrac14\), and the diagonal rate is \(-(\tfrac38+\tfrac14)=-\tfrac58\). At \(n=2\), \(\lambda_2=2(\tfrac12)(\tfrac12)=\tfrac12\), \(\mu_2=\tfrac12\), and the diagonal rate is \(-1\). At \(n=3\), \(\lambda_3=3(\tfrac12)(\tfrac14)=\tfrac38\), \(\mu_3=\tfrac34\), and the diagonal rate is \(-\tfrac98\). At \(n=4\), \(\lambda_4=4(\tfrac12)(0)=0\), \(\mu_4=1\), and the diagonal rate is \(-1\).

$$ Q= \begin{bmatrix} 0&0&0&0&0\\ \tfrac14&-\tfrac58&\tfrac38&0&0\\ 0&\tfrac12&-1&\tfrac12&0\\ 0&0&\tfrac34&-\tfrac98&\tfrac38\\ 0&0&0&1&-1 \end{bmatrix}. $$

Check: The five row sums are respectively \(0\), \(\tfrac14-\tfrac58+\tfrac38=0\), \(\tfrac12-1+\tfrac12=0\), \(\tfrac34-\tfrac98+\tfrac38=0\), and \(1-1=0\). In particular, state \(4\) has no transition to state \(5\), because its per-capita birth hazard is zero.

Rate Translation as a Modeling Rule

The examples use two different sources of population-wide rates. In the first, many individual clocks have the same constant per-capita rate, so the total is the number of individuals times that rate. In the second, pairwise mechanisms contribute a number of clocks proportional to the number of possible pairs. As the population changes, the number of active clocks changes too; this is one way a constant mechanism at the individual or pair level creates a state-dependent population rate.

More generally, to construct rates from a proposed mechanism, first specify the current state and the event that occurs. Then count the eligible individuals, pairs, or other event opportunities; multiply by the hazard per opportunity; and assign the resulting rate to the state transition caused by that event. If multiple mechanisms cause the same transition, add their rates. Finally, set the diagonal generator entry to minus the sum of all off-diagonal rates in that row. This procedure is local: it determines the instantaneous rates at each state, not by itself long-term behavior or the existence of a non-explosive process on an unbounded state space.

Common mistake. If each of four individuals has birth hazard \(\tfrac15\), it is incorrect to set the population birth rate to \(\tfrac15\). That is the hazard for one individual. The population has four birth clocks, so the total rate is \(4(\tfrac15)=\tfrac45\). The same distinction applies to deaths. Also, adding per-capita rates before accounting for the number of individuals can give the wrong answer: in the pairwise example, the interaction rate is based on six pairs, not four individuals each with a full \(\tfrac1{10}\) hazard.

Check Your Understanding

For each question, distinguish the rate per event opportunity from the total population rate, and check any generator row by summing its entries.

  1. If five individuals each give birth at rate \(\tfrac13\), what is the total birth rate? What is the probability the next event is a birth if the total death rate is \(2\)?
  2. At population size \(n\), each individual dies at rate \(d\). Explain why the total death rate is \(nd\), and state what happens at \(n=0\).
  3. At population size \(5\), every unordered pair triggers an event at rate \(c\), and every event removes exactly one individual. What is the total interaction-caused death rate?
  4. For a state with birth rate \(\lambda_n=\tfrac23\) and death rate \(\mu_n=\tfrac13\), write the three nonzero entries of the generator row and verify their sum.
  5. Why must rates from two distinct mechanisms be added when both mechanisms cause the same one-step population change?