Branching with General Lifetimes
In a Galton–Watson process, generations are the natural clock: each individual produces offspring to form the next generation. In many applications, however, the timing of reproduction matters. An individual may reproduce only after living for a random length of time, so two populations with the same mean offspring number can grow at different rates if their lifetime distributions differ. The equation that connects these ingredients is the Euler–Lotka equation.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Each individual has a lifetime \(L\) with distribution \(F\) on \((0,\infty)\), and an offspring count \(\xi\) with mean \(m=\mathbb{E}\xi<\infty\). Assume that the lifetime and offspring count are independent, and that individuals’ pairs \((L,\xi)\) are independent and identically distributed. An individual lives for \(L\) time units, then dies and produces \(\xi\) children, all born at that time. The initial population consists of one newborn at time \(0\). Write \(Z(t)\) for the number alive at time \(t\ge0\); it takes values in \(S=\{0,1,2,\ldots\}\). This is a continuous-time population process, but \(Z(t)\) alone need not be Markov: with general lifetimes, the ages of living individuals can affect their future reproduction.
Let generation \(0\) consist of the initial individual, and define generation \(n+1\) to be the children of generation \(n\). Let \(S_n\) be the birth time of a generation-\(n\) individual along its ancestral line, so \(S_0=0\). The Malthusian parameter describes the exponential weighting at which the expected contributions of successive generations balance. In a supercritical model, it is usually a positive number.
Why the Equation Holds
Suppose a birth cohort grows in an exponential mode with rate \(r\): a cohort born \(s\) time units earlier contributes a factor proportional to \(e^{-rs}\) relative to the current cohort. An individual produces an expected \(m\) children at age \(s\) according to the lifetime law \(F(ds)\). The discounted expected number of children contributed by one individual is therefore \(m\int e^{-rs}F(ds)\). A self-sustaining exponential mode requires this discounted contribution to equal one.
Proof. For generation \(0\), the sum has one term \(e^{-rS_0}=1\), so its expectation is \(1\). Given a generation-\(n\) individual born at time \(s\), its children are all born at time \(s+L\). The conditional expected sum of the children’s weights is $$ \mathbb{E}\left[\xi e^{-r(s+L)}\right] =e^{-rs}\mathbb{E}\left[\xi e^{-rL}\right] =e^{-rs}m\mathbb{E}e^{-rL}, $$ where the last equality uses independence of \(\xi\) and \(L\). Summing this conditional expectation over generation \(n\) and taking expectations shows that the expected weight in generation \(n+1\) is \(m\mathbb{E}e^{-rL}\) times the expected weight in generation \(n\). Starting from \(1\) and iterating gives the displayed formula. The multiplier is one precisely when \(m\mathbb{E}e^{-rL}=1\), which is the Euler–Lotka equation. \(\square\)
This derivation identifies a characteristic exponential rate; it does not, by itself, prove that every population path or every expected population has an asymptotic equivalent proportional to \(e^{rt}\). Such conclusions need additional renewal-theoretic hypotheses. The root nevertheless gives the standard growth-rate calculation, and for some lifetime laws the expected population can be solved exactly.
Proof. Define \(\phi(r)=m\mathbb{E}e^{-rL}\) for \(r\ge0\). Since \(0\le e^{-rL}\le1\), dominated convergence shows that \(\phi\) is continuous on \([0,\infty)\). Also \(\phi(0)=m>1\). Since \(L>0\) almost surely, \(e^{-rL}\) tends to zero almost surely as \(r\) tends to infinity; dominated convergence then gives \(\phi(r)\to0\). The intermediate value theorem gives at least one \(r>0\) with \(\phi(r)=1\). If \(r_2>r_1\ge0\), then \(e^{-r_2L}<e^{-r_1L}\) almost surely, so \(\phi(r_2)<\phi(r_1)\). Thus \(\phi\) is strictly decreasing and can equal one at most once. \(\square\)
If \(m=1\), the equation has the solution \(r=0\), and strict decrease means there is no positive solution. If \(m<1\), then \(m\mathbb{E}e^{-rL}\le m<1\) for every \(r\ge0\), so there is no nonnegative solution. These statements concern nonnegative roots; a negative root would require the corresponding exponential moment of \(L\) to be finite.
Worked Growth-Rate Calculations
Worked Example: A Deterministic Lifetime
Every individual lives exactly \(2\) time units and then has an expected \(4\) children. Find the Malthusian parameter.
Here \(m=4\) and \(F\) places probability one at \(2\), so the Euler–Lotka equation is $$ 4e^{-2r}=1. $$ Dividing by \(4\) gives \(e^{-2r}=\tfrac14\). Taking logarithms gives \(-2r=\log(\tfrac14)=-\log 4\), and hence $$ r=\frac{\log 4}{2}=\log 2. $$
Check: Substituting \(r=\log 2\) gives \(4e^{-2\log 2}=4e^{-\log 4}=4\cdot\tfrac14=1\), as required. There is also a direct generation check: at time \(2n\), the expected size of generation \(n\) is \(4^n\), and \(e^{r(2n)}=e^{2n\log 2}=4^n\). The generations occur at fixed time intervals, which is why the rate is \(\log 4\) divided by the lifetime \(2\).
Worked Example: Exponential Lifetimes
Suppose lifetimes are exponential with rate \(\lambda=3\), and each individual has mean offspring number \(m=2\). Find the Malthusian parameter and verify the expected population directly.
The lifetime density is \(f(s)=3e^{-3s}\) for \(s>0\). Its Laplace transform at \(r\ge0\) is $$ \mathbb{E}e^{-rL}=\int_0^\infty e^{-rs}3e^{-3s}\,ds =3\int_0^\infty e^{-(r+3)s}\,ds =\frac{3}{r+3}. $$ The Euler–Lotka equation becomes \(2\cdot\frac{3}{r+3}=1\). Thus \(6=r+3\) and \(r=3\). More generally, for exponential rate \(\lambda>0\) and \(m>1\), the same calculation gives \(m\lambda/(\lambda+r)=1\), hence \(r=\lambda(m-1)\).
For a direct check, let \(z(t)=\mathbb{E}Z(t)\). The initial individual is alive at time \(t\) with probability \(e^{-3t}\). If it dies at time \(s\le t\), it has mean \(2\) children, each with expected population \(z(t-s)\) at time \(t\). Conditioning on its death time therefore gives $$ z(t)=e^{-3t}+2\int_0^t z(t-s)3e^{-3s}\,ds. $$ Try \(z(t)=e^{3t}\). Substitution into the right-hand side gives $$ e^{-3t}+6\int_0^t e^{3(t-s)}e^{-3s}\,ds =e^{-3t}+6e^{3t}\int_0^t e^{-6s}\,ds =e^{-3t}+e^{3t}(1-e^{-6t}) =e^{3t}. $$ Thus the proposed function satisfies the equation. In this exponential-lifetime model the mean population is exactly \(e^{3t}\), not only an exponential mode. The integral equation also checks the initial condition: at \(t=0\), \(z(0)=1\).
Worked Example: Two Possible Lifetime Lengths
An individual lives for \(1\) or \(2\) time units, each with probability \(\tfrac12\), and has mean offspring number \(2\). Find the positive Malthusian parameter.
The lifetime transform is \(\tfrac12e^{-r}+\tfrac12e^{-2r}\). The Euler–Lotka equation is $$ 2\left(\tfrac12e^{-r}+\tfrac12e^{-2r}\right)=1, $$ so \(e^{-r}+e^{-2r}=1\). Put \(x=e^{-r}\). A positive \(r\) corresponds to \(0<x<1\), and the equation becomes \(x^2+x-1=0\). Its roots are \(\frac{-1+\sqrt5}{2}\) and \(\frac{-1-\sqrt5}{2}\); only the first is positive. Therefore $$ r=-\log\left(\frac{\sqrt5-1}{2}\right) =\log\left(\frac{1+\sqrt5}{2}\right). $$
Check: For \(x=(\sqrt5-1)/2\), direct expansion gives \(x^2=(3-\sqrt5)/2\), so \(x+x^2=(\sqrt5-1+3-\sqrt5)/2=1\). Thus the original Euler–Lotka left side is \(x+x^2=1\). Also \(0<x<1\), so \(r=-\log x>0\), as expected for \(m>1\).
Interpreting the Parameter Carefully
The lifetime distribution enters through its entire Laplace transform, not just through its mean. For a deterministic lifetime, each generation arrives at a fixed delay; for an exponential lifetime, reproduction times are spread continuously. These distinct timing patterns explain why a formula based only on \(m\) and \(\mathbb{E}L\) will not generally work.
Common mistake. A tempting but generally incorrect shortcut is to use \(r=(m-1)/\mathbb{E}L\). In the deterministic example, \(m=4\) and \(\mathbb{E}L=2\), so this shortcut gives \(r=\tfrac32\). But substituting into the actual Euler–Lotka equation gives \(4e^{-2(3/2)}=4e^{-3}\), which is not \(1\). The correct rate is \(\log 2\), obtained by solving the lifetime transform equation. The shortcut happens to agree with the exponential-lifetime formula, but that does not make it valid for general \(F\).
A second caution is to distinguish the root equation from an asymptotic theorem. The Euler–Lotka equation specifies the exponential mode through the discounted expected offspring contribution. Whether a particular expected population has a precise asymptotic form \(C e^{rt}\), or whether sample paths have a corresponding growth limit, requires further assumptions and results. In applications, first calculate the characteristic rate from the stated lifetime law, then check separately what asymptotic conclusion the model and the available renewal results justify.
Check Your Understanding
Use the Euler–Lotka equation in each calculation, and check any proposed root by substitution.
- If every individual lives exactly \(3\) time units and has mean \(8\) children, find the Malthusian parameter.
- For exponential lifetimes of rate \(\lambda\) and mean offspring number \(m>1\), derive the positive root from the lifetime transform.
- If \(m=1\) and lifetimes are strictly positive, explain why there is no positive Malthusian parameter.
- For lifetimes \(1\) and \(3\), each with probability \(\tfrac12\), and \(m=2\), write the equation for the positive parameter and explain why it has a unique solution.
- Why does solving the Euler–Lotka equation alone not prove that every sample path grows asymptotically at rate \(r\)?