Generating-Function Equations for Continuous-Time Branching

From Exponential Events to a Generating-Function Equation

In a discrete-generation branching process, one generation is replaced by the next in a single step. In a continuous-time model, individuals instead experience reproduction events at random times. The exponential waiting-time assumption makes the population a Markov process: conditional on the current population, future events do not depend on the ages of individuals. The probability generating function then evolves according to a differential equation that records what can happen during a short initial interval.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Each individual waits an independent exponential time with rate $\lambda>0$, then is replaced by $\xi$ offspring. The offspring counts for different individuals and events are independent and have a common distribution $\mathbb{P}(\xi=k)=p_k$, $k\in\{0,1,2,\ldots\}$. Assume throughout that the offspring count is bounded: there is a finite $K$ such that $\mathbb{P}(\xi\le K)=1$. We consider the non-explosive process, meaning that only finitely many reproduction events occur in each bounded time interval. Let $Z_t$ be the number of individuals at time $t$, with state space $S=\{0,1,2,\ldots\}$. State $0$ is absorbing. The time index is continuous, $t\ge0$.

Definition: The offspring probability generating function is $G(u)=\mathbb{E}[u^\xi]=\sum_{k=0}^{K}p_k u^k$, for $0\le u\le1$. Starting with one individual, the population probability generating function at time $t$ is $F_t(s)=\mathbb{E}_1[s^{Z_t}]$, for $0\le s\le1$. We use $0^0=1$ in the generating-function convention, so $F_t(0)=\mathbb{P}_1(Z_t=0)$.

If the process starts with $z$ individuals, their descendant populations evolve independently. Consequently, the generating function of $Z_t$ from $z$ ancestors is $F_t(s)^z$. This independence is the source of the composition rule used below. The bounded-offspring and non-explosion assumptions ensure that the model is well-defined; the equation we derive concerns the process started from one ancestor.

The First-Event Equation and the Backward Equation

Let $\tau$ be the first reproduction time of the initial individual. It has exponential density $\lambda e^{-\lambda u}$ and survival probability $\mathbb{P}(\tau>t)=e^{-\lambda t}$. If $\tau>t$, the population at time $t$ is one. If $\tau=u\le t$, the individual is replaced by $\xi$ children, each with an independent descendant process running for $t-u$ time units. Conditional on $\xi$, the generating function of their total population is $F_{t-u}(s)^\xi$. Averaging over $\xi$ gives $G(F_{t-u}(s))$. Thus, for $0\le s\le1$,

F_t(s)=e^{-\lambda t}s+\int_0^t \lambda e^{-\lambda u}G(F_{t-u}(s))\,du.

This integral equation conditions on the first event time. The differential equation below is called the backward equation because one may instead condition on what happens during a short interval at the beginning of the time horizon.

Theorem (Backward differential equation): Under the assumptions above, for every $s\in[0,1]$, $F_t(s)$ satisfies $$ \frac{\partial}{\partial t}F_t(s)=\lambda\bigl(G(F_t(s))-F_t(s)\bigr),\qquad F_0(s)=s. $$ The equation has at most one solution taking values in $[0,1]$ on any bounded time interval.

Proof. First establish the composition identity. For $t,h\ge0$, condition on the population at time $h$. Each of the $Z_h$ individuals has an independent descendant population over the following $t$ units, with generating function $F_t(s)$. Hence $$ F_{t+h}(s)=\mathbb{E}_1\!\left[F_t(s)^{Z_h}\right]=F_h(F_t(s)). $$ For a fixed $u\in[0,1]$, examine $F_h(u)$ as $h$ decreases to zero. The probability of no event by time $h$ is $e^{-\lambda h}=1-\lambda h+o(h)$, and the probability of one event is $\lambda h+o(h)$. The probability of at least two events is $o(h)$: after the first event, there are at most $K$ individuals, so their total event rate is at most $\lambda K$ until the next event. If one event occurs, the offspring generating function evaluated at $u$ is $G(u)$, up to an error $o(h)$ from the possibility of another event. Therefore $$ F_h(u)=u+\lambda h\bigl(G(u)-u\bigr)+o(h). $$ Substitute $u=F_t(s)$ into the composition identity. It follows that the right derivative in $t$ is $\lambda(G(F_t(s))-F_t(s))$. The first-event integral equation, whose integrand is continuous because $G$ is a polynomial and $F_t(s)$ is continuous, gives continuity in $t$; the same right-derivative identity at every $t$ yields the stated differential equation. At $t=0$, $Z_0=1$, so $F_0(s)=s$.

For uniqueness, $G$ is a polynomial and is Lipschitz on $[0,1]$: there is a finite constant $L$ such that $|G(x)-G(y)|\le L|x-y|$ there. If $f$ and $g$ are two solutions with the same initial value, then, since both take values in $[0,1]$, $$ |f(t)-g(t)|\le \lambda(L+1)\int_0^t |f(r)-g(r)|\,dr. $$ Set $H(t)=\int_0^t|f(r)-g(r)|\,dr$. Then $H(0)=0$, $H(t)\ge0$, and $H'(t)\le\lambda(L+1)H(t)$. Differentiating $e^{-\lambda(L+1)t}H(t)$ shows its derivative is nonpositive. It starts at zero and is nonnegative, so it is identically zero. Thus $f=g$ on the interval, proving uniqueness. $\square$

Proposition (Branching composition): For $s\in[0,1]$ and $t,h\ge0$, $F_{t+h}(s)=F_h(F_t(s))$.

Proof. Given $Z_h=z$, the $z$ individuals at time $h$ have independent descendant populations during the next $t$ units. The conditional generating function of their total is therefore $F_t(s)^z$, including $z=0$, when it equals $1$. Taking expectation over $Z_h$ gives $\mathbb{E}_1[F_t(s)^{Z_h}]=F_h(F_t(s))$, which is $F_{t+h}(s)$ by conditioning at time $h$. $\square$

Solving the Equation for Simple Reproduction Laws

Worked Example: Every Event Causes Extinction

Suppose $\xi=0$ with probability $1$, and let $\lambda=2$. Then $G(u)=1$, so the backward equation is $$ F_t'(s)=2(1-F_t(s)),\qquad F_0(s)=s. $$ Set $Y_t=1-F_t(s)$. Then $Y_t'=-2Y_t$, with $Y_0=1-s$, so $Y_t=(1-s)e^{-2t}$. Therefore $$ F_t(s)=1-(1-s)e^{-2t}. $$

Check with exact values. Take $t=\ln 2$ and $s=\tfrac13$. Since $e^{-2\ln 2}=\tfrac14$, $$ F_{\ln 2}\!\left(\tfrac13\right)=1-\left(1-\tfrac13\right)\tfrac14 =1-\tfrac23\cdot\tfrac14=\tfrac56. $$ Directly, the population is one if the first event has not occurred, and zero otherwise. The probability of no event by $\ln 2$ is $e^{-2\ln 2}=\tfrac14$, giving generating function $\tfrac14s+\tfrac34$. At $s=\tfrac13$, this is $\tfrac1{12}+\tfrac34=\tfrac{10}{12}=\tfrac56$, as required.

Worked Example: One Offspring or None

Suppose $\mathbb{P}(\xi=1)=r$ and $\mathbb{P}(\xi=0)=1-r$, where $0\le r\le1$. Then $G(u)=1-r+ru$, and $$ F_t'(s)=\lambda(1-r)(1-F_t(s)),\qquad F_0(s)=s. $$ Solving this linear equation gives $$ F_t(s)=1-(1-s)e^{-\lambda(1-r)t}. $$ When $r=1$, this formula gives $F_t(s)=s$: every event replaces one individual by one individual, so the population remains one. When $r=0$, it reduces to the extinction-on-first-event formula.

Check with exact values. Take $r=\tfrac13$, $\lambda=3$, $t=\ln2$, and $s=\tfrac12$. The exponent is $-3(1-\tfrac13)\ln2=-2\ln2$, so $$ F_{\ln2}\!\left(\tfrac12\right)=1-\left(1-\tfrac12\right)\tfrac14 =1-\tfrac18=\tfrac78. $$ The population is either zero or one. It is one at time $t$ precisely when every event so far produced one offspring. The formula assigns probability $e^{-2t}$ to that outcome, which at $t=\ln2$ is $\tfrac14$. The resulting generating function is $\tfrac34+\tfrac14s$, and at $s=\tfrac12$ it equals $\tfrac34+\tfrac18=\tfrac78$.

Worked Example: Each Event Produces Two Offspring

Suppose $\xi=2$ with probability $1$. Then $G(u)=u^2$ and $$ F_t'(s)=\lambda(F_t(s)^2-F_t(s)),\qquad F_0(s)=s. $$ For $0<s<1$, separate variables: $$ \frac{dF}{F(F-1)}=\lambda\,dt,\qquad \frac{1}{F(F-1)}=\frac{1}{F-1}-\frac{1}{F}. $$ Integrating gives $\log\!\left(\frac{1-F_t(s)}{F_t(s)}\right)=\lambda t+\log\!\left(\frac{1-s}{s}\right)$. Solving for $F_t(s)$ yields $$ F_t(s)=\frac{s}{s+(1-s)e^{\lambda t}}. $$ The displayed formula also applies at $s=0$ and $s=1$: its denominator is positive, and it gives $F_t(0)=0$ and $F_t(1)=1$, respectively.

Check by substitution. Let $D=s+(1-s)e^{\lambda t}$, so $F_t=s/D$. Then $$ F_t'=-\frac{\lambda s(1-s)e^{\lambda t}}{D^2}, \qquad F_t^2-F_t=\frac{s^2-sD}{D^2} =-\frac{s(1-s)e^{\lambda t}}{D^2}. $$ Thus $F_t'=\lambda(F_t^2-F_t)$, and $F_0=s/(s+1-s)=s$. As a numerical check, with $\lambda=1$, $t=\ln2$, and $s=\tfrac12$, the denominator is $\tfrac12+\tfrac12\cdot2=\tfrac32$, giving $F_t(s)=\tfrac{1/2}{3/2}=\tfrac13$. The differential-equation right side is $(\tfrac13)^2-\tfrac13=-\tfrac29$. The derivative formula gives $-(\tfrac12)(\tfrac12)2/(\tfrac32)^2=-\tfrac29$, in agreement.

Interpreting and Checking the Equation

The term $G(F_t(s))$ describes replacement at the event: each offspring starts an independent population whose generating function is $F_t(s)$, and averaging the product over the offspring count gives $G(F_t(s))$. The term $-F_t(s)$ removes the contribution of the individual that has just experienced the event. Multiplication by $\lambda$ accounts for the event rate of one individual. The initial condition is $s$, not $1$, because at time zero the initial population is exactly one.

For any offspring law, $G(1)=1$. Accordingly, $F_t(1)=1$ solves the differential equation and expresses that a probability generating function evaluated at one equals one. Also, $F_t(0)$ has a direct interpretation: it is the probability that the population is zero at time $t$. These are useful checks, but they do not replace solving the equation for the offspring law of interest.

Common mistake: reversing the sign. A tempting but incorrect equation is $F_t'=\lambda(F_t-G(F_t))$. For binary fission, $G(u)=u^2$, and at $t=0$, $s=\tfrac12$, this wrong equation gives $$ F_0'=\tfrac12-\left(\tfrac12\right)^2=\tfrac14. $$ The correct equation gives $$ F_0'=\left(\tfrac12\right)^2-\tfrac12=-\tfrac14. $$ The negative sign is consistent with the event: replacing one individual by two increases the population, and for $0<s<1$, increasing the exponent decreases $s^{Z_t}$. The correct right-hand side is therefore $\lambda(G(F_t)-F_t)$, not its negative.

These equations convert the reproduction rule into a tractable evolution equation for the full population distribution, not only for a mean or a variance. The method is: specify $G$, set $F_0(s)=s$, solve the resulting initial-value problem, and verify both the differential equation and the initial condition. If a closed-form solution is unavailable, the differential equation still specifies the generating function and can be used for numerical analysis.

Check Your Understanding

Use the backward equation and the initial condition to answer each question. In each generating-function evaluation, take $s\in[0,1]$.

If every event produces zero offspring, what are $G(u)$ and the differential equation for $F_t(s)$?
For offspring counts zero and one with probabilities $1-r$ and $r$, derive the coefficient multiplying $1-F_t(s)$ in the differential equation.
For deterministic two-offspring replacement, verify the initial condition of $F_t(s)=s/[s+(1-s)e^{\lambda t}]$ at $t=0$.
Explain why conditioning on the population at time $h$ gives $F_{t+h}(s)=F_h(F_t(s))$.
For binary fission at $s=\tfrac12$ and $t=0$, compute the correct derivative and the derivative predicted by the sign-reversed equation.

Generating-Function Equations for Continuous-Time Branching

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