Extinction Risk Near the Epidemic Threshold

Early Epidemics as Branching Processes

At the start of an epidemic, each infected person may infect several others while nearly everyone they contact is still susceptible. In this early phase, a useful approximation is to treat different infected people’s numbers of new infections as independent draws from the same distribution. The resulting branching model is simple enough to calculate extinction risk, but it also highlights an important feature of epidemics near the threshold: a modest average number of infections per case does not determine what any one outbreak will do.

Consider a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(Z_n\) be the number of infected individuals in generation \(n\), where generation \(0\) consists of the initial infectives and generation \(n+1\) consists of people infected by generation \(n\). The state space is \(S=\{0,1,2,\ldots\}\), and time is discrete, indexed by \(n\ge0\). Let \(\mathcal{F}_n\) record the infections and offspring counts observed through generation \(n\). In the branching approximation, conditional on \(\mathcal{F}_n\), each of the \(Z_n\) individuals independently produces a Poisson number of new infections with mean \(R\). Here \(R\) is the mean number of secondary infections per case in this early-phase model. Once \(Z_n=0\), all later generation sizes are zero.

The Poisson offspring probabilities are \(p_k=e^{-R}R^k/k!\), \(k\ge0\), for \(R>0\). Their total is \(\sum_{k=0}^\infty e^{-R}R^k/k!=e^{-R}e^R=1\), so they form a probability distribution. The probability-generating function, which is the expectation of \(s\) raised to the offspring count, is \(f(s)=\exp(R(s-1))\) for \(0\le s\le1\). The threshold \(R=1\) separates a regime where the mean generation size decreases from one where it increases. It does not, by itself, say that a particular epidemic must die out or take off.

The Extinction Equation and Its Threshold

Starting from one infective, extinction occurs if that infective has no children, or if every child’s lineage eventually becomes extinct. If there are \(k\) children, the latter event has probability \(q^k\), by independence of their lineages. Averaging over the Poisson offspring count gives \(q=f(q)\). This argument suggests the equation, but the smallest-root condition matters: a fixed point of the equation is not automatically the extinction probability.

Proof. Let \(q_n=\mathbb{P}_1(Z_n=0)\), where the subscript indicates one initial individual. Since zero is absorbing, the events \(\{Z_n=0\}\) increase with \(n\). Thus \(q_n\) increases to the probability \(q\) of eventual extinction. Set \(q_0=0\), since the initial individual has not yet produced a generation at time zero. Conditioning on the number of children gives \(q_{n+1}=f(q_n)\): if the initial individual has \(k\) children, each of their lineages must have reached zero by the relevant generation, an event of probability \(q_n^k\); averaging gives \(\sum_k p_kq_n^k=f(q_n)\). Continuity of \(f\) implies \(q=f(q)\). Also, if \(r\in[0,1]\) is any fixed point, then \(q_0=0\le r\), and the fact that \(f\) is increasing gives \(q_n\le r\) for every \(n\), by induction. Taking the limit gives \(q\le r\), so \(q\) is the smallest fixed point.

It remains to identify the fixed points. For \(R=0\), every individual has zero children, so \(q=1\), which is the only fixed point because \(f(s)=1\). For \(R>0\), \(f\) is strictly convex on \([0,1]\), since \(f''(s)=R^2\exp(R(s-1))>0\). If \(0<R<1\), the tangent to \(f\) at \(1\) is \(1+R(s-1)\), and strict convexity gives \(f(s)>1+R(s-1)>s\) for \(0\le s<1\). If \(R=1\), strict convexity gives \(f(s)>1+(s-1)=s\) for \(s<1\). In both cases there is no fixed point below \(1\), while \(f(1)=1\).

If \(R>1\), then \(f(0)=e^{-R}>0\), and \(f(1)-1=0\). The derivative at \(1\) of \(f(s)-s\) is \(R-1>0\), so \(f(s)-s<0\) for \(s<1\) sufficiently close to \(1\). Continuity therefore gives a root of \(f(s)-s\) strictly between \(0\) and \(1\). Strict convexity means there cannot be three distinct roots: if there were, strict convexity at the middle root would put the function strictly below the chord joining the two outer roots, whose value there is zero. Hence there is exactly one root below \(1\), in addition to the root at \(1\). The smallest-root argument identifies the smaller root as \(q\).

Finally, with \(z\) initial individuals, the \(z\) lineages evolve independently. Extinction of the whole process occurs exactly when each lineage becomes extinct, so \(q_z=q^z\) for \(z\ge1\). For \(z=0\), extinction has already occurred, and the formula gives \(q^0=1\). \(\square\)

The theorem says that extinction is certain when \(R\le1\), including exactly at the threshold. When \(R>1\), extinction still has positive probability; the probability of avoiding extinction is \(1-q\), not certainty. The branching approximation describes the early phase. It does not model the later reduction in susceptible people, so its survival event should be read as avoiding early extinction within this approximation, not as a complete description of the final epidemic.

Generation Sizes and Variability

Extinction probability is only one part of the picture. The generation-size mean and variance show how widely outcomes can spread. Suppose \(Z_0=z\), where \(z\) is a fixed nonnegative integer. Given \(\mathcal{F}_n\), \(Z_{n+1}\) is a sum of \(Z_n\) independent Poisson variables with mean \(R\). Therefore its conditional mean and variance are both \(RZ_n\).

Proof. Write \(m_n=\mathbb{E}[Z_n]\) and \(V_n=\operatorname{Var}(Z_n)\). Conditional expectation gives \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=RZ_n\), so taking expectations yields \(m_{n+1}=Rm_n\). Since \(m_0=z\), iteration gives \(m_n=zR^n\). The conditional variance is \(\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)=RZ_n\). The law of total variance then gives $$ V_{n+1} =\mathbb{E}[\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)] +\operatorname{Var}(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]) =Rm_n+R^2V_n. $$ For \(R\ne1\), repeated substitution, starting with \(V_0=0\), yields $$ V_n=z\sum_{j=1}^{n}R^{2(n-j)}R^j =zR^n\sum_{j=0}^{n-1}R^j =zR^n\frac{R^n-1}{R-1}. $$ The geometric-sum identity applies because \(R\ne1\). For \(R=1\), the recurrence becomes \(V_{n+1}=V_n+z\), so \(V_0=0\) gives \(V_n=zn\). These formulas also cover \(z=0\): the process is identically zero, and both its mean and variance are zero. \(\square\)

At criticality, \(R=1\), the expected generation size stays at \(z\), but the variance grows as \(zn\). Thus a constant mean does not mean that every outbreak stays near that mean. Some lineages have already died out, while others can be much larger. When \(R\) is just above one, the survival probability can still be small, so many realizations die out even though the expected generation size increases. The expectation combines these different outcomes; it should not be read as a prediction that each realization follows smooth exponential growth.

How the Risk Changes Near the Threshold

For \(R>1\), the extinction probability is generally defined implicitly rather than by an elementary closed form. Its behavior close to the threshold can nevertheless be described. Write \(R=1+\delta\), with \(\delta>0\), and let \(q=1-x\) be the smaller root, so \(0<x<1\). Taking logarithms of the fixed-point equation gives $$ -\ln(1-x)=(1+\delta)x. $$ For \(0<x<1\), the power series \(-\ln(1-x)=x+x^2/2+x^3/3+\cdots\) yields $$ \delta=\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\cdots. $$ As \(\delta\) decreases to zero, \(x\) must decrease to zero: the right side is positive and increasing in \(x\), and is bounded away from zero whenever \(x\) is bounded away from zero. Dividing the displayed series by \(x\) then gives \(\delta/x\to\tfrac12\), so \(x/\delta\to2\). Consequently, for small positive \(\delta\), the survival probability \(1-q=x\) is approximately \(2\delta\), and the extinction probability is approximately \(1-2\delta\). This is a local approximation, not an exact formula for arbitrary \(R\).

For several initial infections, the distinction between “each lineage” and “the whole outbreak” is important. If one lineage has extinction probability \(q\), then \(z\) initial lineages all become extinct with probability \(q^z\). More initial infections reduce the chance of total extinction when \(0<q<1\), but do not make survival certain for any finite \(z\). Near \(R=1\), \(q\) is close to one on the supercritical side, and extinction can remain likely even though the mean generation size grows.

Interpreting the Approximation

The branching model is appropriate for early epidemic growth when each case has approximately the same opportunity to infect others and susceptible depletion is negligible. Its independence assumption is an approximation: in a real epidemic, people share contacts, local outbreaks can be correlated, and the susceptible population is finite. Those features can change the dynamics once an outbreak is no longer small. Here the model’s purpose is narrower: quantify early extinction risk under a specified offspring law and show why average growth alone can be misleading.

Common pitfall. A common incorrect calculation is: “Since \(R>1\), the epidemic cannot become extinct, so \(q=0\).” For \(R=2\ln2\), substituting \(q=0\) into the required equation gives \(f(0)=e^{-2\ln2}=\tfrac14\), not \(0\). The correct fixed point is \(q=\tfrac12\), since \(f(\tfrac12)=\exp(-\ln2)=\tfrac12\). Supercriticality means positive probability of avoiding extinction, not zero probability of extinction. A second related error is to treat the expected generation size \(R^n\) as the size that will occur in every realization. The variance calculation shows why that interpretation fails, especially near \(R=1\).