Two Interventions, Two Model Parameters
Vaccination and contact reduction can both reduce transmission, but they act on different parts of an SIR model. Perfect vaccination before an outbreak removes people from the susceptible population. Contact reduction lowers the rate at which an infectious person transmits to those who remain susceptible. Translating each intervention into the model makes it possible to compare their effects on early outbreak risk without treating them as interchangeable policies.
Consider a continuous-time stochastic SIR process \((S_t,I_t,R_t)\) on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with finite state space \(E=\{(s,i,r):s+i+r=N\}\). Its filtration is \(\mathcal{F}_t=\sigma((S_u,I_u,R_u):0\le u\le t)\). In a population of size \(N\), infection changes \((s,i,r)\) to \((s-1,i+1,r)\) at rate \(\beta si/N\), and recovery changes it to \((s,i-1,r+1)\) at rate \(\gamma i\), where \(\beta>0\) and \(\gamma>0\). The basic reproduction number in a fully susceptible population is \(R_0=\beta/\gamma\).
Assume vaccination is given before infection is introduced, protects perfectly, and covers a fraction \(v\) of the population, where \(0\le v\le1\). Treat the remaining susceptible fraction as \(1-v\), so that \(S_0/N\) is approximately \(1-v\). Assume contact reduction multiplies the transmission rate by \(c\), where \(0\le c\le1\); recovery rates are unchanged. With both interventions, the infection rate is therefore \(c\beta si/N\). These assumptions describe homogeneous mixing and do not include partial vaccine protection, changing behaviour in response to cases, or waning immunity.
To see why this expression arises, write \(s=S_0/N\) for the initial susceptible fraction. While the susceptible population is treated as approximately fixed, an infectious person generates new infections at rate \(c\beta s\), and recovers at rate \(\gamma\). Thus the expected number of infections generated during that person’s infectious period is \(c\beta s/\gamma\). Substituting \(s\approx1-v\) gives \(R_{\mathrm{eff}}\). The approximation is intended for the early phase, before depletion of susceptibles substantially changes the infection rate.
Extinction and Early Outbreak Risk
The early phase can be approximated by a branching process: each infectious person produces a random number of new infections, and those infections reproduce in the same way. This approximation neglects collisions in which two infectious people infect the same susceptible person. It is most useful when the initial number infected is small relative to the population. In this model, an infectious period is exponential with rate \(\gamma\), and infections occur during it at rate \(\lambda=c\beta s\). Conditional on an infectious period of length \(t\), the number of infections has a Poisson distribution with mean \(\lambda t\).
Proof. Let \(K\) be the number of infections generated by one infectious person, and let \(G(z)=\mathbb{E}[z^K]\), for \(0\le z\le1\), be its probability generating function. Conditional on an infectious period of length \(t\), the Poisson generating function is \(\exp(\lambda t(z-1))\). Averaging over an exponential infectious period gives \(G(z)=\int_0^\infty \gamma e^{-\gamma t}\exp(\lambda t(z-1))\,dt=\gamma/(\gamma+\lambda(1-z))\). The integral is finite because \(\gamma+\lambda(1-z)\ge\gamma>0\). With \(R=\lambda/\gamma\), this is \(G(z)=1/[1+R(1-z)]\).
Let \(q_n\) be the probability that there are no infections in generation \(n\), starting with one infection in generation zero. Then \(q_0=0\), and conditioning on the number of children of the initial infection gives \(q_{n+1}=G(q_n)\). The events that generation \(n\) is empty increase with \(n\), because an empty generation has no descendants. Their union is eventual extinction, so \(q=\lim_{n\to\infty}q_n\). Since \(G\) is continuous on \([0,1]\), taking limits gives \(q=G(q)\). Moreover, if \(a\in[0,1]\) is any fixed point of \(G\), then \(q_0=0\le a\); monotonicity of \(G\) implies inductively \(q_n\le a\) for every \(n\), and hence \(q\le a\). Thus \(q\) is the smallest fixed point in \([0,1]\).
If \(R=0\), then \(G(z)=1\), so the fixed point is \(q=1\). If \(R>0\), the fixed-point equation is \(R q^2-(1+R)q+1=0\), which factors as \((q-1)(Rq-1)=0\). For \(0<R\le1\), the root \(1/R\) is at least \(1\), so the only root in \([0,1]\) is \(1\). For \(R>1\), both roots are in \([0,1]\), and the smaller is \(1/R\); the smallest-fixed-point property gives \(q=1/R\). With \(m\) initial infections, extinction requires all \(m\) independent descendant families to become extinct, which has probability \(q^m\). Its complement is \(1-q^m\). \(\square\)
In this approximation, \(1-q\) is the probability that an outbreak escapes early extinction from one introduction. It is often used as an approximation to the probability of a large outbreak in a large population. It is not the probability of a particular final epidemic size: once many people are infected, susceptible depletion matters, and the branching approximation no longer describes the process accurately. Even when \(R\le1\), a finite chain of infections can occur; the result says that eventual extinction has probability one in the branching model, not that transmission stops immediately.
Intervention Thresholds
Proof. With vaccination alone, set \(c=1\). The condition is \(R_0(1-v)\le1\). Since \(R_0>0\), dividing by \(R_0\) and rearranging gives \(v\ge1-1/R_0\). With contact reduction alone, set \(v=0\); then \(R_0c\le1\), or \(c\le1/R_0\). If \(R_0\le1\), the untreated value already satisfies the condition, and multiplying by either \(1-v\le1\) or \(c\le1\) cannot increase it. Finally, the extinction theorem gives \(q=1\) exactly when \(R_{\mathrm{eff}}\le1\); for any finite \(m\), this implies \(q^m=1\). \(\square\)
The equality case is included: at \(R_{\mathrm{eff}}=1\), extinction occurs with probability one in this branching model, although outbreaks can persist for a long time. To obtain a strictly subcritical value, use a coverage strictly above \(1-1/R_0\), or a contact multiplier strictly below \(1/R_0\), when \(R_0>1\). The risk from one introduction is a nondecreasing function of \(R_{\mathrm{eff}}\): it is zero for values at most one and equals \(1-1/R_{\mathrm{eff}}\) above one. For \(R>1\), increasing \(R\) decreases \(1/R\), so it increases \(1-1/R\).
Worked Example: Vaccination Reduces the Susceptible Pool
Suppose \(R_0=3\), one infectious person is introduced, and perfect vaccination covers \(v=1/3\) of the population. There is no contact reduction, so \(c=1\). The remaining susceptible fraction is \(1-v=2/3\), and
Since \(2>1\), the extinction probability from one introduction is \(q=1/2\), and the probability of avoiding extinction in the early-outbreak approximation is \(1-q=1/2\). For two independent introductions, the extinction probability is \(q^2=(1/2)^2=1/4\), so the probability of avoiding extinction is \(1-1/4=3/4\).
Check: The mean number of early infections per infectious person is \(R_{\mathrm{eff}}=2\). Substituting \(q=1/2\) into \(G(z)=1/[1+2(1-z)]\) gives \(G(1/2)=1/[1+2(1/2)]=1/2=q\), as required.
Worked Example: Matching Vaccination and Contact Reduction
Let \(R_0=3\). Compare vaccinating half the population with no contact reduction against halving the contact rate with no vaccination. For vaccination, \(v=1/2\) and \(c=1\), so \(R_{\mathrm{eff}}=3(1/2)(1)=3/2\). For contact reduction, \(v=0\) and \(c=1/2\), so \(R_{\mathrm{eff}}=3(1)(1/2)=3/2\). Both give the same early-outbreak extinction probability, \(q=1/(3/2)=2/3\), and the same probability \(1-q=1/3\) of avoiding extinction from one introduction.
Check: For either intervention the offspring generating function is \(G(z)=1/[1+(3/2)(1-z)]\). At \(z=2/3\), \(G(2/3)=1/[1+(3/2)(1/3)]=1/(3/2)=2/3\), confirming the extinction probability.
The equality is about this model’s early-outbreak reproduction number and risk, not an assertion that the policies are identical in practice. Vaccination changes who is susceptible; contact reduction changes the rate of exposure. Their feasibility, timing, duration, and effects on particular groups can differ.
Worked Example: Combining Interventions to Cross the Threshold
Suppose \(R_0=3\), vaccination coverage is \(v=2/5\), and contact reduction uses \(c=1/2\). The remaining susceptible fraction is \(1-v=3/5\). Combining the two effects gives
Because \(9/10<1\), the extinction probability is \(q=1\), and the early-outbreak probability of avoiding extinction from one introduction is zero. By contrast, vaccination alone gives \(R_{\mathrm{eff}}=3(3/5)=9/5\), so the risk from one introduction is \(1-1/(9/5)=1-5/9=4/9\). Contact reduction alone gives \(R_{\mathrm{eff}}=3(1/2)=3/2\), so that risk is \(1-1/(3/2)=1-2/3=1/3\).
Check: Substituting \(R=9/10\) into the offspring generating function gives \(G(z)=1/[1+(9/10)(1-z)]\). At \(z=1\), \(G(1)=1\), so \(q=1\) satisfies the fixed-point equation; the extinction theorem establishes it is the only fixed point in \([0,1]\) when \(R<1\).
Assumptions and a Common Calculation Error
The formula \(R_{\mathrm{eff}}=R_0(1-v)c\) relies on the specified intervention mechanisms. It treats vaccination as perfect protection that removes a fraction \(v\) from the initially susceptible pool, and contact reduction as a uniform multiplier \(c\) on transmission. If vaccination is imperfect, the susceptible pool may not simply shrink by \(v\); if contact reduction changes infectious duration or affects some contacts more than others, multiplying the transmission rate by a single \(c\) may not represent the intervention. In a finite population, the actual number vaccinated and the susceptible count can also vary, so the early branching calculation is an approximation rather than an exact finite-population outbreak probability.
Common mistake: subtracting intervention levels from \(R_0\). With \(R_0=3\), \(v=1/3\), and \(c=1/2\), an incorrect additive calculation gives \(3-1/3-1/2=13/6\). The model instead applies the two multipliers: \(3(1-1/3)(1/2)=3(2/3)(1/2)=1\). The incorrect value would suggest a positive early takeoff risk of \(1-1/(13/6)=7/13\). The correct value is critical, \(R_{\mathrm{eff}}=1\), for which the branching-process extinction probability is one. Multiplying the remaining susceptible fraction and contact multiplier, rather than subtracting them from \(R_0\), is essential.
For policy comparisons, the effective reproduction number summarizes the early transmission potential under stated assumptions. It does not by itself describe how quickly an epidemic grows, the distribution of its final size, or which intervention is preferable under practical constraints. Those questions require additional features of the model and data.
Check Your Understanding
Use the early-outbreak branching approximation and show the substitutions in each calculation.
- If \(R_0=4\), what vaccination coverage is the threshold for \(R_{\mathrm{eff}}\le1\) when there is no contact reduction?
- For \(R_0=5/2\), \(v=1/5\), and \(c=1\), calculate \(R_{\mathrm{eff}}\) and the probability of avoiding extinction from one introduction.
- For \(R_{\mathrm{eff}}=2\), calculate the extinction probability from three independent initial infections.
- Why can vaccination and contact reduction give the same early-outbreak risk while representing different changes to the SIR model?
- What does extinction with probability one at \(R_{\mathrm{eff}}=1\) say about the branching model, and what does it not say about the duration of transmission?