R₀ and herd immunity
Tonight's idea
\(R_0\) (say "R nought") is how many people one infected person passes the disease to when everyone around them can still catch it. Measles: 12 to 18; the 1918 flu: about 2. If \(R_0 \gt 1\) each generation of cases is bigger than the last, 1, \(R_0\), \(R_0^2\), \(R_0^3\): exponential growth. If \(R_0 \lt 1\) the outbreak fizzles.
But \(R_0\) assumes everyone is susceptible, which stops being true as people recover. The simplest model splits \(N\) people into three boxes. \(\beta\) is infections per infectious person per day if everyone were susceptible; \(\gamma\) is the fraction of infectious people recovering each day, so \(1/\gamma\) is the infectious period.
With \(\Delta\) meaning "change over one step", one day here (the true model takes vanishingly short steps):
Infectious people only pass it on when they meet a susceptible, so the effective reproduction number is \(R_{\text{eff}} = R_0 \cdot S/N\). The number currently infectious, \(I\), grows while \(R_{\text{eff}} \gt 1\) and peaks when \(S/N = 1/R_0\). Everyone else, \(1 - 1/R_0\), is no longer susceptible by then.
Past the threshold, by infection or vaccination, each case produces fewer than one new case.
Watch one: two days by hand
\(N = 1000\), with \(S = 990\), \(I = 10\), \(R = 0\). Take \(\beta = 0.5\) and \(\gamma = 0.25\) (a four-day infectious period), so \(R_0 = 2\).
Day 1. New infections: \(0.5 \times 990 \times 10 / 1000 = 4.95\). Recoveries: \(0.25 \times 10 = 2.5\). So \(S = 985.05\), \(I = 12.45\), \(R = 2.5\).
Day 2. New infections: \(0.5 \times 985.05 \times 12.45 / 1000 \approx 6.13\). Recoveries \(\approx 3.11\). So \(S \approx 978.92\), \(I \approx 15.47\), \(R \approx 5.61\).
Fractional people are fine: the model tracks averages. \(I\) grows about 25% a day at first, because \(\beta S/N - \gamma \approx 0.25\), and that rate falls as \(S\) falls.
Your turn
- A disease has \(R_0 = 3\) and a generation time of 5 days (one generation is one round of passing it on; it takes one generation time). Starting from one case, how many new cases are there in the fourth generation? How many cases in total by then, and how many days have passed?
- Compute the herd-immunity threshold for \(R_0 = 1.5\), \(2\), \(3\), \(5\) and \(12\). Give each as a fraction and a percentage.
- A vaccine stops infection in 90% of the people who receive it. The disease has \(R_0 = 4\). What fraction of the population must be vaccinated to reach the herd-immunity threshold? Why is the answer bigger than the threshold itself?
- \(R_0 = 2.5\) in a city of 10,000, all susceptible. (a) How many are still susceptible when the number currently infectious peaks? (b) How many have been infected by then? (c) If 40% had been vaccinated beforehand, what is \(R_{\text{eff}}\) on day 0?
Stretch: the overshoot
Herd immunity is where the epidemic stops growing, not where it stops: the infectious people at the peak keep infecting. In the smooth model, started from a vanishingly small seed, the fraction \(s_\infty\) still susceptible at the very end satisfies \(s_\infty = e^{-R_0(1 - s_\infty)}\). For \(R_0 = 2\), start from \(s = 0.5\), compute \(e^{-2(1-s)}\), feed it back in, and repeat until it settles. What fraction is infected in total, against the 50% threshold? Do the one-day steps of "Watch one" end at the same place?
Answers
- Fourth generation: \(3^4 = 81\) new cases. Total \(1 + 3 + 9 + 27 + 81 = 121\). Four generations is 20 days.
- \(1 - 1/R_0\): \(1/3 \approx 33\%\), \(1/2 = 50\%\), \(2/3 \approx 67\%\), \(4/5 = 80\%\), \(11/12 \approx 92\%\).
- Threshold \(1 - 1/4 = 75\%\) must be immune. If a fraction \(v\) is vaccinated, \(0.9v\) are immune, so \(0.9v \ge 0.75\) gives \(v \ge 0.833\): about 83%. Only 90% of each dose counts, so you must over-vaccinate to make up for the 10% in whom it fails.
- (a) \(I\) peaks when \(S/N = 1/2.5 = 0.4\): 4,000 still susceptible. (b) So about 6,000 have been infected by then, some still infectious, most recovered. Daily new cases peak slightly earlier. (c) With 40% vaccinated, \(R_{\text{eff}} = 2.5 \times 0.6 = 1.5\): still an epidemic, but a slower one.
Stretch. From 0.5 the iterates go 0.368, 0.282, 0.238, 0.218, … and settle at \(s_\infty \approx 0.203\). About 80% are infected in total, far more than the 50% threshold: a 30-point overshoot. Vaccination reaches the threshold with \(I\) essentially zero, so nothing drives an overshoot; natural infection reaches it when infections are most numerous. The one-day steps end near \(S \approx 185\) of 1000, not 203: coarse steps and the 1% seed each shave a little off.