5.5 Preventable fractions in a population
For a protective exposure, the counterpart of the PAF is the proportion of the cases that would occur without the exposure which it prevents in the population as a whole.
Proposition 5.5.1. If a proportion \(P_e\) of the population has a protective exposure with risk ratio \(\RR < 1\), the population preventable fraction is \[ \text {PF} = \frac {p_0 - p}{p_0} = P_e(1 - \RR ) . \]
Proof. \(p = P_e p_1 + (1 - P_e)p_0\), so \(p_0 - p = P_e(p_0 - p_1)\), and dividing by \(p_0\) gives \(P_e(1 - p_1/p_0)\). □
Example 5.5.2. In an outbreak, the attack rate is \(2\%\) among vaccinated children and \(10\%\) among unvaccinated children. The vaccine efficacy is \(1 - 0.02/0.10 = 0.80\). With \(70\%\) coverage, the vaccination programme has prevented \(\text {PF} = 0.7 \times 0.8 = 56\%\) of the cases that would have occurred without it. Raising coverage to \(90\%\) would raise this to \(72\%\), if the efficacy and the risk among the unvaccinated stayed the same. In practice they do not, because vaccinated children do not pass infection to unvaccinated ones (herd immunity). The formulas here ignore this, and so understate the benefit of high coverage.
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