5.4 Several exposure levels and several factors
An exposure with several levels
If an exposure has levels \(j = 0, 1, \dots , J\) with risk ratios \(\RR _j\) relative to the reference level \(0\), and a proportion \(P_{cj}\) of cases is at level \(j\), Miettinen’s argument applied level by level gives \[ \text {PAF} = \sum _{j \geq 1} P_{cj}\,\frac {\RR _j - 1}{\RR _j} . \] Collapsing the levels into “exposed” and “unexposed” and using a single average risk ratio gives a different, wrong answer (Problem 5.6.3).
Attributable fractions for several factors do not add
When two exposures both affect a disease, the fraction attributable to each can be computed, but the fractions are not shares of a fixed total. A case caused by the joint action of two exposures would be prevented by removing either, so it is counted in both fractions.
Example 5.4.1. Suppose the risk of a disease is \(p_0 = 0.001\) in people with neither of two exposures, smoking multiplies it by \(10\) and an occupational exposure by \(5\), and the two act multiplicatively, so people with both have risk \(0.05\). Half the population smokes and a tenth has the occupational exposure, independently. The population risk is \[ p = 0.001(0.5 + 0.5 \times 10)(0.9 + 0.1 \times 5) = 0.001 \times 5.5 \times 1.4 = 0.0077 . \] Removing smoking leaves \(0.001 \times 1.4 = 0.0014\), so \(\text {PAF}_{\text {smoking}} = (0.0077 - 0.0014)/0.0077 = 0.818\). Removing the occupational exposure leaves \(0.001 \times 5.5 = 0.0055\), so \(\text {PAF}_{\text {occ}} = 0.286\). The two sum to \(1.10\), more than all the cases. Removing both leaves \(0.001\), so the joint PAF is \(0.870\). Each separate figure is correct for its own question; what is wrong is to add them, or to read them as “\(82\%\) of cases are caused by smoking and the rest by something else”.
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