3.4 Prevalence, incidence and duration
Prevalence depends on how often a disease begins and how long it lasts. The relation is exact in a population in a steady state.
Theorem 3.4.1. In a population of constant size \(N\) in which the incidence rate \(I\), the prevalence \(P\) and the mean duration \(\bar D\) of the disease do not change over time, \[ \frac {P}{1 - P} = I\,\bar D . \] When the prevalence is small, \(P \approx I\bar D\).
Proof. At any time there are \(NP\) cases and \(N(1 - P)\) people at risk. New cases arise at rate \(I\,N(1 - P)\) per unit time. Cases leave the prevalent pool, by recovery or death, at rate \(NP/\bar D\) per unit time: in a steady state the pool of \(NP\) cases turns over once in the mean duration \(\bar D\). A steady state requires inflow to equal outflow, \(I\,N(1 - P) = NP/\bar D\), which rearranges to the result. When \(P\) is small, \(1 - P \approx 1\). □
Two consequences matter for design. A treatment that keeps patients alive without curing them raises prevalence, by increasing \(\bar D\). And any factor associated with longer survival with a disease is over-represented among prevalent cases, so a cross-sectional comparison of cases with non-cases mixes the causes of the disease with the determinants of its duration (Problem 3.6.5).
Example 3.4.2. A chronic disease has incidence \(4\) per \(1000\) person-years and lasts on average \(5\) years. In a steady state the prevalence odds are \(0.004 \times 5 = 0.02\), so \(P = 0.02/1.02 = 0.0196\), about \(2\%\). If a new treatment doubles survival with the disease, the prevalence rises to \(0.04/1.04 = 3.8\%\) with no change at all in how often the disease begins.
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