3.3 Incidence rate and person-time
Definition 3.3.1. The incidence rate of a disease in a population over a period is \[ I = \frac {\text {number of new cases in the period}}{\text {total person-time at risk in the period}} . \]
Each person contributes the time from entering observation (or the start of the period) until the earliest of: developing the disease, leaving observation, or the end of the period. The units are \(\text {time}^{-1}\), conventionally written as cases per \(1000\) person-years. A rate of \(5\) per \(1000\) person-years is the same whether it comes from \(1000\) people followed for one year or \(100\) people for ten.
Example 3.3.2. Eight workers are followed during a five-year study. Times are years from the start of the study.
| Worker | Enters | Leaves | Reason for leaving |
| 1 | 0 | 5 | End of study |
| 2 | 0 | 2.5 | Disease |
| 3 | 1 | 5 | End of study |
| 4 | 0 | 3 | Moved away |
| 5 | 2 | 4.5 | Disease |
| 6 | 0 | 5 | End of study |
| 7 | 0.5 | 1.5 | Died of another cause |
| 8 | 3 | 5 | End of study |
The person-years at risk are \(5 + 2.5 + 4 + 3 + 2.5 + 5 + 1 + 2 = 25\) and there are \(2\) cases, so \(I = 2/25 = 0.08\) per person-year, or \(80\) per \(1000\) person-years. No risk can be computed from these data without further assumptions: only \(4\) workers were observed for all five years, and the others were not observed long enough to say whether they would have become ill.
In an open population whose size \(N(t)\) changes slowly, the person-time over a period of length \(\Delta \) is approximately \(N_{\text {mid}}\Delta \), where \(N_{\text {mid}}\) is the population at the middle of the period. National annual death rates are computed in this way, as the deaths in the year divided by the mid-year population.
The rate as a hazard
The incidence rate estimates a population parameter: the hazard \(\lambda (t)\), the instantaneous rate at which disease occurs among those still at risk, \[ \lambda (t) = \lim _{h \downarrow 0} \frac {P(t < T \leq t + h \mid T > t)}{h} , \] where \(T\) is the time to disease. The hazard determines the risk.
Theorem 3.3.3 (Risk and rate). In the absence of competing risks, the risk over \((0, t]\) is \[ R(t) = 1 - \exp \Bigl (-\int _0^t \lambda (u)\,du\Bigr ) . \] If the hazard is constant, \(R(t) = 1 - e^{-\lambda t}\), and \(R(t) \approx \lambda t\) when \(\lambda t\) is small.
Proof. Let \(S(t) = P(T > t) = 1 - R(t)\). Then \(P(t < T \leq t + h \mid T > t) = \{S(t) - S(t + h)\}/S(t)\), so dividing by \(h\) and letting \(h \to 0\) gives \(\lambda (t) = -S'(t)/S(t) = -\frac {d}{dt}\log S(t)\). Integrating from \(0\) with \(S(0) = 1\), \(\log S(t) = -\int _0^t \lambda (u)\,du\). For constant \(\lambda \) the integral is \(\lambda t\), and \(1 - e^{-x} = x - x^2/2 + \cdots \approx x\) for small \(x\). □
Example 3.3.4. A rate of \(6.23\) per \(1000\) person-years, if constant, gives a ten-year risk of \(1 - e^{-0.0623} = 0.0604\); the approximation \(\lambda t = 0.0623\) is \(3\%\) too high. For a rate of \(50\) per \(1000\) the ten-year risk is \(1 - e^{-0.5} = 0.39\), against the approximation \(0.5\), which is useless.
The Poisson likelihood
The statistical model for a rate is justified by the following result, which is why counts of cases over person-time are analysed as Poisson variables even though nobody’s follow-up was fixed in advance.
Theorem 3.3.5. Suppose \(n\) people have a constant hazard \(\lambda \), and person \(i\) is observed for time \(t_i\) with event indicator \(\delta _i\) (\(1\) if the disease occurred at \(t_i\), \(0\) if observation ended without it), where the end of observation without disease is independent of the time to disease. Let \(D = \sum \delta _i\) and \(T = \sum t_i\). Then the likelihood is \[ L(\lambda ) = \lambda ^D e^{-\lambda T} , \] which is proportional to the likelihood of a Poisson observation \(D\) with mean \(\lambda T\). The maximum likelihood estimate is \(\hat \lambda = D/T\), and the observed information gives \(\text {SE}(\log \hat \lambda ) = 1/\sqrt {D}\).
Proof. With constant hazard, \(T_i\) is exponential with density \(\lambda e^{-\lambda t}\) and survivor function \(e^{-\lambda t}\). A person whose disease is observed at \(t_i\) contributes the density \(\lambda e^{-\lambda t_i}\); a person observed disease-free until \(t_i\) contributes \(P(T_i > t_i) = e^{-\lambda t_i}\). (Independence of the end of observation is what makes these the right contributions.) Both are \(\lambda ^{\delta _i} e^{-\lambda t_i}\), and the product over people is \(\lambda ^D e^{-\lambda T}\). A Poisson variable with mean \(\lambda T\) has probability \((\lambda T)^D e^{-\lambda T}/D!\), which as a function of \(\lambda \) is a constant times the same expression.
The log-likelihood \(D\log \lambda - \lambda T\) has derivative \(D/\lambda - T\), zero at \(\hat \lambda = D/T\), and second derivative \(-D/\lambda ^2\). In terms of \(\theta = \log \lambda \) the log-likelihood is \(D\theta - Te^\theta \), with second derivative \(-Te^\theta = -D\) at the maximum, so the observed information for \(\theta \) is \(D\) and \(\text {SE}(\hat \theta ) = 1/\sqrt D\). □
The approximate interval for the rate is therefore \(\hat \lambda \, e^{\pm z/\sqrt D}\). Its precision depends on the number of cases, not on the number of people or the amount of person-time: a study with \(25\) cases estimates a rate to within a factor of about \(e^{1.96/5} = 1.48\), however large the cohort.
The exact interval for a rate
Treating \(D\) as Poisson with mean \(\mu = \lambda T\), exact limits for \(\mu \) come from the gamma distribution, as the binomial ones came from the beta.
Lemma 3.3.6. If \(X \sim \text {Poisson}(\mu )\) and \(d \geq 1\), then \(P(X \geq d) = P(G \leq \mu )\) where \(G \sim \text {Gamma}(d, 1)\); equivalently \(2G \sim \chi ^2_{2d}\).
Proof. Let events occur in a Poisson process of unit rate on \((0, \infty )\). The number in \((0, \mu ]\) is \(\text {Poisson}(\mu )\), and at least \(d\) events occur in \((0, \mu ]\) exactly when the \(d\)-th event time is at most \(\mu \). The \(d\)-th event time is the sum of \(d\) independent unit exponentials, which is \(\text {Gamma}(d, 1)\), and twice a \(\text {Gamma}(d, 1)\) variable is \(\chi ^2_{2d}\). □
Solving \(P(X \geq d) = 0.025\) and \(P(X \leq d) = 0.025\) for \(\mu \) gives the exact \(95\%\) limits for the rate, \begin {equation} \label {eq:exactrate} \lambda _L = \frac {\chi ^2_{2D,\,0.025}}{2T}, \qquad \lambda _U = \frac {\chi ^2_{2D + 2,\,0.975}}{2T} , \end {equation} where \(\chi ^2_{k,\alpha }\) is the \(\alpha \) quantile of \(\chi ^2_k\) and \(\lambda _L = 0\) when \(D = 0\).
Example 3.3.7. A cohort accumulates \(3850\) person-years and \(24\) cases. The rate is \(24/3850 = 6.23\) per \(1000\) person-years. The log-scale interval is \[ 6.23\, e^{\pm 1.96/\sqrt {24}} = (4.18,\ 9.30) \text { per } 1000 , \] and the exact interval, from \(\chi ^2_{48,\,0.025} = 30.75\) and \(\chi ^2_{50,\,0.975} = 71.42\), is \((3.99,\ 9.28)\) per \(1000\).
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