From a Concentration Profile to Event Risk
A drug concentration usually changes continuously between observations or dose administrations, while event risk is often reported over a fixed horizon such as 5 days or 30 days. To connect these quantities, first use the concentration at each time to determine the instantaneous event hazard, then accumulate that hazard over the horizon. The result is cumulative hazard; transforming it gives the probability of remaining event-free and the probability of experiencing the event.
The calculation is not generally the same as inserting an average concentration into an exposure–hazard equation. If concentration changes, the hazard changes with it, and the cumulative hazard depends on the entire hazard trajectory. This tutorial focuses on that integration step, assuming a single event type and a specified concentration profile.
Hazard has units of inverse time, such as day\(^{-1}\); cumulative hazard is dimensionless. Survival and event risk are probabilities between 0 and 1. These relationships presume that \(h(t\mid C)\) describes the relevant event process for the individual or population being predicted and that the individual is event-free at the start of the prediction interval. If prediction begins at a later time, the integral and survival probability must be defined over that interval and conditional on being event-free at its start.
Why the Integral Gives Survival
The connection follows from the definition of hazard. Over a short interval of length \(dt\), an event-free individual has an event probability approximately \(h(t\mid C)dt\). Thus the change in survival probability is approximately \(-S(t)h(t\mid C)dt\). In the continuous-time limit, this becomes a differential equation.
Because events remove individuals from the event-free group, \(dS(t)/dt=-h(t\mid C)S(t)\), with \(S(0)=1\) for an event-free individual at the start.
Divide by \(S(t)\), then integrate from 0 to \(T\): \(\int_1^{S(T)} dS/S=-\int_0^T h(t\mid C)\,dt\).
The left side is \(\log S(T)\), so \(\log S(T)=-H(T)\), and therefore \(S(T)=\exp[-H(T)]\). Event risk by \(T\) is \(1-S(T)\).
This derivation also shows why hazards are integrated, not added as probabilities. The event-free population remaining at each time affects the number still at risk, which is represented by the factor \(S(t)\) in the differential equation. The cumulative hazard itself is not an event probability: it can exceed 1, whereas \(1-\exp[-H(T)]\) remains a valid probability.
Analytic Integration When the Hazard Has a Simple Form
Some concentration–hazard specifications allow direct integration. For example, under an additive hazard model with a constant baseline component, \(h(t\mid C)=h_0+\gamma C(t)\), where \(h_0\geq 0\) and \(\gamma\geq 0\). If \(C(t)=C_0\exp(-kt)\), with \(k>0\), then the concentration contribution can be integrated exactly. Here \(C_0\) is concentration at time 0, and \(k\) has units of inverse time.
The expression has the required units: \(h_0T\) is dimensionless, and \(\gamma C_0\) has units of inverse time, so \(\gamma C_0/k\) is dimensionless. An additive hazard is one possible model choice, not a universal PK/PD assumption. Its coefficients and nonnegative hazard constraint must be scientifically and statistically appropriate for the application.
Worked Example: Exponentially Declining Concentration
Suppose, hypothetically, that concentration follows \(C(t)=4\exp(-0.5t)\) mg/L for \(0\leq t\leq4\) days. Consider the additive hazard \(h(t\mid C)=0.020+0.010C(t)\) day\(^{-1}\), where the coefficient 0.010 has units L/(mg·day). Calculate cumulative hazard, survival, and event risk by day 4.
The constant baseline contribution is \(0.020\ {\rm day}^{-1}\times4\ {\rm days}=0.0800\). The integrated concentration is \(4\ {\rm mg/L}\times[1-\exp(-0.5\times4)]/(0.5\ {\rm day}^{-1})=6.917\ {\rm mg\cdot day/L}\). Multiplying by 0.010 L/(mg·day) gives a concentration-related cumulative hazard of 0.06917.
Therefore \(H(4)=0.0800+0.06917=0.1492\). Survival is \(S(4)=\exp(-0.1492)=0.8614\), and event risk is \(1-0.8614=0.1386\), or 13.86% by day 4 under this hypothetical model. The risk is not 14.92%; that value is the cumulative hazard expressed as a percentage, not the event probability.
Piecewise Hazards and Time-Varying Baselines
A common practical case is a hazard that is constant within intervals but changes between them. This may represent a piecewise baseline hazard combined with a concentration effect. If the hazard on interval \(j\) is \(h_j\), and the interval length is \(\Delta t_j\), that interval contributes \(h_j\Delta t_j\) to cumulative hazard. Summing the contributions gives the total. The survival probability follows only after the total cumulative hazard has been calculated.
Worked Example: Piecewise-Constant Concentration and Baseline Hazard
For a hypothetical 5-day prediction, suppose concentration is 1 mg/L from day 0 to day 2 and 3 mg/L from day 2 to day 5. Let the log-linear hazard be \(h(t\mid C)=h_{0,j}\exp(\beta C(t))\) within each interval, with \(\beta=0.30\) L/mg. Suppose \(h_{0,1}=0.010\) day\(^{-1}\) on the first interval and \(h_{0,2}=0.020\) day\(^{-1}\) on the second. These hazards are nonnegative, and the concentration and baseline are constant within each interval, so the interval-wise integration is exact under the stated model.
On days 0–2, the hazard is \(0.010\exp(0.30\times1)=0.01350\) day\(^{-1}\). Its cumulative-hazard contribution is \(0.010\exp(0.30\times1)\times2=0.0270\).
On days 2–5, the hazard is \(0.020\exp(0.30\times3)=0.04919\) day\(^{-1}\). Its contribution is \(0.020\exp(0.30\times3)\times3=0.1476\). Thus \(H(5)=0.0270+0.1476=0.1746\). The survival probability is \(S(5)=\exp(-0.1746)\approx0.8398\), so event risk by day 5 is approximately \(1-0.8398=0.1602\), or 16.02%.
The second interval contributes more despite lasting only one day longer because both its baseline hazard and concentration are higher. The calculation accounts for the full duration of each hazard level rather than treating the final concentration as if it applied throughout the 5 days.
Numerical Integration of a Sampled Concentration Profile
Concentration profiles are often available at discrete times, and the hazard may not have a convenient antiderivative. A straightforward option is to calculate hazard at each time point and numerically integrate the resulting hazard curve. With the trapezoidal rule, adjacent hazard values are joined by a straight line; the area over each interval is the interval width times the average of its endpoint hazards. This is an approximation when hazard varies between sample times.
Worked Example: Trapezoidal Integration of a Concentration Profile
Suppose hypothetical concentration measurements are 0, 2, 1, and 0 mg/L at times 0, 1, 2, and 3 days. Let \(h(t\mid C)=0.020\exp(0.20C(t))\) day\(^{-1}\), with the coefficient 0.20 in L/mg. For this calculation, assume concentration changes linearly between measurements; the trapezoidal rule applied to endpoint hazards approximates the integral under that interpolation.
The hazards at days 0, 1, 2, and 3 are, respectively, \(0.0200\), \(0.020\exp(0.4)=0.02984\), \(0.020\exp(0.2)=0.02443\), and \(0.0200\) day\(^{-1}\). Each interval is 1 day. The three trapezoid areas are \((0.0200+0.02984)/2\times1=0.02492\), \((0.02984+0.02443)/2\times1=0.02713\), and \((0.02443+0.0200)/2\times1=0.02221\).
Adding them gives \(H(3)\approx0.02492+0.02713+0.02221=0.07426\). Hence \(S(3)=\exp(-0.07426)\approx0.9284\), and event risk is approximately 0.0716, or 7.16% by day 3. These are numerical approximations conditional on the assumed profile and integration rule; they are not exact merely because the hazard was calculated exactly at the four listed times.
Assumptions, Diagnostics, and Common Pitfalls
A credible event-risk calculation requires more than correct arithmetic. The hazard model must be appropriate for the event and time scale, the concentration profile must represent exposure over the prediction interval, and the link between concentration and hazard must be specified on the intended scale. When concentration is measured with error or predicted from a PK model, uncertainty in that profile can also affect the risk prediction. The examples above condition on the given concentration profile and model parameters; they do not include parameter or exposure uncertainty.
- Check units. A hazard in day\(^{-1}\) must be integrated over days. For a log-linear effect, \(\beta C(t)\) must be dimensionless; if concentration is in mg/L, \(\beta\) is in L/mg.
- Check the time grid. Sparse observations can miss a rapid concentration peak or a rapid hazard change. Refine the grid or use a suitable integration method when the profile changes sharply.
- Check numerical stability. Compare results using a finer time grid or a second quadrature rule. Large changes suggest that the original approximation is insufficiently accurate.
- Check the hazard itself. Predicted hazards must be nonnegative and finite over the interval. Implausibly large hazards can drive survival close to zero and may signal an unsuitable parameter value, scale, or extrapolation.
- Check the interpolation assumption. Linear interpolation of concentration does not imply a linear hazard when the concentration effect is nonlinear. If the model is nonlinear, calculate hazard from the interpolated concentration and then integrate that hazard.
A frequent mistake is to average concentration over the interval and substitute that average into a nonlinear hazard model. For example, with \(h(t)=h_0\exp(\beta C(t))\), in general \(\int_0^T h_0\exp(\beta C(t))dt\) is not equal to \(Th_0\exp(\beta\bar{C})\). The exponential transformation means that high-concentration periods can contribute disproportionately. Integrate the hazard over time, using analytic integration when available or a suitable numerical method otherwise.
A second conceptual distinction is between a time-varying instantaneous effect and a cumulative-exposure effect. In the first, current concentration determines the hazard at that time, such as \(h(t)=h_0(t)\exp[\beta C(t)]\). In the second, the hazard may depend on an exposure summary accumulated up to time \(t\), such as an integral of past concentration. These are different model structures; integrating a current-concentration-dependent hazard to obtain \(H(T)\) does not itself assert that past concentration drives the instantaneous hazard.
Analytic integration is efficient and exact when the model structure permits it, as in the exponential-concentration additive-hazard example. Piecewise integration is exact for a truly piecewise-constant hazard. Numerical quadrature is more flexible for complex concentration profiles and hazard functions, but its accuracy depends on the time grid and approximation method. No integration method corrects a misspecified hazard model, an incorrect concentration profile, or extrapolation beyond the data.
Check Your Understanding
Work each problem using the distinction between hazard, cumulative hazard, survival, and event risk. Solutions are included below the questions.
- A constant hazard is 0.04 day\(^{-1}\) for 5 days. Find cumulative hazard and event risk by day 5.
- Why is cumulative hazard not generally equal to event risk?
- Under \(h(t)=0.01\exp(0.2C(t))\) day\(^{-1}\), explain why using average concentration in the hazard equation may not reproduce the integrated hazard.
- For a hazard of 0.03 day\(^{-1}\) over 2 days followed by 0.01 day\(^{-1}\) over 3 days, find cumulative hazard and survival at day 5.
Solutions. (1) \(H(5)=0.04\times5=0.20\). Thus \(F(5)=1-\exp(-0.20)=0.1813\), or 18.13%; the cumulative hazard is 0.20, not an 20% event probability. (2) Event risk is \(1-\exp[-H(T)]\), whereas cumulative hazard is the integrated hazard and is not bounded above by 1. (3) The exponential is nonlinear, so the average of \(\exp[0.2C(t)]\) generally differs from \(\exp[0.2\bar C]\); integrate the time-varying hazard or use justified numerical quadrature. (4) \(H(5)=0.03\times2+0.01\times3=0.09\). Therefore \(S(5)=\exp(-0.09)=0.9139\). The corresponding event risk is \(1-0.9139=0.0861\), or 8.61%.