Why Model a Saturable Hazard Effect?
A drug concentration may alter the instantaneous rate at which an event occurs. A log-linear concentration effect makes the log hazard ratio change in direct proportion to concentration. That can be a useful approximation over a limited range, but it implies that the effect continues to grow without bound as concentration increases. A saturable Emax model instead represents an effect that approaches a finite limiting value.
Here the concentration is treated as fixed over the time interval of interest. The model therefore describes how a particular concentration changes the hazard at each time relative to the baseline hazard. It does not, by itself, describe delayed drug effects or changes in concentration over time.
The model assumes a multiplicative effect on the hazard: at a given concentration, the hazard is the baseline hazard multiplied by a concentration-specific factor. That factor does not depend on time, so the model is proportional hazards for fixed \(C\). It also assumes the stated Emax curve is an adequate description of the concentration effect and that the concentration is known and held fixed for the calculation. For an observed-data analysis, conclusions additionally rely on an appropriate baseline hazard and censoring assumptions.
A positive \(E_{\max}\) represents an increase in hazard as concentration rises. A negative \(E_{\max}\) represents a decrease. Importantly, \(E_{\max}\) is not itself a hazard ratio: it is the high-concentration limiting change in log hazard ratio. The corresponding limiting hazard ratio is \(\exp(E_{\max})\).
Limits, Monotonicity, and the Half-Effect Concentration
Write the concentration-dependent log hazard ratio as \(g(C)=E_{\max}C/(EC_{50}+C)\). Its derivative is
Since \(EC_{50}>0\) and the denominator is positive, the sign of \(g'(C)\) is the sign of \(E_{\max}\). Because \(HR(C)=\exp(g(C))\), its derivative is \(HR(C)g'(C)\), which has the same sign. Thus the hazard ratio increases with concentration when \(E_{\max}>0\), decreases when \(E_{\max}<0\), and is constant when \(E_{\max}=0\).
At zero concentration, the model gives \(HR(0)=1\): the low-concentration value is attained at \(C=0\). As concentration grows without bound, \(C/(EC_{50}+C)\) approaches 1, so \(HR(C)\) approaches \(\exp(E_{\max})\). For nonzero \(E_{\max}\), this high-concentration limit is approached but not attained at any finite concentration. If \(E_{\max}=0\), the hazard ratio is 1 at every concentration, including the limit.
At \(C=EC_{50}\), the fraction \(C/(EC_{50}+C)\) is \(1/2\). The log hazard-ratio effect is therefore \(E_{\max}/2\), and the hazard ratio is \(\exp(E_{\max}/2)\). This is the meaning of \(EC_{50}\) in this model: it is the concentration at which half of the limiting log hazard-ratio effect has been reached, not necessarily the concentration at which the hazard ratio itself is halfway between 1 and its limiting value.
When \(E_{\max}>0\), the hazard ratio ranges from 1 at \(C=0\) toward \(\exp(E_{\max})>1\), so the limiting value is its supremum and is not attained at finite concentration. When \(E_{\max}<0\), the ratio declines from 1 toward \(\exp(E_{\max})<1\); its maximum over \(C\geq0\) is 1 at zero concentration. These distinctions matter when describing an “increase,” “maximum,” or “plateau.”
Worked Example: Hazard at the Half-Effect Concentration
Suppose, hypothetically, that \(E_{\max}=\log(4)\), \(EC_{50}=2.00\) mg/L, and the baseline hazard at a particular time is \(0.0120\) day\(^{-1}\). Calculate the hazard at \(C=2.00\) mg/L.
The concentration equals \(EC_{50}\), so the fraction in the exponent is \(2.00/(2.00+2.00)=0.5\). Thus
The hazard is \(0.0120\times2=0.0240\) day\(^{-1}\). At this concentration the log hazard-ratio effect is half its limiting value, \(\log(4)\), and the hazard ratio is 2.
From a Hazard Ratio to Survival
The hazard is an instantaneous rate, not a probability of an event over a time interval. To obtain survival, use cumulative hazard. Let \(H_0(t)=\int_0^t\lambda_0(u)\,du\) be the baseline cumulative hazard. Under fixed concentration \(C\), the multiplier \(HR(C)\) is constant over time, so the cumulative hazard is \(H(t\mid C)=HR(C)H_0(t)\). The survival probability is then
This relation allows the baseline hazard to vary over time. Only if the baseline hazard is constant over the interval can it be multiplied by elapsed time to obtain baseline cumulative hazard. The next examples make that distinction explicit.
Worked Example: A Decreasing Hazard Effect and Survival
Suppose, hypothetically, that \(E_{\max}=-\log(3)\), \(EC_{50}=1.50\) mg/L, and the baseline hazard is constant at \(0.0200\) day\(^{-1}\). Find the hazard ratio, hazard, and survival over 10 days at \(C=4.50\) mg/L.
The concentration fraction is \(4.50/(1.50+4.50)=0.75\). Using the unrounded value of \(\log(3)\),
The hazard is \(0.0200\times0.438691\approx0.008774\) day\(^{-1}\). Because the baseline hazard is constant, the baseline cumulative hazard over 10 days is \(0.0200\times10=0.200\). The concentration-specific cumulative hazard is approximately \(0.200\times0.438691=0.087738\), giving
Under these hypothetical model assumptions, estimated survival through 10 days is about 91.60% at this fixed concentration. The survival calculation uses cumulative hazard; the daily hazard itself is not a 0.8774% daily event probability.
Worked Example: Survival With a Time-Varying Baseline Hazard
Suppose, hypothetically, that \(E_{\max}=\log(5)\), \(EC_{50}=3.00\) mg/L, and the baseline cumulative hazard over six days is \(H_0(6)=0.180\). At \(C=6.00\) mg/L, calculate the hazard multiplier and survival. The baseline hazard is assumed constant at \(0.0300\) day\(^{-1}\) in this example, consistent with the stated cumulative hazard.
The concentration fraction is \(6.00/(3.00+6.00)=2/3\). Therefore,
The hazard at this concentration is \(0.0300\times2.9240\approx0.08772\) day\(^{-1}\). The concentration-specific cumulative hazard is \(0.180\times2.9240\approx0.5263\). Hence
The hypothetical model predicts survival of about 59.08% through six days. If the baseline hazard were not constant, the supplied \(H_0(6)\) would still suffice for this fixed-concentration calculation; multiplying a single baseline hazard value by six days would not.
Comparison, Diagnostics, and Limitations
The earlier material on log-linear concentration effects on hazard uses a log hazard ratio that changes linearly with concentration. The Emax model instead makes the log hazard ratio rise or fall toward a finite limit. At concentrations much smaller than \(EC_{50}\), \(C/(EC_{50}+C)\) is approximately \(C/EC_{50}\), so the Emax model is approximately log-linear with slope \(E_{\max}/EC_{50}\) near zero concentration. At high concentrations, the saturable model plateaus rather than continuing to change without bound.
This comparison concerns the assumed concentration-response shape; neither model is automatically correct. The Emax form is useful when saturation is scientifically plausible, but data confined to a narrow concentration range may not distinguish a saturating curve from a log-linear one. If observations do not cover concentrations near and above \(EC_{50}\), the plateau may be weakly informed by data. The separate parameters \(E_{\max}\) and \(EC_{50}\) can also be difficult to estimate precisely when the observed concentrations show little curvature.
For a fitted model, examine whether observed event patterns across concentration are consistent with the predicted direction and curvature, and compare model-based survival predictions with suitable observed-data summaries. Large systematic departures by concentration or time suggest that the dose-response shape, baseline hazard, or proportional-hazards assumption may be inadequate. Predictions that continue to change substantially at the largest observed concentrations are not evidence that the model has established a plateau; the data may simply not reach it.
- Concentration must match the model’s interpretation. The equations above use a fixed concentration. Substituting a time-varying concentration into them as if it were constant does not generally give the correct cumulative hazard.
- The hazard ratio is not a risk ratio. A multiplier of 2 on the hazard does not mean that the event probability over a specified interval doubles. Survival depends on cumulative hazard.
- \(E_{\max}\) is on the log scale. The high-concentration limiting hazard ratio is \(\exp(E_{\max})\), not \(E_{\max}\). For negative \(E_{\max}\), that limit is below 1, and the largest hazard ratio is 1 at \(C=0\).
- A limiting value need not be observed. At finite positive concentrations, the Emax fraction is less than 1. In particular, the high-concentration limit is generally approached rather than attained.
The model also imposes a monotone effect with one saturation scale and no hysteresis or delay. If the observed response is nonmonotone, delayed relative to concentration, or driven by cumulative exposure rather than instantaneous concentration, this fixed-concentration relationship may be inadequate. A good fit within the observed range does not guarantee reliable extrapolation far beyond it.
Check Your Understanding
Answer each question using the fixed-concentration Emax hazard model. Brief solutions follow the questions.
- If \(E_{\max}=\log(2)\), what is the high-concentration limiting hazard ratio, and what is the hazard ratio at \(C=EC_{50}\)?
- For \(E_{\max}<0\), does the hazard ratio increase or decrease with concentration? What is its maximum over \(C\geq0\)?
- If \(E_{\max}=0.8\), \(EC_{50}=4.00\) mg/L, and \(C=4.00\) mg/L, calculate the hazard ratio. If the baseline hazard is \(0.0150\) hour\(^{-1}\), calculate the hazard.
- At fixed concentration, \(HR(C)=1.50\) and \(H_0(t)=0.200\). Calculate survival through time \(t\).
Solutions. (1) The limiting ratio is \(\exp\{\log(2)\}=2\); at \(EC_{50}\), it is \(\exp\{\log(2)/2\}=\sqrt{2}\approx1.414\). (2) It decreases, and its maximum over nonnegative concentrations is \(HR(0)=1\). (3) The concentration fraction is \(4.00/(4.00+4.00)=0.5\), so \(HR=\exp(0.8\times0.5)=\exp(0.4)\approx1.492\). The hazard is \(0.0150\times1.49182\approx0.02238\) hour\(^{-1}\). (4) \(S(t\mid C)=\exp\{-1.50(0.200)\}=\exp(-0.300)\approx0.7408\).