Weibull Hazards with Drug Exposure

Why Use a Weibull Hazard With Drug Exposure?

The risk of a clinical event may change over time even when a patient’s drug exposure is held fixed. For example, event risk could rise as treatment continues, or be highest soon after a clinical transition and then decline. The Weibull model represents these patterns with a baseline hazard whose direction of change is governed by one shape parameter. An exposure multiplier then shifts the hazard up or down.

The resulting model is useful when a simple, interpretable time pattern is plausible. It also has a special property: with fixed exposure, it can be expressed both as a proportional-hazards model and as an accelerated failure time model. These interpretations depend on the assumptions below; the equations do not establish that those assumptions are appropriate for a particular drug or endpoint.

Definition: Let $T$ be a nonnegative event time, $t$ time since a defined origin, $\lambda>0$ the Weibull scale parameter, and $k>0$ the Weibull shape parameter. For a fixed exposure value $C$, let $\eta(C)$ be the exposure effect on the log-hazard scale. The exposure-adjusted Weibull hazard is $$ h(t\mid C)=\frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1}\exp\{\eta(C)\},\qquad t>0. $$

Here, $h(t\mid C)$ is an instantaneous event rate, with units of inverse time. The scale $\lambda$ has units of time, while $k$ and $\eta(C)$ are dimensionless. If a log-linear exposure effect has been specified, one may set $\eta(C)=\beta C$; other exposure functions can be used, provided the chosen function enters the hazard through this log-hazard multiplier. This tutorial treats $C$ as fixed over the time interval being modeled.

From Hazard to Cumulative Hazard and Survival

The cumulative hazard is the integral of the hazard from time zero to $t$. The survival function $S(t\mid C)=P(T>t\mid C)$ is related to cumulative hazard by $S(t\mid C)=\exp\{-H(t\mid C)\}$. These definitions give a direct derivation of the Weibull survival function.

\begin{aligned} H(t\mid C) &=\int_0^t h(u\mid C)\,du\\ &=\exp\{\eta(C)\}\int_0^t \frac{k}{\lambda} \left(\frac{u}{\lambda}\right)^{k-1}\,du\\ &=\exp\{\eta(C)\}\left(\frac{t}{\lambda}\right)^k,\\ S(t\mid C) &=\exp\left[-\exp\{\eta(C)\}\left(\frac{t}{\lambda}\right)^k\right]. \end{aligned}

For the integral, the derivative of $(u/\lambda)^k$ is $(k/\lambda)(u/\lambda)^{k-1}$; its value at $u=0$ is zero because $k>0$. Thus the cumulative hazard is finite at every finite $t$, including when the instantaneous hazard tends to infinity as $t$ approaches zero. Differentiating $S(t\mid C)$ also recovers the stated hazard, since $h(t\mid C)=-d\log S(t\mid C)/dt$.

The baseline functions are obtained by setting $\eta(C)=0$: $H_0(t)=(t/\lambda)^k$ and $S_0(t)=\exp[-(t/\lambda)^k]$. The scale $\lambda$ is the baseline time at which cumulative hazard equals one and survival equals $\exp(-1)$, approximately 0.3679. It is not generally the median event time.

Worked Example: Interpreting a Declining Baseline Hazard

Suppose a hypothetical endpoint follows a Weibull model with $\lambda=10$ days and $k=0.7$, and take the reference exposure so that $\eta(C)=0$. Find the baseline median event time, then calculate baseline hazard and survival at day 6.

The median $t_{0.5}$ satisfies $S_0(t_{0.5})=0.5$. Solving the survival equation gives $$ \exp\left[-\left(\frac{t_{0.5}}{10}\right)^{0.7}\right]=0.5 \quad\Longrightarrow\quad t_{0.5}=10[\log(2)]^{1/0.7}\approx 10(0.5924)=5.924\ \text{days}. $$

At day 6, the baseline hazard is $$ h_0(6)=\frac{0.7}{10}(0.6)^{-0.3} =0.07000(1.1656)\approx 0.08159\ \text{day}^{-1}. $$ The cumulative hazard and survival are $$ H_0(6)=(0.6)^{0.7}\approx 0.6994,\qquad S_0(6)=\exp(-0.6994)\approx 0.4970. $$

Because $k<1$, the baseline hazard declines over time: the event rate is higher near the time origin than later. The survival of approximately 0.497 at day 6 is consistent with the calculated median of about 5.924 days.

What the Shape Parameter Changes

Differentiating the log baseline hazard with respect to time shows how its direction of change depends on $k$: $$ \frac{d}{dt}\log h_0(t)=\frac{k-1}{t}. $$ For $k>1$, the hazard increases with time; for $0<k<1$, it decreases; and for $k=1$, it is constant. At $t$ approaching zero, the hazard tends to zero for $k>1$, tends to infinity for $0<k<1$, and equals $1/\lambda$ for $k=1$. These limits concern the instantaneous hazard; the survival function remains well-defined at zero with $S(0\mid C)=1$.

The median at exposure $C$ follows by setting survival to one half: $$ t_{0.5}(C)=\lambda\left[\frac{\log(2)}{\exp\{\eta(C)\}}\right]^{1/k} =\lambda[\log(2)]^{1/k}\exp\{-\eta(C)/k\}. $$ More generally, the time by which a fraction $q$ of the population has experienced the event, for $0<q<1$, is found by solving $S(t\mid C)=1-q$. The same calculation gives $t_q(C)=\lambda[-\log(1-q)/\exp\{\eta(C)\}]^{1/k}$.

Exposure Effects and Proportional Hazards

At two fixed exposure values $C_1$ and $C_0$, the hazard ratio is $$ \frac{h(t\mid C_1)}{h(t\mid C_0)} =\exp\{\eta(C_1)-\eta(C_0)\}. $$ It does not depend on time. Therefore, a fixed exposure effect that enters only through $\eta(C)$ gives proportional hazards, regardless of whether the baseline Weibull hazard increases, decreases, or stays constant. The shape parameter controls the baseline time pattern; it does not, by itself, make this exposure hazard ratio vary over time.

The same equations yield an accelerated failure time interpretation. Since $$ S(t\mid C)=S_0\left(t\exp\{\eta(C)/k\}\right), $$ the event-time distribution at exposure $C$ is the baseline event-time distribution multiplied by $\exp\{-\eta(C)/k\}$. A positive exposure effect ($\eta(C)>0$) increases the hazard and shortens event times in this distributional sense; a negative effect lowers the hazard and lengthens them. This interpretation describes a model-based distributional shift, not proof that exposure causes the change.

Worked Example: A Protective Exposure Effect

Consider a hypothetical Weibull model with $\lambda=12$ days, $k=2$, and an exposure effect $\eta(C)=-0.25$. Calculate the hazard and survival at day 6, and compare them with the reference exposure where $\eta(C)=0$.

At the reference exposure, the baseline hazard at day 6 is $$ h_0(6)=\frac{2}{12}\left(\frac{6}{12}\right)^{1} =0.08333\ \text{day}^{-1}. $$ The exposure multiplier is $\exp(-0.25)\approx0.7788$, so $$ h(6\mid C)=0.08333(0.7788)\approx0.06490\ \text{day}^{-1}. $$

The baseline cumulative hazard is $H_0(6)=(0.5)^2=0.25$, and the exposure-adjusted cumulative hazard is $H(6\mid C)=0.25(0.7788)=0.1947$. Hence the survival probabilities are $$ S_0(6)=\exp(-0.25)\approx0.7788,\qquad S(6\mid C)=\exp(-0.1947)\approx0.8231. $$

The hazard ratio is $\exp(-0.25)\approx0.7788$ at every time under this fixed-effect model. At day 6, the modeled event rate is lower and survival is higher at the specified exposure than at the reference exposure. These are conditional model comparisons, not causal conclusions.

Worked Example: Shape, Hazard Ratio, and Median Shift

Suppose a hypothetical model has $\lambda=8$ days and $k=2$. A fixed exposure has $\eta(C)=\log(2)$, corresponding to twice the reference hazard. Calculate the hazards at day 4 and the median event times at the two exposure levels.

At day 4, the baseline hazard is $$ h_0(4)=\frac{2}{8}\left(\frac{4}{8}\right)=0.1250\ \text{day}^{-1}. $$ The exposed hazard is $h(4\mid C)=0.1250(2)=0.2500\ \text{day}^{-1}$. The ratio is 2 at day 4 and, by the model equation, at every other time as well.

For the reference exposure, the median is $$ t_{0.5}(0)=8[\log(2)]^{1/2}\approx6.660\ \text{days}. $$ For the exposed group, $$ t_{0.5}(C)=8\left[\frac{\log(2)}{2}\right]^{1/2} \approx4.710\ \text{days}. $$ Equivalently, the exposure multiplies the baseline median by $\exp\{-\log(2)/2\}=1/\sqrt{2}$. The increasing baseline hazard follows from $k=2>1$, while the exposure effect preserves proportional hazards and shifts the event-time distribution earlier.

Assumptions, Diagnostics, and Alternatives

The formulas describe a conditional event-time distribution with a specified time origin, a positive scale and shape, and a Weibull baseline. The exposure value in $\eta(C)$ is fixed for the modeled comparison, and the effect is assumed to multiply the hazard without changing its time shape. Applying the model to censored event-time data additionally requires an appropriate censoring assumption for the intended inference. The hazard and survival equations alone do not resolve issues such as confounding, measurement error, or how exposure was assigned.

Check whether model predictions are compatible with the observed time pattern and whether exposure-specific survival curves show patterns consistent with the fitted model. A persistent mismatch in the timing of events, a baseline hazard that plausibly changes direction, or an exposure contrast that appears to change over time can signal poor fit. Such patterns may indicate a misspecified baseline, nonproportional effects, or a need to represent exposure differently; they are not automatically fixed by changing the Weibull shape parameter.

The exponential model is the special Weibull case $k=1$, with a constant baseline hazard. It is simpler, but cannot represent a monotonically increasing or decreasing baseline hazard. A lognormal event-time model offers a different distributional shape and is not generally a proportional-hazards model. A Cox proportional-hazards model leaves the baseline hazard unspecified, whereas the Weibull model imposes a specific parametric baseline and can provide a direct survival curve. The choice depends on scientific plausibility, fit, and the intended predictions; a flexible-looking formula is not a substitute for checking its implications.

A common mistake is to say that any Weibull model has a constant hazard ratio because it has a shape parameter. The correction is that proportional hazards follows here from the exposure multiplier being constant over time and common to the hazard at each time. If exposure changes over time, or if the model makes its effect interact with time, the ratio need not be constant. Likewise, $k$ determines the baseline hazard trend, not the direction of the exposure effect.

Key takeaway: With fixed exposure entering as $\exp\{\eta(C)\}$, the Weibull cumulative hazard is $\exp\{\eta(C)\}(t/\lambda)^k$ and survival is its negative exponential. The shape $k$ determines whether baseline hazard rises, falls, or remains constant; the exposure contrast is proportional over time under this model.

Check Your Understanding

Answer each question, then compare with the solutions below.

For a Weibull baseline with $k=0.6$, does the hazard increase or decrease over time? What is its limit as time approaches zero?
At fixed exposures $C_1$ and $C_0$, the fitted exposure effects are $\eta(C_1)=0.3$ and $\eta(C_0)=0.1$. What is the hazard ratio, and does it vary with time under this model?
If $\lambda=5$ days, $k=1$, and $\eta(C)=0$, calculate survival at day 5 and identify the baseline hazard.
Explain why a Weibull model with a fixed exposure multiplier can also have an accelerated failure time interpretation.

Solutions. (1) Since $k<1$, the hazard decreases; it tends to infinity as time approaches zero. (2) The hazard ratio is $\exp(0.3-0.1)=\exp(0.2)\approx1.221$, constant over time because the exposure multiplier is fixed. (3) The hazard is $1/\lambda=0.2000\ \text{day}^{-1}$, and $S(5)=\exp[-(5/5)^1]=\exp(-1)\approx0.3679$. (4) The survival function satisfies $S(t\mid C)=S_0(t\exp\{\eta(C)/k\})$, so the exposure-specific event-time distribution is a time-scaled version of the baseline distribution by the factor $\exp\{-\eta(C)/k\}$.