Exposure-Linked Time to Disease Progression

Why Link Exposure to Progression Time?

Disease progression is often recorded as a time-to-event outcome: the event is progression, and a patient who has not progressed by the end of observation is censored. Patients may have different drug exposures over time, so a useful PK/PD question is whether the instantaneous progression hazard changes with exposure. The model must use each patient’s exposure history up to each time, not just a dose label or a single exposure summary that ignores when exposure occurred.

The analysis below relates an exposure history to progression hazard while allowing hazard to change over time even at the same exposure. It also shows how to compare this model with one that has a baseline hazard alone. The fitted relationship is an exposure-response association conditional on the model and data; it does not by itself establish that exposure causes a change in progression.

Model and Assumptions

Definition: Let $T_i$ be patient $i$'s time to progression, $C_i$ the censoring time, and $Y_i=\min(T_i,C_i)$ the observed follow-up time. Let $\delta_i=1$ if progression is observed and $0$ if the patient is censored. Let $x_i(t)$ be the patient's exposure at time $t$, measured in mg/L. A proportional-hazards exposure model is $$ h_i(t\mid x_i(t))=h_0(t)\exp\{\beta x_i(t)\}, $$ where $h_i(t)$ is the instantaneous progression hazard conditional on remaining progression-free until $t$, $h_0(t)$ is the baseline hazard, and $\beta$ is the log hazard ratio per 1 mg/L increase in exposure.

For a concrete parametric baseline, use the Weibull hazard

h_0(t)=\frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1}, \qquad H_0(t)=\left(\frac{t}{\lambda}\right)^k, \qquad S_0(t)=\exp\{-H_0(t)\}.

Here $k>0$ is a dimensionless shape parameter and $\lambda>0$ is a scale measured in the time unit used for follow-up. $H_0(t)$ is cumulative baseline hazard and $S_0(t)$ is baseline survival. If $k=1$, the baseline hazard is constant; if $k>1$, it increases over time; and if $k<1$, it decreases. The exposure model multiplies the baseline hazard by $\exp\{\beta x_i(t)\}$. Thus a positive $\beta$ means higher exposure is associated with a higher progression hazard, conditional on the model.

The principal assumptions are that the event and censoring definitions are consistent across patients; censoring is independent of progression, conditional on modeled information; and the recorded or predicted exposure history is adequate for the exposure effect being modeled. The proportional-hazards form assumes the same $\beta$ applies over time: a one mg/L difference has the same hazard ratio at any time, although the baseline hazard itself can vary. This simple model also assumes no additional lag, cumulative-exposure, or delayed effect unless those features are explicitly added. If exposure is predicted from a PK model, treating it as known ignores uncertainty in those predictions.

Likelihood for Events and Censoring

For a specified exposure history, cumulative hazard through time $t$ is the integral of the hazard over that history. Survival is the probability of remaining progression-free through $t$, which equals the exponential of negative cumulative hazard. The event-time likelihood contribution follows directly from the event density $f_i(t)=h_i(t)S_i(t)$ and the survival probability for a censored observation.

H_i(t)=\int_0^t h_0(u)\exp\{\beta x_i(u)\}\,du, \qquad S_i(t)=\exp\{-H_i(t)\},

L_i= \left[h_i(Y_i)\right]^{\delta_i}\exp\{-H_i(Y_i)\}, \qquad \ell=\sum_i\left[\delta_i\log h_i(Y_i)-H_i(Y_i)\right].

An observed progression contributes its hazard at the event time as well as survival up to that time. A censored patient contributes survival through the censoring time, but no event hazard term. With a changing exposure history, the integral is evaluated over the whole history; using only exposure at the event time would generally give the wrong cumulative hazard.

For the baseline-only model, set $\beta=0$, so $h_i(t)=h_0(t)$. The baseline-only model is nested within the exposure model. Comparing the models therefore asks whether adding the exposure coefficient improves fit enough to justify the additional parameter.

Worked Example: Likelihood Contribution From a Changing Exposure History

Hypothetical illustration. Suppose a patient progresses at month 9. Exposure is 0.8 mg/L from month 0 through month 3, then 1.2 mg/L from month 3 through month 9. Take $\lambda=18$ months, $k=1.2$, and $\beta=\log(1.5)=0.4055$ per mg/L. Compute this patient’s likelihood contribution.

For an interval with constant exposure $x$, cumulative hazard over that interval is $\exp(\beta x)$ times the corresponding increment in $H_0$. Here $H_0(3)=(3/18)^{1.2}=0.1165$ and $H_0(9)=(9/18)^{1.2}=0.4353$. Therefore,

H_i(9)=\exp(0.4055\times0.8)(0.1165) +\exp(0.4055\times1.2)(0.4353-0.1165)

Using $\exp(0.3244)=1.383$ and $\exp(0.4866)=1.627$, this is $1.383(0.1165)+1.627(0.3188)=0.6798$. The hazard at month 9 is

h_i(9)=\frac{1.2}{18}(9/18)^{0.2}\exp(0.4055\times1.2) \approx0.09441\ \text{per month}.

Because progression was observed, the likelihood contribution is the event hazard times survival through month 9: $L_i=0.09441\exp(-0.6798)=0.04784$ per month. The survival factor uses both exposure intervals; the event hazard uses the exposure at the event time.

Estimating and Interpreting the Exposure Effect

Estimate $\lambda$, $k$, and $\beta$ by maximizing the log-likelihood over patients. The resulting coefficient is on the log-hazard scale. For an exposure difference $\Delta x$, the hazard ratio is $\exp(\beta\Delta x)$. It compares instantaneous hazards among patients who have remained progression-free to that time, conditional on the specified model; it is not a ratio of progression probabilities or a direct time ratio.

Worked Example: Hazard Ratio and Model Comparison

Hypothetical illustration. Suppose a Weibull baseline-only model has maximized log-likelihood $-142.6$, with two estimated parameters, $\lambda$ and $k$. Adding exposure gives a maximized log-likelihood of $-137.9$, with three parameters, including $\beta$. Suppose the exposure coefficient estimate is $0.4055$ per mg/L with standard error $0.1600$.

A one mg/L increase has estimated hazard ratio $\exp(0.4055)=1.500$. In this hypothetical fit, that exposure difference is associated with a 50.0% higher instantaneous progression hazard, conditional on the model. A Wald 95% confidence interval on the coefficient scale is $0.4055\pm1.96(0.1600)=[0.0919,0.7191]$ per mg/L. Exponentiating its endpoints gives a hazard-ratio interval of $[1.096,2.052]$ for a one mg/L difference. Under the model and usual large-sample approximation, the interval procedure has 95% coverage over repeated samples; it does not mean there is a 95% probability that the fixed parameter lies in this particular interval.

The likelihood-ratio statistic for testing $H_0:\beta=0$ against $H_a:\beta\ne0$ is $2[-137.9-(-142.6)]=9.4$. With one added parameter and regular likelihood conditions, this is compared with a chi-square distribution with 1 degree of freedom, giving an approximate $p$-value of 0.0022. This is evidence against the no-exposure-effect model under the specified assumptions; it is not the probability that $H_0$ is true.

For AIC, use $2q-2\ell$, where $q$ is the number of estimated parameters. The baseline model has AIC $2(2)-2(-142.6)=4+285.2=289.2$. The exposure model has AIC $2(3)-2(-137.9)=6+275.8=281.8$. The exposure model’s AIC is 7.4 points lower, favoring it among these two fitted models according to this criterion. This is a relative model comparison, not proof that the exposure relationship is causal or that the model is correctly specified.

From Hazard to Progression-Time Predictions

A hazard ratio does not translate directly into a fixed percentage change in progression time. For a constant exposure $x$, the Weibull model gives $S(t\mid x)=\exp[-\exp(\beta x)(t/\lambda)^k]$. The median time solves $S(t\mid x)=0.5$. When exposure changes over time, use the integrated hazard along the specified history and solve $H_i(t)=\log(2)$ for the median, if the history and model imply that this crossing occurs.

Worked Example: Median Time Under a Changing Exposure History

Hypothetical illustration. Use $\lambda=18$ months, $k=1.2$, and $\beta=\log(1.5)$ per mg/L. Consider a prediction scenario with exposure 0 mg/L from month 0 through month 6 and 1 mg/L thereafter. For the first six months, $H_0(6)=(6/18)^{1.2}=0.2676$. After month 6, the exposure multiplier is $\exp(\beta)=1.5$. Thus, for $t>6$,

H_i(t)=0.2676+1.5\left[(t/18)^{1.2}-0.2676\right].

Set this equal to $\log(2)=0.6931$. Then $(t/18)^{1.2}=0.2676+(0.6931-0.2676)/1.5=0.5513$, so $t=18(0.5513)^{1/1.2}=10.96$ months. Under the same baseline model with exposure kept at 0 mg/L, the median is $18(0.6931)^{1/1.2}=13.26$ months. The difference between these model-based medians is about 2.30 months for these specified histories. It is a prediction conditional on the fitted model, not a guaranteed treatment effect for an individual patient.

Diagnostics, Alternatives, and Limitations

Before interpreting estimates, check that the data support the model. Plot estimated survival by exposure group or range and compare it with model-based survival predictions, while recognizing that categorizing a continuous exposure is mainly a diagnostic display. Examine whether predicted survival is calibrated over clinically relevant follow-up, and inspect residual or score-based diagnostics for influential patients and systematic lack of fit. In particular, a time-varying exposure effect can violate proportional hazards: if higher exposure is associated with a larger hazard ratio early than late, a single $\beta$ may conceal that pattern. Check this with exposure-by-time extensions or appropriate residual diagnostics rather than assuming the proportional effect holds automatically.

Also inspect exposure overlap and event counts. If nearly all high-exposure patients progress early, there may be little information about their later risk; sparse events or limited exposure variation can make estimates unstable. Verify the time alignment of exposure and progression, and ensure exposure after progression is not used. Informative censoring, misclassified progression times, exposure measurement error, or changes in treatment prompted by worsening disease can bias the estimated relationship. A time-varying exposure can itself reflect evolving health status, so adjustment and causal interpretation require careful design and assumptions beyond this model.

The Weibull model imposes a particular baseline-hazard shape. A Cox proportional-hazards model instead leaves the baseline hazard unspecified and estimates exposure effects through the partial likelihood; it can reduce baseline-shape risk but does not directly supply the same parametric absolute-time predictions without estimating a baseline survival curve. A baseline-only Weibull model is simpler and useful as a comparator, but cannot represent an exposure association. More flexible parametric baseline hazards can relax the Weibull shape restriction, at the cost of additional parameters and possible instability. Models with lagged exposure or cumulative exposure may be scientifically appropriate when a delayed or accumulated effect is expected, but they answer different questions and need adequate data to estimate their added structure.

Common pitfall: A hazard ratio of 1.5 does not mean progression occurs 50% sooner, nor does it mean a patient’s probability of progression is 50% higher at every time. It is a ratio of instantaneous hazards among patients still at risk. Translate it into survival or time predictions by integrating the fitted hazard over a clearly stated exposure history.

Key takeaway: Exposure-linked time-to-progression analysis combines an event-time likelihood, a baseline hazard, and each patient’s exposure history. Interpret the coefficient on the hazard scale, compare models on the same data, and derive time predictions from cumulative hazard rather than treating a hazard ratio as a time ratio.

Check Your Understanding

Answer each question, then compare with the solutions below.

If $\beta=0.4055$ per mg/L, what hazard ratio corresponds to a 0.5 mg/L exposure difference?
In the example with $\lambda=18$ months and $k=1.2$, is $\lambda$ the baseline median progression time? Explain.
What likelihood contribution does a patient censored at $Y_i$ make under this model?
Why does a lower AIC for the exposure model not establish a causal exposure effect?

Solutions. 1. The ratio is $\exp(0.4055\times0.5)=\exp(0.20275)=1.225$, for a 0.5 mg/L difference. 2. No. At zero exposure, the median solves $(t/18)^{1.2}=\log(2)$, giving $t=18(0.6931)^{1/1.2}=13.26$ months. 3. It contributes survival through censoring, $S_i(Y_i)=\exp\{-H_i(Y_i)\}$, without an event-hazard factor. 4. AIC compares relative fit with a parameter penalty among candidate models; confounding, informative exposure histories, or misspecification can remain, so the comparison alone does not identify a causal effect.