Why trajectory shape matters in link comparisons
A link-sensitivity analysis compares ways to connect a subject’s latent biomarker trajectory to their event hazard. That comparison is informative only if the trajectory model supplies reasonable estimates of the quantities being linked. If the true biomarker path curves but the longitudinal model forces it to be linear, estimated current values and slopes can be wrong in systematic, time-dependent ways. A difference between fitted hazard links may then reflect the trajectory approximation as well as the scientific association structure.
This concern is distinct from ordinary measurement noise. A flexible joint model can account for noisy repeated measurements while still imposing the wrong mean trajectory. The issue is especially important for a slope link: differentiating a misspecified mean function does not recover the subject’s actual rate of change.
For orientation, a current-value model can be written as
where \(h_i(t\mid b_i)\) is the hazard conditional on the random effects, \(h_0(t)\) is the baseline hazard, \(w_i\) is a vector of covariates, and \(\alpha\) is the association parameter. A slope-link alternative replaces or supplements \(m_i(t)\) with \(m_i'(t)\). The joint likelihood accounts for the repeated measurements and event-time information while integrating over the random effects. These formulations rely on an adequate longitudinal mean and random-effects distribution, a correctly specified association form and baseline hazard, and censoring that is independent conditional on the modeled information. Here the focus is the mean trajectory: changing the hazard link does not repair a poor trajectory model.
A curved path and its linear approximation
A simple way to see the problem is to compare a quadratic path with the best-fitting straight line over the same observation window. The line may summarize the average trend reasonably while failing to represent local values or rates of change. Since a slope link uses a derivative, its error can be more consequential than the error in the fitted biomarker level.
Worked Example: A Linear Fit Misses a Curved Biomarker Path
Suppose a hypothetical subject’s latent biomarker trajectory, in units per year, is \(m(t)=2+1.2t-0.15t^2\), observed over years \(t=0,1,2,3,4\). For these five equally spaced times, fit the least-squares straight line to the latent values and compare its levels and slopes with the quadratic trajectory.
The quadratic gives the values \(2.00, 3.05, 3.80, 4.25,\) and \(4.40\). Their mean is \(3.50\), and the mean observation time is \(2\). The fitted linear slope is \[ \frac{\sum (t-\bar t)\{m(t)-\bar m\}}{\sum (t-\bar t)^2} =\frac{6.00}{10}=0.600. \] The fitted intercept is \(3.50-(0.600)(2)=2.30\), so the linear approximation is \(\tilde m(t)=2.30+0.600t\).
At years 0 through 4, that line predicts \(2.30, 2.90, 3.50, 4.10,\) and \(4.70\). Thus the discrepancies \(m(t)-\tilde m(t)\) are \(-0.30, 0.15, 0.30, 0.15,\) and \(-0.30\) biomarker units. The errors change sign over time: they are not simply a constant offset.
The true derivative is \(m'(t)=1.2-0.30t\), which takes values \(1.20, 0.90, 0.60, 0.30,\) and \(0.00\) biomarker units per year. The fitted line instead has derivative \(0.600\) at every time. In particular, at year 4 it overstates the rate by \(0.600\) units per year, while at year 0 it understates it by the same amount. A constant estimated slope has erased the slowing and eventual flattening.
This calculation is about the trajectory for one hypothetical subject, not an estimated joint-model coefficient. In an actual analysis, repeated noisy measurements and the random-effects distribution determine the fitted subject-specific paths. The lesson carries over: a linear model can produce time-varying value errors and a derivative that is structurally unable to vary with time.
How misspecification can look like link sensitivity
Suppose the scientific hazard association is with current biomarker value, but a linear longitudinal mean is used. At time \(t\), the fitted current-value contribution is \(\alpha\tilde m_i(t)\) rather than \(\alpha m_i(t)\). The error in the log hazard contribution is exactly \(\alpha\{\tilde m_i(t)-m_i(t)\}\), conditional on the same covariates and baseline hazard. For a slope link, the corresponding error is \(\eta\{\tilde m_i'(t)-m_i'(t)\}\), where \(\eta\) is the slope association parameter. Neither error is generally constant over time or across subjects.
A comparison of current-value and slope links can therefore mix two questions: which trajectory feature is associated with risk, and whether the longitudinal model estimates that feature adequately. In particular, a linear trajectory forces the subject-specific slope to be constant. If the true path bends, the fitted slope can average over periods of rapid change and periods of little change. A slope-link coefficient may compensate for that averaging, and its value need not be directly comparable to the coefficient under a better-fitting curved trajectory.
Worked Example: The Same Curved Subject Under Two Links
Use the quadratic trajectory from the previous example at year 3. Suppose, hypothetically, a current-value association parameter is \(\alpha=0.35\) per biomarker unit. The true value is \(m(3)=4.25\), whereas the linear approximation gives \(\tilde m(3)=4.10\). Holding all other hazard terms fixed, the fitted log-hazard contribution differs from the true one by \[ 0.35(4.10-4.25)=-0.0525. \] The ratio of the fitted to true hazard contributions on the multiplicative scale is \(\exp(-0.0525)=0.9489\). This is a 5.11% lower hazard contribution at that time, attributable here to the value discrepancy.
Now consider a slope link with hypothetical parameter \(\eta=0.50\) per biomarker unit per year. At year 3 the true slope is \(0.300\), while the linear approximation gives \(0.600\). The fitted log-hazard contribution is too high by \(0.50(0.600-0.300)=0.150\); its hazard ratio relative to the contribution from the true slope is \(\exp(0.150)=1.162\). This is a 16.2% higher contribution. These calculations do not establish which link is scientifically correct. They show that a curved path creates different errors for level and slope links, so the apparent contrast between links can be amplified or changed by the trajectory approximation.
A useful justification for why coefficient estimates can shift comes from a simple projection calculation. At a fixed time, suppose the true log hazard, after removing the baseline and other covariates, is \(\alpha m_i\). A fitted current-value model instead uses \(\tilde m_i\). In a weighted least-squares projection of the true log hazard onto an intercept and \(\tilde m_i\), the projected slope is
The second equality follows by writing \(m_i=\tilde m_i+(m_i-\tilde m_i)\) and applying covariance additivity. Thus, when the trajectory discrepancy is associated with the fitted values in the event-weighted risk set, the projected association coefficient differs from \(\alpha\). The weights and risk sets in a survival likelihood make its exact behavior more complicated, so this projection is an explanation of a possible distortion, not a formula for the joint-model maximum-likelihood estimate.
Worked Example: An Omitted Curvature Component Alters a Projected Association
Consider four hypothetical subjects in a fixed-time risk set. Their fitted current values are \(\tilde m_i=(2,3,4,5)\), and their true latent values are \(m_i=(1.4,2.8,4.2,5.6)\). Assume equal weights and a true current-value log-hazard coefficient \(\alpha=0.40\) per biomarker unit.
The discrepancies \(m_i-\tilde m_i\) are \((-0.6,-0.2,0.2,0.6)\). Both sets of values have mean \(3.5\) for \(\tilde m_i\) and \(3.5\) for \(m_i\). The centered fitted values are \((-1.5,-0.5,0.5,1.5)\). Their squared deviations sum to \(2.25+0.25+0.25+2.25=5.00\). The cross-products of those centered values with the discrepancies sum to \[ (-1.5)(-0.6)+(-0.5)(-0.2)+(0.5)(0.2)+(1.5)(0.6)=2.00. \] The discrepancy-to-fitted-value covariance ratio in the projection formula is therefore \(2.00/5.00=0.400\), giving \[ \alpha^*=0.40(1+0.400)=0.560 \] per biomarker unit. The true hazard ratio for a one-unit increase is \(\exp(0.40)=1.492\); the projected coefficient implies \(\exp(0.56)=1.751\). This difference arises because the omitted component increases along with the fitted value in this constructed risk set. It illustrates a mechanism for coefficient distortion; it is not a claimed output from a survival-model fit.
Assessing whether the trajectory is driving the result
A practical assessment changes the longitudinal mean structure in a controlled way while holding the association comparisons as comparable as possible. For example, compare a linear trajectory with a scientifically plausible quadratic or spline-based mean, and then re-estimate the candidate hazard links under each. This is not a search for whichever model gives the preferred association. It tests whether the reported link contrast depends on forcing the biomarker to follow a straight path.
Plot repeated measurements and fitted subject-specific or population trajectories against time. Look for systematic bends in residuals, such as positive residuals in the middle of follow-up and negative residuals at both ends.
For a current-value link, inspect fitted values over the event-risk period. For a slope link, inspect estimated derivatives and their uncertainty. A fit that looks acceptable in levels can still imply implausible or overly rigid derivatives.
Keep the analysis choices not under assessment aligned, and compare the resulting fitted trajectories, association estimates, uncertainty, and hazard implications. Large changes point to dependence on trajectory structure, not solely to the hazard link.
State which trajectory forms were considered and whether the substantive link comparison persisted. If the result changes markedly, describe it as sensitive to longitudinal mean specification.
Residual plots are useful but not decisive: observations may be sparse near event times, and apparent curvature can be difficult to separate from noise. Compare plausible mean structures using longitudinal fit, scientific knowledge, and the stability of the joint-model conclusions. Avoid adding curvature unsupported by the observation schedule or data, since an overcomplicated trajectory can also make subject-specific derivatives unstable.
Alternatives, limits, and a common mistake
The main alternative to treating a linear trajectory as adequate is to allow curvature in the longitudinal mean, for example through a quadratic time term or a spline. A second option is to keep a simpler mean only when diagnostics and the scientific time scale support it. These choices address the longitudinal representation; they are separate from choosing a current-value, slope, or other hazard association. A fair link comparison evaluates candidate links under a trajectory structure that is credible for each, rather than letting different trajectory assumptions silently accompany different links.
There are limits to what this sensitivity check can establish. Better longitudinal fit does not prove that a particular hazard link is true. When visits are sparse, slopes can be weakly identified even under a curved mean. Flexible curves may also extrapolate poorly beyond observed visits, and the random-effects distribution or event process may be misspecified for other reasons. The exact decomposition of apparent sensitivity into trajectory and association components is generally not identifiable from a simple coefficient comparison alone.
- Common mistake: “The slope-link coefficient differs from the current-value coefficient, so the data show that rate of change has a different biological association with risk.”
- Correction: First check whether the fitted trajectory can represent the observed curvature and whether its derivative is credible. A linear mean makes the slope constant by construction; differences between link results may partly be a consequence of that constraint.
- Another pitfall: Adding a quadratic term and treating any coefficient change as proof that the earlier model was wrong. A changed estimate signals sensitivity, but interpretation also requires checking fit, uncertainty, and whether the added shape is plausible.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- A fitted linear trajectory has slope \(0.4\) biomarker units per year. The actual path is known to flatten late in follow-up. Why is a slope-link result especially worth checking?
- For the quadratic path \(m(t)=2+1.2t-0.15t^2\), calculate the true slope at year 2 and compare it with the least-squares linear slope from the worked example.
- In the projection example, what feature of the discrepancy caused the projected coefficient to exceed the true coefficient?
- Name one diagnostic pattern that could indicate an inadequate linear longitudinal mean, and one limitation of that diagnostic.
Solutions. (1) The linear model imposes the same slope at every time, so it cannot represent late flattening; the estimated derivative used by the hazard link may be wrong. (2) \(m'(2)=1.2-0.30(2)=0.600\) units per year, equal to the linear slope \(0.600\) at that particular time, though not at other times. (3) The discrepancy increased with the fitted value, giving a positive covariance term of \(0.400\) relative to its variance and raising the projected slope from \(0.40\) to \(0.56\). (4) A curved residual pattern, such as positive residuals in the middle and negative ones at both ends, is a warning; sparse or noisy visits can make such a pattern hard to detect or distinguish from random variation.