Why the Time at Which Slope Is Evaluated Matters
A slope-based association relates the rate of change in a longitudinal marker to the event hazard. For a curved marker trajectory, however, the rate of change need not be constant. A hazard model can use the slope at the time the hazard is evaluated, or it can use the slope fixed at baseline. These are different association structures: one updates as the trajectory changes, while the other keeps the baseline rate as a subject-specific summary.
The distinction matters when trajectories accelerate or decelerate. Two subjects can have one ordering of baseline slopes and the opposite ordering of their slopes later. A hazard ratio based on the baseline slopes may therefore differ in magnitude or even direction from a hazard ratio based on slopes at the event-risk time.
In a shared-random-effects joint model, the longitudinal trajectory is commonly specified conditional on subject-specific random effects, and the event hazard is also conditional on those effects. A slope association can be written using either slope definition:
for an event-time slope link, or
for a baseline-slope link. Here \(h_i(t\mid b_i)\) is the subject-specific hazard conditional on the random effects \(b_i\), \(h_0(t)\) is the baseline hazard, \(w_i\) is a vector of other covariates, and \(\gamma\) is its coefficient vector. The parameters \(\alpha_E\) and \(\alpha_B\) are slope-association parameters. They need not be equal: the two links are distinct models, and each parameter is estimated under its own model.
If time is measured in years and the marker in units, a slope has units of marker-units per year. The corresponding \(\alpha\) has units of years per marker-unit, so that the exponent is dimensionless. A positive \(\alpha_E\) means that, holding other terms fixed, a higher instantaneous slope at time \(t\) is associated with a higher conditional hazard at \(t\). It does not by itself imply a causal effect.
What Curvature Does to the Slope
For a quadratic trajectory, the distinction can be derived directly. Suppose
Differentiating with respect to time gives the slope at time \(t\); setting \(t=0\) gives the baseline slope:
Thus the event-time slope differs from the baseline slope by \(2\beta_{2i}t\). If \(\beta_{2i}>0\), the slope increases over time; if \(\beta_{2i}<0\), it decreases. The two slopes agree at baseline, and they also agree at later times if the trajectory is linear (\(\beta_{2i}=0\)). This calculation assumes the trajectory is on the time scale and marker scale used in the hazard association. Changing the time origin changes which instant is called baseline.
Worked Hazard Calculations
Worked Example: One Curved Trajectory at Two Evaluation Times
Consider a hypothetical subject whose marker trajectory is \(m(t)=2+0.4t+0.1t^2\), with time in years and marker values in units. Assume a slope-only hazard association, equal other covariates, and association parameter \(\alpha=0.5\) years per marker-unit. Calculate the slope at baseline and at year 3, then compare the two model-implied hazard multipliers relative to a subject with slope zero.
Baseline slope. Differentiation gives \(m'(t)=0.4+0.2t\), in marker-units per year. Therefore \(m'(0)=0.4\) marker-units per year. The baseline-slope model gives the multiplier \(\exp\{0.5(0.4)\}=\exp(0.20)=1.221\).
Slope at year 3. Substitution gives \(m'(3)=0.4+0.2(3)=1.0\) marker-unit per year. The event-time slope model gives the multiplier \(\exp\{0.5(1.0)\}=\exp(0.50)=1.649\).
These are illustrative model-implied multipliers relative to slope zero, not a comparison of two observed subjects. If the same numerical association parameter is used solely to compare the two links, the event-time multiplier divided by the baseline-slope multiplier is \(\exp\{0.5(1.0-0.4)\}=\exp(0.30)=1.350\). Thus the event-time link gives a 1.350-fold multiplier relative to the baseline-slope link at year 3 under this illustration. In separately fitted models, the estimated association parameters can differ, so this ratio is not generally a direct empirical comparison.
Worked Example: Curvature Reverses a Between-Subject Comparison
Two hypothetical subjects have trajectories \(m_A(t)=1+0.3t+0.08t^2\) and \(m_B(t)=1.2+0.5t+0.02t^2\), with years and marker units as above. At year 2, compare subject A’s hazard with subject B’s under each slope link. Assume the same other covariates and, for transparent arithmetic, use \(\alpha=1.2\) years per marker-unit in both illustrative calculations.
Their slopes are \(m_A'(t)=0.3+0.16t\) and \(m_B'(t)=0.5+0.04t\). At baseline, the slopes are \(0.3\) and \(0.5\) marker-units per year, so the baseline-slope difference (A minus B) is \(0.3-0.5=-0.2\). The baseline-slope hazard ratio is \(\exp\{1.2(-0.2)\}=\exp(-0.24)=0.7866\). Under that link, A’s conditional hazard is about 0.787 times B’s, at the same time and conditional on equal other covariates.
At year 2, \(m_A'(2)=0.3+0.16(2)=0.62\), while \(m_B'(2)=0.5+0.04(2)=0.58\) marker-units per year. The event-time slope difference is \(0.62-0.58=0.04\), so the event-time hazard ratio is \(\exp\{1.2(0.04)\}=\exp(0.048)=1.049\). At year 2, the event-time link associates A’s slightly higher slope with a hazard about 1.049 times B’s. The ordering reverses because the two trajectories have different curvature. The stated direction follows from the signed differences; an unsigned difference alone would not determine a hazard ratio.
Worked Example: When the Two Links Give the Same Comparison
Suppose two hypothetical trajectories are \(m_C(t)=3+0.6t\) and \(m_D(t)=2+0.2t\). Both are linear, so their slopes are \(0.6\) and \(0.2\) marker-units per year at baseline and at every later time. With equal other covariates and \(\alpha=0.7\) years per marker-unit, the signed slope difference C minus D is \(0.6-0.2=0.4\). Under either link, the hazard ratio is \(\exp\{0.7(0.4)\}=\exp(0.28)=1.323\).
The agreement here follows from constant slopes, not from the two model definitions being identical in general. If both trajectories had the same nonzero curvature, their individual slopes would change over time, but their slope difference would remain constant; with the same association parameter and the same covariates, the between-subject hazard ratio would then also remain the same across evaluation times.
Interpreting and Checking the Association
For two subjects \(i\) and \(j\) evaluated at the same time \(t\), an event-time slope model implies, when other covariates are equal,
The corresponding baseline-slope ratio replaces both slopes at \(t\) with their baseline values and uses \(\alpha_B\). The difference must be signed: if subject \(i\)'s slope is lower, the difference \(m_i'-m_j'\) is negative. A hazard ratio greater than 1 means a higher conditional hazard for the numerator subject at that time, under the stated model and covariate comparison; it is not a probability of experiencing the event.
Confirm which instant is \(t=0\), and use the time and marker scales specified by the model.
For an event-time link, evaluate the derivative at the hazard time \(t\). For a baseline link, evaluate it at \(0\).
Use the numerator slope minus the denominator slope, then multiply by the association parameter.
Compute the exponential of the signed product and state the evaluation time, units, and covariate conditions.
The interpretation relies on several modeling conditions. The latent trajectory must be differentiable at the evaluation time, and its functional form must adequately describe the marker process in the relevant time range. The hazard link must specify the intended slope—baseline or event-time—and the conditional hazard model must be appropriate for the risk process. A derivative calculated from a fitted latent trajectory is not the same as a raw difference between two noisy measurements. Measurement timing and noise can make an observed finite difference unstable; it should not silently replace the model-defined derivative.
Useful checks include plotting fitted trajectories and their implied slopes over the risk period, checking whether estimated curvature is plausible, and examining whether slope estimates depend strongly on sparse observations or extrapolation. A polynomial can fit observed values yet imply implausible slopes outside the observed time range. Also check that the time origin and units are consistent: re-expressing years as months changes numerical slope values and requires a corresponding change in the association coefficient for the linear predictor to retain its meaning.
How This Link Differs From Alternatives
| Association structure | Quantity entering the hazard | What changes over time? |
|---|---|---|
| Baseline slope | \(m_i'(0)\) | The subject’s slope contribution is fixed after baseline. |
| Event-time slope | \(m_i'(t)\) | The contribution can evolve with the trajectory’s curvature. |
| Current value | \(m_i(t)\) | The marker level, rather than its rate of change, evolves with time. |
A current-value association answers a different question from either slope link: it relates the hazard to the latent marker level at the evaluation time. A baseline-slope link can be attractive when the scientific target is a fixed early rate or when updating the slope is not desired. An event-time slope link is appropriate when the instantaneous rate of change is the intended time-varying association. Neither choice is automatically preferable; the target interpretation and the plausibility of the trajectory model should guide the choice.
Common Mistakes and Limitations
- Using the baseline derivative by habit. In a curved trajectory, \(m_i'(0)\) is not generally \(m_i'(t)\). Differentiate and substitute the required time.
- Dropping the sign of a slope difference. A magnitude alone cannot identify the direction of a hazard ratio. State which subject has the greater slope.
- Calling the event-time slope an observed instantaneous measurement. It is typically a derivative of the model-based latent trajectory, not a directly observed quantity.
- Assuming equal association parameters across links. The event-time and baseline-slope specifications are different models; \(\alpha_E\) and \(\alpha_B\) are generally estimated separately.
- Ignoring the time origin or units. Baseline slope depends on the chosen origin, while numerical slopes depend on time units. Report the scale used.
Both links also inherit uncertainty from estimating the longitudinal trajectory and its random effects. In practice, a slope association should be interpreted as part of the joint model, not as if each latent slope were known without error. A simple polynomial derivative can be sensitive to model misspecification, especially near the edges of observed follow-up. These limitations do not erase the distinction: they make it more important to specify precisely which slope the hazard uses.
Check Your Understanding
Answer each question, then compare with the concise solutions below.
- For \(m(t)=4+0.5t-0.03t^2\), find the baseline slope and the slope at year 5, in marker-units per year.
- At year 5, subject A’s slope is 0.8 and subject B’s is 0.5 marker-units per year. If \(\alpha_E=0.6\) years per marker-unit, calculate the event-time hazard ratio A relative to B, holding other covariates equal.
- When do the event-time and baseline slopes agree for a quadratic trajectory?
- Why is a signed slope difference needed to calculate a hazard ratio?
- For two trajectories with the same quadratic coefficient, what happens to their slope difference over time?
Solutions. 1. \(m'(t)=0.5-0.06t\), so \(m'(0)=0.5\) and \(m'(5)=0.5-0.30=0.20\) marker-units per year. 2. The signed difference is \(0.8-0.5=0.3\); the ratio is \(\exp\{0.6(0.3)\}=\exp(0.18)=1.197\). A’s conditional hazard is 1.197 times B’s at that time. 3. They agree at \(t=0\), or at all times if the quadratic coefficient is zero. 4. The exponential uses the signed contrast; reversing its direction changes the sign in the exponent and reciprocates the hazard ratio. 5. If the quadratic coefficients match, the time-dependent curvature terms cancel in the slope difference, so the difference is constant.