Why Link the Hazard to a Latent Current Value?
In a joint model, the longitudinal submodel describes how a marker changes over time, while the event submodel describes the instantaneous event rate. A current-value link connects these components by letting the event hazard at time \(t\) depend on the marker’s underlying expected value at that same time. This is different from using the most recent recorded marker: a recorded value includes measurement error and may have been collected before \(t\).
The link is useful when the scientific model says that the subject’s current underlying marker level is relevant to their event rate. It also makes the shared structure explicit: the longitudinal trajectory is estimated with subject-specific random effects, and its value at \(t\) becomes a predictor in the event model. This is a modeling choice, not a claim that an observed marker value causes an event.
Constructing the Link From the Longitudinal Model
A common longitudinal specification is a linear mixed model:
The vectors \(x_i(t)\) and \(z_i(t)\) contain the fixed- and random-effects design terms; \(\beta\) is the fixed-effects coefficient vector; and \(\epsilon_{ij}\) is measurement error. A standard mean-zero error assumption gives \(E\{\epsilon_{ij}\mid b_i,x_i(t_{ij}),z_i(t_{ij})\}=0\), so \(m_i(t_{ij})\) is the conditional expectation of the marker measurement at that time. The random effects allow different subjects to have different latent trajectories.
To construct the current-value link, evaluate the longitudinal mean function at the time \(t\) used in the hazard. Do not select a measurement index \(j\) and insert \(Y_{ij}\) instead. The observed measurement is \(m_i(t_{ij})+\epsilon_{ij}\); the current-value term is the modeled mean \(m_i(t)\), including the subject-specific random effects. It is defined at times when no measurement was taken as well as at visit times.
This construction rests on substantive and statistical assumptions. The longitudinal mean structure must represent the marker trajectory adequately over the relevant time range. The event model assumes that, conditional on its covariates and the shared random effects, the marker influences the hazard through the specified current value and that the log hazard is linear in that value. The model also assumes that the measurement error is not itself an additional event predictor. Conditional independence assumptions concern the remaining dependence between the longitudinal and event processes; they do not make the current-value link automatically correct.
Worked Example: Evaluating a Random-Intercept-and-Slope Trajectory
Suppose a hypothetical marker is measured in units per liter, and a subject’s fitted latent trajectory is \(m_i(t)=2.4+0.35t\), where \(t\) is in years. At year 4, the subject has an observed measurement of 4.6 units per liter. The event model uses \(\alpha=0.25\) per marker unit. Find the latent current value and its contribution to the hazard multiplier.
Evaluate the trajectory at \(t=4\):
The current-value term in the log hazard is \(\alpha m_i(4)=0.25(3.8)=0.95\). Thus its multiplicative contribution is
The observed value, 4.6, is not the value inserted into the current-value link. It contains measurement error relative to the modeled trajectory. Substituting it would give \(\exp\{0.25(4.6)\}=\exp(1.15)\approx3.158\), a different multiplier driven partly by that measurement error. The model uses 3.8 because it is the fitted conditional expected value at year 4.
Why the Expected Value, Not the Recorded Measurement?
The distinction follows directly from the longitudinal model. At an observation time, the measurement is a sum of the trajectory and an error term. Under the conditional mean-zero assumption, taking the expectation removes that term:
The event model therefore uses the trajectory component as the latent marker level. If instead the hazard predictor used \(Y_{ij}\), the predictor would inherit the measurement error and would only be available at recorded visit times without an extra rule for handling the intervals between visits. The current-value specification avoids making the hazard depend directly on that noisy observation.
This does not mean that \(m_i(t)\) is an exactly observed biological truth. It is a model-based conditional mean. Its quality depends on the longitudinal model, the available measurements, and the assumptions used to estimate the subject’s trajectory. The distinction is between a modeled latent mean and a noisy recorded value, not between an estimate and an infallible measurement of biology.
Worked Example: Combining the Current Value With an Event Covariate
Consider a hypothetical event model with baseline hazard \(\lambda_0(t)=0.010\) per month at a particular time. A subject is in treatment group \(w_i=1\), with treatment coefficient \(\gamma=0.30\). Their fitted trajectory is \(m_i(t)=1.1+0.20t+b_{0i}-0.05t\), where \(t\) is in months and \(b_{0i}=0.40\). Let \(\alpha=0.20\) per marker unit. Calculate the subject-specific hazard at month 6.
First evaluate the latent trajectory:
The linear predictor is \(\gamma w_i+\alpha m_i(6)=0.30(1)+0.20(2.40)=0.78\). Therefore,
This is a conditional instantaneous hazard at month 6 under the stated model, not a probability of an event during the month. The calculation uses both the event covariate and the latent current value; it does not replace the latter with a visit-specific observed measurement.
Scale, Centering, and the Baseline Hazard
The marker scale matters because \(\alpha m_i(t)\) must be dimensionless for the exponential link. If the marker is rescaled, its coefficient must be expressed on the corresponding new scale. Centering can make a trajectory easier to describe without changing the fitted hazard, provided the baseline hazard is adjusted consistently. For a constant \(c\), define \(m_i^c(t)=m_i(t)-c\). Then
Thus the centered formulation has the same hazard if its baseline hazard is \(\lambda_0^c(t)=\lambda_0(t)\exp(\alpha c)\). Centering changes the reference level represented by the baseline; it does not change the subject-specific hazard predictions when the transformation is applied consistently.
Worked Example: Centering a Marker Without Changing the Hazard
Suppose at a particular time \(\lambda_0(t)=0.008\) per day, the latent marker value is 54 units, and \(\alpha=0.03\) per unit. Center the marker at \(c=50\) units and verify that the hazard is unchanged.
Using the uncentered value, the hazard contribution from the marker is \(\exp\{0.03(54)\}=\exp(1.62)\approx5.053\). The hazard is therefore \(0.008(5.053)\approx0.04042\) per day.
The centered value is \(m_i^c(t)=54-50=4\) units. The centered baseline hazard is
The centered hazard is \(0.03585\exp\{0.03(4)\}=0.03585\exp(0.12)\approx0.04042\) per day, agreeing with the uncentered calculation up to rounding.
Alternatives, Diagnostics, and Limitations
The current-value link is not the only way to connect a longitudinal trajectory to an event. The alternatives encode different scientific assumptions. A shared-random-effects link puts one or more random effects directly in the event predictor; a slope link uses the trajectory’s rate of change; and a cumulative-value link uses information accumulated over an interval. These are not interchangeable descriptions of the same effect.
| Link choice | Event predictor uses | Use when the model is intended to represent |
|---|---|---|
| Current value | \(m_i(t)\) | The marker’s underlying level at time \(t\) |
| Shared random effects | Selected components of \(b_i\) | Association through subject-specific trajectory features represented by those effects |
| Current slope | \(m_i'(t)\) | The marker’s rate of change at time \(t\) |
| Cumulative value | An integral of \(m_i(s)\) over a specified interval | Association with accumulated marker exposure |
A current-value model can be inadequate if risk is associated primarily with rapid change or accumulated exposure rather than the current level. It can also be sensitive to misspecification of the longitudinal trajectory, especially where measurements are sparse or the model must extrapolate to event times far from observed visits. A linear trajectory may be an unsuitable representation of a marker with curvature; a more flexible mean function may be needed, but flexibility does not remove the need to check the event link itself.
Useful checks include examining whether the fitted longitudinal trajectories follow the observed marker patterns, looking for systematic residual patterns over time, and assessing whether the current-value relationship is plausible over the marker range actually supported by the data. Check whether conclusions depend strongly on the assumed trajectory shape or on sparse regions of follow-up. These checks can reveal problems, but no diagnostic proves that the latent marker is measured without error or that the chosen link is scientifically correct.
A frequent mistake is to insert the latest observed marker directly into the hazard because it is the most recent available number. That reasoning confuses a noisy measurement with the latent current value and can make the hazard predictor depend on measurement error and visit timing. The correction is to evaluate \(m_i(t)\) from the longitudinal model at time \(t\); the observed measurement informs estimation of the trajectory but is not itself the current-value term.
Check Your Understanding
Answer each question, then compare with the solutions below.
- If \(m_i(t)=3+0.2t+b_{0i}\), \(b_{0i}=0.5\), and \(t=5\), what current value enters the hazard?
- Why does a mean-zero measurement error assumption support using \(m_i(t)\), rather than \(Y_{ij}\), as the current-value term?
- If the marker is centered at \(c=10\), how must the baseline hazard change so that the hazard remains the same?
- Name one scientific situation where a current-slope link might be more suitable than a current-value link.
Solutions. (1) \(m_i(5)=3+0.2(5)+0.5=4.5\) marker units. (2) Conditional on the random effects and covariates, \(Y_{ij}=m_i(t_{ij})+\epsilon_{ij}\) and the conditional expectation of \(\epsilon_{ij}\) is zero, so the conditional expected measurement is \(m_i(t_{ij})\); the recorded value still contains error. (3) If \(m_i^c(t)=m_i(t)-10\), use \(\lambda_0^c(t)=\lambda_0(t)\exp(10\alpha)\). (4) A current-slope link may be more suitable when the event rate is hypothesized to depend on how rapidly the marker is changing at that time rather than its level.