Evaluating the Latent Marker at an Event Time

From a Longitudinal Trajectory to an Event-Time Hazard

A shared random-effects joint model connects a subject’s longitudinal marker trajectory with that subject’s event hazard. The trajectory is not just a line through recorded measurements: it is a model-based representation of the subject’s underlying marker level. In a current-value association model, the hazard at time $t$ depends on the value of that latent trajectory at that same time.

This tutorial focuses on the calculation at a specified event time. Given a mixed-model trajectory, the time, covariates, random effects, association coefficient, and baseline hazard, we first evaluate the latent marker and then calculate the conditional hazard. The result is conditional on the specified subject-specific random effects and covariates; it is not an average over a population of subjects.

Definition: In a current-value joint model, the latent current marker for subject $i$ at time $t$, written $m_i(t)$, is the longitudinal model’s conditional mean trajectory at that time, given the subject’s random effects and covariates. A common linear mixed-model form is $$ m_i(t)=\mathbf{x}_i(t)^\top\boldsymbol{\beta}+\mathbf{z}_i(t)^\top\mathbf{b}_i, $$ where $\mathbf{x}_i(t)$ and $\mathbf{z}_i(t)$ are the fixed- and random-effects design vectors, $\boldsymbol{\beta}$ is the fixed-effects vector, and $\mathbf{b}_i$ is subject $i$’s random-effects vector.

For a random-intercept-and-slope model, this can be written as $m_i(t)=\beta_0+\beta_1t+b_{0i}+b_{1i}t$, when there are no additional covariates in the longitudinal mean. Here $\beta_0$ and $\beta_1$ describe the population mean intercept and slope, while $b_{0i}$ and $b_{1i}$ are this subject’s deviations from them. The time $t$ must use the same origin and units used to define the model.

A longitudinal observation is often represented as $Y_{ij}=m_i(t_{ij})+\epsilon_{ij}$, where $\epsilon_{ij}$ is measurement error. Thus, $m_i(t)$ is the model’s underlying mean at time $t$, not the noisy measurement that might be observed at that time. This distinction matters even when the calculation is made at a time when a marker measurement is available.

The Conditional Current-Value Hazard

A common association structure uses the latent marker value as a predictor in a proportional-hazards model:

h_i(t\mid \mathbf{b}_i,\mathbf{w}_i) = h_0(t)\exp\left\{\boldsymbol{\gamma}^{\top}\mathbf{w}_i+\alpha m_i(t)\right\}.

Here $h_i(t\mid \mathbf{b}_i,\mathbf{w}_i)$ is subject $i$’s hazard at time $t$, conditional on the subject-specific random effects and the event-submodel covariates $\mathbf{w}_i$. The baseline hazard $h_0(t)$ is the hazard when the event-submodel covariate contribution and the marker contribution are both zero. The vector $\boldsymbol{\gamma}$ contains covariate coefficients, and $\alpha$ is the current-value association coefficient.

The hazard has units of inverse time, such as events per year. The exponent must be dimensionless. Consequently, $\alpha$ has units inverse to the units of the marker, while $\boldsymbol{\gamma}^{\top}\mathbf{w}_i$ is dimensionless under the chosen covariate coding. The exponential converts the linear predictor into a positive multiplicative factor on the baseline hazard.

Calculation rule: At a specified event time $t_i$, evaluate $m_i(t_i)$ using the subject’s trajectory, calculate $\eta_i(t_i)=\boldsymbol{\gamma}^{\top}\mathbf{w}_i+\alpha m_i(t_i)$, and then calculate $h_i(t_i\mid\mathbf{b}_i,\mathbf{w}_i)=h_0(t_i)\exp\{\eta_i(t_i)\}$.

Worked Evaluations

Worked Example: Random Intercept and Slope at the Event Time

Suppose a hypothetical subject has the longitudinal trajectory $m_i(t)=1.00+0.20t+0.20+0.00t$, with the last two terms representing $b_{0i}=0.20$ and $b_{1i}=0.00$. At the event time $t_i=4.5$ years, suppose the baseline hazard is $h_0(4.5)=0.12$ per year, the event covariate is $w_i=1$ with coefficient $\gamma=0.6315$, and the association coefficient is $\alpha=0.35$ per marker unit. Treat these model quantities as specified for this calculation.

Evaluate the latent marker. Substitute the time and subject’s random effects into the trajectory:

$$ m_i(4.5)=1.00+(0.20)(4.5)+0.20+(0.00)(4.5) =1.00+0.90+0.20+0.00=2.10. $$

Calculate the hazard predictor. The covariate contribution is $(0.6315)(1)=0.6315$, and the marker contribution is $(0.35)(2.10)=0.7350$. Thus $\eta_i(4.5)=0.6315+0.7350=1.3665$.

Calculate the conditional hazard.

h_i(4.5\mid b_i,w_i) =0.12\exp(1.3665) \approx 0.12(3.9216) \approx 0.47059 \approx 0.4706\ \text{per year}.

The subject’s hazard at 4.5 years, conditional on the specified random effects and covariate, is approximately 0.4706 per year. The calculation assumes that the supplied trajectory and hazard model are the intended models, that the time is measured in years as specified, and that the supplied baseline hazard is evaluated on that same time scale.

Worked Example: A Different Subject-Specific Slope

Consider a second hypothetical subject with $m_i(t)=1.00+0.25t+0.20-0.05t$. The event occurs at $t_i=4$ years. Let the event covariate be treatment status $w_i=1$, with $\gamma=0.40$, let $\alpha=0.25$ per marker unit, and suppose $h_0(4)=0.08$ per year.

The fixed-effects trajectory at year 4 is $1.00+(0.25)(4)=2.00$. The random-effects contribution is $0.20+(-0.05)(4)=0.00$. Therefore,

$$ m_i(4)=1.00+1.00+0.20-0.20=2.00. $$

The hazard predictor is $\eta_i(4)=(0.40)(1)+(0.25)(2.00)=0.40+0.50=0.90$. The resulting conditional hazard is

h_i(4\mid b_i,w_i)=0.08\exp(0.90) \approx 0.08(2.4596) =0.196768 \approx 0.1968\ \text{per year}.

The negative random slope reduces the subject’s trajectory slope relative to the fixed slope, but it does not make the current marker contribution negative: the marker value at year 4 is 2.00. The hazard calculation uses that value at the specified time, not the slope alone.

Worked Example: Evaluating a Time-Varying Baseline Hazard

For a third hypothetical subject, suppose $m_i(t)=0.80+0.30t-0.10+0.02t$. Let $t_i=3$ years, $\alpha=0.40$ per marker unit, and $\boldsymbol{\gamma}^{\top}\mathbf{w}_i=0$. Suppose the baseline hazard is Weibull, $h_0(t)=\lambda\kappa t^{\kappa-1}$, with $\lambda=0.02$ per year squared and $\kappa=2$.

First evaluate the trajectory:

$$ m_i(3)=0.80+(0.30)(3)-0.10+(0.02)(3) =0.80+0.90-0.10+0.06=1.66. $$

Next evaluate the baseline hazard at three years. With $\kappa=2$, $h_0(3)=(0.02)(2)(3)^{1}=0.12$ per year. The marker predictor is $\eta_i(3)=(0.40)(1.66)=0.664$. Hence,

h_i(3\mid b_i,w_i) =0.12\exp(0.664) \approx 0.12(1.9426) =0.233112 \approx 0.2331\ \text{per year}.

This example makes clear that the baseline hazard must be evaluated at the event time before applying the exponential multiplier. Replacing $h_0(3)$ with a constant unrelated to the specified Weibull function would give a different, incorrect conditional hazard.

Why the Substitution Works

The current-value model defines the conditional hazard through its baseline and exponential predictor. Once the longitudinal model supplies $m_i(t)$, the event-model predictor at a fixed time is completely determined by the specified covariates and parameters. Substitution therefore gives the calculation directly; no approximation is needed for evaluating the model at that time.

The hazard itself is an instantaneous rate, not the probability that an event occurs at exactly one time point. More precisely, for a small positive interval $\Delta t$, the conditional probability of an event in $[t,t+\Delta t)$, given survival just before $t$, is approximately $h_i(t\mid\mathbf{b}_i,\mathbf{w}_i)\Delta t$ when the hazard changes little over that interval. This explains why the result is reported per year in the examples, rather than as a probability.

Set the time and units.
Use the specified event time and the time origin, time scale, and units used by both the trajectory and baseline hazard.

Evaluate the latent trajectory.
Substitute that time, the fixed effects, relevant covariates, and the subject’s random effects into $m_i(t)$.

Form the event-model predictor.
Calculate $\boldsymbol{\gamma}^{\top}\mathbf{w}_i+\alpha m_i(t)$, checking that each term uses the specified coding and units.

Apply the baseline hazard.
Evaluate $h_0(t)$ at the same time, multiply by the exponential of the predictor, and report the hazard in inverse-time units.

Assumptions, Checks, and Alternatives

These calculations rely on a specified longitudinal model and a specified event model. The longitudinal model must define a meaningful trajectory at the requested time, including the correct fixed-effect terms, random-effect terms, and time coding. The event model must use the stated current-value link and baseline hazard, with the event covariates correctly coded. Conditional on the random effects and covariates, the hazard is defined by the displayed model. The numerical exercise treats model quantities as given; it does not account for uncertainty in estimated parameters or random effects.

Before trusting a numerical result, check that the marker value is plausible on the marker’s scale, that the random slope has been multiplied by time, and that covariate contributions have been included exactly once. Also check that the baseline hazard is positive and expressed per unit of the same time scale used for the event time. A conversion from years to months changes the numerical rate scale and must be applied consistently to the model, not just to the final answer.

A useful comparison is between a current-value model and an event model with no longitudinal-marker association. In the latter, $\alpha=0$, so $\exp\{\alpha m_i(t)\}=1$ and the marker does not multiply the baseline hazard. A current-value model instead lets subjects with different latent marker values have different conditional hazards, holding other included quantities fixed. A recorded marker value is not interchangeable with the latent value in this calculation: the longitudinal model separates the trajectory from measurement error. Which model is scientifically and statistically appropriate depends on the intended association structure and the data; a single conditional hazard calculation cannot establish that one structure fits better.

The result is conditional on the specified random effects. It should not be described as a population-average hazard, nor should the hazard value be read as an event probability. The calculation also inherits limitations of the specified models: a misspecified trajectory, unsuitable time scale, or poor representation of the baseline hazard can make a precisely executed substitution scientifically misleading. When a time is outside the range supported by the specified trajectory, the arithmetic may still be possible, but the model-based value may be less reliable.

Common Mistakes

Using the wrong marker value. The current-value link takes $m_i(t)$, not a noisy observation $Y_{ij}$. Evaluate the model trajectory at the event time.
Omitting the random slope’s time multiplier. In a random-intercept-and-slope trajectory, the random contribution is $b_{0i}+b_{1i}t$, not $b_{0i}+b_{1i}$.
Exponentiating only part of the predictor. Both the covariate term and the marker term belong inside the same exponential.
Rounding too early. Retain enough digits in the exponential and product, then round the final hazard. For example, $\exp(1.3665)\approx3.9216$, and $0.12(3.9216)\approx0.47059$, or 0.4706 per year.
Calling a hazard a probability. A hazard is an instantaneous rate. A probability over an interval requires an interval calculation, not just the hazard at one time.

Key takeaway: Evaluate the subject-specific latent trajectory at the event time first; then place that value, with the correctly coded covariates, inside the conditional hazard’s exponential predictor and multiply by the baseline hazard at that same time.

Check Your Understanding

Work each calculation using the stated models. Solutions are provided below the questions.

A trajectory is $m_i(t)=2+0.1t-0.2+0.03t$. Find $m_i(5)$.
At a time when $m_i(t)=1.5$, suppose $h_0(t)=0.10$ per year, $\alpha=0.20$ per marker unit, and the covariate contribution is 0.30. Find the conditional hazard.
If $\alpha=0$, does the current marker value change the hazard multiplier? Explain.
A student calculates $m_i(t)=\beta_0+\beta_1t+b_{0i}+b_{1i}$ for a random-intercept-and-slope model. Identify the error.

Solutions. (1) Substitute $t=5$: $m_i(5)=2+0.5-0.2+0.15=2.45$. (2) The predictor is $0.30+(0.20)(1.5)=0.60$, so $h_i(t)=0.10\exp(0.60)\approx0.10(1.8221)=0.1822$ per year. (3) No. If $\alpha=0$, then $\exp\{\alpha m_i(t)\}=\exp(0)=1$, so the marker contributes no multiplier. (4) The random slope must be multiplied by time: the random-effects contribution is $b_{0i}+b_{1i}t$.