A Random-Intercept Joint Model Worked Example

Why Share a Random Intercept?

Repeated marker measurements on the same person are related: some people tend to have higher marker values throughout follow-up than others. A random-intercept model represents this persistent subject-to-subject variation with a latent quantity \(b_i\). A joint model can then use that same \(b_i\) to connect the marker trajectory to the event hazard. This is important when the underlying marker level is associated with event risk and the observed measurements are noisy.

The focus here is a current-value association: conditional on a subject’s random intercept, the hazard at time \(t\) depends on the latent marker value at that time. The derivation makes explicit how the shared latent effect enters both submodels and why the subject-level likelihood averages over its distribution rather than treating it as observed.

Model and Assumptions

Definition: For subject \(i\), let \(Y_{ij}\) be the marker measurement at time \(t_{ij}\), and let \(b_i\) be a subject-specific random intercept. A random-intercept longitudinal model is

Y_{ij}=m_i(t_{ij})+\epsilon_{ij},\qquad m_i(t)=\beta_0+\beta_1t+b_i.

Here \(m_i(t)\) is the subject’s latent marker trajectory, \(\epsilon_{ij}\sim N(0,\sigma^2)\) are conditionally independent measurement errors, and \(b_i\sim N(0,\tau^2)\). The random intercept and measurement errors are independent. In the current-value survival submodel,

h_i(t\mid b_i)=h_0(t)\exp\{\gamma^\mathsf{T}w_i+\alpha m_i(t)\},

where \(h_0(t)\) is the baseline hazard, \(w_i\) is a vector of observed covariates, \(\gamma\) is its coefficient vector, and \(\alpha\) is the marker-association parameter.

The hazard is conditional on the latent intercept, not on the noisy observed value \(Y_{ij}\). For fixed covariates and \(b_i\), a one-unit increase in the latent marker value multiplies the hazard by \(\exp(\alpha)\). This is a conditional association; it is not, by itself, a causal effect.

The model assumes that the specified longitudinal mean and error distribution are adequate, and that the random intercept follows the stated normal distribution. Conditional on \(b_i\) and the observed covariates, the event process has the hazard shown above. The marker errors and event process are conditionally independent given \(b_i\) and covariates, so their likelihood contributions can be multiplied before integrating over \(b_i\). The observation schedule must also be ignorable for the likelihood used here: informally, the timing or presence of marker measurements must not carry additional information about the unobserved marker or event process beyond what the model accounts for. Censoring is assumed conditionally independent under the event-time model. These conditions connect to the earlier material on conditional independent censoring and assumptions about the observation schedule; they are modeling assumptions, not consequences of choosing a random intercept.

Conditional Hazard and Event-Time Contribution

Write \(T_i\) for the observed event or censoring time and \(\delta_i=1\) if the event was observed, with \(\delta_i=0\) for right censoring. Given \(b_i\), the cumulative hazard through time \(t\) is

H_i(t\mid b_i)=\int_0^t h_i(u\mid b_i)\,du.

The conditional survival function is \(S_i(t\mid b_i)=\exp\{-H_i(t\mid b_i)\}\). Thus the event-time contribution, including right censoring, is \(h_i(T_i\mid b_i)^{\delta_i}S_i(T_i\mid b_i)\). An observed event contributes a density; censoring contributes the probability of surviving to the censoring time.

Substituting the random-intercept trajectory into the hazard gives a useful explicit form:

h_i(t\mid b_i)=h_0(t)\exp\{\gamma^\mathsf{T}w_i+\alpha(\beta_0+\beta_1t+b_i)\}.

The random intercept therefore acts as a multiplicative factor \(\exp(\alpha b_i)\) on the hazard, while the fixed time trend contributes \(\exp(\alpha\beta_1t)\). Unless \(\beta_1=0\) or the baseline hazard has a special form, the cumulative hazard requires integrating this time-dependent expression; it is not generally \(t\) times the hazard at the endpoint.

Worked Example: A Constant Latent Marker Trajectory

Suppose \(h_0(t)=0.1\) per year, \(\gamma^\mathsf{T}w_i=0\), and the marker trajectory for a subject is constant at \(m_i(t)=0.8\). Let \(\alpha=0.25\). Find the conditional hazard, and the event-time contributions for an event at two years and censoring at two years.

The conditional hazard is constant: \(h_i(t\mid b_i)=0.1\exp(0.25\times0.8)=0.1\exp(0.2)=0.1221403\) per year. Hence \(H_i(2\mid b_i)=2\{0.1\exp(0.2)\}\approx0.2443\).

For an observed event at two years, the conditional contribution is \(h_i(2\mid b_i)\exp\{-H_i(2\mid b_i)\} =0.1\exp(0.2)\exp\{-2(0.1\exp(0.2))\}\approx0.0957\) per year. For censoring at two years, it is \(\exp\{-2(0.1\exp(0.2))\}\approx0.7833\), a survival probability. The two contributions have different interpretations and units: an event density is per year, whereas a censoring contribution is dimensionless.

Constructing the Subject-Level Likelihood

For a subject with measurements \(Y_{i1},\ldots,Y_{in_i}\), conditional independence gives the longitudinal density

p(\mathbf{Y}_i\mid b_i) =\prod_{j=1}^{n_i}\frac{1}{\sqrt{2\pi\sigma^2}} \exp\left\{-\frac{[Y_{ij}-m_i(t_{ij})]^2}{2\sigma^2}\right\}.

Multiplying this by the conditional event-time contribution and the random-effect density gives the joint density conditional on \(b_i\), up to the common parameter notation:

p(\mathbf{Y}_i,T_i,\delta_i,b_i) =p(\mathbf{Y}_i\mid b_i)\, h_i(T_i\mid b_i)^{\delta_i} \exp\{-H_i(T_i\mid b_i)\}\, \phi(b_i;0,\tau^2).

The observed-data subject likelihood does not condition on an observed random intercept, because \(b_i\) is latent. It integrates the conditional likelihood over its normal distribution:

L_i =\int_{-\infty}^{\infty} p(\mathbf{Y}_i\mid b)\, h_i(T_i\mid b)^{\delta_i} \exp\{-H_i(T_i\mid b)\}\, \phi(b;0,\tau^2)\,db.

This is the shared-random-effects construction: the same value of \(b\) determines the expected marker measurements and the hazard, so the two data components inform the same latent quantity. The integral is generally not available in closed form because the survival contribution contains an exponential of an integrated hazard. Numerical integration, often Gaussian quadrature, is therefore used in estimation.

Worked Example: Conditional Likelihood at a Given Intercept

Consider a hypothetical subject with \(m_i(t)=1+0.2t+b_i\), measurement-error standard deviation \(\sigma=0.5\), and \(b_i=0.3\). The subject has measurements \(Y_{i1}=1.4\) at \(t_{i1}=0\) and \(Y_{i2}=1.6\) at \(t_{i2}=1\). Let the baseline hazard be \(h_0(t)=0.08\) per year, \(\alpha=0.5\), and no survival covariates. The event occurs at \(T_i=2\). Calculate the conditional joint contribution of the marker data and event time, given \(b_i=0.3\).

The fitted latent means at the measurement times are \(1.3\) and \(1.5\), so each residual is \(0.1\). Each marker density is \((0.5\sqrt{2\pi})^{-1}\exp\{-0.1^2/(2\times0.5^2)\} =0.7979\exp(-0.02)\approx0.7821\). The longitudinal contribution is therefore \(0.7821^2\approx0.6117\).

At the event time, \(m_i(2)=1+0.2(2)+0.3=1.7\), giving \(h_i(2\mid b_i)=0.08\exp(0.5\times1.7)\approx0.1872\) per year. Across follow-up, the hazard is \(0.08\exp\{0.5(1+0.2t+0.3)\}=0.08\exp(0.65+0.1t)\). Thus

H_i(2\mid b_i) =0.08\exp(0.65)\frac{\exp(0.2)-1}{0.1} \approx0.3393.

The conditional event contribution is \(0.1872\exp(-0.3395)\approx0.1333\) per year. Multiplying by the marker contribution gives \(0.6117(0.1333)\approx0.0816\) per year per squared marker unit. This is conditional on the specified intercept; it is not yet the observed-data likelihood, which must also average over possible intercepts.

Marginalizing Over the Shared Intercept

The integral in \(L_i\) is an expectation under the random-intercept distribution. For \(b_i\sim N(0,\tau^2)\), a change of variable \(b=\tau z\), where \(z\sim N(0,1)\), writes it as an expectation over a standard normal variable. In practice, quadrature approximates this expectation by a weighted sum. Each quadrature point must be used in both submodels: evaluating the marker density at one intercept and the survival contribution at another would break the shared-effects model.

The following calculation illustrates that integration. It uses a three-point Gaussian quadrature approximation; the result is approximate because the integrand is not a low-degree polynomial.

Worked Example: A Quadrature Approximation to the Subject Likelihood

Suppose one measurement \(Y_i=1.2\) is recorded at time zero, with \(m_i(0)=1+b_i\), \(\sigma=0.4\), and \(b_i\sim N(0,0.25)\). Let \(h_0(t)=0.1\) per year, \(\alpha=0.4\), no survival covariates, and an observed event at \(T_i=2\). Three-point standard-normal quadrature uses nodes \(z=0,\pm\sqrt{3}\), with weights \(2/3,1/6,1/6\); here the corresponding intercepts are \(b=0,\pm0.8660\).

At each node, multiply the marker density by the event density \(h_i(2\mid b)\exp\{-2h_i(2\mid b)\}\), since the marker trajectory is constant in time in this example. At \(b=0\), the marker density is \(0.9974\exp(-0.125)\approx0.8802\), and the event contribution is \(0.1\exp(0.4)\exp\{-0.2\exp(0.4)\}\approx0.1107\). Their product is approximately \(0.0975\).

At \(b=0.8660\), the marker density is approximately \(0.2494\); the event contribution is \(0.1\exp(0.4(1+0.8660))\exp\{-0.2\exp(0.4(1+0.8660))\}\approx0.1383\). The product is about \(0.0345\). At \(b=-0.8660\), the corresponding density and event contribution are about \(0.02862\) and \(0.08545\), giving a product of \(0.002445\).

Weighting and adding the three products gives \((2/3)(0.0975)+(1/6)(0.0345)+(1/6)(0.002444)\approx0.0711\) per year per marker unit. The weights sum to one, and the calculation uses the same intercept at each node for both data components. A fitted likelihood would use more accurate numerical integration as needed and evaluate the approximation over all subjects.

Diagnostics, Alternatives, and Limitations

A valid formula does not guarantee an adequate model. Check whether the longitudinal mean is compatible with the observed time trends and whether residual variation appears roughly constant and plausibly Gaussian after accounting for the fitted trajectory. Examine whether the random-intercept distribution is a reasonable description of between-subject heterogeneity. On the event side, assess the chosen baseline hazard and whether the current-value link is scientifically and empirically plausible. Numerical integration should be checked by increasing the number of quadrature points and verifying that log-likelihoods and estimates remain stable.

Warning signs include unstable estimates, strong sensitivity to the integration settings, implausible variance estimates, and systematic residual patterns. These may indicate a poorly specified longitudinal mean, an inadequate random-effects distribution, an unsuitable hazard specification, or insufficient event information. A random-intercept model also imposes a common time trend \(\beta_1\) across subjects: subjects differ in level but not in slope. A random-intercept-and-slope model relaxes that restriction, but is a distinct model.

A separate longitudinal and survival analysis does not use the shared latent intercept to represent their dependence. Plugging noisy observed marker values directly into the hazard is also not equivalent: it treats measurement error as if it were absent and does not integrate uncertainty in \(b_i\). Conversely, the shared-random-intercept model has its own limits. Its conclusions depend on the distributional and conditional-independence assumptions, and the conditional hazard interpretation should not be confused with a marginal population hazard. Alternative baseline hazard specifications, such as Weibull or piecewise-constant forms, change the hazard calculation but not the basic multiply-then-integrate likelihood construction.

Common pitfall: It is tempting to estimate \(b_i\) from the marker observations, substitute that estimate into the hazard, and call the result the joint likelihood. That yields a plug-in conditional calculation, not the observed-data likelihood. The latter averages the product of the longitudinal and survival contributions over the distribution of \(b_i\), as in the integral for \(L_i\).

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If \(\alpha=0\), what happens to the dependence of the conditional hazard on the latent marker trajectory?
For a constant hazard of \(0.2\) per year, what are the event contribution at \(T=1\) and the right-censoring contribution at \(T=1\)?
Why must the marker and event-time contributions be evaluated at the same random-intercept value before integration?
In the random-intercept model, what feature of individual trajectories is constrained to be the same across subjects?

Solutions. (1) The marker term becomes \(\exp\{0m_i(t)\}=1\), so the conditional hazard no longer depends on the latent marker. (2) For an event, \(0.2\exp(-0.2\times1)\approx0.1637\) per year; for censoring, \(\exp(-0.2)\approx0.8187\). (3) The random intercept is shared: it jointly determines the marker mean and hazard, so the conditional joint density is their product at one common value of \(b_i\), subsequently averaged over its distribution. (4) The fixed time slope \(\beta_1\) is common; subjects may have different levels through \(b_i\), but this model does not allow their slopes to vary.