The Zero-Association Limit

What the Zero-Association Limit Means

A shared random-effects joint model links a subject’s longitudinal trajectory to their event hazard. The current-value association coefficient controls how the latent marker value enters that hazard. Setting this coefficient to zero gives a useful benchmark: it removes the current latent marker value from the event model, while leaving the longitudinal model itself intact.

The claim is precise, but narrower than “the marker and event are unrelated.” It concerns the event process conditional on modeled covariates and the specified joint model. Covariates can still connect marker levels and event risk, and other shared latent terms could create dependence if they were included.

Definition: Let \(Y_{ij}\) be the observed marker for subject \(i\) at time \(t_{ij}\), with latent trajectory \(m_i(t)=x_i(t)^\mathsf{T}\beta+z_i(t)^\mathsf{T}b_i\). A current-value joint model specifies \(h_i(t\mid b_i,w_i)=h_0(t)\exp\{w_i^\mathsf{T}\gamma+\alpha m_i(t)\}\). The zero-association restriction is \(\alpha=0\). Under that restriction, the event hazard does not depend on the subject’s latent trajectory or its random effects, conditional on the modeled survival covariates \(w_i\).

Here, \(h_0(t)\) is the baseline hazard, \(w_i\) is a vector of survival covariates, \(\gamma\) is its coefficient vector, and \(b_i\) is the vector of subject-specific random effects. A typical longitudinal observation model is \(Y_{ij}=m_i(t_{ij})+\epsilon_{ij}\), with measurement errors independent of \(b_i\) and of the event process conditional on the model. The coefficient \(\alpha\) has units inverse to those of the marker, so that \(\alpha m_i(t)\) is dimensionless.

The interpretation relies on the model having no other path from \(b_i\) or the latent trajectory into the event hazard. It also relies on the stated conditional event model being adequate. In examples below, covariates are treated as fixed, the event hazard has no additional shared frailty, and the observation schedule and censoring mechanism are handled under the course’s stated assumptions.

Deriving Conditional Independence

The hazard determines the event-time survival function conditional on the random effects. For a continuous event time, integrating the hazard gives the conditional survival probability. When \(\alpha=0\), the integrand no longer depends on \(b_i\).

\begin{aligned} S_i(t\mid b_i,w_i) &=\exp\left\{-\int_0^t h_i(s\mid b_i,w_i)\,ds\right\}\\ \alpha=0\quad &\Longrightarrow\quad h_i(s\mid b_i,w_i)=h_0(s)\exp(w_i^\mathsf{T}\gamma),\\ S_i(t\mid b_i,w_i) &=\exp\left\{-\exp(w_i^\mathsf{T}\gamma) \int_0^t h_0(s)\,ds\right\}. \end{aligned}

Thus the conditional survival distribution is the same for every value of \(b_i\), given \(w_i\). The event-time density, \(f_i(t\mid b_i,w_i)=h_i(t\mid b_i,w_i)S_i(t\mid b_i,w_i)\), is also unchanged by \(b_i\). Since both the event density and survival probability are invariant to the random effects, the event time \(T_i\) and \(b_i\) are conditionally independent given \(w_i\) in this model.

If the longitudinal measurement errors are also independent of the event process given the random effects and covariates, then the event time is conditionally independent of the full longitudinal process given the modeled covariates. This follows because the process is generated from \(b_i\) and measurement errors, while the event-time distribution at \(\alpha=0\) depends on neither. This argument concerns the underlying process; it does not say that a dataset’s number of recorded measurements is independent of event time. An event can end follow-up and thereby determine how many measurements were collected.

Likelihood Factorization at Zero

Let \(U_i=\min(T_i,C_i)\) be the observed follow-up time, where \(C_i\) is the censoring time, and let \(\delta_i=I(T_i\le C_i)\) indicate an observed event. Given \(b_i\), the event-time contribution is the event density when \(\delta_i=1\), or the survival probability when \(\delta_i=0\). Under conditionally independent censoring, its usual form is \(h_i(U_i\mid b_i,w_i)^{\delta_i}S_i(U_i\mid b_i,w_i)\).

For marker observations collected up to \(U_i\), write their conditional likelihood as \(L_{Y_i}(b_i)\). Conditional on the modeled observation schedule, the joint contribution integrated over the random effects is

L_i = \int L_{Y_i}(b_i) \left\{h_i(U_i\mid b_i,w_i)^{\delta_i} S_i(U_i\mid b_i,w_i)\right\} p(b_i)\,db_i.

At \(\alpha=0\), the event contribution no longer contains \(b_i\), so it can be taken outside the integral. This gives the factorization below.

\begin{aligned} L_i\big|_{\alpha=0} &= \left\{h_0(U_i)\exp(w_i^\mathsf{T}\gamma)\right\}^{\delta_i} \exp\left\{-\exp(w_i^\mathsf{T}\gamma) \int_0^{U_i}h_0(s)\,ds\right\} \int L_{Y_i}(b_i)p(b_i)\,db_i\\ &=L_{T_i}(U_i,\delta_i\mid w_i)\,L_{Y_i}^{\mathrm{marg}}. \end{aligned}

The last expression separates the event contribution from the marker likelihood marginalized over the random effects. This is a consequence of the zero restriction and the conditional model assumptions, not a general property of joint models. When \(\alpha\ne 0\), the event contribution depends on \(b_i\), so the integral generally cannot be split this way.

Worked Example: Equal Event Risk for Different Trajectories

Suppose two hypothetical subjects have random-intercept trajectories \(m_i(t)=8+b_i\), measured in marker units, with \(b_1=2\) and \(b_2=-1\). Let the baseline hazard be constant at \(0.04\) per year, let \(w_i=0\), and set \(\alpha=0\). Assume no other random effect enters the event model.

Their latent marker levels are \(m_1(t)=10\) and \(m_2(t)=7\). But for either subject the hazard is \(h_i(t)=0.04\exp(0+0\times m_i(t))=0.04\) per year.

For three years of follow-up, each subject’s event-free probability is \(S_i(3)=\exp(-0.04\times3)=\exp(-0.12)=0.8869\). The corresponding event probability by three years is \(1-0.8869=0.1131\) for each. The model permits different marker trajectories but, at \(\alpha=0\), assigns them the same event-time distribution conditional on the stated covariates.

This conclusion depends on the specified hazard containing no other subject-specific term. If a random effect were added directly to the hazard, equal current-value association at zero would not by itself remove that additional dependence.

Worked Example: Factoring a Marker Density and Event Contribution

Consider a hypothetical subject with \(b_i\sim N(0,1)\), one marker observation satisfying \(Y_i\mid b_i\sim N(2+b_i,0.5^2)\), and a constant event hazard of \(0.10\) per year. Set \(\alpha=0\), with no survival covariates. Suppose an event is observed at \(U_i=2\) years. The schedule and censoring assumptions are taken as given.

After integrating out \(b_i\), the marker has distribution \(Y_i\sim N(2,1+0.5^2)=N(2,1.25)\). Its density at the hypothetical observation \(Y_i=2\) is \[ f_Y(2)=\frac{1}{\sqrt{2\pi(1.25)}}=0.3568 \] per marker unit.

The event-time density contribution is \[ h_i(2)S_i(2)=0.10\exp(-0.10\times2) =0.10\exp(-0.2)=0.08187 \] per year. Because \(\alpha=0\), this contribution does not depend on \(b_i\). The joint density for the marker value and event time is therefore \(0.3568\times0.08187=0.02921\), per marker unit per year.

This is a joint density, not a probability for observing the exact continuous values. The calculation illustrates that the marker density is marginalized over \(b_i\) separately from the event contribution.

What Zero Association Does Not Say

The restriction removes the latent current marker value from the hazard. It does not force all marker-related covariates out of the survival model. If a covariate helps determine both the marker mean and event risk, the marker and event can be associated marginally even when \(\alpha=0\). Conditional independence is always relative to the covariates and latent terms included in the model.

Worked Example: A Shared Covariate Can Still Link Marker and Risk

Suppose a hypothetical binary covariate \(w_i\) determines both marker means and event risk. Let \(m_i(t)=5+2w_i+b_i\), let \(h_0(t)=0.02\) per year, let \(\gamma=\log 2\), and set \(\alpha=0\). Assume \(b_i\) does not enter the hazard through any other term.

For \(w_i=0\), the marker trajectory is \(5+b_i\) and the hazard is \(0.02\exp(0)=0.020\) per year. For \(w_i=1\), the trajectory is \(7+b_i\) and the hazard is \(0.02\exp(\log 2)=0.040\) per year. These calculations show a relationship between the covariate-defined marker means and event hazards, but no effect of the within-group random deviation \(b_i\) on the hazard. The zero current-value coefficient removes only the latter pathway in this model.

Diagnostics, Alternatives, and Limitations

The zero-association model is a restricted version of the model with a freely estimated \(\alpha\). A useful diagnostic is to compare the restricted fit with the unrestricted fit and examine whether the estimated association and its uncertainty are compatible with zero. A likelihood-ratio test can be used under its regularity conditions when the models are nested and zero is an interior parameter value; a one-parameter test then uses a chi-squared reference distribution with one degree of freedom. A profile likelihood can also reveal whether the data distinguish zero from practically meaningful nonzero values.

Raw correlations between marker summaries and event times are not a direct test of this restriction. Shared covariates can create marginal association, and event-related follow-up can leave early-event subjects with fewer measurements. Conversely, a near-zero estimate may reflect weak information about the association rather than compelling evidence that the latent trajectory is irrelevant. Check model fit for both components, the adequacy of the covariates and baseline hazard, and whether the assumed random-effects structure is reasonable.

The main alternatives answer different questions. A nonzero current-value model lets the hazard depend on the latent marker level; a model with a separate shared frailty can make event risk depend on latent subject characteristics through another term. A separate longitudinal and survival analysis may be appropriate when the conditional-independence assumptions are defensible, but it does not automatically account for dependence if those assumptions fail. None of these specifications establishes a causal effect of changing the marker.

Common pitfall: “The fitted current-value coefficient is zero, so the event time and every observed marker value are unconditionally independent.” This is too broad. The zero restriction gives conditional independence from the latent trajectory under the specified model, given modeled covariates and in the absence of other shared latent terms. Covariate pathways and event-dependent observation patterns can still produce observed-data associations.

Check Your Understanding

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

In a model with \(h_i(t\mid b_i,w_i)=h_0(t)\exp\{w_i^\mathsf{T}\gamma+\alpha m_i(t)\}\), what changes in the hazard when \(\alpha\) is set to zero?
With \(\alpha=0\), a constant baseline hazard of \(0.05\) per year, and no survival covariates, calculate the probability of remaining event-free for four years.
Why might marker level and event risk still be associated across subjects when \(\alpha=0\)?
Does the zero-association restriction imply that the number of recorded marker measurements is independent of event time? Explain.

Solutions. (1) The hazard becomes \(h_0(t)\exp(w_i^\mathsf{T}\gamma)\), with no dependence on \(m_i(t)\) or \(b_i\) through this current-value term. (2) The survival probability is \(\exp(-0.05\times4)=\exp(-0.20)=0.8187\). (3) Shared modeled covariates can affect both marker means and event risk; other shared latent terms could also do so if included. (4) No. An event can end follow-up and therefore limit the number of measurements, even though the event hazard does not depend on the latent trajectory under the zero restriction.

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