Treating Observed Biomarkers as Error-Free

Why an Observed Reading Is Not the Trajectory

A dynamic prediction uses a patient’s biomarker history to update the estimated risk of a future event. The measurements in that history are not usually the underlying biological values themselves. They contain measurement error, while the latent trajectory represents the evolving marker level that the joint model relates to the event hazard. Treating a noisy reading as exact can therefore change a patient’s predicted risk, sometimes substantially.

The issue is not whether the observed measurement is useful: it is the evidence used to learn about the latent trajectory. The issue is whether the prediction accounts for the uncertainty that remains after observing it. This distinction is especially important when measurements are sparse or noisy, and when the event hazard changes sharply with the marker.

Definition: Let \(Y_i(t_{ij})\) be the observed biomarker for patient \(i\) at measurement time \(t_{ij}\), and let \(m_i(t)\) be the patient’s latent marker trajectory. A common measurement model is \(Y_i(t_{ij})=m_i(t_{ij})+\epsilon_{ij}\), where the measurement errors have mean zero conditional on the latent trajectory and are independent of the event process conditional on the latent trajectory and model covariates. An error-free plug-in prediction replaces a latent marker value used in the hazard with an observed value as though they were equal.

The measurement model is an assumption, not a universal law. It may need to include a nonzero measurement bias, time-varying measurement variance, or other structure. The comparison below assumes the standard zero-mean error model and a correctly specified joint model. For clarity, numerical examples treat model parameters as known; practical predictions can also include parameter uncertainty.

Landmark Risk With Latent-Marker Uncertainty

Let \(s\) be the prediction landmark, \(u\) the prediction horizon, \(T_i\) the event time, and \(H_i(s)\) the biomarker measurements available by \(s\). The target is the probability of an event in \((s,s+u]\), conditional on being event-free at \(s\), the observed history, and any specified covariates. In a joint model with random effects \(b_i\), the event hazard may depend on the latent trajectory:

h_i(t\mid b_i)=h_0(t)\exp\{\gamma^\top w_i+\alpha m_i(t;b_i)\}.

Here \(h_0(t)\) is the baseline hazard, \(w_i\) denotes covariates, and \(\alpha\) describes the association between the latent marker and the hazard. Given \(b_i\), the event-free probability from \(s\) to \(s+u\) is the exponential of minus the cumulative hazard over that interval. A prediction that accounts for marker uncertainty averages this conditional event-free probability over the random-effects distribution given the available evidence:

R_i(s,u)=1-\int \exp\left\{-\int_s^{s+u}h_i(v\mid b_i)\,dv\right\} p\{b_i\mid H_i(s),T_i>s;\theta\}\,db_i.

The posterior distribution in this expression is conditioned both on the biomarker history and on survival to the landmark. In particular, the observed measurements inform the latent trajectory through the longitudinal model; they do not replace it. The survival condition also matters because the fact that a patient remained event-free through \(s\) provides information about their event-related random effects.

The contrast is the plug-in calculation: insert the latest observed value \(Y_i(s)\) wherever the hazard model calls for the marker, then calculate the risk as if this value were the relevant latent value and known without error. If the hazard depends on the marker’s future path, substituting a single reading does not by itself specify that path; the plug-in procedure must make an additional assumption about how the reading represents future marker values.

Why Integration and Plugging In Differ

The difference follows from the nonlinearity of survival probabilities. For a given latent trajectory, risk is \(1-\exp\{-\text{cumulative hazard}\}\). Averaging that risk across plausible trajectories is generally not the same as calculating risk at one observed value or even at the posterior mean trajectory. The prediction target is an average over uncertainty, not simply a transformation of an average marker.

A short derivation makes this explicit. Suppose the marker is constant over the horizon at a latent value \(m\), covariates are fixed, and the cumulative baseline hazard over the horizon is \(c\). Absorb any fixed covariate contribution into \(c\), and write the risk as \(g(m)\). Then:

g(m)=1-\exp\{-c\exp(\alpha m)\}, \qquad R=E\{g(M)\mid H_i(s),T_i>s\}.

The second derivative is

g''(m)=\alpha^2 x\exp(-x)(1-x), \qquad x=c\exp(\alpha m).

Thus, for a positive association \(\alpha\), risk is locally convex where \(x<1\) and locally concave where \(x>1\). When uncertainty is small, a second-order Taylor expansion around the posterior mean \(\mu_M\) gives \(E\{g(M)\}\) approximately equal to \(g(\mu_M)+\tfrac12g''(\mu_M)\operatorname{Var}(M)\). This is an approximation, not an identity. It shows why uncertainty may raise risk in one range and lower it in another. In addition, the observed value can lie above or below the latent value inferred from the full history, so a plug-in prediction can err in either direction.

Worked Example: A Noisy Reading Produces a Higher Plug-In Risk

Suppose the prediction is for the next interval after landmark \(s\). For illustration, assume the marker is constant over that interval, the association parameter is \(\alpha=\log(2)\), and the cumulative baseline hazard is \(c=0.10\). The longitudinal and survival information together give a two-point posterior distribution for the latent marker: \(M=0\) with probability \(0.5\), and \(M=2\) with probability \(0.5\). This distribution is already conditional on the observed history and survival to \(s\). The latest observed reading is \(Y_i(s)=2\).

Integrated prediction. At \(M=0\), the cumulative hazard is \(0.10\exp\{(\log 2)0\}=0.10\), so the risk is \(1-\exp(-0.10)=0.09516\). At \(M=2\), it is \(0.10\exp\{2\log 2\}=0.40\), so the risk is \(1-\exp(-0.40)=0.3297\). Averaging over the posterior gives

$$ R=0.5(0.09516)+0.5(0.3297)=0.2124. $$

Error-free plug-in prediction. Treating the observed value \(2\) as the exact marker gives \(1-\exp\{-0.10\exp(2\log 2)\}=1-\exp(-0.40)=0.3297\). It exceeds the integrated prediction by \(0.1173\), or 11.73 percentage points. This discrepancy is possible because the plug-in calculation discards the posterior possibility that the latent marker is 0.

Worked Example: A Local Approximation Shows the Role of Variance

Consider the same constant-marker risk function, now with \(\alpha=0.8\), \(c=0.10\), and a posterior distribution for \(M\) with mean \(0.5\) and variance \(0.25\). At the posterior mean, \(x=0.10\exp(0.8\times0.5)=0.1492\), and the plug-in risk is \(g(0.5)=1-\exp(-0.1492)=0.1386\).

To approximate the effect of posterior variance, use the second-order expression. At \(m=0.5\), the curvature is

g''(0.5)=0.8^2(0.1492)\exp(-0.1492)(1-0.1492)\approx 0.0700.

Therefore the approximation to integrated risk is \(0.1386+\tfrac12(0.0700)(0.25)=0.1473\). The approximation is about 0.00875 higher than risk at the posterior mean. Since \(x<1\), the risk function is locally convex at this marker value, consistent with the positive correction. This is a local approximation; the exact integrated risk would require the full posterior distribution, not only its mean and variance.

Worked Example: Equal Posterior Means Need Not Give Equal Risk

Use \(\alpha=\log(2)\) and \(c=0.10\) again. Compare two posterior distributions, each with mean marker value 1. In the first, \(M=1\) with probability 1. In the second, \(M=0\) or \(M=2\), each with probability 0.5. The first distribution gives risk \(1-\exp\{-0.10\exp(\log 2)\}=1-\exp(-0.20)=0.1813\).

For the second distribution, the integrated risk is \(0.5(1-\exp(-0.10))+0.5(1-\exp(-0.40))=0.2124\), as calculated above. The two predictions differ by \(0.03115\), or 3.115 percentage points, despite having the same posterior mean. The posterior spread matters because risk is nonlinear in the marker. In applications, replacing the posterior distribution by its mean alone can therefore also miss relevant uncertainty.

Assumptions, Diagnostics, and Alternatives

The integrated prediction is meaningful only relative to its model. The longitudinal model must represent the relationship between observed readings and latent trajectories adequately, and the event model must relate the hazard to the appropriate latent marker quantity. The measurement-error assumptions must be plausible, and the conditional distribution of random effects must account for the observed history and survival to the landmark. If the hazard depends on future marker values, those values must be specified or integrated over according to the prediction target and model. A constant-marker calculation is useful for illustration, but is not a general substitute for modeling a time-varying trajectory.

Useful checks focus on whether the measurement and event models support the intended prediction:

Compare observed measurements with fitted longitudinal trajectories and inspect residual patterns over time. Persistent trends or changing residual spread may indicate that the measurement model is inadequate.
Check whether the estimated measurement-error variance is plausible in the context of the measurement process. When measurement error is appreciable, a single observed reading may be a particularly poor stand-in for the latent value.
Inspect the posterior distribution of the latent trajectory at and after the landmark. A wide or skewed posterior indicates that a point substitution may hide meaningful uncertainty.
Check the event-model form and the implied risk over the marker range. Strong curvature or rapidly changing risk makes differences between plug-in and integrated predictions more consequential.
Recalculate predictions under plausible changes to measurement-error assumptions and compare them. Large changes signal that conclusions rely on those assumptions.

Three approaches should be kept distinct. An observed-value plug-in treats the reading as the latent value and is easiest to calculate, but ignores measurement error. A posterior-mean plug-in uses an estimated latent value, which can be more reasonable than using the raw reading but still ignores uncertainty and generally does not equal integrated risk. A posterior integration averages conditional risk over plausible latent trajectories given the history and survival to the landmark. It is the natural model-based approach when the prediction target concerns risk given the observed history, but it depends on the joint-model assumptions and may require numerical integration.

The fact that a model accounts for measurement error does not guarantee a correct prediction. A misspecified trajectory, event hazard, or measurement process can still produce misleading results. Nor is integration always numerically simple: the earlier material on Gauss-Hermite quadrature, Laplace approximation, and Monte Carlo integration provides general tools for evaluating expectations over latent variables. The central point here is which quantity to integrate, not which numerical method to choose.

Common Pitfall

A frequent error is to argue: “The latest biomarker is the best available value, so I can put it directly into the hazard and treat the resulting risk as the patient’s risk.” The reading is evidence about the patient’s latent marker, but its difference from that marker is not known to be zero. Moreover, because the hazard-to-risk transformation is nonlinear, even substituting the posterior mean latent value need not reproduce the average risk across plausible latent values.

Key takeaway: For a prediction conditional on observed history, use the history to obtain the distribution of latent trajectories, condition that distribution on survival to the landmark, and average the resulting conditional risks. A raw observed biomarker is not generally an error-free replacement for the latent marker.

Check Your Understanding

Answer each question before reading its solution.

In the constant-marker model, let \(c=0.10\), \(\alpha=\log(2)\), and let the posterior probability be 0.5 at each of \(M=0\) and \(M=2\). What is the integrated risk?

Solution: The risks are \(1-\exp(-0.10)=0.09516\) and \(1-\exp(-0.40)=0.3297\). Their average is \(0.5(0.09516)+0.5(0.3297)=0.2124\), or 21.24%.
For \(g(m)=1-\exp\{-c\exp(\alpha m)\}\), when is the risk function locally convex as a function of \(m\)?

Solution: Its second derivative is \(\alpha^2x\exp(-x)(1-x)\), where \(x=c\exp(\alpha m)\). For nonzero \(\alpha\) and positive \(x\), it is positive when \(x<1\), so the function is locally convex there.
Why can two posterior distributions with the same mean marker value lead to different predictions?

Solution: Risk is a nonlinear function of the marker. The integrated prediction averages that function over the entire posterior distribution, so changing the posterior spread or shape can change the average risk even when the mean stays fixed.
A patient’s latest observed marker is below the posterior mean of the latent marker. Must the observed-value plug-in risk be lower than the integrated risk?

Solution: No. The integrated risk depends on the full posterior distribution and on the curvature of the risk function. Even risk at the posterior mean is not generally equal to integrated risk; the observed value’s position alone does not determine the comparison.