Why the Observed Trajectory May Not Be the Population Trajectory
A longitudinal data set records outcomes when measurements are made, not necessarily at times chosen independently of patients’ health. If worsening symptoms prompt a biomarker test, for example, the measurements collected at a given time may disproportionately represent patients with higher biomarker values. A regression through the recorded values can then describe the trajectory among visits rather than the average trajectory among the population of patients.
The distinction is about the observation schedule, not simply missing measurements. A useful first calculation is to ask: at a fixed time, what outcome distribution would we see if we sampled a visit rather than a patient? Once visits are more likely at some outcome values than others, those two distributions need not agree.
For the calculations below, assume patients are under observation and eligible for a visit at the time being considered, the outcome is defined for each eligible patient, and the stated intensity correctly describes how visits arise. We compare the population mean at a fixed time with the mean outcome attached to visits at that time. This comparison isolates selection by the visit process; it does not by itself address dropout, measurement error, or terminal events.
How Outcome-Triggered Visits Tilt the Mean
At a fixed time \(t\), suppose the visit intensity is \(\lambda(t\mid Y)\), with \(Y\) denoting the outcome across eligible patients. A patient with outcome \(Y=y\) contributes visits in proportion to \(\lambda(t\mid y)\). Thus the outcome distribution among visits is the population outcome distribution reweighted by the intensity. Its mean is
Subtracting the population mean \(\mu(t)=E\{Y(t)\}\) gives a useful expression for the difference:
To verify this, write \(E(Y\lambda)=E(Y)E(\lambda)+\operatorname{Cov}(Y,\lambda)\), divide by \(E(\lambda)\), and subtract \(E(Y)\). When higher outcomes correspond to higher visit intensity, the covariance is positive and the mean among visits exceeds the population mean. The formula also shows that a visit-sampled mean equals the population mean if the intensity is constant across outcome values, or more generally if the covariance is zero.
Worked Example: A Linear Intensity Shifts the Mean at One Time
Suppose that at a particular time the population consists of equal numbers of patients with outcomes 3 and 7 units. Let the visit intensity, in visits per standardized time interval, be \(\lambda=1+0.1Y\). This intensity is positive for both possible outcomes and increases with the outcome, as required for the proposed triggering mechanism.
The population mean is \(\mu=(3+7)/2=5\) units, and the variance is \(\sigma^2=4\) square units. The mean intensity is \(1+0.1(5)=1.5\). The intensity-weighted mean is
The visit-sampled mean is about 0.267 units above the population mean. Equivalently, for \(\lambda=1+cY\), the covariance formula gives \(\mu_{\mathrm{visit}}=\mu+c\sigma^2/(1+c\mu)\), which here is \(5+0.1(4)/1.5=5.267\) units. Both calculations show that the distortion comes from unequal visit rates, not from a change in the population mean.
A Worked Linear-Trajectory Distortion
Consider a hypothetical cohort with two equally sized patient groups. Their outcomes follow exact linear trajectories, with time \(t\) measured in years and the outcome measured in biomarker units. Group A follows \(Y_A(t)=2t\); Group B follows \(Y_B(t)=10\). The population mean trajectory is therefore \(\{2t+10\}/2=5+t\) units per patient. Its intercept is 5 units, and its slope is 1 unit per year.
Suppose a patient has probability \(0.1(1+Y/10)\) of receiving a measurement in a small, standardized observation window. This probability is positive for the outcomes considered and increases with the current outcome. Assume 100 patients in each group. For a transparent illustration, take the realized visit counts to equal the expected counts at \(t=0\) and \(t=1\).
Worked Example: Outcome-Triggered Visits Change the Estimated Slope
At \(t=0\), Group A has outcome 0 and visit probability 0.10, giving 10 visits among 100 patients. Group B has outcome 10 and visit probability 0.20, giving 20 visits. The observed mean among visits is
At \(t=1\), Group A has outcome 2 and visit probability 0.12, giving 12 visits. Group B still has outcome 10 and probability 0.20, giving 20 visits. The visit-sampled mean is
With just these two time points, the slope from an ordinary regression of the recorded outcomes on time is the difference between the two visit-sampled means divided by the one-year time difference: \((7.000-6.667)/(1-0)=0.333\) units per year. The population slope is 1 unit per year. The naive estimate is therefore lower by about 0.667 units per year, while the visit-sampled levels at both times are above their population counterparts, 5 and 6 units.
This calculation describes the point estimate from treating these records as the trajectory sample. It does not establish that the same numerical bias occurs with every mixed-effects model or every visit schedule. Its purpose is to show directly how changing representation across time can alter both the level and the apparent trend.
Why the Slope Distortion Is Not Automatic
Outcome-dependent sampling need not change a trajectory’s slope in every setting. In particular, suppose patients have parallel linear trajectories \(Y_i(t)=a_i+\beta t\), where \(a_i\) varies across patients but the slope \(\beta\) is common. If the intensity is \(\lambda_i(t)=\exp\{\gamma Y_i(t)\}\), then at a fixed time it can be written as
The first factor is the same for all patients at that time, so it cancels when calculating the visit-sampled mean. The distribution of the \(a_i\)'s among visits is tilted by \(\exp(\gamma a_i)\), but that tilt does not vary with time. The visit-sampled mean is consequently a shifted intercept plus \(\beta t\): the level is distorted, but the common slope is preserved. This argument relies on parallel trajectories and this separable intensity; it is not a general guarantee for outcome-triggered observation.
Worked Example: A Shifted Level with an Unchanged Common Slope
Take equal numbers of patients with \(a=0\) and \(a=10\), and suppose each has trajectory \(Y(t)=a+2t\) units, with time in years. Let \(\lambda=\exp\{\gamma Y(t)\}\), where \(\gamma=\log(2)/10\) per biomarker unit. At each fixed time, the high-intercept group’s intensity is twice the low-intercept group’s intensity, because their outcomes differ by 10 units.
The population mean is \(5+2t\) units. The visit-sampled mean weights the two intercepts in the ratio 1 to 2, so its intercept is \(\{1(0)+2(10)\}/3=6.667\) units. At \(t=0\), the visit mean is 6.667 units; at \(t=1\), it is \(6.667+2=8.667\) units. The visit-sampled slope is \((8.667-6.667)/(1-0)=2.000\) units per year, equal to the population slope, even though the estimated level is higher by 1.667 units.
The example is a limiting case, not a reason to ignore visit timing. If slopes vary across patients, or if the way visits depend on outcome changes over time, the composition of the visit sample can change across time and the slope can change as well.
Interpreting the Fit and Checking Its Weak Points
An ordinary regression or mixed-effects model for outcomes can be useful, but its interpretation depends on how observations enter the analysis and on the assumptions connecting observed records to the target population. If visit times are independent of outcomes, conditional on the variables represented in the model, the observed measurements may represent the intended population trajectory under suitable modeling conditions. If the visit intensity still depends on the current or latent outcome, the records may instead overrepresent some parts of the trajectory. A random-effects model for outcomes does not, by itself, remove selection caused by an unmodeled visit process.
In an actual analysis, consider whether visits cluster after unusual or worsening outcomes, and whether patients with different outcome levels contribute different numbers of records. Compare the measurement schedule and the target estimand: an average across patients at a specified time is not automatically the same quantity as an average across visits. These observations can flag a possible problem, but they cannot prove that the visit process is noninformative or identify its dependence on unobserved outcomes.
The main alternatives address different aspects of the problem. A design based on scheduled measurements limits outcome-driven differences in measurement opportunities. Inverse-intensity weighting uses a model for the visit process to reweight observed records toward the target population, subject to correct specification and adequate positivity. A joint model represents the outcome and visit processes together, potentially linking them through shared latent quantities. These approaches are not interchangeable guarantees: each requires assumptions, and results can be sensitive to how the observation mechanism is represented.
The central limitation is that the visit-sampled distribution alone may not reveal how many measurements would have been made for patients who were observed less often, particularly when visit dependence involves an unobserved current outcome. Consequently, a fitted trajectory can be precisely estimated for the recorded visits yet still differ from the population trajectory of interest. The distinction is about the estimand and the observation mechanism, not simply sample size.
Check Your Understanding
Try each problem first, then compare your reasoning with the solutions.
- At a fixed time, \(E(Y)=4\), \(\operatorname{Var}(Y)=9\), and the visit intensity is \(\lambda=1+0.2Y\). Calculate the mean among visits.
- The visit intensity is constant across patients at each fixed time. What is the visit-sampled mean, and why?
- Two groups have equal population sizes. At \(t=0\), their outcomes are 0 and 8, and their visit counts are 5 and 15. At \(t=1\), their outcomes are 2 and 8, and their visit counts are 10 and 15. Calculate the visit means at both times and the two-time visit-sampled slope.
- Under what conditions can outcome-triggered visits shift the estimated level but leave a common linear slope unchanged?
Solutions. (1) The mean intensity is \(1+0.2(4)=1.8\). Using the linear-intensity formula, the visit mean is \(4+0.2(9)/1.8=5.000\) units. (2) If \(\lambda\) is constant across patients, then \(E(Y\lambda)/E(\lambda)=E(Y)\lambda/\lambda=E(Y)\); there is no outcome-based reweighting at that time. (3) At \(t=0\), the visit mean is \(\{5(0)+15(8)\}/20=6.000\) units. At \(t=1\), it is \(\{10(2)+15(8)\}/25=5.600\) units. The visit-sampled slope is \((5.600-6.000)/(1-0)=-0.400\) units per year. (4) For example, if trajectories are parallel, \(Y_i(t)=a_i+\beta t\), and the intensity is exponential in the current outcome, the time factor is common to all patients and cancels from the relative visit weights. The selected intercept can differ, while the common slope remains \(\beta\).