Why Model Visits Through a Latent Trajectory?
A patient’s disease severity can change between clinical assessments. If worsening health prompts extra contact, then visits may become more frequent precisely when the underlying condition is changing, even if that condition is not measured at every visit. The observed schedule is therefore potentially informative about a disease trajectory that is only partially seen.
The key distinction is between a state that is observed at a visit and an underlying state that evolves continuously or frequently over follow-up. The earlier material on current-state-triggered visit intensities treats a visit rate as depending on a current state available to the model. Here the trigger is a latent trajectory: the disease state at time \(t\) may affect visits even when no measurement is taken at that time. A latent state is not automatically the same as the most recent recorded measurement.
A common model for patient \(i\)'s instantaneous visit rate is
Here \(X_i\) is a vector of covariates, \(\lambda_0^V(t)\) is a baseline visit intensity, and \(\alpha\) describes how the latent state is associated with the visit rate. The intensity is a rate per unit time, conditional on the specified trajectory and covariates. If \(\alpha>0\), higher latent values correspond to higher conditional visit rates; if \(\alpha<0\), they correspond to lower rates. This is an association in the model, not by itself evidence that changing the disease state would cause a change in visits.
Assumptions and Interpretation
An intensity model needs a well-defined follow-up interval and a visit-counting process. The latent trajectory must be specified on that interval, including times between recorded visits, and the model must yield a nonnegative, integrable intensity. In the log-linear form above, the exponential ensures nonnegativity; integrability over a finite interval is still needed for a finite expected number of visits. The trajectory and intensity must also be defined using information available up to time \(t\), so the rate at \(t\) does not depend on future values in a way that would make it non-predictable.
For a one-unit increase in \(M_i(t)\), holding time and covariates fixed, the conditional visit-rate ratio is \(\exp(\alpha)\). If \(M_i(t)\) is measured in a meaningful unit, this ratio is per one-unit increase in that latent scale. The interpretation depends on how the latent state is scaled and oriented: reversing which direction represents worse disease changes the sign of \(\alpha\).
The conditional intensity describes the rate given the latent trajectory. It does not imply that the trajectory is observed, nor does it by itself specify the full distribution of visit times. In particular, an intensity alone should not be treated as proof that visit counts have independent increments. When a calculation uses a conditional inhomogeneous Poisson process, that is an additional assumption and will be stated explicitly.
Worked Example: Comparing Conditional Visit Rates
Suppose, hypothetically, that visits follow a conditional Poisson process with rate \(\lambda_i^V(t)=0.12\exp\{0.7M_i(t)\}\) visits per week. Assume the latent state is constant over a two-week period. Compare a patient with \(M_i(t)=0.5\) to one with \(M_i(t)=2\).
For the first patient, the conditional rate is \(\lambda^V=0.12e^{0.7(0.5)}=0.12e^{0.35}\approx0.1703\) visits per week. Over two weeks, the integrated intensity is \(\Lambda^V=2(0.12e^{0.35})\approx0.3406\).
For the second patient, the rate is \(\lambda^V=0.12e^{0.7(2)}=0.12e^{1.4}\approx0.4866\) visits per week, giving \(\Lambda^V=2(0.12e^{1.4})\approx0.9732\).
The rate ratio for a 1.5-unit state difference is \(e^{0.7(2-0.5)}=e^{1.05}\approx2.858\). Thus the second patient's conditional rate is about 2.86 times the first patient's rate, under the model.
Because this example assumes a conditional Poisson process, the probability of at least one visit over the interval is \(1-e^{-\Lambda^V}\). The probabilities are \(1-e^{-0.3406}\approx0.2886\) and \(1-e^{-0.9732}\approx0.6221\), respectively. These probabilities depend on the stated Poisson assumption as well as the intensity model.
From a Trajectory to an Integrated Visit Rate
For a fixed latent path \(M_i(t)\), the expected number of visits over an interval \([a,b]\), conditional on that path and the model covariates, is the integrated intensity
This quantity accumulates the rate over time. A high rate for a short period and a moderate rate over a longer period can produce the same integrated intensity. If, additionally, visits form a conditional inhomogeneous Poisson process, then the conditional visit count on \([a,b]\) is Poisson with mean \(\Lambda_i^V(a,b\mid M_i,X_i)\). In that case, the probability of no visits is \(e^{-\Lambda_i^V}\), and the probability of at least one is \(1-e^{-\Lambda_i^V}\).
The no-visit formula follows directly from the Poisson probability at count zero. For \(N_i^V(b)-N_i^V(a)\) with conditional mean \(\Lambda_i^V\),
Thus \(P\{N_i^V(b)-N_i^V(a)\geq1\mid M_i,X_i\}=1-e^{-\Lambda_i^V}\). This probability calculation is not valid from an intensity alone if the conditional Poisson assumption has not been made.
Worked Example: A Trajectory That Worsens Midway
Consider a hypothetical four-week follow-up. Assume a conditional Poisson visit process with \(\lambda^V(t)=0.10e^{0.6M(t)}\) visits per week. Let the latent state be \(0.5\) during weeks 0–2 and \(1.5\) during weeks 2–4. Assume the baseline rate and model coefficient remain constant throughout.
During the first two weeks, the rate is \(0.10e^{0.6(0.5)}=0.10e^{0.3}\approx0.1350\) visits per week. The integrated intensity for that segment is \(2(0.10e^{0.3})\approx0.2700\).
During the second two weeks, the rate is \(0.10e^{0.6(1.5)}=0.10e^{0.9}\approx0.2460\) visits per week. The segment's integrated intensity is \(2(0.10e^{0.9})\approx0.4919\).
Adding the two contributions gives \(\Lambda^V=2(0.10e^{0.3})+2(0.10e^{0.9})\approx0.7619\) expected visits over four weeks, conditional on this path. Under the conditional Poisson assumption, the probability of at least one visit is \(1-e^{-0.7619}\approx0.5332\). This uses the entire piecewise trajectory rather than a single rate evaluated at one assessment time.
Why Replacing the Latent State by Its Average Can Fail
When the trajectory is unobserved, it can be tempting to substitute its mean into the rate function. In a nonlinear model, this generally does not produce the average rate. For the log-linear model, conditional on covariates, \(E[\lambda_i^V(t)\mid X_i]=\lambda_0^V(t)e^{\gamma^\top X_i}E[e^{\alpha M_i(t)}\mid X_i]\), which is generally different from \(\lambda_0^V(t)e^{\gamma^\top X_i}e^{\alpha E[M_i(t)\mid X_i]}\).
The difference matters because visit patterns reflect both the relationship between visits and the latent state and the distribution of latent states in the population. A latent value at an observed visit does not reveal the whole trajectory for that patient, and people with different latent paths can contribute different amounts of follow-up at different rates.
Worked Example: A Mixture of Two Latent States
Suppose, hypothetically, that each patient has a latent state that remains either \(1\) or \(3\) over a one-week interval, with equal probability. Visits follow a conditional Poisson process with \(\lambda^V=0.08e^{0.5M}\) per week. The state is not observed during the interval.
For \(M=1\), the rate is \(0.08e^{0.5}\approx0.1319\) visits per week. For \(M=3\), it is \(0.08e^{1.5}\approx0.3585\). Averaging these conditional rates gives \(\{0.08e^{0.5}+0.08e^{1.5}\}/2\approx0.2452\) visits per week.
The mean latent state is \(2\). Substituting that mean into the rate model instead gives \(0.08e^{0.5(2)}=0.08e^1\approx0.2175\), not \(0.2452\). The average rate is therefore not the rate evaluated at the average state.
The same issue appears in visit probabilities. Averaging the two conditional probabilities gives \(\{(1-e^{-0.1319})+(1-e^{-0.3585})\}/2\approx(0.1236+0.3013)/2=0.2124\). Evaluating at the mean state gives \(1-e^{-0.2175}\approx0.1954\). These differ because the exponential rate and the probability transformation are nonlinear.
Model Checks, Alternatives, and Limitations
A latent-trajectory visit model is useful when a plausible disease process may influence contact even between measurements. It also raises a challenge: the key predictor is not fully observed, so a fitted relationship can depend on how the trajectory is represented and inferred. A model that fits recorded measurements well may still describe the unobserved path poorly.
Useful checks include comparing observed visit counts and timing with patterns predicted across follow-up, examining whether discrepancies are concentrated in periods or patient groups, and assessing whether conclusions change under reasonable alternative trajectory specifications. Check that the modeled trajectory scale has a clinically interpretable direction and that fitted rates are plausible. Systematic underprediction of visits during apparent deterioration, for example, can indicate that the latent path or its association with visits is inadequate. Because the trajectory itself is partly unobserved, such checks cannot conclusively verify the trigger mechanism.
The main alternatives answer different questions. An externally scheduled visit model is appropriate when the schedule is fixed independently of evolving health, while a model based on an observed current state uses information actually available at that time. A latent-trajectory model allows the underlying health state to matter between recorded assessments, but requires assumptions about that unobserved state. If its assumptions are implausible or weakly supported by the data, a simpler visit model and sensitivity analysis may be more defensible than a highly detailed latent specification.
- Do not treat the latest recorded value as the true state throughout the gap. That substitution is an additional approximation, not a consequence of a latent-trajectory model.
- Do not infer a causal effect from \(\alpha\). A nonzero trajectory coefficient describes conditional dependence in the specified model; unmeasured factors may influence both disease progression and visits.
- Do not translate intensity into a visit probability without assumptions. The conditional Poisson formula requires a conditional Poisson process, not just a stated conditional intensity.
- Do not assume the latent trajectory is identified by visit times alone. Limited measurements or few visits may leave many plausible combinations of trajectory and visit dependence.
Check Your Understanding
Try each problem, then compare your work with the solutions below.
- In \(\lambda^V(t)=0.05e^{0.4M(t)}\) visits per day, what is the conditional rate ratio for a two-unit increase in the latent state?
- A conditional Poisson visit process has constant rate \(0.20\) visits per week for three weeks. Find its integrated intensity and the probability of at least one visit.
- In a log-linear rate model, explain why \(e^{\alpha E[M(t)]}\) generally differs from \(E[e^{\alpha M(t)}]\).
- State one assumption needed to use \(1-e^{-\Lambda^V}\) as the probability of at least one visit.
Solutions. (1) The rate ratio is \(e^{0.4(2)}=e^{0.8}\approx2.226\). (2) The integrated intensity is \(3(0.20)=0.600\); under the conditional Poisson assumption, the probability is \(1-e^{-0.600}\approx0.4512\). (3) The exponential is nonlinear, so averaging its value is generally not the same as applying it to the average; the two can coincide in special cases, such as a non-varying state or \(\alpha=0\). (4) The visit process must be conditionally Poisson over the interval, given the trajectory and covariates; intensity alone is insufficient.