Why shared data require explicit accounting
An integrated analysis can connect several models without bringing in several independent sources of evidence. A population PK model may use concentration measurements that later inform an exposure-response model; a clinical outcome analysis may reuse patients or trial summaries already represented in another component. If the same observations influence two model components and both contributions are treated as independent, the integrated analysis can become overconfident. Its estimates may be pulled too strongly toward the shared data, and its uncertainty may be understated.
The central task is therefore not merely to notice that two components use related variables. It is to trace which observations, participants, and derived estimates enter each component, then express their joint contribution correctly. Reusing a result can be valid: the error occurs when evidence already represented in that result is added again as if it were new, independent evidence.
A practical audit follows data lineage, not just model names. For each component, record the data source, participant and record identifiers, outcomes, time points, transformations, and estimates passed to or from another component. Two files with different names may contain the same records; two estimates may summarize the same underlying measurements. Conversely, two outcomes from the same participants need not be duplicates, although their dependence may still matter under the joint data-generating model.
Likelihood accounting and a simple derivation
Let \(D_1\) and \(D_2\) denote data used by two components, and let \(\theta\) denote the parameters relevant to the integrated analysis. If the data are conditionally independent given \(\theta\), their joint likelihood factors as \(L(\theta;D_1,D_2)=L_1(\theta;D_1)L_2(\theta;D_2)\). That factorization is justified by the conditional-independence assumption, not by the fact that the data were stored in separate files or analyzed by separate teams. If the components share observations, the correct joint likelihood must account for that overlap and its dependence; multiplying marginal likelihoods as if they were independent is generally not justified.
A particularly clear failure occurs when a posterior based on data is reused as a prior and the same data likelihood is then applied again. By Bayes' rule, the posterior based on prior density \(p(\theta)\) and data \(D\) is proportional to \(p(\theta)L(\theta;D)\). Treating that posterior as the new prior and multiplying by the same likelihood gives:
The likelihood is squared: the data have been represented twice. The result is not the posterior from one use of the data, but a different analysis that implicitly gives those observations an extra contribution. This derivation assumes the second step uses the same likelihood and the same parameterization; the underlying warning also applies when shared data are passed between more complicated linked components.
Worked Example: Reusing One Observation Twice
Suppose a hypothetical model has a prior \(\theta\sim N(0,100)\). One measurement is \(y=10\), with \(y\mid\theta\sim N(\theta,4)\). The variance is known, and the normal prior and observation model are the assumptions for this calculation.
For one measurement, the posterior precision is the prior precision plus the observation precision: \(1/100+1/4=0.01+0.25=0.26\). Thus the posterior variance is \(1/0.26=3.846\), and the posterior mean is \((0/100+10/4)/0.26=2.5/0.26=9.615\). After one use of the measurement, the result is \(\theta\mid y\sim N(9.615,3.846)\).
If this posterior is incorrectly treated as a new prior and the same measurement is used again, the updated precision is \(1/3.846+1/4=0.26+0.25=0.51\), giving variance \(1/0.51=1.961\). The updated mean is \((9.615/3.846+10/4)/0.51=(2.5+2.5)/0.51=9.804\).
The mean has moved closer to the single observed value, and the variance has fallen from 3.846 to 1.961, even though no new measurement was collected. The second update has counted the same measurement again. With one use, the posterior is \(N(9.615,3.846)\); the narrower \(N(9.804,1.961)\) result is not justified by these data alone.
Representing overlap rather than ignoring it
There are several defensible strategies, chosen according to what information is available and what the analysis needs. A joint model can include the shared observations once and specify their contribution to all relevant components. A sequential analysis can pass a result forward when the downstream data are genuinely new, or when the handoff is explicitly constructed to preserve the relevant joint uncertainty without reintroducing the original evidence. When only summary estimates are passed, a summary likelihood can incorporate their covariance if that covariance and a suitable sampling model are available.
For example, two estimates based on overlapping participants should not automatically be combined as independent estimates. If their sampling covariance is known, a multivariate likelihood can represent the pair. For a vector of estimates \(z\) with model mean \(m(\theta)\) and covariance matrix \(V\), a normal summary likelihood is proportional to \(\exp[-\tfrac{1}{2}(z-m(\theta))^\mathsf{T}V^{-1}(z-m(\theta))]\). The off-diagonal elements of \(V\) encode dependence. Setting them to zero is an additional independence assumption, not a neutral default.
Worked Example: Two Contrasts Sharing a Comparator
In a hypothetical three-arm trial, treatment arms A and B share a control arm C. Suppose the estimated contrasts are \(d_A=2\) and \(d_B=3\) outcome units. Assume independent arm-mean estimation errors, each with variance 1. Each contrast subtracts the same estimated control mean, so each contrast has variance \(1+1=2\), and the covariance between contrasts is the control-mean variance, 1. Thus:
Suppose the target is a common contrast \(\delta\) estimated from \(z\), with \(z\sim N(\delta\mathbf{1},V)\). The generalized least-squares estimate is \((\mathbf{1}^\mathsf{T}V^{-1}z)/(\mathbf{1}^\mathsf{T}V^{-1}\mathbf{1})\), and its variance is \(1/(\mathbf{1}^\mathsf{T}V^{-1}\mathbf{1})\). Here, \(V^{-1}=\tfrac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\). Therefore \(\mathbf{1}^\mathsf{T}V^{-1}z=5/3\) and \(\mathbf{1}^\mathsf{T}V^{-1}\mathbf{1}=2/3\). The estimate is \((5/3)/(2/3)=2.5\) outcome units, with variance \(1/(2/3)=1.5\) and standard error \(\sqrt{1.5}=1.225\) outcome units.
If the covariance were incorrectly set to zero, the two variances of 2 would yield an equal-weight estimate of 2.5 with variance \(2/2=1\), or standard error 1 outcome unit. The point estimate happens to be unchanged here, but the naive analysis claims greater precision. The shared control explains why the contrast estimates are not independent.
Worked Example: Quantifying Participant Overlap
Two hypothetical estimates are sample means for the same outcome in the same target population. Each uses 50 participants, and 20 participants appear in both samples. Assume independent participants, an outcome variance of \(100\) units squared, and that the two estimates use the same outcome definition and measurement occasion. Under these assumptions, each sample mean has variance \(100/50=2\) units squared.
The covariance of the two sample means is the number of shared participants multiplied by the population variance and divided by the product of the sample sizes: \(\operatorname{Cov}(\bar X_1,\bar X_2)=20(100)/(50\cdot50)=0.8\) units squared. Their correlation is \(0.8/\sqrt{2\cdot2}=0.4\). The overlap therefore creates positive dependence.
If the target is the average of the two estimates, \((\bar X_1+\bar X_2)/2\), its variance is \[ \frac{1}{4}\{2+2+2(0.8)\}=1.4 \] units squared, so its standard error is \(\sqrt{1.4}=1.183\) units. Treating the estimates as independent would instead give variance \((2+2)/4=1\) and standard error 1 unit. In this example, the independence assumption understates uncertainty because it ignores the 20 shared participants.
A practical audit for integrated analyses
Before fitting or updating a linked analysis, make the evidence flow explicit. The purpose is to identify duplicate contributions and determine whether dependence is already represented, not to discard valid information merely because components are connected.
List source datasets and their participants, records, outcomes, and time points. Mark reused records and any estimates derived from them.
For every quantity passed between components, identify the data and fitted results that produced it. Record whether it is a raw observation, a point estimate, a posterior distribution, or a derived prediction.
Determine whether the likelihood factorization is justified conditional on the model parameters. If not, specify a joint likelihood or an appropriate dependence structure.
Use shared observations once in a joint analysis, or pass evidence forward without adding the same likelihood again. For summaries, retain their covariance when it is part of the sampling model.
Compare the chosen analysis with a clearly labeled independence-based sensitivity analysis where useful. Investigate unexpected precision gains, repeated likelihood contributions, and changes driven by assumptions about overlap.
Diagnostics, alternatives, and limitations
Double counting can be difficult to spot from model output alone: a plausible estimate and a small standard error do not establish that evidence was counted correctly. Useful diagnostics include matching participant or record identifiers across inputs, reviewing code and data-flow documentation, and tracing whether a downstream likelihood uses raw data already included in an upstream posterior. A sharp increase in precision after adding a component that contains no new observations is a warning sign, though precision can also change legitimately when the model structure changes.
The main alternatives differ in how they represent evidence. A joint likelihood is generally appropriate when a coherent joint data model can be specified and the relevant observations are available. A sequential handoff can be efficient when data entering the downstream step are new or the transferred uncertainty is adequate for its purpose. A covariance-aware summary likelihood can be practical when only estimates and their sampling covariance are available. Treating components as independent is acceptable only when the relevant conditional-independence assumptions are defensible. These approaches are not interchangeable merely because they give similar point estimates.
Limitations arise when data lineage is incomplete, identifiers cannot be reconciled, or only point estimates have been retained. A point estimate alone usually does not reveal the likelihood or covariance needed to reconstruct its evidence contribution. In that case, a precise correction for overlap may be impossible without additional information; the analysis should state the limitation and assess sensitivity to plausible dependence assumptions rather than silently claiming independence. As in the earlier material on preserving dependence in integrated simulations, the general principle is to retain dependence that is relevant to the target inference, but here the specific concern is repeated evidence in the likelihood or handoff.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A posterior based on data \(D\) is used as a prior, and the same likelihood for \(D\) is applied again. What likelihood contribution results relative to the original prior?
- Two contrast estimates share a control group. What feature of their covariance should be examined before combining them?
- Two estimates each have variance 4 and covariance 1. What is the variance of their average?
- Why is a point estimate passed between components generally insufficient to establish that no evidence has been counted twice?
Solutions. (1) The resulting density is proportional to the original prior multiplied by \(L(\theta;D)^2\), so the data likelihood has been applied twice. (2) Examine the off-diagonal covariance induced by the shared control estimate; do not assume it is zero solely because the contrasts have different treatment arms. (3) The variance is \(\{4+4+2(1)\}/4=10/4=2.5\) in squared units. (4) A point estimate does not generally specify which observations produced it, its likelihood, or its dependence on other estimates. Without that information, the evidence flow and any correction for overlap cannot be fully assessed.