Why an Assay Result May Be an Interval
An assay record does not always contain an exact concentration or merely indicate that a result fell below the lower limit of quantification (LLOQ). A laboratory may report a range, such as “between 0.80 and 1.00 mg/L,” because the output is rounded, categorized, or otherwise only resolved to an interval. That record contains more information than a simple below-LLOQ flag, but less than an exact numerical result. The likelihood must represent precisely what was observed: the assay result fell within the reported bounds.
The interval is applied to the random assay observation described by the observation model. In an NLME model, that observation is related to a model-predicted concentration and may vary with the subject’s random effects. The interval likelihood is therefore a probability under the observation model, not a density evaluated at an invented midpoint. As in the earlier material on censored likelihoods, the observation mechanism determines the contribution to the likelihood.
For a continuous observation model, including or excluding either endpoint does not change this probability: a single point has probability zero. The interval limits must nevertheless be recorded consistently, particularly when data are assembled from assay categories. If the lower limit is zero, the lower CDF term is \(G_{ij}(0\mid\eta_i)\), not automatically zero; it is zero only when the observation model assigns no probability at or below zero.
Deriving the Contribution
The derivation follows directly from the event recorded by the assay. If the report says the result lies within the interval, the observed event is \(a_{ij}<Y_{ij}\le b_{ij}\). The probability of this event is the integral of the density over the range. By the definition of a CDF, that integral is the CDF at the upper endpoint minus the CDF at the lower endpoint.
This form covers a general continuous observation model, provided its density and CDF are defined for the assay observation. It does not require the concentration prediction itself to be normally distributed. For example, under an additive normal residual model with model prediction \(f_{ij}(\theta,\eta_i)\) and residual standard deviation \(\sigma\), suppose \(Y_{ij}\mid\eta_i\sim N(f_{ij}(\theta,\eta_i),\sigma^2)\). Then standardizing the endpoints gives
where \(\Phi\) is the standard normal CDF. This expression assumes the stated normal observation model and the same conditional prediction and residual standard deviation across the interval. It is a probability, so it is dimensionless and lies between zero and one. In contrast, an exact continuous observation \(y_{ij}\) contributes a density \(g_{ij}(y_{ij}\mid\eta_i)\), which has inverse-concentration units when \(Y_{ij}\) is measured in concentration units.
Worked Example: Probability for a Reported Concentration Range
Suppose a hypothetical assay reports that a result lies in \((0.80,1.00]\) mg/L. Conditional on the subject’s random effects, the model predicts \(1.20\) mg/L, and the additive normal residual standard deviation is \(0.20\) mg/L. Assume the assay observation follows this normal model.
The standardized lower and upper endpoints are \(\bigl(0.80-1.20\bigr)/0.20=-2.00\) and \(\bigl(1.00-1.20\bigr)/0.20=-1.00\). Therefore the interval contribution is
Thus, under the specified conditional model, the probability of a result in the reported range is about 0.1359. The reported interval is not replaced by its midpoint, 0.90 mg/L: the observation is the event of falling in the range, and its likelihood contribution is the probability of that event.
From an Assay Interval to an NLME Likelihood
Let \(\theta\) denote the population parameters and let \(\eta_i\) denote the random effects for subject \(i\). Conditional on \(\eta_i\), the model specifies the distribution of each assay observation. If observations are conditionally independent given the random effects and model parameters, the subject’s conditional likelihood is the product of the contributions for that subject’s records. An exact result contributes its density, an interval report contributes a probability, and a below-LLOQ indicator contributes the probability of the event below the LLOQ.
For example, with an exact result \(y_{ij}\), an interval \((a_{ik},b_{ik}]\), and a below-LLOQ record with limit \(q_{im}\), the corresponding factors are respectively \(g_{ij}(y_{ij}\mid\eta_i)\), \(G_{ik}(b_{ik}\mid\eta_i)-G_{ik}(a_{ik}\mid\eta_i)\), and \(G_{im}(q_{im}\mid\eta_i)\). The last expression assumes the reported event is \(Y_{im}\le q_{im}\). These are different observations and must not be encoded as if they were all exact concentrations.
Here \(E_i\), \(I_i\), and \(B_i\) index the subject’s exact, interval-reported, and below-LLOQ records. This product is conditional on the subject’s random effects. To obtain the marginal likelihood contribution for that subject, integrate over the random-effects distribution. If \(\eta_i\sim N(0,\Omega)\), for example, then
where \(\varphi(\eta_i;0,\Omega)\) is the random-effects density. In general, the probability for an interval is computed at each value of \(\eta_i\) inside this integral. Substituting a population-average prediction for the subject-specific prediction, or taking the interval probability only at \(\eta_i=0\), does not generally give the marginal likelihood.
Worked Example: An Interval Record Alongside an Exact Result
Consider a hypothetical subject with random effects fixed at \(\eta_i=0\) for this conditional calculation. The model predicts \(1.00\) mg/L at two observation times, and the residual standard deviation is \(0.20\) mg/L. The first record is an exact result of \(1.10\) mg/L. The second is reported as \((0.80,1.00]\) mg/L. Assume conditional independence of residual errors.
For the exact observation, the normal density contribution is
For the interval record, the standardized endpoints are \(-1.00\) and \(0\), so its contribution is
Conditional independence means the two contributions multiply. The subject’s conditional likelihood for these two records is therefore \(1.7603\times0.3413447\approx0.6009\ {\rm (mg/L)}^{-1}\). The units are inverse concentration because one factor is a density and the interval factor is a dimensionless probability. In an NLME fit, this conditional product would then be integrated over the subject’s random-effects distribution rather than treated as the subject’s final marginal likelihood.
Assumptions, Checks, and Common Mistakes
The interval formula is simple, but its validity depends on matching the probability calculation to the data-generating and reporting process. Before using it, verify the following:
- The stored lower and upper endpoints describe the assay’s reported interval on the same scale as the observation model. A range on a log scale must not be inserted as though it were a range on the original concentration scale.
- The observation model’s conditional density or CDF is appropriate for the assay result, with the model prediction and residual parameters evaluated for that record.
- The interval endpoints are ordered, \(a_{ij}<b_{ij}\), and correspond to the category actually reported. If categories are adjacent, their boundary convention should not create gaps or overlaps in the represented outcomes.
- The conditional independence assumption used to multiply records is appropriate. If the model specifies dependence among residuals, the joint probability for the observed intervals and exact values is required instead of a product of separate marginal contributions.
A useful basic diagnostic is to check that every computed interval probability is between zero and one and that increasing the interval width cannot decrease its probability when the interval is expanded to contain the original one. A probability near zero can be valid when the interval is far from the model prediction; it can also reveal a mismatched scale, incorrect endpoints, or an incorrect CDF calculation. Compare the model-implied probabilities with the frequencies of reported categories across records where such a comparison is meaningful. This does not establish that the observation model is correct, but substantial discrepancies merit investigation.
Common mistake: replacing the interval with its midpoint. A tempting approach is to treat the report \((a,b]\) as an exact value \((a+b)/2\) and use the density at that value. This changes the observed event and generally changes the likelihood. The correct contribution is the integral of the density across the interval. For a sufficiently narrow interval and a smooth density, the approximation \(G(b)-G(a)\approx g(c)(b-a)\) for some \(c\in(a,b)\) explains why a density near the midpoint may sometimes be a rough numerical approximation after multiplication by interval width. It is not the exact interval likelihood, and using the midpoint density alone also has different units.
Another error is to use only the upper-tail or lower-tail probability. The interval event requires subtracting the lower-endpoint CDF from the upper-endpoint CDF. Using \(G(b)\) alone would include all observations below \(a\), which the report does not say occurred. Similarly, if an interval ends at the LLOQ but begins above it, its contribution is not the same as a below-LLOQ contribution.
How the Interval Likelihood Relates to Other Records
The interval contribution sits between two familiar forms of information. An exact continuous assay value contributes a density at one point. A left-censored below-LLOQ record contributes the probability of the entire region below a threshold. An interval report contributes the probability of a bounded region. The distinction is about what the assay records, not about which value is most convenient to enter into a data file.
| Recorded information | Observed event | Conditional contribution |
|---|---|---|
| Exact assay value \(y\) | The observation is recorded as \(y\) | \(g(y\mid\eta_i)\) |
| Below-LLOQ indicator with limit \(q\) | \(Y\le q\) | \(G(q\mid\eta_i)\) |
| Reported range \((a,b]\) | \(a<Y\le b\) | \(G(b\mid\eta_i)-G(a\mid\eta_i)\) |
A limitation is that interval reporting discards information within the range: it does not reveal where the assay result fell between the endpoints. Narrower reported intervals generally preserve more information than wider intervals, but they still do not provide exact observations. If assay reporting boundaries are chosen based on information not represented in the model, or if the interval itself is recorded incorrectly, the likelihood can be misspecified. Finally, numerical implementations must evaluate differences of CDF values accurately, especially when both values are close to zero or both are close to one; stable tail-probability calculations may be needed in those cases.
Check Your Understanding
For each problem, identify the appropriate likelihood contribution and compare your result with the solution.
- A hypothetical assay reports \((1.4,1.8]\) mg/L. Under a normal observation model with conditional mean \(1.6\) mg/L and standard deviation \(0.2\) mg/L, calculate the interval probability.
- A hypothetical result is reported as \((0,0.5]\) mg/L. The conditional normal mean is \(0.4\) mg/L and the standard deviation is \(0.1\) mg/L. Write and evaluate the contribution.
- An instrument rounds to the nearest \(0.1\) mg/L and reports \(0.7\) mg/L. Under a normal model with mean \(0.70\) mg/L and standard deviation \(0.10\) mg/L, calculate the appropriate interval probability.
- For conditionally independent records, an interval contributes probability \(0.20\), and an exact result contributes density \(1.5\ {\rm (mg/L)}^{-1}\). What is their joint conditional contribution, and what are its units?
Solutions. (1) The standardized endpoints are \(-1\) and \(1\). Thus, \(\Phi(1)-\Phi(-1)\approx0.8413447-0.1586553=0.6826894\approx0.6827\). (2) The standardized endpoints are \(-4\) and \(1\), so the contribution is \(\Phi(1)-\Phi(-4)\approx0.8413447-0.0000317=0.8413130\approx0.8413\). (3) Rounding to the nearest tenth means the underlying reported assay result lies in \([0.65,0.75)\) mg/L. The standardized endpoints are \(-0.5\) and \(0.5\), giving \(\Phi(0.5)-\Phi(-0.5)\approx0.6914625-0.3085375=0.3829250\approx0.3829\). (4) Multiply the factors: \(0.20\times1.5=0.30\ {\rm (mg/L)}^{-1}\). The probability factor is dimensionless, so the joint contribution retains the density’s inverse-concentration units.