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Optimal Sampling for Individual PK Models · Tutorial 56 of 497

Designing Around Below-Quantification Concentrations

You will learn to evaluate late PK samples using a likelihood that treats below-quantification results as censored observations rather than as zeroes or missing data.

Advanced 12 min read

What You'll Learn

  • Define a left-censored assay result and distinguish it from a missing concentration.
  • Derive expected Fisher information for a normal observation subject to a lower quantification limit.
  • Calculate how a sample’s expected information changes as its predicted concentration approaches or falls below the limit.
  • Compare candidate PK sampling times using both information per sample and model sensitivity.
  • Identify when normal-error calculations are unsuitable and why common substitutions for censored values can mislead.

Why the Quantification Limit Changes a Sampling Design

A late PK sample can be valuable because its concentration may respond strongly to a parameter such as clearance or terminal elimination rate. But if the predicted concentration is near an assay’s lower limit of quantification (LLOQ), the design must account for the possibility that the laboratory will report only “below quantification.” That result is not an exact concentration, but it is not devoid of information: it tells us the response fell on one side of a known threshold.

For design, the relevant question is not simply whether a candidate time has a high predicted concentration or a large model sensitivity. It is how much information the assay’s possible outcomes are expected to provide at that time. The calculations below use a normal observation model to make that comparison explicit. They concern expected information over possible results, not a guarantee that any individual sample will be quantified.

Definition: A result is left-censored at the LLOQ when the assay reports that the response is at or below a known threshold \(L\), without reporting its exact value. A result above \(L\) is reported quantitatively. A below-quantification result therefore records the event \(Y\leq L\); it is not an exact value of zero, \(L\), or \(L/2\).

Observation Model and Expected Information

Let \(C(t;\boldsymbol\theta)\) be the model-predicted concentration at time \(t\), where \(\boldsymbol\theta\) is the parameter vector. For a single candidate sample, suppose the assay response \(Y\), measured in concentration units, follows

$$ Y\mid\boldsymbol\theta \sim N\!\left(\mu,\sigma^2\right), \qquad \mu=C(t;\boldsymbol\theta), $$

with known LLOQ \(L\) and known residual standard deviation \(\sigma\). The observation mechanism is assumed to report the exact value \(y\) if \(y>L\), and only the event \(Y\leq L\) otherwise. For multiple candidate samples, assume conditional independence given \(\boldsymbol\theta\). These assumptions make the information calculation tractable; they should not be used automatically when the residual model, assay threshold, or dependence structure is different.

Write \(\Phi\) and \(\phi\) for the standard normal cumulative distribution function and density, and define the standardized threshold

$$ a=\frac{L-\mu}{\sigma}. $$

The probability of a below-quantification result is \(\Phi(a)\), and the probability of a quantified result is \(1-\Phi(a)\). If the result is quantified, its likelihood contribution is the normal density at the reported value. If it is censored, the likelihood contribution is the probability \(\Phi(a)\). The expected Fisher information for the mean \(\mu\), including both kinds of possible outcome, is

$$ \mathcal I_\mu = \frac{1}{\sigma^2} \left[ a\phi(a)+1-\Phi(a) +\frac{\phi(a)^2}{\Phi(a)} \right]. $$

The first two terms in brackets arise from the possible exact observations above the threshold; the last term arises from the possible censored event. The bracketed quantity is a fraction of the information \(1/\sigma^2\) that an always-quantified normal observation would provide about its mean. It approaches 1 when the threshold is far below the mean, and decreases when the threshold moves above the mean.

Why the Formula Has Two Contributions

A short derivation clarifies why a censored result must not be treated as either an exact number or as no observation. Let \(Z=(Y-\mu)/\sigma\), so \(Z\sim N(0,1)\) and the reporting threshold is \(Z>a\). For a quantified value, the score for \(\mu\) is \((Y-\mu)/\sigma^2=Z/\sigma\). Its expected squared score over the quantified outcomes is

$$ \int_{a}^{\infty}\frac{z^2}{\sigma^2}\phi(z)\,dz = \frac{a\phi(a)+1-\Phi(a)}{\sigma^2}, $$

where \(\int_a^\infty z^2\phi(z)\,dz=a\phi(a)+1-\Phi(a)\), obtained by integration by parts using \(\phi'(z)=-z\phi(z)\). For a censored result, the likelihood is \(\Phi(a)\). Since \(da/d\mu=-1/\sigma\), its score is \(-\phi(a)/[\sigma\Phi(a)]\). This outcome occurs with probability \(\Phi(a)\), so its contribution to expected squared score is

$$ \Phi(a)\left(\frac{\phi(a)}{\sigma\Phi(a)}\right)^2 = \frac{\phi(a)^2}{\sigma^2\Phi(a)}. $$

Adding the two outcome contributions gives the stated formula. This is expected Fisher information: it averages over the chance of obtaining either a quantified value or a censored result. It is not the information of a particular result considered in isolation.

Worked Example: A Threshold Two Standard Deviations Below the Mean

Suppose a hypothetical sample has \(\mu=1.00\) mg/L, \(\sigma=0.25\) mg/L, and \(L=0.50\) mg/L. Calculate its expected information about \(\mu\) relative to an always-quantified observation.

First, \(a=(0.50-1.00)/0.25=-2\). Using \(\Phi(-2)=0.0227501\), \(\phi(-2)=0.0539910\), and \(1-\Phi(-2)=0.977250\), the information fraction is

$$ \frac{\sigma^2\mathcal I_\mu}{1} = \frac{\phi(-2)^2}{\Phi(-2)} +(-2)\phi(-2)+1-\Phi(-2) $$
$$ \approx \frac{0.0539910^2}{0.0227501} -2(0.0539910)+0.977250 =0.9974. $$

Thus \(\mathcal I_\mu=0.9974/(0.25\ \text{mg/L})^2=15.96\ (\text{mg/L})^{-2}\), compared with \(1/0.25^2=16.00\ (\text{mg/L})^{-2}\) if the observation were always quantified. The chance of censoring is only about 2.275%, so the expected information loss is small.

Worked Example: The Mean at the LLOQ

Now suppose a hypothetical sample has \(\mu=L=0.50\) mg/L and \(\sigma=0.20\) mg/L. Here \(a=0\), \(\Phi(0)=0.5\), and \(\phi(0)=0.398942\). Its information fraction is

$$ \frac{\sigma^2\mathcal I_\mu}{1} = 0.5+\frac{0.398942^2}{0.5} =0.8183. $$

The expected information is therefore \(0.8183/0.20^2=20.46\ (\text{mg/L})^{-2}\), rather than \(25.00\ (\text{mg/L})^{-2}\) for an always-quantified observation. Although a censored result still informs us that \(Y\leq L\), the exact value is unavailable for roughly half the observations under this model. The expected information is consequently lower.

Using the Calculation to Compare Candidate Times

For a vector parameter, let \(\mathbf g(t)=\partial C(t;\boldsymbol\theta)/\partial\boldsymbol\theta\) be the sensitivity vector evaluated at the nominal parameter values used for design. If \(\sigma\) is fixed and the model depends on the parameters through the mean, the expected information matrix from one sample at time \(t\) is

$$ \mathcal I_{\boldsymbol\theta}(t) = \mathcal I_\mu(t)\,\mathbf g(t)\mathbf g(t)^{T}. $$

Under conditional independence, the expected information matrices from candidate samples add. This expression makes two ingredients visible: the assay-adjusted information about the mean and the sensitivity of the predicted concentration to the parameters. A sample can be informative about the mean yet unhelpful for a particular parameter if its sensitivity to that parameter is small. Conversely, a sensitive late sample can lose value if censoring is likely.

Worked Example: Comparing Two Hypothetical Late Samples

Consider one scalar parameter \(\theta\), with candidate A having predicted concentration \(0.80\) mg/L and sensitivity \(g_A=\partial C/\partial\theta=-0.30\) mg/L per unit of \(\theta\). Candidate B has predicted concentration \(0.40\) mg/L and sensitivity \(g_B=-0.10\) mg/L per unit of \(\theta\). For both, let \(L=0.50\) mg/L and \(\sigma=0.20\) mg/L. Assume the normal model and fixed residual standard deviation stated above.

For A, \(a_A=(0.50-0.80)/0.20=-1.5\). Using \(\Phi(-1.5)=0.0668072\), \(\phi(-1.5)=0.129518\), and \(1-\Phi(-1.5)=0.933193\), the information fraction is

$$ r_A=(-1.5)(0.129518)+0.933193+ \frac{0.129518^2}{0.0668072} \approx 0.9900. $$

For B, \(a_B=(0.50-0.40)/0.20=0.5\). With \(\Phi(0.5)=0.691462\), \(\phi(0.5)=0.352065\), and \(1-\Phi(0.5)=0.308538\),

$$ r_B=(0.5)(0.352065)+0.308538+ \frac{0.352065^2}{0.691462} \approx 0.6639. $$

The scalar parameter information is \(g^2r/\sigma^2\). Thus candidate A contributes \((0.30)^2(0.9900)/(0.20)^2=2.228\) per squared unit of \(\theta\); candidate B contributes \((0.10)^2(0.6639)/(0.20)^2=0.1660\) per squared unit of \(\theta\). A is more informative under these assumptions: it has both the larger sensitivity magnitude and the higher information fraction. Even so, B is not equivalent to no observation; its censored outcomes retain some information.

Design Choices, Diagnostics, and Limitations

The censoring likelihood is the principled choice when the assay reports “below LLOQ” and the assumed response model is appropriate. By contrast, dropping every censored sample treats the event as if it had never been observed. Replacing it by zero, \(L/2\), or \(L\) treats an interval of possible responses as an exact point. These substitutions alter the likelihood and can bias parameter estimates or understate uncertainty. They may also distort a design comparison: the apparent information can depend on an arbitrary replacement value rather than on the actual reporting process.

A useful design diagnostic is the predicted probability of censoring, \(\Phi(a)\), at every candidate time, considered alongside the assay-adjusted information and model sensitivities. If \(\Phi(a)\) is large, do not infer that the sample is useless; calculate its censored-data information. But if the information fraction is very small and the relevant parameter sensitivity is also weak, that time may contribute little. Check that the planned LLOQ and residual error are on the same measurement scale as the model prediction, and that the nominal predictions used in the calculation are plausible.

  • Check the reporting rule. The formula assumes exact results above a fixed threshold and a censoring event at or below it. It does not describe assay failures, missing samples, or a threshold that changes unpredictably between samples.
  • Check the residual model. A normal model on the concentration scale can assign probability to negative responses. If that is untenable, a model on a transformed scale or another suitable observation distribution may be needed; the threshold and likelihood must then be expressed on that scale.
  • Check independence and parameter dependence. The additive information calculation assumes conditionally independent samples. If residual variability depends on parameters, the information includes additional terms not represented by \(\mathcal I_\mu\mathbf g\mathbf g^T\).
  • Distinguish expected from realized information. The design formula averages across possible outcomes. Once a result is observed, its likelihood contribution depends on whether it was quantified and, if so, on its actual value.

A common mistake is to say, “The predicted concentration is below the LLOQ, so this sample provides zero information.” That confuses absence of an exact measurement with absence of an observation. A censored result supplies the inequality \(Y\leq L\), whose probability changes with the model prediction. The opposite mistake is to count a censored sample as though its exact value were known. The correction in both cases is to use the likelihood for the actual reporting process, then evaluate its expected information under the candidate design.

Key takeaway: A below-quantification result is a threshold event, not a measured concentration and not automatically a discarded sample. Compare candidate PK times using both the probability of censoring and the information in the censored-data likelihood, together with the model sensitivities relevant to the design objective.

Check Your Understanding

Answer each question, then check the solutions below.

  1. Under the normal model, what does a report “below LLOQ” contribute to the likelihood?
  2. If \(a=0\), what is the approximate expected information fraction for the mean, and what does it mean?
  3. For a candidate time with a high probability of censoring, should its expected information automatically be set to zero? Explain.
  4. Why is replacing each below-quantification result by \(L/2\) not equivalent to using the censoring likelihood?

Solutions. (1) It contributes the event probability \(\Phi(a)=P(Y\leq L)\), not a density evaluated at zero or at \(L\). (2) The fraction is \(0.8183\). Under the stated model, the expected information about the mean is about 81.83% of the information from an always-quantified normal observation with the same \(\sigma\). (3) No. The censored event still informs the model because its probability depends on the predicted mean; calculate its contribution with the censoring likelihood and combine it with the possible quantified outcomes. (4) \(L/2\) asserts an exact response value, whereas the report only establishes that the response lies at or below \(L\). The substitution therefore uses a different likelihood and can change estimates, uncertainty, and design comparisons.

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