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Below-Quantification-Limit Observations · Tutorial 302 of 797

Censoring Under Proportional Residual Error

Learn how proportional residual error determines the probability and likelihood contribution of a BQL observation, and why that probability behaves differently from additive error near zero concentration.

Advanced 10 min read

What You'll Learn

  • Express proportional residual error as a conditional observation distribution with a concentration-dependent standard deviation.
  • Derive the probability that an observation falls below the LLOQ.
  • Construct the likelihood contribution for BQL and quantified observations.
  • Explain why likelihoods in an NLME model must account for individual random effects.
  • Compare proportional and additive error as predicted concentrations approach zero.
  • Identify positivity, distributional, and implementation pitfalls.

Why Proportional Error Changes BQL Probabilities

A below-quantification-limit (BQL) result is an observation known to fall below an assay threshold, not a measured concentration that should be treated as an exact value. Its likelihood contribution therefore depends on the probability assigned to the region below the threshold by the observation model. Under proportional residual error, the residual standard deviation changes with the model-predicted concentration. Consequently, the BQL probability is not calculated in the same way as it is under additive error.

Let \(f_{ij}=f(t_{ij},\theta,\eta_i)\) denote the model-predicted concentration for individual \(i\) at observation time \(t_{ij}\), conditional on population parameters \(\theta\) and individual random effects \(\eta_i\). Let \(L_{ij}>0\) be the known LLOQ applying to that observation. The derivation below is conditional on \(\eta_i\); the population likelihood must also account for the distribution of the random effects.

Definition: Under proportional normal residual error, an observation is modeled as \(Y_{ij}=f_{ij}(1+\epsilon_{ij})\), where \(\epsilon_{ij}\sim N(0,\sigma^2)\). For \(f_{ij}>0\) and \(\sigma>0\), this is equivalent to \(Y_{ij}\mid\eta_i\sim N(f_{ij},\sigma^2 f_{ij}^2)\). Thus the conditional mean is \(f_{ij}\), the conditional standard deviation is \(\sigma f_{ij}\), and \(\sigma\) is the residual coefficient of variation.

This model assumes that the residual errors have the specified normal distribution and, when the usual conditionally independent observation likelihood is used, are independent across observations given the individual effects and model predictions. The prediction must be positive for the standard deviation \(\sigma f_{ij}\) to be positive. The model also uses one relative error scale \(\sigma\); it does not describe data whose relative variability changes systematically with concentration unless that feature is modeled separately.

Deriving the Conditional BQL Probability

A BQL result records the event \(Y_{ij}<L_{ij}\). Under the proportional model, divide both sides by the positive prediction \(f_{ij}\), then standardize the residual error. The resulting probability is a normal cumulative probability.

$$ \begin{aligned} P(Y_{ij}<L_{ij}\mid\eta_i) &=P\left(f_{ij}(1+\epsilon_{ij})<L_{ij}\mid\eta_i\right)\\ &=P\left(\epsilon_{ij}<\frac{L_{ij}}{f_{ij}}-1\mid\eta_i\right)\\ &=\Phi\left(\frac{L_{ij}/f_{ij}-1}{\sigma}\right)\\ &=\Phi\left(\frac{L_{ij}-f_{ij}}{\sigma f_{ij}}\right). \end{aligned} $$

Here \(\Phi\) is the standard normal cumulative distribution function. The last expression is also obtained by taking the conditional normal distribution of \(Y_{ij}\) and standardizing its threshold using its standard deviation, \(\sigma f_{ij}\). With a continuous normal distribution, using \(Y_{ij}\le L_{ij}\) instead of \(Y_{ij}<L_{ij}\) gives the same probability.

The formula is conditional: it gives the BQL probability at a particular value of \(\eta_i\). In a nonlinear mixed-effects model, the individual effects are generally not known. The subject-level likelihood therefore averages the conditional likelihood over their distribution; substituting a population-average prediction into the conditional formula is not generally equivalent.

Worked Example: BQL Probability at a Given Prediction

Suppose a hypothetical sample has a conditional prediction of \(2.00\ \text{mg/L}\), an LLOQ of \(0.500\ \text{mg/L}\), and proportional error with \(\sigma=0.250\). Calculate the conditional probability of a BQL result.

The proportional standard deviation is \(\sigma f=0.250(2.00)=0.500\ \text{mg/L}\). Therefore,

$$ z=\frac{L-f}{\sigma f} =\frac{0.500-2.00}{0.500} =-3.00, \qquad P(Y<L\mid\eta)=\Phi(-3.00)=0.001350. $$

The model assigns a \(0.1350\%\) conditional probability to a BQL result at these values. The negative standardized threshold is consistent with the prediction being well above the LLOQ. This is a model-based probability, not a statement about the chance that the particular prediction or parameter is correct.

Likelihood Contributions and Random Effects

A quantified result contributes a density, whereas a BQL result contributes the probability of the censored region. For a quantified concentration \(y_{ij}\), the conditional density under proportional error is

$$ g(y_{ij}\mid\eta_i) = \frac{1}{\sigma f_{ij}\sqrt{2\pi}} \exp\left[-\frac{(y_{ij}-f_{ij})^2}{2\sigma^2 f_{ij}^2}\right]. $$

If \(b_{ij}=1\) denotes a BQL record and \(b_{ij}=0\) a quantified record, the conditional contribution can be written as

$$ \ell_{ij}(\eta_i)= \begin{cases} \Phi\left(\dfrac{L_{ij}-f_{ij}}{\sigma f_{ij}}\right), & b_{ij}=1,\\[6pt] g(y_{ij}\mid\eta_i), & b_{ij}=0. \end{cases} $$

The subject’s conditional likelihood is the product of these contributions over that subject’s observations when residual errors are conditionally independent. If \(g_\eta(\eta_i\mid\Omega)\) is the random-effects distribution, the marginal subject likelihood is the integral of that conditional likelihood against \(g_\eta\). This distinction matters: an observed BQL record contributes its conditional probability at each possible individual effect, averaged according to the model, rather than a probability evaluated only at an average effect.

Worked Example: A BQL Probability and a Quantified Density

For a hypothetical observation, let \(f=0.400\ \text{mg/L}\), \(\sigma=0.300\), and \(L=0.500\ \text{mg/L}\). First calculate the BQL contribution. Then, if a separate quantified result is \(y=0.620\ \text{mg/L}\) at the same prediction, calculate its conditional density.

The conditional standard deviation is \(0.300(0.400)=0.120\ \text{mg/L}\). For the BQL result,

$$ z=\frac{0.500-0.400}{0.120}=0.8333, \qquad P(Y<L\mid\eta)=\Phi(0.8333)\approx0.7977. $$

Thus, the BQL likelihood contribution is approximately \(0.7977\). For the quantified value, the standardized residual is \((0.620-0.400)/0.120=1.833\), and

$$ g(0.620\mid\eta) = \frac{1}{0.120\sqrt{2\pi}} \exp\left[-\frac{(0.620-0.400)^2}{2(0.120)^2}\right] \approx 0.6188\ (\text{mg/L})^{-1}. $$

The BQL contribution is a probability and has no units. The quantified contribution is a density, with units reciprocal to concentration; it is not the probability of observing exactly \(0.620\ \text{mg/L}\).

Worked Example: Averaging Over Individual Effects

Consider a simple hypothetical random-effects distribution that gives equal probability to two individual predictions, \(f=0.200\ \text{mg/L}\) and \(f=0.800\ \text{mg/L}\). Let \(L=0.500\ \text{mg/L}\) and \(\sigma=0.250\). Calculate the marginal BQL probability under this two-point distribution, and compare it with evaluating the formula at the average prediction.

At \(f=0.200\), the standard deviation is \(0.0500\ \text{mg/L}\), so \(z=(0.500-0.200)/0.0500=6.00\) and the BQL probability is approximately \(0.999999999\). At \(f=0.800\), the standard deviation is \(0.200\ \text{mg/L}\), so \(z=(0.500-0.800)/0.200=-1.50\) and the probability is approximately \(0.06681\). Averaging over the two equally likely effects gives

$$ P(\mathrm{BQL})= \frac{0.999999999+0.06681}{2} \approx 0.5334. $$

The average prediction is \(0.500\ \text{mg/L}\). Evaluating the conditional formula at that prediction gives \(z=0\) and a BQL probability of \(0.5000\), not \(0.5334\). This finite example demonstrates why averaging the conditional likelihood over random effects is generally different from calculating it at an average prediction. In an actual NLME model, the integral uses the specified random-effects distribution rather than this illustrative two-point distribution.

Low-Concentration Behavior and Comparison with Additive Error

For proportional error and a positive LLOQ, the standardized threshold can be written as \(z=(L/f-1)/\sigma\). As \(f\) approaches zero from above, \(L/f\) grows without bound, so \(z\) approaches infinity and the BQL probability approaches one. This follows from the normal CDF limit \(\Phi(z)\to1\) as \(z\to\infty\). At \(f=L\), the probability is exactly \(0.5\); for a fixed positive \(L\), it decreases as \(f\) increases.

Under additive normal error with standard deviation \(\sigma_a>0\), the observation model is \(Y\mid\eta\sim N(f,\sigma_a^2)\). Its BQL probability is \(\Phi((L-f)/\sigma_a)\). As \(f\) approaches zero, this tends to \(\Phi(L/\sigma_a)\), which is less than one for finite positive \(L\) and \(\sigma_a\). Thus additive error retains a fixed absolute spread at low predictions, while proportional error’s spread shrinks with the prediction.

FeatureProportional normal errorAdditive normal error
Conditional standard deviation\(\sigma f\), with \(\sigma\) dimensionless\(\sigma_a\), in concentration units
Conditional BQL probability\(\Phi((L-f)/(\sigma f))\)\(\Phi((L-f)/\sigma_a)\)
Limit as \(f\to0^+\), for \(L>0\)1\(\Phi(L/\sigma_a)\)
Main low-concentration featureAbsolute variability shrinks with the predictionAbsolute variability stays constant

Worked Example: Contrasting Low-Concentration Predictions

Suppose a hypothetical assay has \(L=0.500\ \text{mg/L}\). Compare the BQL probabilities at \(f=0.0500\ \text{mg/L}\) under proportional error with \(\sigma=0.300\) and additive error with \(\sigma_a=0.200\ \text{mg/L}\).

Under proportional error, the standard deviation is \(0.300(0.0500)=0.0150\ \text{mg/L}\), giving

$$ z_p=\frac{0.500-0.0500}{0.0150}=30.0, \qquad P_p(\mathrm{BQL})=\Phi(30.0)\approx1. $$

Under additive error, the standard deviation remains \(0.200\ \text{mg/L}\), giving

$$ z_a=\frac{0.500-0.0500}{0.200}=2.25, \qquad P_a(\mathrm{BQL})=\Phi(2.25)\approx0.9878. $$

Both models assign a high BQL probability, but the proportional model assigns one that is essentially one at this prediction because its absolute standard deviation is very small. These calculations illustrate the models’ different implications; they do not establish which residual model is appropriate for a particular assay.

Conditions, Diagnostics, and Common Mistakes

The proportional normal model is mathematically a normal distribution for any finite positive \(f\) and \(\sigma\), but its suitability for concentration data requires attention. A normal distribution has support over the entire real line. Under the proportional model, the probability of a negative observation is

$$ P(Y<0\mid\eta)=\Phi\left(-\frac{1}{\sigma}\right), $$

which does not depend on \(f\). For example, with \(\sigma=0.300\), this probability is \(\Phi(-3.333)\approx0.000429\). If the fitted relative error is large enough that negative values are not negligible, a normal proportional model may be a poor description of a nonnegative assay measurement. Also, at exactly \(f=0\), the model \(Y=f(1+\epsilon)\) collapses to a point mass at zero: it no longer has the ordinary positive-variance normal density used for quantified observations. The standard likelihood calculations therefore require strictly positive predictions for observations modeled with a continuous proportional normal distribution.

Check that the prediction is positive at every observation and that the LLOQ supplied to the likelihood is the threshold that actually applies to that record. Examine whether a single relative error scale is plausible across the concentration range and whether the implied negative-value probability is acceptable. A fit or implementation that produces nonpositive predictions, a zero residual scale, or an invalid standard deviation cannot use the stated density and BQL formula as written.

A common mistake is to calculate the proportional-error probability as \(\Phi((L-f)/\sigma)\). That expression uses \(\sigma\) as though it were an additive standard deviation in concentration units. Under proportional error, the standard deviation is \(\sigma f\), so the correct expression is \(\Phi((L-f)/(\sigma f))\). A related mistake is to treat the LLOQ itself as the observed concentration; the BQL contribution is the probability of the entire event below the LLOQ, not a density evaluated at the threshold.

The main alternative considered here is additive residual error. It is appropriate to compare the implied absolute variability and BQL probabilities over the concentration range, while recognizing that the two models make different assumptions about how variability scales. Neither model should be selected solely because one gives a preferred BQL rate at a single prediction. Proportional error can be unsuitable when absolute assay variability does not shrink at low concentrations; additive error can be unsuitable when variability is primarily relative. The error model is an assumption about the conditional observation distribution, not a device for changing the observed censoring status.

Key takeaway: Under proportional normal error, the conditional BQL probability is \(\Phi((L-f)/(\sigma f))\), provided \(f>0\) and \(\sigma>0\). Its standard deviation shrinks with \(f\), so for a fixed positive LLOQ its BQL probability tends to one as the prediction approaches zero from above. The likelihood contribution must be averaged over individual random effects in the NLME model.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A hypothetical observation has \(f=1.20\ \text{mg/L}\), \(L=0.600\ \text{mg/L}\), and \(\sigma=0.200\). What is its conditional BQL probability?
  2. Why is \(\Phi((L-f)/\sigma)\) not the BQL probability under proportional error?
  3. For proportional error with \(\sigma=0.250\), what is \(P(Y<0\mid\eta)\), and does it depend on positive \(f\)?
  4. For a positive finite LLOQ, state the limiting BQL probability as \(f\to0^+\) under proportional error and under additive normal error.

Solutions. (1) The standard deviation is \(0.200(1.20)=0.240\ \text{mg/L}\), so \(z=(0.600-1.20)/0.240=-2.50\), and \(P(\mathrm{BQL}\mid\eta)=\Phi(-2.50)\approx0.006210\). (2) The proportional standard deviation is \(\sigma f\), not \(\sigma\); the correct standardized threshold is \((L-f)/(\sigma f)\). (3) \(P(Y<0\mid\eta)=\Phi(-1/0.250)=\Phi(-4.00)\approx0.00003167\), independent of positive \(f\). (4) Under proportional error the limit is one. Under additive error with standard deviation \(\sigma_a>0\), it is \(\Phi(L/\sigma_a)\), which is less than one for finite positive \(L\) and \(\sigma_a\).