Tutorials › Nonlinear Mixed-Effects Models (NLME) › Residual Error and Assay Precision in Design Information

Study Design and Optimal Sampling for NLME · Tutorial 776 of 797

Residual Error and Assay Precision in Design Information

Quantify how residual-error assumptions and assay precision change the Fisher information contributed by each observation.

Advanced 12 min read

What You'll Learn

  • Distinguish additive, proportional, and combined residual-error models.
  • Calculate observation-level Fisher information with known or parameter-dependent variance.
  • Compare information across concentrations under combined error.
  • Quantify the effect of improving assay precision and adding technical replicates.
  • Recognize when conditional observation information does not equal marginal NLME information.
  • Diagnose error-model choices that can mislead design calculations.

Why residual error changes design information

A sampling time can be highly informative when the predicted concentration changes sharply with a parameter. But sensitivity alone is not enough: a measurement is less informative when its residual variability is large. In design calculations, residual error therefore determines how much each observation contributes, and concentration-dependent error can change the relative value of different sampling times.

The calculations here describe information from a normally distributed observation, conditional on the model quantities stated. In an NLME model, this may mean conditioning on an individual’s random effects. If those effects are integrated out, observations can be dependent and the marginal information must account for that dependence; the conditional contributions cannot simply be treated as the full marginal information.

Definition: For an observation \(Y_i\) with model mean \(m_i(\boldsymbol{\theta})\) and variance \(v_i(\boldsymbol{\theta})\), the Fisher information describes the expected curvature of the log likelihood with respect to the parameter vector \(\boldsymbol{\theta}\). When observations are conditionally independent, their conditional Fisher information matrices add. The sensitivity of the mean to parameter \(\theta_a\) is \(\partial m_i/\partial\theta_a\).

Specify the residual-error model first

A common combined-error model assumes that, conditional on the model prediction, the observation is normal with variance \(v_i=\sigma_{\mathrm{add}}^2+(\sigma_{\mathrm{prop}}m_i)^2\). Here \(\sigma_{\mathrm{add}}\) is an additive residual standard deviation in concentration units, while \(\sigma_{\mathrm{prop}}\) is a dimensionless proportional-error coefficient. The corresponding residual standard deviation is the square root of this variance. This tutorial uses the sum of the two variance components; it is not the alternative model that adds standard deviations.

With additive error alone, \(v_i=\sigma_{\mathrm{add}}^2\), so the absolute residual spread is constant across concentrations. With proportional error alone, \(v_i=(\sigma_{\mathrm{prop}}m_i)^2\), so the residual standard deviation is a constant fraction of the predicted concentration. In the combined model, the two terms are equal when \(m_i=\sigma_{\mathrm{add}}/\sigma_{\mathrm{prop}}\), for positive predicted concentrations. Below this crossover, additive error tends to dominate; above it, proportional error tends to dominate.

These models assume a correctly specified conditional mean, a normal conditional residual distribution, and the stated variance structure. The calculations also require the variance to be positive. A proportional-only model needs positive predictions to describe positive concentrations sensibly; it is not a remedy for zeros or values near zero. Whether residuals are independent conditional on the model is another assumption to check, not something guaranteed by the variance formula.

Key distinction: If the variance is known and does not depend on the parameter being evaluated, mean sensitivity contributes information in proportion to \(1/v_i\). If \(v_i\) itself depends on that parameter, the normal likelihood also contains information through the change in variance. A design calculation must state which convention it uses.

Information from a normal observation

Suppose \(Y_i\mid\boldsymbol{\theta}\sim N(m_i(\boldsymbol{\theta}),v_i(\boldsymbol{\theta}))\), where the variance is positive and both functions are differentiable. Write \(m_{ia}=\partial m_i/\partial\theta_a\) and \(v_{ia}=\partial v_i/\partial\theta_a\). The observation’s Fisher information for components \(\theta_a\) and \(\theta_b\) is

$$ \mathcal{I}_{i,ab} = \frac{m_{ia}m_{ib}}{v_i} + \frac{1}{2}\frac{v_{ia}v_{ib}}{v_i^2}. $$

The first term is information carried by changes in the mean relative to residual variance. The second term is information carried by changes in the variance. When \(v_i\) is fixed with respect to the parameters of interest, \(v_{ia}=0\), leaving only the first term. For independent observations, sum the observation matrices to obtain the conditional total information. In vector notation, the mean term is \(\mathbf{J}_i^\mathsf{T}\mathbf{J}_i/v_i\), where \(\mathbf{J}_i\) is the row of mean sensitivities.

Why the variance term appears

For one observation, let \(r=Y_i-m_i\). The log likelihood, apart from constants, is \(-\tfrac12\log(v_i)-r^2/(2v_i)\). Differentiating with respect to \(\theta_a\) gives the score

$$ S_a = \frac{m_{ia}r}{v_i} + \frac{v_{ia}}{2} \left(\frac{r^2}{v_i^2}-\frac{1}{v_i}\right). $$

Under the assumed normal model, \(E(r)=0\), \(E(r^2)=v_i\), and \(E(r^4)=3v_i^2\). Thus the two score components have zero expected cross-product: their product contains only terms proportional to \(E(r)\) or \(E(r^3)\), both of which are zero for a centered normal variable. The mean-score product has expectation \(m_{ia}m_{ib}/v_i\). For the variance-score product, the centered quantity \(r^2/v_i^2-1/v_i\) has variance \(2/v_i^2\), giving \(v_{ia}v_{ib}/(2v_i^2)\). Since Fisher information is the expected product of score components, the displayed formula follows.

In the combined-error model, if \(\sigma_{\mathrm{add}}\) and \(\sigma_{\mathrm{prop}}\) are known and \(m_i\) depends on \(\theta_a\), then \(\partial v_i/\partial\theta_a=2\sigma_{\mathrm{prop}}^2m_i\,\partial m_i/\partial\theta_a\). Some design calculations evaluate the variance at nominal predictions and hold it fixed while differentiating the mean. That gives the mean-information term only. It is a deliberate approximation or convention, not the exact Fisher information for a likelihood whose variance changes with the parameter.

Worked Example: Mean and variance information at one concentration

Suppose a hypothetical observation has predicted concentration \(m=10\) mg/L, and a scalar parameter \(\theta\) has sensitivity \(\partial m/\partial\theta=2\) mg/L per unit of \(\theta\). Let the known combined-error coefficients be \(\sigma_{\mathrm{add}}=1\) mg/L and \(\sigma_{\mathrm{prop}}=0.10\). Calculate its information for \(\theta\), including the variance term.

The variance is \(v=1^2+(0.10\times10)^2=2\;(\mathrm{mg/L})^2\). The mean-information contribution is \((\partial m/\partial\theta)^2/v=2^2/2=2\) per squared unit of \(\theta\).

Because the variance depends on the mean, \(\partial v/\partial\theta=2(0.10)^2(10)(2)=0.40\;(\mathrm{mg/L})^2\) per unit of \(\theta\). The variance-information contribution is \(\tfrac12(0.40)^2/2^2=0.0200\) per squared unit of \(\theta\).

The exact total is \(2+0.0200=2.020\) per squared unit of \(\theta\). If the design convention freezes the variance at its nominal value, it reports 2 instead. The variance term is small here, but its size depends on the model and should not be silently assumed to be zero.

Concentration changes the value of an observation

Under the mean-information term, an observation is valuable when its parameter sensitivity is large relative to its residual standard deviation. Under pure proportional error, for example, that term for a scalar parameter is \((\partial m_i/\partial\theta)^2/(\sigma_{\mathrm{prop}}m_i)^2\). Equivalently, it is the squared relative sensitivity \((\partial m_i/\partial\theta)/m_i\), divided by the squared proportional-error coefficient. A large concentration by itself does not guarantee high information: what matters is sensitivity relative to noise.

With combined error, increasing concentration can move an observation from the additive-dominated to the proportional-dominated region. If the sensitivity does not grow enough to offset that increase in variance, its mean-information contribution falls. This is why a design that ranks times using sensitivities alone can favor noisy observations and give a misleading picture of their value.

Worked Example: Comparing low and high concentrations

Consider two hypothetical candidate observations for the same scalar parameter. At each time the predicted concentration has sensitivity \(1\) mg/L per unit of \(\theta\). Use \(\sigma_{\mathrm{add}}=1\) mg/L and \(\sigma_{\mathrm{prop}}=0.10\), and calculate exact information at predictions of 2 mg/L and 20 mg/L.

At 2 mg/L, \(v=1^2+(0.10\times2)^2=1.04\;(\mathrm{mg/L})^2\), and \(\partial v/\partial\theta=2(0.10)^2(2)(1)=0.040\;(\mathrm{mg/L})^2\) per unit of \(\theta\). Therefore, \(\mathcal{I}=1/1.04+\tfrac12(0.040)^2/(1.04)^2=0.9615+0.000739=0.9623\) per squared unit of \(\theta\).

At 20 mg/L, \(v=1^2+(0.10\times20)^2=5\;(\mathrm{mg/L})^2\), and \(\partial v/\partial\theta=2(0.10)^2(20)(1)=0.40\;(\mathrm{mg/L})^2\) per unit of \(\theta\). Thus, \(\mathcal{I}=1/5+\tfrac12(0.40)^2/5^2=0.2000+0.003200=0.2032\) per squared unit of \(\theta\).

The low-concentration observation contributes about \(0.9623/0.2032=4.74\) times as much information in this calculation, despite equal mean sensitivities. With additive error alone and the same sensitivity, each would contribute \(1/1^2=1\). This comparison is conditional on the stated error model and sensitivities; it does not establish that low-concentration sampling is generally preferable.

Assay precision and technical replication

Assay precision enters the information calculation through the relevant variance component. Improving a known additive assay standard deviation reduces its variance contribution by the square of the improvement factor. For instance, halving an additive standard deviation quarters its variance. If proportional or other residual variability remains, however, the total variance does not generally fall by that same factor.

Technical replicates can provide repeated measurements of the same underlying concentration. If their assay errors are conditionally independent, have common known variance \(v\), and share the same mean sensitivity, their mean-information contributions add: \(r\) replicates contribute \(r(\partial m/\partial\theta)^2/v\), where \(r\) is the number of replicates. This calculation applies to the independently replicated error component. If the model also has shared biological or occasion-level variability, treating all replicate measurements as independent draws from the entire variance model can overstate information.

Worked Example: Improving additive assay precision

At a hypothetical time, the predicted concentration is \(8\) mg/L and its sensitivity is \(1.5\) mg/L per unit of \(\theta\). The proportional coefficient is \(0.05\). Compare known additive assay standard deviations of \(0.8\) mg/L and \(0.4\) mg/L, including the variance-information term.

With additive standard deviation \(0.8\) mg/L, \(v=0.8^2+(0.05\times8)^2=0.64+0.16=0.80\;(\mathrm{mg/L})^2\). The variance derivative is \(2(0.05)^2(8)(1.5)=0.060\;(\mathrm{mg/L})^2\) per unit of \(\theta\). Therefore, \(\mathcal{I}=1.5^2/0.80+\tfrac12(0.060)^2/(0.80)^2=2.8125+0.002813=2.815\) per squared unit of \(\theta\).

With additive standard deviation \(0.4\) mg/L, \(v=0.4^2+0.16=0.32\;(\mathrm{mg/L})^2\). The variance derivative remains \(0.060\), because only the additive component changed and it is constant with respect to \(\theta\). Now \(\mathcal{I}=1.5^2/0.32+\tfrac12(0.060)^2/(0.32)^2=7.0313+0.01758=7.049\) per squared unit of \(\theta\). The exact information is about \(7.049/2.815=2.50\) times as large, not four times as large: the proportional component remains, and the variance-information term also changes.

Checks, alternatives, and limitations

A useful design check is to calculate the predicted residual standard deviation and the mean sensitivity at each candidate time, then compare their ratio through the mean-information term. Inspect the additive and proportional variance components separately. If the high-concentration variance is unexpectedly dominant, verify the units and scale of the proportional coefficient; a coefficient intended as a fraction should not be entered as a percentage without conversion.

Residual diagnostics from suitable fitted models can reveal patterns inconsistent with the design assumption: residual spread increasing with prediction challenges an additive-only model, while a nearly constant absolute spread may challenge a proportional-only model. Curvature, outliers, censoring, and serial dependence can also make a normal independent-error calculation inadequate. Diagnostics cannot prove the model correct, and the assumed error structure should be justified for the assay and measurement process.

ApproachHow information is weightedImportant limitation
Additive normal errorUses a constant absolute varianceMay understate high-concentration variability
Proportional normal errorUses variance proportional to predicted concentration squaredCan be unsuitable near zero and assumes constant relative spread
Combined normal errorAdds additive and proportional variance componentsRequires defensible estimates or assumptions for both components
Log-scale error modelRepresents variability on a logarithmic scaleChanges the distributional and mean interpretation; it is not interchangeable with a normal model on the original scale

For a population design, conditional information given random effects is not automatically the information in the marginal likelihood. Marginalization can induce dependence among observations and changes the relevant mean and covariance structure. Likewise, if residual-error parameters are estimated rather than known, their uncertainty must be included as nuisance-parameter information when assessing precision for other parameters. The computations above isolate the effect of the specified residual model; they do not replace those broader calculations.

Common pitfall: “This time has the largest concentration and a large sensitivity, so it must be the most informative.” That reasoning ignores the variance. Compare sensitivity relative to residual standard deviation, and state whether variance derivatives are included. A larger prediction can carry less information when its noise rises faster than its sensitivity.

Check Your Understanding

Answer each question using the stated assumptions. Brief solutions follow.

  1. With additive standard deviation \(2\) mg/L and proportional coefficient \(0.10\), at what positive predicted concentration are the two variance components equal?
  2. A parameter-independent variance is \(9\;(\mathrm{mg/L})^2\), and the mean sensitivity is \(3\) mg/L per unit of a scalar parameter. What is the observation’s Fisher information?
  3. Why can a proportional-only normal error model be problematic near zero predicted concentration?
  4. Under what conditions does doubling the number of technical replicates double the mean-information contribution?
  5. Why is conditional information given random effects not necessarily the complete population information in an NLME design?

Solutions. (1) Equality requires \(\sigma_{\mathrm{add}}^2=(\sigma_{\mathrm{prop}}m)^2\), so \(m=2/0.10=20\) mg/L. (2) Since the variance is parameter-independent, the variance term is zero and the information is \(3^2/9=1\) per squared unit of the parameter. (3) Its standard deviation is proportional to the prediction, so it tends to zero as the prediction tends to zero; that may not represent additive assay noise or the behavior of measurements near a detection limit. (4) The replicate errors must be conditionally independent, share the same mean sensitivity and variance, and the replication must apply to the error component being modeled. (5) Integrating out random effects can induce dependence and change the marginal likelihood, so conditional contributions alone need not equal marginal Fisher information.