Power Error Models and Concentration-Dependent Precision

Why Let Residual Precision Change With Concentration?

A residual-error model describes the difference between an observed concentration and the concentration predicted by the structural model. In some assays, the typical absolute size of that difference changes systematically with the predicted concentration. A power error model represents this change with a single exponent: it specifies how quickly the residual standard deviation grows or shrinks as the prediction changes.

The exponent is useful because it governs both absolute precision, measured by the residual standard deviation, and relative precision, measured by the coefficient of variation. These are not the same: an absolute error can grow with concentration even while the error as a percentage of the prediction declines.

Model and Assumptions

Definition: For observation $j$ on individual $i$, let $Y_{ij}$ be the observed concentration and $f_{ij}$ the model-predicted concentration. The power residual-error model is $$ Y_{ij}=f_{ij}+f_{ij}^{\,p}\epsilon_{ij}, $$ where $p$ is the power parameter. Conditional on the predictions, assume $E(\epsilon_{ij}\mid f_{ij})=0$, $\operatorname{Var}(\epsilon_{ij}\mid f_{ij})=\sigma^2$, and, when a normal observation model is used, $\epsilon_{ij}\mid f_{ij}\sim N(0,\sigma^2)$. The errors are assumed independent across observations unless a separate correlation structure is specified. The usual interpretation below assumes $f_{ij}>0$.

The model says that the residual $Y_{ij}-f_{ij}$ is the prediction raised to the power $p$, multiplied by a mean-zero error with common variance $\sigma^2$. Thus, the conditional mean remains $f_{ij}$. The variance and precision calculations are conditional on the predicted concentration; they do not include between-subject or between-occasion variation in the predictions.

For concentrations, $f_{ij}$ and $Y_{ij}$ have concentration units. Since $f_{ij}^{\,p}\epsilon_{ij}$ must also have concentration units, $\epsilon_{ij}$ and $\sigma$ have units of concentration$^{1-p}$. The numerical value and units of $\sigma$ therefore depend on the concentration scale and on $p$; it is not generally a dimensionless coefficient of variation.

Deriving Variance, Standard Deviation, and CV

Given $f_{ij}$, the prediction is fixed, so subtracting it from the observation leaves $f_{ij}^{\,p}\epsilon_{ij}$. Its conditional mean is zero. Its conditional variance follows by pulling the fixed multiplier outside the variance:

\begin{aligned} \operatorname{Var}(Y_{ij}\mid f_{ij}) &=\operatorname{Var}(f_{ij}^{\,p}\epsilon_{ij}\mid f_{ij})\\ &=f_{ij}^{\,2p}\operatorname{Var}(\epsilon_{ij}\mid f_{ij})\\ &=\sigma^2 f_{ij}^{\,2p}. \end{aligned}

Consequently, for positive $f_{ij}$, the conditional residual standard deviation is $\sigma f_{ij}^{\,p}$. The coefficient of variation (CV), defined here as conditional standard deviation divided by conditional mean, is

\operatorname{CV}(Y_{ij}\mid f_{ij}) =\frac{\sigma f_{ij}^{\,p}}{f_{ij}} =\sigma f_{ij}^{\,p-1}.

This distinction is central: $p$ is the exponent for the standard deviation, whereas $2p$ is the exponent for the variance. The CV depends on $p-1$, not $p$.

Interpretation: If $p=0$, the standard deviation is constant and the model is additive-error-like. If $p=1$, the standard deviation is proportional to the prediction and the CV is constant. If $0<p<1$, absolute error grows with concentration while the CV declines. If $p>1$, both absolute error and CV grow with concentration. If $p<0$, the model implies larger absolute residual variation at smaller positive predictions.

How the Exponent Controls Changes Across Concentrations

A ratio makes the interpretation of $p$ especially clear. For two positive predictions $f_1$ and $f_2$, with the same $\sigma$, the ratio of their residual standard deviations is $(f_2/f_1)^p$. Taking logarithms shows that $p$ is the slope relating log standard deviation to log prediction. The corresponding slope for log variance is $2p$, and the slope for log CV is $p-1$.

\frac{\operatorname{SD}(Y\mid f_2)}{\operatorname{SD}(Y\mid f_1)} =\left(\frac{f_2}{f_1}\right)^p, \qquad \log\{\operatorname{SD}(Y\mid f)\} =\log(\sigma)+p\log(f).

This ratio interpretation assumes the same power model and the same $\sigma$ apply at both predictions. It is not a claim that two individual observations will have residuals in this exact ratio: it compares their model-implied standard deviations.

Worked Example: A Square-Root Power and Changing CV

Suppose hypothetical concentrations are modeled with $p=0.5$ and $\sigma=0.12$ in units of $(\text{mg/L})^{0.5}$. Calculate the conditional standard deviation, variance, and CV at predictions of 4 mg/L and 16 mg/L.

At $f=4$ mg/L, $f^p=4^{0.5}=2$. Therefore, the standard deviation is $0.12(2)=0.240$ mg/L, and the variance is $0.240^2=0.0576$ $(\text{mg/L})^2$. The CV is $0.240/4=0.0600$, or 6.00%.

At $f=16$ mg/L, $f^p=16^{0.5}=4$. The standard deviation is $0.12(4)=0.480$ mg/L, and the variance is $0.480^2=0.2304$ $(\text{mg/L})^2$. The CV is $0.480/16=0.0300$, or 3.00%.

The prediction increased fourfold. The standard deviation doubled, as $4^{0.5}=2$, while the CV halved. This matches $p=0.5$: absolute residual variation increases with concentration, but relative variation decreases.

Worked Example: A Power Greater Than One

For a second hypothetical model, let $p=1.2$ and $\sigma=0.080$ in units of $(\text{mg/L})^{-0.2}$. Find the standard deviation and CV at predictions of 2 mg/L and 8 mg/L.

At $f=2$ mg/L, $2^{1.2}\approx2.297$. The standard deviation is $0.080(2.297)=0.1838$ mg/L. The CV is $0.080(2^{0.2})\approx0.0919$, or 9.19%. Equivalently, $0.1838/2\approx0.0919$.

At $f=8$ mg/L, $8^{1.2}\approx12.126$. The standard deviation is $0.080(12.126)=0.9701$ mg/L. The CV is $0.080(8^{0.2})\approx0.1213$, or 12.13%. The standard deviation increased by a factor of $4^{1.2}\approx5.278$ between the two predictions, and the CV increased by a factor of $4^{0.2}\approx1.320$.

Because $p>1$, both absolute and relative residual variability increase with prediction. The standard deviation is still less than the predicted concentration at both points, but this is not guaranteed by the model for every prediction.

Learning the Power From Hypothetical Replicate Summaries

The ratio formula also gives a simple way to interpret an exponent estimated from data. If replicate observations at two predictions provide reliable estimates of residual standard deviation, their ratio can illustrate the power needed to describe the change. In an actual NLME analysis, uncertainty and all available observations must be considered; two sample standard deviations alone do not establish a precise value of $p$.

Worked Example: Interpreting Replicate Residual Spreads

Suppose a hypothetical assay has replicate residual standard deviations of 0.12 mg/L at a prediction of 1 mg/L and 0.24 mg/L at a prediction of 4 mg/L. Under one common power model, what exponent matches these two spreads, and what does it imply about CV?

The ratio relation gives $0.24/0.12=(4/1)^p$, so $2=4^p$. Taking logarithms, $p=\log(2)/\log(4)=0.500$. With the 1 mg/L prediction, the model standard deviation equals $\sigma(1)^{0.5}$, so the matching scale is $\sigma=0.12$ in units of $(\text{mg/L})^{0.5}$.

The implied CV is 12.0% at 1 mg/L and $0.24/4=0.0600$, or 6.00%, at 4 mg/L. The fitted exponent describes a doubling in absolute spread over a fourfold increase in prediction, while relative spread falls by half. These two summaries illustrate the algebra; replicate estimates are subject to sampling variability and should not be treated as exact population standard deviations.

Comparison, Diagnostics, and Limitations

The power model includes familiar error structures as special cases, but its flexibility has costs. The table compares the main interpretations. Here, “standard deviation” means conditional residual standard deviation.

Model or case	Standard deviation	Implication
Power model, $p=0$	$\sigma$	Constant absolute precision; additive-error-like.
Power model, $p=1$	$\sigma f$	Constant CV of $\sigma$; proportional-error-like.
Power model, general $p$	$\sigma f^p$	Absolute and relative precision change according to $p$.
Combined additive and proportional error	Typically based on an additive variance term plus a term proportional to $f^2$	Allows a nonzero additive component as well as concentration-scaled variation.

The earlier material on lognormal residual error describes a different observation model. A normal power-error model is additive on the concentration scale and can assign positive probability to negative observations. A lognormal model instead keeps observations positive and models variation on the log scale; it should not be treated as merely another value of $p$.

A basic diagnostic is to examine whether residual spread changes with prediction in the way the fitted power implies. Under the model, residuals divided by $f_{ij}^{\,p}$ should have approximately constant variance $\sigma^2$; equivalently, residuals divided by $\sigma f_{ij}^{\,p}$ should have approximately unit variance. A remaining trend in spread across predictions suggests that the chosen power or the observation model may be inadequate. This check does not, by itself, identify the cause of a mismatch.

The exponent is difficult to distinguish from the scale parameter when predictions cover only a narrow range: several combinations of $p$ and $\sigma$ may imply similar standard deviations there. Interpretation also requires positive predictions for noninteger powers. At predictions near zero, $p>0$ forces the modeled standard deviation toward zero, whereas $p<0$ makes it grow without bound. Such behavior may be implausible for the measurement process. For a normal error model, a large CV can also imply a material probability of negative concentrations. These are model limitations to assess in context, not reasons to assume one exponent is universally appropriate.

Common Mistake

Common mistake: “The power is $p=0.5$, so the variance is proportional to the square root of concentration and the CV is constant.” This confuses standard deviation, variance, and CV. The standard deviation is proportional to $f^{0.5}$, the variance to $f^{1.0}$, and the CV to $f^{-0.5}$. The CV is constant only when $p=1$.

Always derive the quantity needed from the model rather than transferring the exponent unchanged. Squaring the standard deviation gives variance; dividing it by the mean gives CV.

Check Your Understanding

Answer each question, then compare with the solutions below.

For $p=0.25$, by what factor does the residual standard deviation change when the prediction increases eightfold?
For $p=0.25$, does the CV rise or fall as prediction increases? Give its concentration exponent.
A model has $p=1$ and $\sigma=0.07$. What is its CV, and what is the standard deviation at a prediction of 5 mg/L?
Why can a power model with a noninteger exponent be problematic at zero or negative predictions?
What should be examined if normalized residual spread still changes systematically with prediction?

Solutions. (1) The standard-deviation ratio is $8^{0.25}\approx1.682$, so it increases by a factor of about 1.682. (2) The CV is proportional to $f^{p-1}=f^{-0.75}$, so it falls as prediction rises. (3) With $p=1$, CV is $\sigma=0.070$, or 7.00%; the standard deviation is $0.07(5)=0.350$ mg/L. (4) A negative prediction may not have a real-valued fractional power, and zero can make the model imply zero or unbounded variability depending on the exponent. (5) The fitted power or observation model may not describe the residual spread adequately; examine the model and its assumptions rather than interpreting the remaining trend as evidence for a particular cause on its own.

Model or case	Standard deviation	Implication
Power model, \(p=0\)	\(\sigma\)	Constant absolute precision; additive-error-like.
Power model, \(p=1\)	\(\sigma f\)	Constant CV of \(\sigma\); proportional-error-like.
Power model, general \(p\)	\(\sigma f^p\)	Absolute and relative precision change according to \(p\).
Combined additive and proportional error	Typically based on an additive variance term plus a term proportional to \(f^2\)	Allows a nonzero additive component as well as concentration-scaled variation.