Combined Additive and Proportional Error

Why Combine Additive and Proportional Error?

A single residual-error pattern may not describe measurements across the full concentration range. At low concentrations, an approximately fixed amount of measurement error can be important; at higher concentrations, the absolute error may grow with the predicted concentration. A combined additive and proportional error model represents both patterns in one variance function.

The key question is not simply whether the model has two parameters. It is how much each parameter contributes at a given predicted concentration. The relative contributions change with concentration, and their equality defines a useful crossover point.

Definition: Let $C_{ij}$ be the measured concentration for observation $j$ from subject $i$, and let $f_{ij}$ be its model-predicted concentration, conditional on the subject and occasion effects included in the prediction. A combined additive and proportional residual-error model is $$ C_{ij}=f_{ij}+\epsilon_{ij}, \qquad \epsilon_{ij}\mid f_{ij}\sim N\left(0,\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2 f_{ij}^2\right). $$ Here $\sigma_{\mathrm{add}}$ has concentration units, while $\sigma_{\mathrm{prop}}$ is dimensionless. The variance contributions are $\sigma_{\mathrm{add}}^2$ and $\sigma_{\mathrm{prop}}^2 f_{ij}^2$, respectively.

This specification assumes that the conditional residual mean is zero and that the two variance contributions add. A common construction is $\epsilon_{ij}=\epsilon_{\mathrm{add},ij}+f_{ij}\epsilon_{\mathrm{prop},ij}$, where the two terms are independent, have mean zero, and have variances $\sigma_{\mathrm{add}}^2$ and $\sigma_{\mathrm{prop}}^2$. Residuals from different observations are usually taken to be conditionally independent in this basic model. These assumptions concern residual error after accounting for the prediction; they do not say that between-subject or between-occasion variability is absent.

Deriving the Variance Function

The variance function follows directly from the independent error components. For independent random variables, the variance of a sum is the sum of their variances. Multiplying a random variable by a fixed value $f_{ij}$ multiplies its variance by $f_{ij}^2$.

\begin{aligned} \operatorname{Var}(\epsilon_{ij}\mid f_{ij}) &=\operatorname{Var}(\epsilon_{\mathrm{add},ij} +f_{ij}\epsilon_{\mathrm{prop},ij}\mid f_{ij})\\ &=\operatorname{Var}(\epsilon_{\mathrm{add},ij}) +f_{ij}^2\operatorname{Var}(\epsilon_{\mathrm{prop},ij})\\ &=\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2. \end{aligned}

The addition depends on the independence assumption. If the components were correlated, the variance would also contain a covariance term: $2f_{ij}\operatorname{Cov}(\epsilon_{\mathrm{add},ij},\epsilon_{\mathrm{prop},ij})$. The standard combined model above omits that term by assuming the components are independent. Its conditional standard deviation is the square root of the total variance, not the sum of the two standard deviations.

The additive component contributes a constant variance as $f_{ij}$ changes. The proportional component contributes variance that grows with $f_{ij}^2$; its standard deviation is $\sigma_{\mathrm{prop}}|f_{ij}|$. For nonnegative predicted concentrations, this is $\sigma_{\mathrm{prop}}f_{ij}$. Thus, the model permits a fixed absolute-error floor at low concentrations and increasing absolute error at higher concentrations.

Worked Example: Constructing the Variance at a Given Prediction

Suppose, hypothetically, that a concentration model has $\sigma_{\mathrm{add}}=0.20$ mg/L and $\sigma_{\mathrm{prop}}=0.15$. For an observation with predicted concentration $f_{ij}=10$ mg/L, calculate the additive and proportional variance contributions, the total variance, and the residual standard deviation.

The additive contribution is $(0.20\text{ mg/L})^2=0.0400\;(\text{mg/L})^2$. The proportional contribution is $(0.15)^2(10\text{ mg/L})^2=0.0225(100)=2.25\;(\text{mg/L})^2$.

Therefore, $$ \operatorname{Var}(\epsilon_{ij}\mid f_{ij}=10) =0.0400+2.25=2.29\;(\text{mg/L})^2. $$ The conditional residual standard deviation is $\sqrt{2.29}=1.513$ mg/L. The proportional contribution is much larger at this prediction, even though the additive term is not zero.

The Crossover Concentration

To identify where the components contribute equally, set their variance contributions equal. For positive $\sigma_{\mathrm{add}}$ and $\sigma_{\mathrm{prop}}$, and a nonnegative predicted concentration, the equality is

\sigma_{\mathrm{add}}^2 =\sigma_{\mathrm{prop}}^2 f_*^2 \quad\Longrightarrow\quad f_*=\frac{\sigma_{\mathrm{add}}}{\sigma_{\mathrm{prop}}}.

The ratio has concentration units, as it must: the numerator has concentration units and the denominator is dimensionless. Below $f_*$, the additive variance contribution is larger; above $f_*$, the proportional contribution is larger. At $f_*$, each contributes half of the total variance. This comparison concerns variance contributions, not a claim that one error source is the only source of variability on either side.

The same conclusion follows by comparing the component standard deviations, $\sigma_{\mathrm{add}}$ and $\sigma_{\mathrm{prop}}f_{ij}$, since both are nonnegative for nonnegative predictions and squaring preserves their ordering. If $\sigma_{\mathrm{add}}=0$, there is no additive component; if $\sigma_{\mathrm{prop}}=0$, there is no proportional component and no finite crossover. If both are zero, the model specifies zero residual variance.

Worked Example: Finding the Range Where Each Component Dominates

In a hypothetical assay model, suppose $\sigma_{\mathrm{add}}=0.12$ mg/L and $\sigma_{\mathrm{prop}}=0.08$. Find the crossover and determine the dominant contribution at predictions of 0.75 mg/L and 3.0 mg/L.

The crossover is $$ f_*=\frac{0.12\text{ mg/L}}{0.08}=1.50\text{ mg/L}. $$ Thus, the additive contribution is larger below 1.50 mg/L, and the proportional contribution is larger above 1.50 mg/L.

At $f_{ij}=0.75$ mg/L, the additive variance is $0.12^2=0.0144\;(\text{mg/L})^2$, and the proportional variance is $0.08^2(0.75)^2=0.0036\;(\text{mg/L})^2$. The total is $0.0180\;(\text{mg/L})^2$. The additive share is $0.0144/0.0180=0.80$, or 80%; the proportional share is 20%.

At $f_{ij}=3.0$ mg/L, the additive variance remains $0.0144\;(\text{mg/L})^2$, while the proportional variance is $0.08^2(3.0)^2=0.0576\;(\text{mg/L})^2$. The total is $0.0720\;(\text{mg/L})^2$, so the proportional share is $0.0576/0.0720=0.80$, or 80%. The calculated dominance agrees with the crossover rule.

Using the Model Across a Concentration Range

A fitted variance function can be examined at any predicted concentration by calculating both contributions. Their proportions of the total variance are

\text{additive share} =\frac{\sigma_{\mathrm{add}}^2} {\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2 f_{ij}^2}, \qquad \text{proportional share} =\frac{\sigma_{\mathrm{prop}}^2 f_{ij}^2} {\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2 f_{ij}^2}.

The shares sum to one. They can communicate the transition more precisely than the word “dominates,” especially near the crossover, where neither contribution is negligible. If $f_{ij}>0$, the combined model's conditional coefficient of variation is

\frac{\sqrt{\operatorname{Var}(\epsilon_{ij}\mid f_{ij})}}{f_{ij}} = \sqrt{\left(\frac{\sigma_{\mathrm{add}}}{f_{ij}}\right)^2+ \sigma_{\mathrm{prop}}^2}.

Consequently, the combined model does not have a constant coefficient of variation across all concentrations: the additive term makes the relative error larger at low positive predictions, while the proportional contribution approaches a constant relative-error level as predictions increase.

Worked Example: Comparing Component Shares Below and Above the Crossover

Consider hypothetical estimates $\sigma_{\mathrm{add}}=0.30$ mg/L and $\sigma_{\mathrm{prop}}=0.10$. The crossover is $0.30/0.10=3.00$ mg/L. Compare the components at 2.0 mg/L and 6.0 mg/L.

At 2.0 mg/L, the additive variance is $0.30^2=0.0900\;(\text{mg/L})^2$, and the proportional variance is $0.10^2(2.0)^2=0.0400\;(\text{mg/L})^2$. The total is $0.130\;(\text{mg/L})^2$, giving a standard deviation of $\sqrt{0.130}=0.3606$ mg/L. The additive share is $0.0900/0.130=0.6923$, or 69.2%, so the additive component dominates.

At 6.0 mg/L, the additive variance is still $0.0900\;(\text{mg/L})^2$, while the proportional variance is $0.10^2(6.0)^2=0.360\;(\text{mg/L})^2$. The total is $0.450\;(\text{mg/L})^2$, and the standard deviation is $\sqrt{0.450}=0.6708$ mg/L. The proportional share is $0.360/0.450=0.800$, or 80.0%, so the proportional component dominates.

Assumptions, Checks, and Limitations

The crossover calculation is algebraic, but its interpretation depends on the model being appropriate. The prediction $f_{ij}$ must be on the same concentration scale as the measured $C_{ij}$. The additive parameter must therefore be in concentration units, and the proportional parameter must be dimensionless. The variance function also presumes that its parameters are nonnegative and that the conditional residual mean is represented by $f_{ij}$.

A useful check is to compare the fitted variance with the observed spread of residuals across predicted concentrations. Under the stated model, residuals should have approximately zero conditional mean, and their scale should be compatible with $\sqrt{\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2}$. A repeated increase or decrease in spread beyond what this curve predicts suggests that the chosen variance function may not describe the data well. Systematic residual patterns can also indicate that the prediction itself is inadequate; they should not automatically be attributed to residual variance.

The concentration range in the data matters for estimating both parameters. If nearly all predictions lie far below the crossover, the proportional component may be weakly informed. If nearly all lie far above it, the additive component may be difficult to distinguish from a small contribution to the total variance. A numerically calculated crossover can then be outside the observed range and should not be presented as a well-supported empirical transition.

The pure additive model sets $\sigma_{\mathrm{prop}}=0$, giving constant residual variance. The pure proportional model sets $\sigma_{\mathrm{add}}=0$, giving variance proportional to $f_{ij}^2$. The combined model is more flexible, but flexibility alone does not establish that both components are needed. It also retains a Gaussian residual assumption; when predictions are near zero, a symmetric Gaussian model can assign probability to negative measured concentrations. The combined function does not resolve that limitation by itself.

Common Mistakes

Adding standard deviations. The model adds variances: $\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2$. Taking the square root gives the total standard deviation.
Using the wrong crossover ratio. Equating the variance contributions gives $f_*=\sigma_{\mathrm{add}}/\sigma_{\mathrm{prop}}$, not the inverse. The units provide a check: the answer must have concentration units.
Calling one component absent when it is smaller. Below the crossover, the proportional contribution is still present; it is simply smaller than the additive contribution.
Interpreting a crossover outside the data range as observed evidence. The ratio defines the model's mathematical crossover, but data concentrated on one side may provide little information about the transition.

Key takeaway: The combined residual variance is $\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2$. When both components are positive, they contribute equally at $f_*=\sigma_{\mathrm{add}}/\sigma_{\mathrm{prop}}$; predictions below and above this value are additive- and proportional-dominated, respectively.

Check Your Understanding

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

If $\sigma_{\mathrm{add}}=0.18$ mg/L and $\sigma_{\mathrm{prop}}=0.06$, what is the crossover concentration?
For $\sigma_{\mathrm{add}}=0.10$ mg/L, $\sigma_{\mathrm{prop}}=0.20$, and $f_{ij}=1.0$ mg/L, calculate the total variance and identify the larger contribution.
At the crossover, what fraction of the total variance is contributed by each component?
Why is the sum of component standard deviations not the combined standard deviation under the independent-error model?

Solutions. (1) $f_*=0.18/0.06=3.00$ mg/L. (2) The additive variance is $0.10^2=0.0100\;(\text{mg/L})^2$, and the proportional variance is $0.20^2(1.0)^2=0.0400\;(\text{mg/L})^2$. The total is $0.0500\;(\text{mg/L})^2$, and the proportional contribution is larger. (3) The contributions are equal, so each is one-half, or 50%, of the total variance. (4) Independent variances add, so the combined standard deviation is the square root of the sum of the variances, $\sqrt{\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2}$, not the sum of the standard deviations.