Introduction
A power model expresses how a typical pharmacokinetic parameter changes as a positive continuous covariate changes. It is useful when the relationship is naturally described by proportional rather than fixed-unit changes: for example, a 10% increase in a covariate may correspond to a similar percentage change in clearance across different parts of the covariate range. The reference value anchors the prediction, while the power exponent controls the direction and strength of the relationship.
The key calculation is an exact ratio. For two covariate values, the predicted parameter ratio is the ratio of the covariate values raised to the exponent. This is not generally the same as multiplying the exponent by the covariate’s percentage change; that product is only a small-change approximation.
The subscript “typ” indicates the prediction from the covariate relationship at the typical individual level, rather than a prediction that includes an individual random effect. In a model that also has a multiplicative individual random effect, setting that effect to zero gives this typical prediction. This distinction matters: the power model quantifies the covariate component, not all between-individual variation.
At \(x=x_{\mathrm{ref}}\), the covariate ratio is 1, so \(P_{\mathrm{typ}}(x_{\mathrm{ref}})=P_{\mathrm{ref}}\). If \(\theta_P=0\), the predicted parameter does not vary with \(x\). A positive exponent means the predicted parameter increases as \(x\) increases; a negative exponent means it decreases. These directional statements apply to the specified model and covariate range, not necessarily to a causal effect.
Exact Changes and the Meaning of the Exponent
For covariate values \(x_1\) and \(x_2\), use the same parameter relationship at both values and divide the predictions. The reference parameter cancels, leaving a ratio that depends only on the covariate ratio and the exponent.
The exact percentage change from \(x_1\) to \(x_2\) is therefore \(\left[(x_2/x_1)^{\theta_P}-1\right]\times 100\%\). A percentage change in the parameter is a relative change; it does not depend on whether the parameter is recorded in, for example, litres per hour or millilitres per minute. The predicted parameter value itself does depend on its units.
The exponent also has a local interpretation. Taking logarithms makes the model linear in log covariate and log parameter. The slope on that scale is the exponent, so a small proportional change in \(x\) corresponds approximately to \(\theta_P\) times that proportional change in \(P\).
For a finite change, use the exact ratio \((x_2/x_1)^{\theta_P}\), not its small-change approximation \(1+\theta_P[(x_2-x_1)/x_1]\). To see why the approximation is only local, write \(x_2/x_1=1+\delta\), where \(\delta=(x_2-x_1)/x_1\). A first-order Taylor expansion around \(\delta=0\) gives \((1+\delta)^{\theta_P}\approx1+\theta_P\delta\). This is useful when \(\delta\) is small, but the exact power is straightforward to calculate and should be preferred for specified finite contrasts.
Worked Example: Clearance Across a 20% Weight Increase
Suppose a hypothetical model predicts typical clearance of \(5.00\ \mathrm{L/h}\) at a reference weight of \(70\ \mathrm{kg}\), with weight exponent \(\theta_{CL}=0.75\). Calculate the predicted clearance at \(84\ \mathrm{kg}\) and the percentage change from \(70\ \mathrm{kg}\).
The covariate ratio is \(84/70=1.20\). The exact clearance ratio is \[ (84/70)^{0.75}=1.20^{0.75} =\exp[0.75\log(1.20)] \approx\exp(0.136741) \approx1.147. \] Thus the predicted clearance at \(84\ \mathrm{kg}\) is \(5.00\times1.147\approx5.73\ \mathrm{L/h}\). The percentage change is \((1.147-1)\times100\%\approx14.7\%\).
The covariate increased by 20%, while predicted clearance increased by about 14.7%. The small-change approximation would give \(0.75\times20\%=15.0\%\), close in this case but not exact. The model predicts an increase because the exponent is positive.
Worked Calculations for Different Contrasts
A reference value is useful for reporting an interpretable typical parameter, but it does not change the predicted relationship. Comparisons between any two covariate values can be calculated directly from their ratio. The following examples illustrate a negative exponent and the choice of reference.
Worked Example: Volume With a Negative Covariate Exponent
Consider a hypothetical volume model with \(V_{\mathrm{ref}}=40.0\ \mathrm{L}\) at an eGFR reference value of \(90\ \mathrm{mL/min/1.73\,m^2}\), and exponent \(\theta_V=-0.30\). Calculate the typical volume at eGFR \(60\ \mathrm{mL/min/1.73\,m^2}\) and its percentage change from the reference prediction.
The covariate ratio is \(60/90=2/3\), so \[ \frac{V_{\mathrm{typ}}(60)}{40.0} =(2/3)^{-0.30} =\exp[-0.30\log(2/3)] =\exp(0.1216) \approx1.129. \] The predicted volume is \(40.0\times1.129\approx45.2\ \mathrm{L}\). Relative to the reference prediction, the percentage change is \((1.129-1)\times100\%\approx12.9\%\).
Although the exponent is negative, the covariate decreased from 90 to 60, so the predicted volume increased. The sign of the exponent describes what happens when the covariate increases; it does not by itself state the direction of change for every contrast.
Worked Example: Changing the Reference Without Changing the Relationship
A hypothetical clearance model has \(CL_{\mathrm{ref}}=5.00\ \mathrm{L/h}\) at \(x_{\mathrm{ref}}=70\ \mathrm{kg}\), with \(\theta_{CL}=0.75\). Find the prediction at \(100\ \mathrm{kg}\). Then re-express the model using \(80\ \mathrm{kg}\) as the reference and verify that the prediction at \(100\ \mathrm{kg}\) is unchanged.
Using the original reference, \[ CL_{\mathrm{typ}}(100) =5.00(100/70)^{0.75} =5.00\exp[0.75\log(1.4286)] \approx5.00(1.307) \approx6.53\ \mathrm{L/h}. \]
The typical clearance at \(80\ \mathrm{kg}\), which becomes the new reference parameter, is \[ CL_{\mathrm{ref,new}} =5.00(80/70)^{0.75} \approx5.00(1.105) \approx5.53\ \mathrm{L/h}. \] Using this new reference, the prediction at \(100\ \mathrm{kg}\) is \[ 5.53(100/80)^{0.75} \approx5.53(1.182) \approx6.54\ \mathrm{L/h}. \] The predictions agree up to rounding. Changing the reference changes the reported reference parameter, but not the exponent or the model’s predictions, provided the reference parameter is recalculated consistently.
Assumptions, Diagnostics, and Alternatives
A power model is interpretable when the covariate is positive, its scale and units are correctly specified, and the power relationship is a reasonable description over the range being interpreted. It implies a straight-line relationship between \(\log P_{\mathrm{typ}}\) and \(\log x\); it does not imply that the parameter changes by a fixed amount for every one-unit increase in \(x\). The exponent is constant across the modeled range, so the relative response to a proportional covariate change is also constant.
Before calculating a contrast, check that both covariate values use the same units and definition, that the reference value is positive, and that the parameter prediction and contrast are within a range supported by the data. A reference value may be chosen for convenience or interpretability, but it should be stated explicitly. Extrapolating far beyond observed covariate values can produce numerically valid predictions that are poorly supported by the model data.
Useful diagnostics include plotting the fitted typical parameter relationship against the covariate on both the original and log scales, and examining whether systematic departures remain across the observed range. Check for influential observations, sparse regions, implausible predictions, and coding errors such as mixing kilograms with grams. A power model cannot be evaluated on the log-covariate scale when \(x\leq0\). If zero or negative values are meaningful in the data, this form is not directly applicable without a scientifically justified redefinition or a different model.
The main alternatives encode different shapes. An additive linear relationship predicts a fixed parameter increment per covariate unit and can predict nonpositive parameter values unless constrained. A log-linear relationship such as \(P_{\mathrm{typ}}(x)=P_{\mathrm{ref}}\exp[\beta(x-x_{\mathrm{ref}})]\) has a constant change in log parameter per covariate unit, rather than per proportional covariate change. A power model instead has a constant log-log slope. These forms are not interchangeable: their predictions can differ substantially across a wide covariate range, so the choice should reflect the data and the intended interpretation.
| Model form | What is constant | Key interpretation |
|---|---|---|
| Power: \(P_{\mathrm{ref}}(x/x_{\mathrm{ref}})^{\theta_P}\) | Change in log parameter per change in log covariate | \(\theta_P\) is the local elasticity |
| Additive linear: \(P_{\mathrm{ref}}+\beta(x-x_{\mathrm{ref}})\) | Parameter increment per covariate unit | \(\beta\) has parameter-units per covariate-unit |
| Log-linear: \(P_{\mathrm{ref}}\exp[\beta(x-x_{\mathrm{ref}})]\) | Change in log parameter per covariate unit | \(\beta\) has inverse covariate-units |
The power model’s compact interpretation should not be mistaken for evidence that the covariate causes the parameter to change. It describes a modeled association conditional on the model specification and other included covariates. Nor does a single exponent establish that the form is adequate for every subgroup or range. Predictions are only as reliable as the model and data supporting them.
Common Pitfall
A frequent error is to use \(\theta_P\) times the covariate’s percentage change as though it were the exact parameter percentage change. For example, with exponent \(0.75\) and a 20% increase in \(x\), this shortcut gives 15%. The exact parameter change is \((1.20^{0.75}-1)\times100\%\), about 14.7%. The approximation is justified only as a first-order result for a small proportional change; for a specified finite contrast, calculate the power ratio first.
Another error is to infer the direction of a contrast from the exponent alone. The exponent gives the direction of parameter change as \(x\) increases. If \(x\) decreases, the contrast direction reverses for a nonzero exponent. Always evaluate \((x_2/x_1)^{\theta_P}\) with the actual starting and ending covariate values.
Check Your Understanding
Try each calculation, then compare with the solutions below.
- A parameter has exponent \(0.50\). What is its predicted ratio when the covariate rises from \(100\) to \(121\)?
- Clearance is \(4.00\ \mathrm{L/h}\) at the reference covariate value. If the covariate ratio is \(0.80\) and the exponent is \(-0.25\), calculate predicted clearance and percentage change from reference.
- For exponent \(0.60\), calculate the exact parameter percentage change for a 10% covariate increase, and compare it with the first-order approximation.
- Why can the power model not be applied directly to a covariate value of zero?
Solutions. (1) The covariate ratio is \(121/100=1.21\), so the parameter ratio is \(1.21^{0.50}=1.10\): a 10.0% increase. (2) The parameter ratio is \(0.80^{-0.25}=\exp[-0.25\log(0.80)]\approx1.057\). Predicted clearance is \(4.00\times1.057\approx4.23\ \mathrm{L/h}\), an increase of about 5.74%. (3) The exact change is \((1.10^{0.60}-1)\times100\%\approx5.89\%\). The first-order approximation is \(0.60\times10\%=6.00\%\), close but not exact. (4) The model uses \(x/x_{\mathrm{ref}}\) raised to a power and its log-log interpretation requires positive \(x\); \(\log(0)\) is undefined, and the prediction may be undefined or degenerate depending on the exponent.