Why Use a Power-Law Effect?
A continuous covariate may change a pharmacokinetic parameter by a relative amount rather than by a fixed additive amount. For example, a parameter might increase as body weight increases, with the size of the increase depending on how large the weight change is relative to a reference weight. A power law expresses that relationship directly and gives its exponent a useful interpretation.
The model below describes the covariate effect on a typical parameter value. It does not, by itself, specify the random-effects distribution or residual-error model. Those parts of an NLME model are held fixed when interpreting this covariate relationship.
At \(x=x_{\mathrm{ref}}\), the ratio is 1, so \(\theta_{\mathrm{typ}}(x_{\mathrm{ref}})=\theta_{\mathrm{ref}}\), whatever the value of \(\beta\). If \(\beta>0\), the typical parameter increases with \(x\); if \(\beta<0\), it decreases; and if \(\beta=0\), it is constant with respect to this covariate. These directions describe the modelled relationship, not a causal effect.
Log-Ratio Interpretation
Taking logarithms of the model gives a linear relationship between the log parameter and the log covariate ratio. This is the source of the phrase log-log slope.
Because \(x_{\mathrm{ref}}\) is fixed, the derivative of the log parameter with respect to the log covariate is
Thus, \(\beta\) is the slope on a graph of \(\log\theta_{\mathrm{typ}}\) against \(\log x\). Equivalently, for a small relative change in \(x\), the relative change in the parameter is approximately \(\beta\) times as large. The exact finite-change relationship is more useful when changes are not small.
The exact percentage change depends on the covariate ratio, not on the absolute difference \(x_2-x_1\). For a 1% increase in \(x\), the exact parameter ratio is \(1.01^\beta\); if \(\beta=0\), this is \(1.01^0=1\), so the parameter changes by 0%. For small changes, the percentage change is approximately \(100\beta\) times the covariate percentage change, but this is an approximation rather than the exact finite-change rule.
Worked Predictions
Worked Example: Predicting Clearance at Several Weights
Suppose a hypothetical NLME model describes typical clearance, in L/h, by body weight \(x\), in kg: \[ CL_{\mathrm{typ}}(x)=5.00\left(\frac{x}{70}\right)^{0.75}. \] The reference clearance is 5.00 L/h at 70 kg. Calculate the typical clearance at 50 kg, 70 kg, and 100 kg.
At 50 kg, the covariate ratio is \(50/70=0.7143\). Therefore, \[ CL_{\mathrm{typ}}(50)=5.00(0.7143)^{0.75} =5.00\exp\{0.75\log(0.7143)\} \approx 5.00(0.7770)=3.885\text{ L/h}. \]
At the reference weight, \(CL_{\mathrm{typ}}(70)=5.00(70/70)^{0.75}=5.00\) L/h. At 100 kg, \[ CL_{\mathrm{typ}}(100)=5.00(100/70)^{0.75} =5.00\exp\{0.75\log(1.4286)\} \approx 5.00(1.3067)=6.534\text{ L/h}. \]
The predictions increase with weight because the exponent is positive. The predictions also obey the direct ratio rule: the 100-kg to 50-kg clearance ratio is \((100/50)^{0.75}=2^{0.75}\approx1.682\), consistent with \(6.534/3.885\approx1.682\).
Worked Example: Interpreting a Negative Exponent
In a hypothetical model, a positive parameter has reference value 8.00 units at \(x_{\mathrm{ref}}=40\) units of a positive covariate, with exponent \(\beta=-0.30\). Predict the parameter when the covariate doubles to 80 units, and calculate the percentage change.
The ratio is \(80/40=2\), so \[ \frac{\theta_{\mathrm{typ}}(80)}{\theta_{\mathrm{typ}}(40)} =2^{-0.30} =\exp\{-0.30\log 2\} \approx \exp(-0.2079) \approx0.8123. \] The predicted parameter is \(8.00(0.8123)\approx6.498\) units. Its percentage change is \(100(0.8123-1)\approx-18.77\%\), a decrease of about 18.77%. Doubling the covariate does not halve the parameter: the size of the change is set by the exponent.
Worked Example: A 10% Covariate Increase
Suppose a hypothetical positive parameter is 2.50 L/h at a covariate reference value of 50 units, and \(\beta=0.40\). Predict the parameter at 55 units and find the exact percentage increase.
The covariate ratio is \(55/50=1.10\). The parameter ratio is \[ 1.10^{0.40} =\exp\{0.40\log(1.10)\} \approx\exp(0.03812) \approx1.0389. \] Therefore, the prediction is \(2.50(1.0389)\approx2.597\) L/h. The exact percentage increase is \(100(1.0389-1)\approx3.89\%\). It is less than 10% because the exponent is positive but less than 1.
Reference Values, Assumptions, and Justification
The reference values make the model interpretable: \(\theta_{\mathrm{ref}}\) is the typical parameter at \(x_{\mathrm{ref}}\). Changing the reference value does not change the underlying power-law curve or its exponent. If the reference is changed to \(x_0>0\), the parameter value at the new reference must be \[ \theta_0=\theta_{\mathrm{ref}}\left(\frac{x_0}{x_{\mathrm{ref}}}\right)^\beta. \] Then the same curve can be written as \(\theta_{\mathrm{typ}}(x)=\theta_0(x/x_0)^\beta\). Choosing a clinically meaningful reference can make the reference parameter easier to interpret, but it does not make an unsuitable functional form appropriate.
The log-log interpretation follows directly from the model, rather than from an additional statistical assumption. Write \(u=\log(x/x_{\mathrm{ref}})\). Then \(\log\theta_{\mathrm{typ}}=\log\theta_{\mathrm{ref}}+\beta u\), a straight line in \(u\) with slope \(\beta\). Also, for any two positive covariate values, dividing their model predictions cancels \(\theta_{\mathrm{ref}}\), yielding the exact ratio formula used above.
In an NLME model, the power law is commonly applied to the typical parameter, while individual random effects and the rest of the model are specified separately. The exponent describes how the typical parameter changes with the covariate when those other model components are held fixed. It should not be interpreted as a change in every individual’s parameter by precisely the same amount if additional individual-level effects are present.
Comparison, Diagnostics, and Limitations
A power law is useful when a relative change in the covariate is naturally associated with a relative change in the parameter. Earlier material on exponential effects of continuous covariates uses a different functional form: a power law is linear in log parameter versus log covariate, whereas an exponential effect is linear in log parameter versus the covariate itself. An additive linear effect instead represents a fixed parameter change per covariate unit. These forms imply different predictions, especially far from the reference value; they should be compared on the same data and in the same model context.
| Feature | Power-law effect | Practical implication |
|---|---|---|
| Covariate domain | \(x>0\) | A zero or negative covariate cannot be used in the ratio form. |
| Parameter prediction | Always positive when \(\theta_{\mathrm{ref}}>0\) | The model cannot predict a zero or negative typical parameter. |
| Relative change | Depends on \((x_2/x_1)^\beta\) | Equal covariate ratios imply equal parameter ratios. |
| Shape | One constant log-log slope \(\beta\) | Changing slopes across the range may indicate poor fit. |
Check whether the model describes the covariate range that matters clinically, and inspect residual diagnostics against the covariate for systematic patterns. A persistent trend, curvature, or distinct pattern in part of the range can indicate that a single exponent is inadequate. Also check how many observations support the covariate extremes: apparent lack of fit or a steep exponent can be driven by sparse data there. Compare predicted parameter trends with the data-supported trend, and examine uncertainty in the exponent and its association with other model parameters.
The power law is not automatically appropriate because the parameter and covariate are positive. It imposes a particular shape and may be unrealistic near the boundaries or when the true relationship changes across the range. It cannot directly represent an additive offset in the parameter, and it does not establish a causal covariate effect. If a different shape is scientifically plausible, compare alternatives using the same observations and a justified evaluation strategy rather than selecting a form solely because its exponent is easy to interpret.
Common Mistakes
- Reading \(\beta\) as an exact percentage change. The exponent is a log-log slope. For a finite covariate change, calculate \(100[(x_2/x_1)^\beta-1]\%\); \(100\beta\%\) is not generally the exact change for a 1% covariate increase.
- Using an absolute difference instead of a ratio. The power model uses \(x/x_{\mathrm{ref}}\). A change from 50 to 55 is a 10% increase, not a change whose effect can be determined from the difference of 5 alone.
- Ignoring the sign or size of the exponent. A negative exponent predicts a decrease as \(x\) increases; an exponent between 0 and 1 predicts an increase smaller than the covariate’s percentage increase for small changes.
- Applying the formula outside its domain or data support. The covariate must be positive, and predictions far beyond observed values are extrapolations even when the calculation is straightforward.
Check Your Understanding
Try each problem, then compare your work with the solutions below. All values are hypothetical.
- A parameter is 8.0 units at \(x_{\mathrm{ref}}=40\) units, with \(\beta=0.5\). Predict it at \(x=90\).
- For \(\beta=-0.30\), what is the parameter ratio when the covariate doubles, and what is the percentage change?
- For \(\beta=0.8\), calculate the exact percentage change in the parameter when the covariate increases by 5%.
- What parameter ratio does the model predict for a 1% covariate increase if \(\beta=0\)? Interpret it.
Solutions. (1) The covariate ratio is \(90/40=2.25\), so \(\theta_{\mathrm{typ}}(90)=8.0(2.25)^{0.5}=8.0(1.5)=12.0\) units. (2) The ratio is \(2^{-0.3}=\exp\{-0.3\log 2\}\approx0.8123\), so the parameter decreases by \(100(1-0.8123)\approx18.77\%\). (3) The ratio is \(1.05^{0.8}=\exp\{0.8\log(1.05)\}\approx1.0398\), giving an increase of \(100(1.05^{0.8}-1)\approx3.98\%\). (4) The ratio is \(1.01^0=1\), so the parameter does not change.