Worked Allometric Prediction With Uncertainty

From Animal Clearances to a Human Prediction

Allometric scaling relates a quantity such as drug clearance to body mass through a power law. Its appeal is that a power relationship becomes linear after taking logarithms, so a small multispecies data set can be used to estimate both the scaling exponent and the relationship’s scatter. But a human prediction is not just the fitted value at human body mass: it also depends on uncertainty in the fitted relationship and on the variability of a new response around that relationship.

The calculation below follows one hypothetical clearance data set from regression through prediction. Its purpose is to make the ingredients and assumptions visible, not to suggest that four species are enough to establish a reliable biological model. In particular, the human body mass lies beyond the range of the animal data, so the prediction is an extrapolation.

Definition: Let $W_i$ be the body mass, in kilograms, and $CL_i$ the clearance, in litres per hour, for species $i$. The log-linear allometric model is $$ x_i=\log_{10}(W_i), \qquad y_i=\log_{10}(CL_i), \qquad y_i=\beta_0+\beta_1 x_i+\varepsilon_i. $$ Here $\beta_0$ is the intercept, $\beta_1$ is the allometric exponent, and $\varepsilon_i$ is the species-level residual. On the original scale, the exponentiated fitted log-scale relationship is $CL=10^{\beta_0}W^{\beta_1}$, which under the normal log-residual model is the conditional median clearance, not the conditional arithmetic mean. The model assumes that the log clearances follow a linear relationship with log body mass and, for the usual regression intervals, that the residuals are independent, normally distributed with mean zero and common variance $\sigma^2$, conditional on the body masses.

The power-law interpretation follows directly from the log rules: if $\log_{10}(CL)=\beta_0+\beta_1\log_{10}(W)$, then $CL=10^{\beta_0}W^{\beta_1}$. Thus a one-unit increase in log-base-10 mass corresponds to an estimated $\beta_1$-unit increase in log-base-10 clearance. The exponent is not the percentage change in clearance for each kilogram; it describes how clearance scales multiplicatively with mass.

Fit the Relationship and Check Its Basic Quantities

Consider four hypothetical animal species with the following masses and measured clearances. The log values are rounded for display; the calculations use the displayed values.

Species	Mass $W_i$ (kg)	Clearance $CL_i$ (L/h)	$x_i=\log_{10}(W_i)$	$y_i=\log_{10}(CL_i)$
A	0.01	0.00355	-2	-2.45
B	0.1	0.0158	-1	-1.80
C	1	0.0891	0	-1.05
D	10	0.631	1	-0.20

Worked Example: Fit the Four-Species Allometric Model

For ordinary least squares, the estimated slope is $b_1=S_{xy}/S_{xx}$, where $S_{xx}=\sum_i(x_i-\bar{x})^2$ and $S_{xy}=\sum_i(x_i-\bar{x})(y_i-\bar{y})$. The intercept is $b_0=\bar{y}-b_1\bar{x}$. From the table, $\bar{x}=-0.5$ and $\bar{y}=-1.375$.

The centered log masses are $-1.5,-0.5,0.5,1.5$, so $S_{xx}=2.25+0.25+0.25+2.25=5$. The centered log clearances are $-1.075,-0.425,0.325,1.175$, giving $S_{xy}=1.6125+0.2125+0.1625+1.7625=3.75$. Therefore,

b_1=\frac{3.75}{5}=0.75,\qquad b_0=-1.375-(0.75)(-0.5)=-1.00.

The fitted equation is $\widehat{y}=-1.00+0.75x$, or, on the original scale, $\widehat{CL}=10^{-1}W^{0.75}$ L/h. The fitted log clearances at $x=-2,-1,0,1$ are $-2.50,-1.75,-1.00,-0.25$. Subtracting these from the observed values gives residuals $0.05,-0.05,-0.05,0.05$. Their squared sum is $SSE=0.01$; with $n=4$, the residual degrees of freedom are $df=n-2=2$. Thus

s=\sqrt{\frac{SSE}{n-2}} =\sqrt{\frac{0.01}{2}} =0.07071\text{ log}_{10}\text{ clearance units}.

The slope estimate says clearance scales approximately as mass to the $0.75$ power in this fitted data set. It is an estimate, not a known biological constant. With only two residual degrees of freedom, the small residual standard deviation does not by itself establish that the model is precise or appropriate for humans.

Prediction at Human Body Mass

For a hypothetical 70-kg human, $x_0=\log_{10}(70)=1.8451$, which exceeds the largest observed log mass, 1. This is extrapolation. The fitted log clearance is $b_0+b_1x_0$, and exponentiating gives the fitted clearance on the original scale. That fitted value describes the model’s estimated mean log clearance at this mass; predicting a new clearance also requires allowing for a new residual.

Prediction result: Under the stated linear, independent, normal, equal-variance residual assumptions, a two-sided $100(1-\alpha)\%$ prediction interval for a new log response at fixed $x_0$ is $$ b_0+b_1x_0 \ \pm\ t_{1-\alpha/2,\,n-2}\ s \sqrt{1+\frac{1}{n}+\frac{(x_0-\bar{x})^2}{S_{xx}}}. $$ The $1$ inside the square root accounts for the new response’s residual variance. The remaining terms account for uncertainty in the estimated regression mean, including uncertainty in the slope. Exponentiating both endpoints yields an interval in clearance units.

Worked Example: Predict Human Clearance With a 95% Prediction Interval

At 70 kg, the fitted log clearance is $ -1.00+(0.75)(1.8451)=0.3838$. The exponentiated fitted mean log clearance is $10^{0.3838}=2.420$ L/h, the conditional median under the stated normal log-residual model. To calculate its prediction uncertainty, first find the leverage term:

h_0=\frac{1}{4}+\frac{(1.8451-(-0.5))^2}{5} =0.2500+\frac{5.4995}{5} =1.3499.

The prediction standard error on the log scale is $0.07071\sqrt{1+1.3499}=0.1084$. For a 95% interval, $t_{0.975,2}=4.303$, so the margin is $(4.303)(0.1084)=0.4664$ log units. The log-scale interval is $0.3838\pm0.4664=[-0.0826,0.8502]$. Exponentiating the endpoints gives

[10^{-0.0826},10^{0.8502}] =[0.8268,7.083]\text{ L/h}.

Under the model, this procedure has 95% repeated-sampling coverage for a new response when both the training data and the new response are generated according to the model. It is not a statement that there is a 95% probability that a fixed, unknown human clearance lies in this particular interval. Nor does the interval account automatically for model misspecification, differences in biology between species, or uncertainty in whether the animal-to-human relationship applies.

Where the Prediction Uncertainty Comes From

The prediction formula separates three contributions on the log scale: $1/n$ from estimating the regression level, $(x_0-\bar{x})^2/S_{xx}$ from estimating the slope at a target away from the center of the data, and $1$ from the new response’s residual. In the human example these terms are $0.2500$, $1.0999$, and $1$, respectively. The extrapolation-related slope term is the largest component of the estimated mean uncertainty. The new-response term remains even if the regression line were known exactly.

For comparison, an interval for the mean log clearance at $x_0$, rather than for a new response, omits the residual term $1$. Its standard error here is $s\sqrt{h_0}=0.08215$. The 95% interval is $0.3838\pm(4.303)(0.08215)=[0.0303,0.7373]$, or approximately $[1.072,5.461]$ L/h after exponentiation. It is narrower because it concerns the model mean, not a new species-level response. These intervals answer different questions and should not be interchanged.

Worked Example: Assess Exponent Uncertainty

For this simple regression, the estimated standard error of the slope is $SE(b_1)=s/\sqrt{S_{xx}}$. Here, $SE(b_1)=0.07071/\sqrt{5}=0.03162$ log-clearance units per log-mass unit. With $df=2$, a 95% $t$-interval for the exponent is

0.75\pm(4.303)(0.03162) =0.75\pm0.1361 =[0.6139,0.8861].

This interval is wide relative to the point estimate despite the visually close fitted values, because the data contain only four species. Exponent uncertainty also matters more when the target is far from the mean observed log mass: a change in slope changes the predicted log clearance by $x_0$ times that change, relative to the fitted line’s center. The prediction interval already includes this regression-parameter uncertainty through its leverage term; adding a separate exponent interval to the same calculation without a defined joint procedure would risk double-counting uncertainty.

Sensitivity to Species Inclusion

A fitted allometric relationship can be sensitive to which species are included. This is especially important with few species: removing one observation changes the slope, intercept, residual variance, degrees of freedom, and the center and spread of the retained log masses. Any refit must calculate its summaries from the retained data, rather than carrying over $\bar{x}$ or $S_{xx}$ from the full data.

Worked Example: Refit Without the Largest Species

Suppose species D, the 10-kg species, is excluded as a sensitivity analysis. The retained log masses are $-2,-1,0$, with $\bar{x}=-1$ and $S_{xx}=2$. The retained log clearances have $\bar{y}=-1.7667$, and $S_{xy}=1.4000$. Thus $b_1=1.4000/2=0.7000$ and $b_0=-1.7667-(0.7000)(-1)=-1.0667$.

The fitted residuals are approximately $0.01667,-0.03333,0.01667$, so $SSE=0.001667$. With $n=3$, $df=1$ and $s=\sqrt{0.001667}=0.04082$ log units. At 70 kg, the new fitted log clearance is $-1.0667+(0.7000)(1.8451)=0.2249$, or $10^{0.2249}=1.679$ L/h.

For this refit, the target leverage is $1/3+(1.8451-(-1))^2/2=4.3806$. The prediction standard error is $0.04082\sqrt{1+4.3806}=0.09469$. Since $t_{0.975,1}=12.706$, the margin is $1.203$ log units. The log interval is approximately $[-0.978,1.428]$, which back-transforms to about $[0.105,26.8]$ L/h. The fitted point prediction changed, and the interval became extremely broad. The very large critical value reflects the single residual degree of freedom; excluding a species has not made the prediction more secure simply because the residual standard deviation decreased.

Diagnostics, Alternatives, and Limitations

Before treating a computed interval as informative, check whether the model is a defensible description of the data and target. With a small number of species, diagnostics have limited power, so they cannot certify the assumptions. Useful checks include:

Log-linearity: plot log clearance against log mass and look for curvature or a pattern in residuals. A straight fitted line does not guarantee that a power law is biologically adequate.
Residual behavior: unusually large residuals, changing residual spread, or dependence among related species can undermine the normal, constant-variance, independent-error assumptions used for the intervals.
Leverage and extrapolation: record the target’s distance from the observed mass range and inspect how much the prediction interval depends on the slope. A target well beyond the data can be highly sensitive even when the fitted residuals are small.
Species influence: refit after omitting a species and report changes in slope, prediction, and interval. A large change signals sensitivity, not a reason to select whichever fit is preferred.

A common alternative is to fix the exponent at a prespecified value, such as $0.75$, and estimate only the scale. This can reduce estimation burden when the exponent is justified independently, but it suppresses exponent uncertainty and can bias predictions if the fixed value does not describe the data. Estimating the exponent from the multispecies data, as above, exposes that uncertainty but is unstable with few species or a narrow mass range. Neither approach removes the need to assess biological plausibility and domain differences.

The prediction interval here is conditional on the chosen allometric form, the specified residual assumptions, and the measured body masses. It does not incorporate every source of translational uncertainty: for example, uncertainty about model structure, species-specific biology, or whether human clearance follows the same relationship. A narrow interval under the regression model can therefore still be misleading if the model is an inadequate account of translation. Conversely, a wide interval can be driven by small sample size and extrapolation even when the observed data appear close to a line.

Key takeaway: An allometric human prediction combines the fitted scaling relationship with uncertainty in its estimated parameters and, for a new response, residual variability. Report the target’s position relative to the animal data, the prediction interval in original clearance units, and the sensitivity to species inclusion. Interpret its nominal coverage under the stated model—not as a probability assigned to a fixed human clearance.

Check Your Understanding

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

In the log-linear model, what does $\beta_1$ represent on the original clearance scale?
Why does the prediction interval formula include a $1$ inside its square root?
Why is the 70-kg prediction in the worked example particularly uncertain?
If one species is removed, which quantities must be recalculated before making a new prediction?
Does a nominal 95% prediction interval mean that the fixed human clearance has a 95% probability of lying in the interval? Explain.

Solutions. (1) It is the estimated change in log-base-10 clearance per one-unit increase in log-base-10 body mass, and is the allometric exponent in $CL\propto W^{\beta_1}$. (2) A new species has its own residual deviation from the fitted relationship, with variance $\sigma^2$; the $1$ represents that new-response variability. (3) The target is beyond the observed mass range, only four species were used, and $df=2$ gives a large $t$ critical value; the interval also includes new-response residual variability. (4) Recalculate the retained-data means, $S_{xx}$, $S_{xy}$, fitted coefficients, residual variance, degrees of freedom, and target leverage. In particular, $S_{xx}$ is computed around the mean of the retained species, not the full data. (5) No. Under the model, 95% of intervals constructed this way cover a new response when repetitions include both the training data and the new response generated according to the model. This does not assign a probability to a fixed clearance, and coverage is not generally 95% if only training data are repeatedly sampled while that fixed target is held unchanged.