Why Ratios and Contrasts Need Their Own Uncertainty Calculation
Population pharmacokinetic results are often interpreted through combinations of parameters rather than through a single estimate. A clearance ratio can express the relative clearance between two groups; a difference in typical clearance can express an absolute change in units such as L/h. Each quantity has its own sampling uncertainty. The standard errors of the component estimates alone are not enough: their covariance also matters.
The calculations below treat parameter estimates as a vector with an estimated covariance matrix. They rely on the estimates being approximately jointly normal in the region relevant to the calculation, and on the covariance matrix representing their joint sampling uncertainty. For a nonlinear quantity such as a ratio, a first-order approximation is also needed. These conditions should be considered before interpreting a resulting standard error or interval.
Contrasts: The Covariance Term in a Difference
Let \(\hat{\theta}_1\) and \(\hat{\theta}_2\) estimate two population parameters. Write their estimated variances as \(\widehat{\operatorname{Var}}(\hat{\theta}_1)\) and \(\widehat{\operatorname{Var}}(\hat{\theta}_2)\), and their estimated covariance as \(\widehat{\operatorname{Cov}}(\hat{\theta}_1,\hat{\theta}_2)\). The estimated contrast is \(\hat{\theta}_1-\hat{\theta}_2\). The variance rule for a difference gives
The covariance is subtracted because the second estimate has a negative coefficient in the contrast. Positive covariance means the estimates tend to move together, so their difference tends to vary less than it would if they varied independently. Negative covariance has the opposite effect. Omitting the covariance is justified only when the estimates are uncorrelated, or when an independence assumption is otherwise warranted.
Worked Example: Difference Between Two Typical Clearances
Suppose a hypothetical model estimates typical clearance as \(6.8\) L/h for one group and \(5.9\) L/h for another. The estimated standard errors are \(0.70\) L/h and \(0.60\) L/h, and the estimated covariance is \(0.21\) (L/h)\(^2\). Find the standard error of the difference, defined as the first group minus the second.
The estimated variances are \(0.70^2=0.49\) (L/h)\(^2\) and \(0.60^2=0.36\) (L/h)\(^2\). The estimated difference is \(6.8-5.9=0.90\) L/h. Therefore,
For orientation, a normal-approximation 95% interval is \(0.90\pm1.96(0.6557)\), or approximately \([-0.385,\ 2.185]\) L/h. This interval includes zero, so it does not establish a nonzero difference at the corresponding two-sided 5% level under the normal approximation. It does not show that the group clearances are equal; it shows that this estimate is imprecise enough to include both zero and a range of positive differences.
The covariance is plausible for these inputs: the implied correlation is \(0.21/(0.70\times0.60)=0.50\), which lies between \(-1\) and \(1\). The covariance has the required squared units, and the resulting variance is positive.
Ratios on the Original Scale
For a ratio \(R=\theta_1/\theta_2\), the parameter estimates are transformed nonlinearly. The delta method approximates the variance of a smooth function of estimates by evaluating its gradient at the estimates and combining that gradient with the covariance matrix. For the ratio, the gradient with respect to \((\theta_1,\theta_2)\) is \((1/\theta_2,-\theta_1/\theta_2^2)\). Thus, for \(\hat{R}=\hat{\theta}_1/\hat{\theta}_2\),
The result is an approximation because the ratio is nonlinear. It is most dependable when the denominator is well away from zero relative to its uncertainty and the estimates are reasonably described by a joint normal approximation near their estimated values. If the component parameters have the same units, the ratio and its standard error are dimensionless. If their units differ, the ratio’s units are the numerator’s units divided by the denominator’s units.
Worked Example: Clearance Ratio With Correlated Estimates
In a hypothetical analysis, two typical clearances are estimated as \(\hat{\theta}_1=5.2\) L/h and \(\hat{\theta}_2=4.0\) L/h. Their estimated variances are \(0.36\) and \(0.25\) (L/h)\(^2\), and their covariance is \(0.18\) (L/h)\(^2\). Calculate the ratio and its approximate standard error.
The ratio estimate is \(5.2/4.0=1.30\). Substituting the estimates and covariance into the ratio variance formula gives
Therefore, \(SE(\hat{R})=\sqrt{0.01965625}\approx0.1402\). The estimated ratio is \(1.30\), with an approximate standard error of \(0.1402\). A simple normal-approximation interval on the original ratio scale is \(1.30\pm1.96(0.1402)\), approximately \([1.025,\ 1.575]\). This symmetric interval is a local approximation, not a guarantee that all possible ratio intervals will be symmetric or remain positive.
The covariance term is substantial here: the positive covariance reduces the estimated variance of the ratio. The inputs are internally plausible because the implied correlation is \(0.18/(0.60\times0.50)=0.60\). A correlation outside \([-1,1]\), or a negative variance produced by the calculation, would signal inconsistent inputs or a calculation error.
Ratios on the Log Scale
For positive parameters, a ratio can instead be expressed as a difference of logarithms: \(\log(\theta_1/\theta_2)=\log(\theta_1)-\log(\theta_2)\). This is especially convenient when the model estimates log parameters or when a multiplicative comparison is the scientific target. If \(\hat{\eta}_1\) and \(\hat{\eta}_2\) estimate \(\log(\theta_1)\) and \(\log(\theta_2)\), respectively, then \(\hat{\eta}_1-\hat{\eta}_2\) estimates the log ratio. Its variance is the ordinary variance of a difference:
If the log-scale estimates are approximately jointly normal, an interval for the log ratio can be calculated on that scale and then exponentiated. Exponentiation turns the two endpoints into an interval for the ratio. This produces an interval that is positive and typically asymmetric around the ratio estimate. It does not remove uncertainty or make a weakly determined ratio precise.
Worked Example: Interval for a Geometric Clearance Ratio
Suppose a hypothetical analysis estimates a clearance ratio of \(1.30\), so the estimated log ratio is \(\log(1.30)\approx0.2624\). On the log scale, the estimated variances of the two group parameters are \(0.0081\) and \(0.0064\), and their covariance is \(0.0012\). Construct an approximate 95% interval for the ratio.
The variance of the log ratio is
Using the normal critical value \(1.96\), the interval on the log scale is
Exponentiating the endpoints gives \([\exp(0.0468),\exp(0.4780)]\), approximately \([1.048,\ 1.613]\). Under the approximate joint-normal model on the log scale, this procedure has approximately 95% repeated-sampling coverage for the ratio. It is not a statement that there is a 95% probability that this particular fixed ratio lies in the calculated interval.
Derivation and Assumptions
The formula for a ratio follows from the first-order Taylor approximation. Let \(g(\theta_1,\theta_2)=\theta_1/\theta_2\). Near the estimated parameter vector, a small estimation error changes \(g\) approximately by its gradient multiplied by the parameter error:
For any vector of estimates with covariance matrix \(\Sigma\), a linear combination with coefficient vector \(a\) has variance \(a^T\Sigma a\). Applying that rule to the gradient gives the first-order variance approximation \(\nabla g^T\Sigma\nabla g\). Since the gradient is evaluated using estimated parameters in practice, the calculated result is itself an estimate of the variance. This derivation explains both the covariance term and the approximation: higher-order changes in the nonlinear function are omitted.
Before using these formulas, check that the parameter estimates refer to the same model fit and that the variances and covariance come from the same estimated covariance matrix. A covariance matrix must be symmetric and positive semidefinite; in particular, for two estimates, the absolute covariance cannot exceed the product of their standard errors. The ratio approximation also requires a denominator sufficiently far from zero for a local linear approximation to be useful. For a log ratio, both underlying parameters must be positive, and the log-scale estimates must have a suitable approximate sampling distribution.
Diagnostics, Alternatives, and Limitations
A few checks catch many practical failures:
- Check scale and units. Variances have squared units, covariance has the product of the two units, and the final standard error has the units of the contrast or ratio.
- Check covariance signs and magnitude. For a difference or ratio, a positive covariance often reduces the variance because the second parameter enters with a negative derivative. The covariance must still be compatible with the two variances.
- Check whether a ratio is locally well behaved. If the denominator is near zero relative to its standard error, the ratio can be highly skewed or poorly approximated by a normal distribution. A symmetric interval from the delta method can then be misleading.
- Recompute the arithmetic. In ratio calculations, the powers of the denominator differ across terms. Write out each term separately before summing, and verify that the final variance is nonnegative.
The raw-scale delta method is direct and useful when the denominator is stable and the ratio estimate is reasonably close to normal. A log-scale interval is often attractive for positive parameters and multiplicative comparisons, particularly when the estimated log parameters are approximately normal. The two approaches need not yield identical intervals: they use different approximations and express uncertainty on different scales. When a denominator may approach zero, a Fieller-type interval can better reflect the resulting ratio uncertainty and may be unbounded. Resampling methods can also reflect nonlinear behavior, but depend on a suitable resampling scheme and successful model refitting.
Common mistake: “The covariance is positive, so I should add its contribution to the variance of a ratio.” This is wrong for the usual ratio \(\hat{\theta}_1/\hat{\theta}_2\): its derivative with respect to the denominator is negative, so the covariance contribution is \(-2\hat{\theta}_1\widehat{\operatorname{Cov}}(\hat{\theta}_1,\hat{\theta}_2)/\hat{\theta}_2^3\). With positive estimates and positive covariance, that term is negative. The sign follows from the function’s derivatives, not from a rule that every variance component must be added.
Check Your Understanding
Use the stated assumptions and show how covariance affects each calculation. Solutions follow the questions.
- Two clearance estimates have standard errors \(0.40\) and \(0.30\) L/h and covariance \(0.06\) (L/h)\(^2\). What is the standard error of their difference?
- For estimates \(\hat{\theta}_1=3.0\) and \(\hat{\theta}_2=2.0\), suppose their variances are \(0.16\) and \(0.09\), and their covariance is zero. Find the delta-method standard error of their ratio.
- Log-scale estimates have variances \(0.010\) and \(0.014\), with covariance \(0.004\). Find the standard error of their log ratio.
- What concern arises if the denominator estimate is \(0.20\) and its standard error is \(0.25\)? Name one alternative to relying on a symmetric raw-scale delta interval.
Solutions. (1) The difference variance is \(0.40^2+0.30^2-2(0.06)=0.16+0.09-0.12=0.13\) (L/h)\(^2\), so its standard error is \(\sqrt{0.13}=0.3606\) L/h. (2) The ratio estimate is \(3.0/2.0=1.50\). Its approximate variance is \(0.16/2.0^2+3.0^2(0.09)/2.0^4=0.0400+0.050625=0.090625\), so its standard error is \(0.3010\). (3) The log-ratio variance is \(0.010+0.014-2(0.004)=0.016\), giving a standard error of \(\sqrt{0.016}=0.1265\). (4) The denominator is small relative to its uncertainty, so the ratio’s distribution may be strongly skewed and the local approximation unreliable. A Fieller-type interval is one alternative; it can be unbounded when the denominator is not determined away from zero.