Back-Transforming Intervals for Positive Parameters

Why Work on the Log Scale?

Many NLME parameters are constrained to be positive: clearance, a variance, or a rate constant cannot be negative. A symmetric confidence interval formed directly on the parameter scale can nevertheless extend below zero, especially when uncertainty is large relative to the estimate. Working with the natural logarithm offers one way to respect positivity: an interval on the log scale becomes a positive interval when its endpoints are exponentiated.

This transformation does not by itself guarantee that an interval is well calibrated. Its coverage depends on the quality of the interval on the log scale. The central idea is to construct an interval for the log parameter using an appropriate uncertainty estimate, then transform the entire interval—not just its midpoint—back to the parameter scale.

Definition: Let \(\theta\) be a parameter that is strictly positive, and let \(\theta_{\mathrm{ref}}>0\) be a fixed reference value with the same units as \(\theta\). Define the dimensionless log parameter \(\phi=\log(\theta/\theta_{\mathrm{ref}})\). If a confidence interval for \(\phi\) is \([L,U]\), the corresponding interval for \(\theta\) is \([\theta_{\mathrm{ref}}\exp(L),\theta_{\mathrm{ref}}\exp(U)]\). With the reference value chosen as one unit, the numerical calculation is often written simply as exponentiating the log-scale endpoints.

The reference value clarifies that a logarithm is taken of a dimensionless ratio. In practice, if clearance is recorded in L/h, one can use \(\theta_{\mathrm{ref}}=1\) L/h and take the logarithm of the numerical clearance value expressed in L/h. Changing the unit changes the numerical log value and its center, but a correctly transformed interval in the new unit describes the same physical range.

Constructing and Transforming the Interval

Suppose the fitted model provides an estimate \(\hat{\phi}\) of the log parameter and an estimated standard error \(SE(\hat{\phi})\). If the sampling distribution of \(\hat{\phi}\) is adequately approximated by a normal distribution, a two-sided Wald interval with confidence level \(1-\alpha\) is \([\hat{\phi}-z_{1-\alpha/2}SE(\hat{\phi}),\hat{\phi}+z_{1-\alpha/2}SE(\hat{\phi})]\), where \(z_{1-\alpha/2}\) is the corresponding standard normal quantile. For a 95% interval, this critical value is approximately 1.96.

Exponentiation is a strictly increasing function. Therefore, the lower endpoint remains the lower endpoint and the upper endpoint remains the upper endpoint. This monotonicity also explains why the transformed interval retains the coverage of the interval from which it was obtained.

\begin{aligned} \hat{\phi} &= \log(\hat{\theta}/\theta_{\mathrm{ref}}), \\ [L,U] &= [\hat{\phi}-z_{1-\alpha/2}SE(\hat{\phi}),\ \hat{\phi}+z_{1-\alpha/2}SE(\hat{\phi})], \\ \text{interval for }\theta &= [\theta_{\mathrm{ref}}\exp(L),\ \theta_{\mathrm{ref}}\exp(U)]. \end{aligned}

Result: If \([L,U]\) has coverage \(1-\alpha\) for \(\phi=\log(\theta/\theta_{\mathrm{ref}})\), then \([\theta_{\mathrm{ref}}\exp(L),\theta_{\mathrm{ref}}\exp(U)]\) has the same coverage for \(\theta\). This follows because the events \(L\leq\phi\leq U\) and \(\theta_{\mathrm{ref}}\exp(L)\leq\theta\leq\theta_{\mathrm{ref}}\exp(U)\) are equivalent. If the original interval is only an approximate Wald interval, the transformed interval has that same approximate—not automatically exact—coverage.

A log-scale Wald interval is symmetric around \(\hat{\phi}\), but its back-transformed endpoints are generally not equally far from \(\hat{\theta}\) on the natural scale. If the point estimate is \(\hat{\theta}\) and the log-scale standard error is \(s_\phi\), its endpoints can be written as \(\hat{\theta}\exp(-c s_\phi)\) and \(\hat{\theta}\exp(c s_\phi)\), where \(c=z_{1-\alpha/2}\). The endpoint ratio is \(\exp(2cs_\phi)\), and the geometric mean of the endpoints is \(\hat{\theta}\). These are multiplicative rather than additive relationships.

Check the parameter and scale.
Confirm that the target parameter is strictly positive and specify the units or reference value used for its logarithm.

Obtain the log-scale estimate and uncertainty.
Use the estimated log parameter and its standard error from the fitted model, with the covariance calculation appropriate to that model.

Form the interval on the log scale.
Use the chosen critical value and verify that a normal approximation for the log-scale estimate is reasonable for this purpose.

Exponentiate both endpoints.
Apply the exponential function to each endpoint and express the results in the parameter's original units.

Worked Example: A Clearance Interval

Suppose a hypothetical NLME analysis estimates clearance as \(0.0900\) L/h and reports a standard error of \(0.200\) for the log-clearance estimate. Use a 95% Wald interval on the log scale, with \(1\) L/h as the reference value. Assume the log-scale estimate is sufficiently close to normal for this approximation and that the reported standard error is appropriate.

The estimated log clearance is the logarithm of the numerical estimate expressed in L/h:

\hat{\phi}=\log\left(\frac{0.0900\ \mathrm{L/h}}{1\ \mathrm{L/h}}\right) =\log(0.0900)=-2.408.

The 95% interval on the log scale is \(-2.408\pm1.96(0.200)=-2.408\pm0.392\), giving endpoints \(-2.800\) and \(-2.016\). Back-transforming each endpoint gives \(\exp(-2.800)=0.0608\) and \(\exp(-2.016)=0.133\). The resulting interval is approximately \([0.0608,0.133]\) L/h.

Under the stated approximation, this is a 95% confidence interval for population clearance in the model. It is not an assertion that there is a 95% probability that the fixed clearance parameter lies in this particular interval. The interval is asymmetric around \(0.0900\) L/h: its lower distance is \(0.0292\) L/h and its upper distance is \(0.0430\) L/h. Its positive limits reflect the transformation, not proof that the uncertainty approximation is accurate.

Variance Parameters and the Meaning of the Log

Variance components are also positive parameters, so the same construction can be applied to a variance itself. The target must be explicit: taking the logarithm of a variance and exponentiating the resulting interval returns an interval for that variance. If instead the model estimates the log standard deviation, exponentiating the interval returns an interval for the standard deviation. These are different quantities; converting between a variance and a standard deviation requires squaring or taking a square root.

Worked Example: An Interval for a Positive Variance

Consider a hypothetical estimated residual variance of \(0.0400\) \((\mathrm{mg/L})^2\), with standard error \(0.250\) for its log estimate. Use \(1\ (\mathrm{mg/L})^2\) as the reference value. Assume the log-variance estimate is adequately approximated by a normal distribution and the standard error is reliable.

The log estimate is \(\log(0.0400)=-3.219\). For a 95% interval, the log-scale half-width is \(1.96(0.250)=0.490\). Thus, the interval on the log scale is \([-3.709,-2.729]\). Exponentiating:

\exp(-3.709)=0.0245,\qquad \exp(-2.729)=0.0653.

The 95% interval for the variance is therefore approximately \([0.0245,0.0653]\) \((\mathrm{mg/L})^2\). It is a variance interval, not an interval for a residual standard deviation. To report the corresponding interval for the standard deviation, take square roots of both limits: approximately \([0.1565,0.256]\) mg/L. This last conversion is valid because square root is increasing for positive values.

Interpreting Asymmetry and Comparing Approaches

A back-transformed interval is often described as asymmetric because its endpoints do not fall equally far from the point estimate in the parameter's original units. This does not mean the calculation was inconsistent: symmetry was imposed on the log scale, where equal distances represent equal multiplicative changes. For example, moving upward by \(d\) on the log scale multiplies the parameter by \(\exp(d)\); moving downward by \(d\) divides it by the same factor.

A direct Wald interval on the natural scale instead adds and subtracts a multiple of the natural-scale standard error. It is symmetric in the parameter units and can have a negative lower endpoint for a parameter that must be positive. A log-scale interval avoids negative limits, but it is not automatically superior: it relies on adequate normal approximation on the log scale and can still be misleading when the estimate or uncertainty is poorly determined. The delta-method intervals for derived pharmacokinetic parameters address uncertainty in functions of estimated parameters; the present construction concerns a positive parameter represented on a log scale. These procedures should not be treated as interchangeable without checking what parameter is being estimated and on which scale.

Worked Example: Reading a Multiplicative Interval

Suppose a hypothetical analysis estimates clearance as \(1.20\) L/h, with a standard error of \(0.150\) for log clearance. Use a 95% log-scale Wald interval and assume its normal approximation is adequate.

The log estimate is \(\log(1.20) = 0.1823\). The half-width is \(1.96(0.150)=0.294\), so the log-scale interval is \([-0.1117,0.4763]\). Exponentiating gives \([\exp(-0.1117),\exp(0.4763)] = [0.894,1.610]\) L/h.

The lower and upper limits are not the same additive distance from \(1.20\) L/h. Instead, relative to the estimate, the endpoints are approximately \(1/1.342\) and \(1.342\) times the estimate. This illustrates the multiplicative interpretation of a log-scale interval. The confidence statement concerns the model's clearance parameter, not the probability that a particular future individual's clearance lies in this range.

Assumptions, Diagnostics, and Common Pitfalls

The interval construction requires a positive target parameter, a meaningful estimate and standard error for its log, and a suitable sampling-distribution approximation on the log scale. For the usual Wald calculation, the estimate should be sufficiently well behaved that its standardized uncertainty is approximately normal. The method also presumes that the fitted NLME model and the covariance estimate used for the standard error are adequate for the intended inference. Exponentiation preserves the coverage of the starting interval; it cannot repair model misspecification or a poor uncertainty estimate.

Before reporting the result, check that the endpoints are ordered and positive, the original units are restored, and the interval is for the intended parameter (variance versus standard deviation, for example). Compare the interval with the fitted estimate and examine whether small changes in the fit or covariance estimate would materially change its width. Strong asymmetry on the original scale is expected in many cases; it is not by itself evidence of a failure. Concern is warranted if the log-scale estimate is extremely uncertain, the model estimate is close to a boundary, or the normal approximation is implausible. In such cases, transformation alone does not validate the interval; alternative uncertainty methods may be needed.

Exponentiating only the point estimate. The estimate \(\exp(\hat{\phi})\) is not an interval. Calculate the two log-scale limits first, then exponentiate each one.
Reporting a symmetric interval after transformation. The transformed endpoints generally are not equally distant from the estimate. Report the actual limits rather than forcing symmetry.
Taking the logarithm of a quantity without clarifying its units. Specify the parameter's units or reference value and report the back-transformed interval in the original units.
Confusing a log variance with a log standard deviation. Identify the modeled target before transforming; exponentiation returns the quantity whose logarithm was used.
Treating positivity as proof of validity. Positive limits are guaranteed by exponentiation, but appropriate coverage still depends on the log-scale approximation and model uncertainty calculation.

Key takeaway: For a positive parameter, form the confidence interval on the log scale using a suitable estimate and standard error, then exponentiate both endpoints. The result is positive and typically asymmetric on the original scale; its inferential quality is inherited from the interval on the log scale.

Check Your Understanding

Answer each question and then check the brief solutions.

A positive parameter has estimated log value \(0.300\) and log-scale standard error \(0.100\). Find its 95% interval on the log scale and the interval for the parameter relative to a reference value of 1.
Why does exponentiating both endpoints preserve the coverage event?
A model reports an interval for the log standard deviation. What quantity do you obtain by exponentiating both endpoints?
Does a positive back-transformed interval establish that the log-scale normal approximation is adequate? Explain.
What is the geometric mean of the endpoints of a log-scale Wald interval centered at \(\hat{\phi}\), when expressed relative to its reference value?

Solutions. (1) The half-width is \(1.96(0.100)=0.196\), so the log interval is \([0.104,0.496]\). Exponentiation gives approximately \([1.110,1.642]\) times the reference value. (2) Because the exponential function is strictly increasing, \(L\leq\phi\leq U\) if and only if \(\exp(L)\leq\exp(\phi)\leq\exp(U)\). (3) An interval for the standard deviation, not the variance. (4) No. Exponentiation guarantees positive limits but does not establish that the starting interval has appropriate coverage. (5) The geometric mean is \(\exp(\hat{\phi})\), the estimated parameter relative to its reference value.