Relative Uncertainty on the Log Scale

Why Uncertainty Changes Shape When Exponentiated

Many pharmacokinetic parameters are positive: clearance, volume, an absorption rate constant, or a variance parameter. A model may estimate the logarithm of such a parameter, then exponentiate the estimate to report the parameter in its usual units. This is useful because exponentiation guarantees a positive reported value. It also changes how uncertainty looks: equal upward and downward changes on the log scale become unequal additive changes on the natural scale.

The key is to keep track of which scale the standard error belongs to. A standard error for a log parameter is an absolute measure on the log scale. It is not, in general, the same thing as a standard error in the parameter’s original units. Its exponentiation instead gives a multiplicative factor. For small log-scale standard errors, that factor leads to a useful approximation: the log-scale standard error, expressed as a percentage, is approximately the parameter’s relative standard error.

Definition: Let \(\theta>0\) be a parameter and let \(\eta=\log(\theta)\). If \(\hat{\eta}\) is an estimate of \(\eta\) with standard error \(SE_{\eta}\), the exponentiated estimate is \(\hat{\theta}=\exp(\hat{\eta})\). A one-standard-error change on the log scale corresponds to the natural-scale values \(\exp(\hat{\eta}-SE_{\eta})\) and \(\exp(\hat{\eta}+SE_{\eta})\), or equivalently \(\hat{\theta}\exp(-SE_{\eta})\) and \(\hat{\theta}\exp(SE_{\eta})\).

These two natural-scale values are a way to display the effect of one standard error on the log scale. They are not automatically a confidence interval. In particular, this display alone does not specify a confidence level or account for any uncertainty beyond the standard error being used.

From a Log-Scale Change to a Multiplicative Change

Suppose the estimate on the log scale changes by an amount \(\delta\). The corresponding natural-scale value is \(\exp(\hat{\eta}+\delta)=\hat{\theta}\exp(\delta)\). Thus the change is multiplicative, and the change relative to \(\hat{\theta}\) is \[ \frac{\exp(\hat{\eta}+\delta)-\hat{\theta}}{\hat{\theta}}=\exp(\delta)-1. \] For a one-standard-error change, set \(\delta=SE_{\eta}\) for the upward change and \(\delta=-SE_{\eta}\) for the downward change.

Writing \(s=SE_{\eta}\), the upper value is \(\hat{\theta}\exp(s)\), and the lower value is \(\hat{\theta}\exp(-s)\). They are equally distant from \(\hat{\theta}\) on the log scale because their logarithms differ from \(\hat{\eta}\) by \(+s\) and \(-s\). They are not equally distant on the natural scale. The upward relative change is \(\exp(s)-1\); the downward relative change, measured as a positive fraction of the estimate, is \(1-\exp(-s)\).

\begin{aligned} \text{upward relative change} &= \exp(s)-1,\\ \text{downward relative change} &= 1-\exp(-s). \end{aligned}

For \(s>0\), the upward relative change is greater than the downward relative change. This follows because \(\exp(s)+\exp(-s)>2\), so \(\exp(s)-1>1-\exp(-s)\). The two changes are approximately equal only when \(s\) is small.

Why Log-Scale SE Can Approximate Relative SE

A conventional relative standard error (RSE) for a natural-scale estimate is the natural-scale standard error divided by the estimate, often multiplied by 100 to express it as a percentage. If uncertainty is small, a log-scale standard error gives a close approximation to this relative quantity. To see why, consider a small change \(\delta\) in the log estimate:

\frac{\exp(\hat{\eta}+\delta)-\exp(\hat{\eta})}{\exp(\hat{\eta})} =\exp(\delta)-1 \approx \delta

The approximation uses the expansion \(\exp(\delta)=1+\delta+\delta^2/2+\cdots\), retaining only the first-order term when \(|\delta|\) is small. If the standard error on the log scale is small, the corresponding relative standard error is therefore approximately \(SE_{\eta}\), or \(100SE_{\eta}\%\). This is an approximation, not an identity. As \(SE_{\eta}\) grows, the upward and downward relative changes become more different, and treating \(100SE_{\eta}\%\) as an exact, symmetric percentage error becomes misleading.

This reasoning is about transforming uncertainty in an estimate; it does not say that an estimate’s sampling distribution is exactly normal on either scale. A symmetric one-standard-error displacement on the log scale is a useful summary when that scale is appropriate, but its interpretation depends on how the estimate and its standard error were obtained.

Worked Example: Clearance With Modest Log-Scale Uncertainty

Suppose a hypothetical population pharmacokinetic analysis estimates log clearance as \(\hat{\eta}=\log(12)\), with \(SE_{\eta}=0.18\). Clearance is reported in L/h, so the natural-scale estimate is \(\hat{\theta}=12\) L/h. Calculate the natural-scale values one log-scale standard error below and above the estimate, and compare the resulting relative changes with the small-error RSE approximation.

The lower and upper values are

\begin{aligned} \hat{\theta}_{\mathrm{lower}} &= 12\exp(-0.18) =12(0.835270)\approx 10.023\ \text{L/h},\\ \hat{\theta}_{\mathrm{upper}} &= 12\exp(0.18) =12(1.197217)\approx 14.367\ \text{L/h}. \end{aligned}

The downward change is \(12-10.023\approx1.977\) L/h, or \(1-\exp(-0.18)\approx0.1647\), about 16.47% of the estimate. The upward change is \(14.367-12\approx2.367\) L/h, or \(\exp(0.18)-1\approx0.1972\), about 19.72%. The small-error approximation gives \(100(0.18)=18.0\%\), between the two directional changes. These are one-standard-error displacements, not a confidence interval.

Worked Examples: Reading the Asymmetry

Worked Example: A Larger Standard Error for an Elimination Rate Constant

Suppose a hypothetical analysis reports an elimination rate constant estimate of \(0.80\ \mathrm{h}^{-1}\), with log-scale standard error \(s=0.50\). Compute the one-standard-error natural-scale values and explain why a symmetric additive display would be a poor description of this uncertainty.

Exponentiating the log-scale changes gives

\begin{aligned} \hat{\theta}_{\mathrm{lower}} &=0.80\exp(-0.50) =0.80(0.606531)\approx0.4852\ \mathrm{h}^{-1},\\ \hat{\theta}_{\mathrm{upper}} &=0.80\exp(0.50) =0.80(1.648721)\approx1.319\ \mathrm{h}^{-1}. \end{aligned}

The downward additive change, calculated from the unrounded transformed value, is \(0.80[1-\exp(-0.50)]\approx0.3148\ \mathrm{h}^{-1}\). The upward additive change is \(0.80[\exp(0.50)-1]\approx0.5190\ \mathrm{h}^{-1}\). The corresponding relative changes are approximately 39.35% downward and 64.87% upward. Although \(100s=50\%\) is sometimes used as a rough RSE approximation, it does not represent either directional change exactly here. The asymmetry is large enough that a single symmetric percentage description hides important information.

Worked Example: Volume and Consistent Rounding

Suppose a hypothetical volume estimate is \(40.0\) L, with a log-scale standard error of \(0.200\). Find the one-standard-error natural-scale values and their distances from the estimate. Use unrounded transformed values to calculate the distances.

The transformed values are

\begin{aligned} \hat{\theta}_{\mathrm{lower}} &=40.0\exp(-0.200) \approx32.7492\ \text{L},\\ \hat{\theta}_{\mathrm{upper}} &=40.0\exp(0.200) \approx48.8561\ \text{L}. \end{aligned}

The downward distance is \(40.0-32.7492\approx7.251\) L. Equivalently, it is \(40.0[1-\exp(-0.200)]\approx7.251\) L. The upward distance is \(48.8561-40.0\approx8.856\) L, also given by \(40.0[\exp(0.200)-1]\). The two values are symmetric in log space but not in liters. Rounding the transformed endpoints before subtracting can make the displayed distances differ slightly; calculate with unrounded values, then round the final results consistently.

Assumptions, Checks, and Alternatives

The transformation is directly interpretable when the reported natural-scale parameter is positive and is defined as the exponential of a log-scale estimate. A log-scale standard error must refer to that same estimated quantity. If the model instead estimates a parameter directly on the natural scale, its natural-scale standard error cannot generally be converted to a log-scale standard error exactly by dividing by the estimate. That division is a small-relative-error approximation and can be poor when uncertainty is large.

Before interpreting a transformed uncertainty summary, check that:

The parameter is positive and the reported estimate is indeed obtained by exponentiating its log estimate.
The standard error is on the log scale, rather than on the natural scale or on another transformed scale.
The lower value is calculated with \(\exp(-s)\) and the upper value with \(\exp(s)\), so their order is correct.
Units are attached to the natural-scale estimate and transformed values. The log-scale change \(s\) is dimensionless.
The size of \(s\) is considered before replacing the two directional changes with one approximate percentage.

A direct-scale summary such as an estimate plus or minus its standard error is a common alternative. When uncertainty is small and the estimate is well away from zero, that symmetric approximation may be adequate. For positive parameters with appreciable log-scale uncertainty, it can be misleading and may even produce an impossible negative lower value. The exponentiated log-scale display preserves positivity and makes multiplicative uncertainty explicit.

Other approaches, such as profile-likelihood or bootstrap summaries, can be useful when the uncertainty is strongly asymmetric or the log-scale approximation is questionable. They answer related but not identical questions and depend on their own assumptions and calculations. A transformed one-standard-error display should not be treated as a replacement for those approaches when a more complete uncertainty analysis is needed.

Common Mistakes

Calling log-scale SE a natural-scale SE. A value such as \(0.20\) is a change in log units; it is not \(0.20\) L or \(0.20\ \mathrm{h}^{-1}\). Exponentiate the log-scale displacement to obtain natural-scale values.
Assuming the natural-scale changes are symmetric. Equal changes of \(+s\) and \(-s\) on the log scale correspond to factors \(\exp(s)\) and \(\exp(-s)\), not equal additive changes.
Reporting \(100s\%\) as an exact RSE. This is a small-error approximation. For larger \(s\), calculate the upward and downward relative changes separately.
Calling the one-standard-error display a confidence interval. It is a transformed summary of a one-SE displacement. It does not by itself establish a confidence level.
Rounding too early. Compute transformed endpoints and differences from unrounded values, then round the reported results. Otherwise, displayed arithmetic may not agree.

Key takeaway: For a positive estimate \(\hat{\theta}=\exp(\hat{\eta})\), log-scale uncertainty \(s\) acts multiplicatively: the one-SE values are \(\hat{\theta}\exp(-s)\) and \(\hat{\theta}\exp(s)\). Their natural-scale changes are asymmetric; \(100s\%\) is only an approximation to relative standard error when \(s\) is small.

Check Your Understanding

Answer each question, then compare with the solutions below.

A positive parameter is estimated as \(6.0\) mg/L with log-scale standard error \(0.10\). What are the one-SE transformed values, and what are the approximate upward and downward relative changes?
Why do equal positive and negative changes on the log scale not give equal additive changes on the natural scale?
A colleague says that a log-scale standard error of \(0.40\) means the natural-scale RSE is exactly 40%. Is this correct? Explain.
Can an estimate plus or minus one natural-scale standard error be assumed to equal the exponentiated one-SE display? State when they might be close.

Solutions. (1) The transformed values are \(6.0\exp(-0.10)\approx5.429\) mg/L and \(6.0\exp(0.10)\approx6.631\) mg/L. The downward relative change is \(1-\exp(-0.10)\approx9.52\%\); the upward change is \(\exp(0.10)-1\approx10.52\%\). (2) The factors are \(\exp(-s)\) and \(\exp(s)\), whose distances from 1 differ for \(s>0\); the natural-scale changes are multiplicative, not additive. (3) No. Forty percent is the small-error approximation \(100s\%\); at \(s=0.40\), the downward change is about 32.97% and the upward change about 49.18%. (4) Not in general. The two summaries use different scales and need not match. They can be close when relative uncertainty is small and a first-order approximation is adequate.