Converting Between Log-Scale SD and Within-Subject CV

Why Convert Between These Variability Scales?

Within-subject variability in bioavailability and bioequivalence analyses is often estimated on the natural-log scale, while variability is also communicated as a coefficient of variation (CV) on the original scale. These are two ways of expressing variability, not interchangeable numbers. In particular, a log-scale standard deviation of 0.30 does not mean a CV of 30%. The conversion is nonlinear, so the difference matters when a CV is compared with a threshold.

This tutorial gives the conversion in both directions and explains what it does, and does not, tell us. The calculation concerns a within-subject log-scale standard deviation, denoted here by \(\sigma_w\), and the corresponding within-subject CV, denoted by \(CV_w\). It does not estimate variability from replicate observations; it converts a given log-scale SD or estimate into another scale.

Definition: Suppose a positive measurement \(Y\), under the relevant within-subject model, is log-normal: \(\ln(Y)\sim N(\mu,\sigma_w^2)\). Its within-subject coefficient of variation is \(CV_w=SD(Y)/E(Y)\), expressed as a proportion or multiplied by 100 to express it as a percentage. Under this log-normal model, \(CV_w=\sqrt{\exp(\sigma_w^2)-1}\).

The Log-Normal Conversion

The key assumptions are that the measurement is positive, the logarithm is the natural logarithm, and the within-subject log-scale variation is represented by a normal distribution with standard deviation \(\sigma_w\). In a model with subject and other effects, \(\sigma_w\) must refer to the relevant within-subject component, not a total SD that combines within- and between-subject variation. The introductory material on within-subject reference variance establishes the role of that variance component; the present calculation translates its SD into a CV scale.

To derive the relationship, let \(X=\ln(Y)\), with \(X\sim N(\mu,\sigma_w^2)\), so that \(Y=\exp(X)\). For a log-normal variable, \(E(Y)=\exp(\mu+\sigma_w^2/2)\) and \(Var(Y)=[\exp(\sigma_w^2)-1]\exp(2\mu+\sigma_w^2)\). The coefficient of variation is the standard deviation divided by the mean. Substituting these expressions gives:

\begin{aligned} CV_w &=\frac{\sqrt{Var(Y)}}{E(Y)}\\ &=\frac{\sqrt{[\exp(\sigma_w^2)-1]\exp(2\mu+\sigma_w^2)}} {\exp(\mu+\sigma_w^2/2)}\\ &=\sqrt{\exp(\sigma_w^2)-1}. \end{aligned}

The location parameter \(\mu\) cancels, so the CV depends on the log-scale SD but not on the mean. To express the result as a percentage, multiply the proportion by 100. Conversely, starting with \(CV_w\) as a proportion and solving for \(\sigma_w\) gives the inverse conversion.

CV_w(\%)=100\sqrt{\exp(\sigma_w^2)-1}, \qquad \sigma_w=\sqrt{\ln\left(1+CV_w^2\right)}.

In the inverse formula, \(CV_w\) must be entered as a proportion: for example, 30% is entered as 0.30, not 30. If the CV is supplied as a percentage \(c\), use \(\sigma_w=\sqrt{\ln(1+(c/100)^2)}\). Both \(\sigma_w\) and \(CV_w\) are dimensionless. The original measurement may have units, but a ratio of standard deviation to mean does not.

Worked Example: Convert a Log-Scale SD to CV

Suppose a hypothetical analysis gives a within-subject log-scale SD of \(\sigma_w=0.25\). Convert it to a within-subject CV.

Use the log-normal conversion, keeping the result as a proportion until the last step:

\begin{aligned} CV_w &=\sqrt{\exp(\sigma_w^2)-1}\\ &=\sqrt{\exp(0.25^2)-1}\\ &=\sqrt{\exp(0.0625)-1}\\ &=\sqrt{1.06449-1}\\ &=\sqrt{0.0644945}\\ &=0.25396. \end{aligned}

Thus \(CV_w=100(0.25396)=25.40\%\). Under the stated log-normal model, the within-subject SD on the original measurement scale is about 25.40% of its arithmetic mean. It would be incorrect to report 25%, or 0.25, simply by multiplying the log-scale SD by 100; that shortcut does not apply to this conversion.

Converting Back and Checking the Scale

The reverse calculation is useful when variability is reported as a CV and the analysis requires a log-scale SD. The same model assumptions apply: the CV must describe the same within-subject variability and the log-normal relationship must be an appropriate representation. The algebra follows by squaring the forward relationship, adding 1, taking the natural logarithm, and taking the nonnegative square root. The SD is nonnegative by definition.

CV_w^2=\exp(\sigma_w^2)-1 \quad\Longrightarrow\quad \ln(1+CV_w^2)=\sigma_w^2 \quad\Longrightarrow\quad \sigma_w=\sqrt{\ln(1+CV_w^2)}.

Worked Example: Convert a CV to Log-Scale SD

Suppose a hypothetical report gives a within-subject CV of 32%. Find the corresponding log-scale SD.

First convert the percentage to a proportion: \(CV_w=32/100=0.32\). Then substitute that proportion into the inverse formula:

\begin{aligned} \sigma_w &=\sqrt{\ln(1+CV_w^2)}\\ &=\sqrt{\ln(1+0.32^2)}\\ &=\sqrt{\ln(1.1024)}\\ &=\sqrt{0.09749}\\ &=0.3122. \end{aligned}

The corresponding within-subject log-scale SD is approximately 0.3122. The intermediate value 0.09749 is the log-scale variance, not the SD; taking its square root is necessary. Entering 32 rather than 0.32 would produce a grossly incorrect result because the formula uses a proportion.

Interpreting Values Near a CV Threshold

A 30% within-subject CV is a commonly encountered variability threshold in discussions of scaling, although the applicable rule depends on the relevant regulatory framework, endpoint, and study context. The equivalent log-scale SD is obtained by converting 0.30, not by treating 0.30 as an SD:

\sigma_w=\sqrt{\ln(1+0.30^2)} =\sqrt{\ln(1.09)} =\sqrt{0.08618} =0.2936.

Thus, under the log-normal relationship, a CV of 30% corresponds to a log-scale SD of about 0.2936. This is a scale conversion, not a decision about whether any particular study qualifies for a scaling procedure. A threshold must be applied according to the rule that governs the analysis, using the relevant estimate and required precision.

Worked Example: A Log-Scale SD Close to the 30% CV Equivalent

Consider a hypothetical within-subject log-scale SD of 0.293. It is close to the value 0.2936 that corresponds to a 30% CV. Calculate the CV before comparing it with the threshold.

\begin{aligned} CV_w &=100\sqrt{\exp(0.293^2)-1}\\ &=100\sqrt{\exp(0.085849)-1}\\ &=100\sqrt{1.08964-1}\\ &=100\sqrt{0.08964}\\ &\approx 29.94\%. \end{aligned}

The calculated CV is slightly below 30%, even though the SD is close to its equivalent threshold. Rounding the SD to 0.29 before converting, or reporting only “about 30%,” could obscure which side of the threshold the unrounded calculation falls on. A threshold-based conclusion should therefore use unrounded inputs and calculations, then report enough digits to make the comparison transparent.

For contrast, a log-scale SD of 0.30 gives:

CV_w =100\sqrt{\exp(0.30^2)-1} =100\sqrt{\exp(0.09)-1} \approx 30.69\%.

The result is above 30%, not exactly 30%. This illustrates why comparing the numerical SD directly with a percentage threshold is invalid: the quantities are on different scales and have different meanings.

Assumptions, Diagnostics, and Limitations

The formula is exact for a log-normal distribution, but its use in an analysis depends on what the supplied SD or CV represents. A within-subject log-scale SD is typically a model-based component or estimate of variability among repeated measurements within subjects under specified conditions. It is not automatically equal to the SD of all logged observations pooled together, which may also reflect differences between subjects or other sources of variation. Nor is a raw sample CV necessarily the same quantity as a model-based within-subject CV.

Before converting, check the following:

Confirm the scale. The formula uses an SD on the natural-log scale. If a value was calculated on a base-10 log scale, convert it first: an SD on the natural-log scale equals \(\ln(10)\) times the SD on the base-10 log scale.
Confirm the component. Use the within-subject SD relevant to the intended interpretation, not a total SD that combines within- and between-subject sources.
Confirm the CV format. The inverse formula takes a proportion, such as 0.30 for 30%.
Retain precision near a cutoff. Convert using unrounded values and round the reported result only after the comparison has been made.

The conversion itself cannot verify that the data are log-normal, identify the correct variance component, or quantify uncertainty in an estimated SD. If \(\sigma_w\) is estimated, the converted CV is a plug-in estimate; the conversion alone does not provide a confidence interval for that CV. Distributional features such as strong departures from log-normality can also make the log-normal CV an imperfect description of the original-scale data. These are reasons to examine the model and data, not reasons to substitute an unconverted SD for a CV.

Useful Approximation and Common Pitfall

For small \(\sigma_w\), the exponential expansion \(\exp(\sigma_w^2)\approx 1+\sigma_w^2\) implies \(CV_w\approx\sigma_w\) as proportions. This can be a quick rough check at low variability, but it is not an exact conversion. As variability increases, the difference becomes more apparent: at \(\sigma_w=0.30\), the CV is about 30.69%, rather than 30.00%.

Common pitfall: Treating a log-scale SD of 0.30 as a 30% CV, or entering a CV of 30% as 30 in the inverse formula. The correction is to use \(CV_w=\sqrt{\exp(\sigma_w^2)-1}\) for the forward conversion and enter 0.30 for a 30% CV in the reverse conversion. Near a threshold, keep full precision through the calculation.

A direct raw-scale CV, calculated as a sample SD divided by a sample mean, is an alternative descriptive summary when those raw observations are the intended data. It need not match a CV obtained by converting a within-subject log-scale variance component, especially when the observations include between-subject variation or when the log-normal model is not suitable. The two summaries answer different questions; one should not be substituted for the other without checking the estimand and model.

Check Your Understanding

Attempt each conversion or interpretation, then compare with the solutions.

A hypothetical within-subject log-scale SD is 0.20. What is the corresponding CV?
A hypothetical within-subject CV is 25%. What is the corresponding log-scale SD?
What log-scale SD corresponds to a 30% CV, and why is it not 0.30?
A colleague rounds \(\sigma_w=0.293\) to 0.29, labels it “about 30%,” and concludes it is at a 30% CV threshold. Identify the problem.
A report gives an SD calculated using base-10 logarithms. What must be done before using the natural-log conversion formula?

Solutions. (1) \(CV_w=100\sqrt{\exp(0.20^2)-1}=100\sqrt{\exp(0.04)-1}=100\sqrt{0.04081}\approx20.20\%\). (2) Enter 0.25 as a proportion: \(\sigma_w=\sqrt{\ln(1+0.25^2)}=\sqrt{\ln(1.0625)}\approx\sqrt{0.06062}=0.2462\). (3) \(\sigma_w=\sqrt{\ln(1.09)}\approx0.2936\); the conversion is nonlinear, so a 30% CV does not correspond to an SD of 0.30. (4) The unrounded conversion is about 29.94%, just below 30%; rounding first and calling it “about 30%” does not establish that it meets a threshold. (5) Multiply the base-10 log SD by \(\ln(10)\) to obtain the natural-log SD, then apply the formula.