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Sample Size and Power · Tutorial 124 of 399

Converting Coefficients of Variation to Log-Scale Variance

Learn to convert a reported coefficient of variation to log-scale variance and select the variance quantity that matches a crossover sample-size formula.

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What You'll Learn

  • Derive the lognormal conversion from arithmetic CV to natural-log variance.
  • Distinguish arithmetic CV from geometric CV and use the matching conversion.
  • Identify the within-subject residual variance used in crossover planning.
  • Relate paired log-difference variance to within-subject residual variance.
  • Diagnose variance inputs that are on the wrong scale or represent the wrong source of variability.

Why the Conversion Matters

Average bioequivalence analyses commonly use natural-log-transformed pharmacokinetic endpoints. A planning report, however, may give variability as a coefficient of variation on the original scale. To use that information correctly, the coefficient of variation must be translated into variance on the log scale. A second decision is just as important: the variance supplied to a crossover sample-size formula must represent the within-subject comparison that the design uses, not total variability among all observed concentrations or participants.

The conversion depends on what the report means by “coefficient of variation.” The usual arithmetic CV and a geometric CV are not numerically identical. The convention must be identified before calculating; otherwise, two calculations can both be arithmetically correct yet describe different variability.

Arithmetic CV and Log-Scale Variance

Definition: Let \(X\) be a positive lognormally distributed measurement, so that \(\ln(X)\sim N(\mu,\sigma^2)\). Its arithmetic coefficient of variation is \(CV=SD(X)/E(X)\), expressed as a proportion rather than a percentage. For this convention, \(CV=\sqrt{\exp(\sigma^2)-1}\), and therefore \(\sigma^2=\ln(1+CV^2)\). In a crossover setting, \(\sigma_w^2\) denotes the within-subject residual variance of the natural-log endpoint; the corresponding arithmetic within-subject CV is \(CV_w=\sqrt{\exp(\sigma_w^2)-1}\).

The equation follows from the moments of a lognormal variable. If \(\ln(X)\sim N(\mu,\sigma^2)\), then \(E(X)=\exp(\mu+\sigma^2/2)\) and \(\operatorname{Var}(X)=[\exp(\sigma^2)-1]\exp(2\mu+\sigma^2)\). Dividing the variance by the squared mean removes the terms involving \(\mu\):

$$ \frac{\operatorname{Var}(X)}{[E(X)]^2} = \frac{[\exp(\sigma^2)-1]\exp(2\mu+\sigma^2)} {\exp(2\mu+\sigma^2)} = \exp(\sigma^2)-1. $$

Taking the positive square root gives \(CV=\sqrt{\exp(\sigma^2)-1}\). Squaring both sides and taking natural logarithms gives \(\sigma^2=\ln(1+CV^2)\). This is the standard conversion when a reported percentage is the usual arithmetic CV of a lognormal measurement.

A percentage must first be divided by 100. For example, an arithmetic CV of 25% is \(0.25\), not \(25\), in the formula. The resulting \(\sigma_w^2\) is a variance in squared natural-log units. Its square root, \(\sigma_w\), is a standard deviation in natural-log units.

Worked Example: Converting an Arithmetic CV

Suppose a hypothetical planning report gives a within-subject arithmetic CV of 30% for a positive endpoint assumed lognormal. Convert it to within-subject log-scale variance and standard deviation.

Convert the percentage to a proportion: \(CV_w=30/100=0.30\). Under the arithmetic-CV convention,

$$ \sigma_w^2=\ln(1+CV_w^2) =\ln(1+0.30^2) =\ln(1.09) \approx 0.08618. $$

The log-scale standard deviation is \(\sigma_w=\sqrt{0.0861777\ldots}\approx0.2936\). As a check, \(\sqrt{\exp(0.0861777\ldots)-1}=\sqrt{0.09}=0.30\), recovering the original arithmetic CV. The variance is \(0.08618\) squared natural-log units; it is not a variance of 30% or of 0.30 on the original scale.

Geometric CV Is a Different Convention

Some reports use “geometric CV” to mean \(GCV=\exp(\sigma)-1\), where \(\sigma\) is the standard deviation of the natural-log values. Under this definition, the conversion is \(\sigma=\ln(1+GCV)\), followed by \(\sigma^2=[\ln(1+GCV)]^2\). This convention should not be silently substituted for the arithmetic-CV formula above.

For the same lognormal distribution, the two CV summaries are related but differ: \(CV=\sqrt{\exp(\sigma^2)-1}\), whereas \(GCV=\exp(\sigma)-1\). Consequently, a report that says only “CV” should be checked for its definition, calculation method, or reporting standard. If the underlying log-scale standard deviation or variance is supplied directly, use it rather than reconverting a rounded CV.

Worked Example: Converting a Geometric CV

A hypothetical source explicitly defines its within-subject geometric CV as \(GCV_w=\exp(\sigma_w)-1\) and reports 30%. Find the corresponding log-scale variance and compare it with the arithmetic-CV result.

First convert the percentage to \(0.30\). Apply the geometric-CV definition and keep full precision through the squaring step:

$$ \sigma_w=\ln(1+GCV_w)=\ln(1.30)\approx0.2623643, \qquad \sigma_w^2=[\ln(1.30)]^2\approx0.0688349\approx0.06883. $$

This variance is smaller than the \(0.08618\) obtained when 30% is interpreted as an arithmetic CV. The difference is not a rounding discrepancy; it reflects two different definitions. As a further check, the arithmetic CV implied by this geometric-CV value is \(\sqrt{\exp(0.0688349\ldots)-1}\approx0.2669\), or about 26.69%, not 30%.

Which Variance Belongs in a Crossover Formula?

A crossover compares treatments within participants. The relevant planning quantity is therefore the within-subject residual variance of the log-transformed endpoint, denoted \(\sigma_w^2\), unless the particular formula is explicitly written in terms of a different but related quantity. Between-subject differences can be large without making the within-person treatment contrast equally variable: a participant may tend to have high values under both treatments, and that stable tendency is removed by a within-person comparison.

This distinction can be seen in a simple two-period crossover model. Let \(Y_{ij}\) be the natural logarithm of the endpoint for participant \(i\) in period \(j\). Suppose the model has participant effects and fixed treatment and period effects, with independent residual errors having variance \(\sigma_w^2\). In a participant’s treatment-minus-reference contrast, the stable participant effect cancels. If period effects are absent or properly accounted for, the residual part of that contrast is the difference of two independent errors.

$$ \operatorname{Var}(\varepsilon_{iT}-\varepsilon_{iR}) = \operatorname{Var}(\varepsilon_{iT})+\operatorname{Var}(\varepsilon_{iR}) -2\operatorname{Cov}(\varepsilon_{iT},\varepsilon_{iR}) = \sigma_w^2+\sigma_w^2 = 2\sigma_w^2. $$

Thus, two related quantities can appear in valid planning calculations: the within-subject residual variance \(\sigma_w^2\), and the variance of a two-treatment within-participant log difference, \(2\sigma_w^2\), under the stated assumptions. They are not interchangeable without the factor of two. Check the formula’s definition of its variance input before inserting a number.

In practice, an estimate of \(\sigma_w^2\) may come directly from the residual mean square of an appropriate log-scale crossover analysis. Alternatively, a suitable within-subject arithmetic CV can be converted using \(\ln(1+CV_w^2)\). If the only available estimate is the sample variance of paired log differences, that estimates the difference variance under the simple model; dividing by two gives an estimate of \(\sigma_w^2\). The data structure and fitted model matter: an unadjusted variance of paired differences can also contain period-related variation if the period effect has not been handled.

Worked Example: Recovering Residual Variance From Paired Differences

In a hypothetical, balanced two-period crossover analysis, the sample standard deviation of the appropriately period-adjusted participant-level log treatment differences is \(s_D=0.350\). Assume the differences have been formed and adjusted so that the independent, equal-variance residual model applies. Estimate the within-subject residual variance and the corresponding arithmetic within-subject CV.

The estimated variance of the log differences is

$$ s_D^2=(0.350)^2=0.1225. $$

Because the difference variance is \(2\sigma_w^2\) under this model, the within-subject residual variance estimate is

$$ \widehat{\sigma_w^2}=\frac{s_D^2}{2} =\frac{0.1225}{2} =0.06125. $$

Convert this residual variance to the arithmetic CV convention:

$$ \widehat{CV_w} =\sqrt{\exp(0.06125)-1} \approx\sqrt{0.06316} \approx0.2513. $$

The estimated arithmetic within-subject CV is therefore about 25.13%. If a sample-size formula instead defines its input as the variance of the paired log difference, its input would be \(0.1225\), not \(0.06125\). The reported quantity must be matched to the formula’s definition.

Choosing and Checking the Input

Before using a variability estimate in crossover planning, verify the endpoint, scale, source, and definition of the estimate. A variance from log-transformed AUC is not automatically suitable for a different endpoint, and variability from one population or design may not represent another. A total variance that includes between-subject variation is not a substitute for the within-subject residual variance merely because it is also reported on a log scale.

Useful diagnostics include checking whether a reported CV is a proportion or a percentage, whether it is arithmetic or geometric, and whether its source analysis used log-transformed data. When a residual mean square is available, compare the CV-derived variance with that estimate. Large disagreement may point to a different CV convention, a different endpoint or population, rounding, or a mismatch between total and within-subject variability. It should be resolved rather than averaged away.

The conversion also relies on a lognormal interpretation when an original-scale arithmetic CV is used to infer a log-scale variance. If the endpoint is markedly inconsistent with that model, or if the CV summarizes a heterogeneous mixture of populations, the algebraic conversion may not describe the residual variance needed by the planned model. In that case, a variance estimate from relevant log-scale data is preferable. A point estimate also carries sampling uncertainty; conversion does not make a weak or poorly matched estimate more reliable.

The principal alternatives are to use a directly reported log-scale residual variance, convert a clearly defined arithmetic CV, convert a clearly defined geometric CV using its own formula, or recover residual variance from an appropriate paired-difference variance. The direct log-scale estimate is simplest when it matches the intended endpoint and design. CV conversion is convenient when only an original-scale summary is reported, but depends on the stated convention and lognormal model. Paired-difference conversion is useful only when the difference variance is properly estimated and the factor of two is handled correctly.

Common Mistakes

  • Using the percentage as the input. Substituting 30 instead of 0.30 into \(\ln(1+CV^2)\) produces a nonsensical scale. Convert percentages to proportions first.
  • Mixing CV definitions. The formulas \(\ln(1+CV^2)\) and \([\ln(1+GCV)]^2\) refer to different CV conventions. Confirm the definition rather than choosing the formula that seems familiar.
  • Using total variance for a within-subject comparison. Between-subject heterogeneity does not equal the residual variability of a participant’s treatment contrast.
  • Forgetting the factor of two. Under the simple two-period model, paired-difference variance is \(2\sigma_w^2\). A formula using \(\sigma_w^2\) requires half the paired-difference variance, while a formula explicitly using difference variance requires the full quantity.
  • Rounding too early. Keep the unrounded logarithm or variance through the calculation, then round the final result. Squaring a rounded standard deviation can produce a slightly different displayed variance.
Key takeaway: For an arithmetic within-subject CV, use \(\sigma_w^2=\ln(1+CV_w^2)\). For a geometric CV defined as \(\exp(\sigma_w)-1\), use \(\sigma_w^2=[\ln(1+GCV_w)]^2\). In a crossover formula, use the within-subject residual variance or the paired-difference variance exactly as the formula defines it.

Check Your Understanding

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

  1. A hypothetical report gives an arithmetic within-subject CV of 20%. What is the corresponding log-scale variance?
  2. A source defines geometric CV as \(\exp(\sigma_w)-1\) and reports 20%. What is \(\sigma_w^2\)?
  3. If the within-subject residual variance is \(0.0500\), what is the variance of the treatment-minus-reference log difference under independent, equal-variance residual errors?
  4. A paired log-difference variance is \(0.100\). Under the same assumptions, what is the within-subject residual variance?
  5. Why should a total log-scale variance that includes between-subject variability not automatically be used as the within-subject variance in crossover planning?

Solutions. (1) Convert \(20\%\) to \(0.20\): \(\sigma_w^2=\ln(1+0.20^2)=\ln(1.04)\approx0.03922\). (2) Use the stated geometric convention: \(\sigma_w^2=[\ln(1+0.20)]^2=[\ln(1.20)]^2\approx0.03324\), retaining the unrounded logarithm when squaring. (3) The difference variance is \(2\sigma_w^2=2(0.0500)=0.100\). (4) Divide the paired-difference variance by two: \(\widehat{\sigma_w^2}=0.100/2=0.0500\). (5) Total variance can include stable differences among participants, whereas a crossover treatment contrast removes the stable participant effect; the planning input must represent the within-subject residual variation relevant to that contrast.