Expressing Ratio Margins on the Log Scale

Why Express Ratio Margins on the Log Scale?

In bioavailability and bioequivalence analyses, treatment effects are often expressed as ratios: for example, a test-to-reference ratio of exposure measures. A ratio describes a proportional comparison, while a model for log-transformed outcomes commonly estimates a difference on the log scale. Expressing the equivalence limits on that same scale makes the parameter and its acceptable range directly comparable.

The earlier material on equivalence hypotheses as two one-sided tests establishes the logic of TOST: equivalence is assessed by ruling out effects at or beyond either boundary. Here the task is narrower: translate positive ratio boundaries into log units and state the corresponding hypotheses. The transformation itself does not decide whether equivalence has been demonstrated; that requires an estimate and its uncertainty.

Definition: Let \(R\) be a positive ratio estimand, such as a test-to-reference ratio, and let \(L\) and \(U\) be its lower and upper equivalence limits, with \(0<L<U\). Define the log-ratio parameter \(\theta=\ln(R)\). The equivalent log-scale bounds are \(\ln(L)\) and \(\ln(U)\), and the equivalence region is \(\ln(L)<\theta<\ln(U)\).

Transforming the Bounds and the Hypotheses

The natural logarithm is strictly increasing for positive arguments. Therefore, applying it to an inequality preserves the order: \(L<R<U\) holds exactly when \(\ln(L)<\ln(R)<\ln(U)\). This is the complete algebraic reason the ratio limits transform as they do. The limits must be positive, and the ratio must have the same orientation throughout the analysis, such as test divided by reference.

L<R<U \quad\Longleftrightarrow\quad \ln(L)<\ln(R)<\ln(U) \quad\Longleftrightarrow\quad \ln(L)<\theta<\ln(U).

Thus the equivalence alternative is that the log-ratio parameter lies strictly between the transformed limits. Its null is the complement: the parameter is at or below the lower bound, or at or above the upper bound. TOST separates that union into two one-sided tests. The inequalities at the boundaries are part of the null hypotheses, not the equivalence alternative.

H_a:\ \ln(L)<\theta<\ln(U)

H_0:\ \theta\leq \ln(L)\ \text{or}\ \theta\geq \ln(U).

The component tests are \(H_{01}:\theta\leq\ln(L)\) against \(H_{a1}:\theta>\ln(L)\), and \(H_{02}:\theta\geq\ln(U)\) against \(H_{a2}:\theta<\ln(U)\). The TOST alternative requires both one-sided nulls to be rejected. These are hypotheses about a parameter, not a claim that a sample estimate must itself fall between the limits. The later inferential calculation accounts for sampling uncertainty.

The log-ratio is dimensionless when it is the logarithm of a ratio of quantities in the same units. Its value is in natural-log units, not in the original measurement units. A positive \(\theta\) corresponds to \(R>1\), a negative \(\theta\) to \(R<1\), and \(\theta=0\) to \(R=1\).

Worked Example: Reciprocal Limits of 0.80 and 1.25

Suppose a hypothetical analysis defines the acceptable test-to-reference ratio range as \(L=0.80\) to \(U=1.25\). Transform each boundary using the natural logarithm.

Lower bound. \(\ln(L)=\ln(0.80)=-0.22314\) (rounded).

Upper bound. \(\ln(U)=\ln(1.25)=0.22314\) (rounded).

The log-scale equivalence region is therefore \(-0.22314<\theta<0.22314\). The TOST null is \(\theta\leq-0.22314\) or \(\theta\geq0.22314\), with component null boundaries at \(-0.22314\) and \(0.22314\). These calculations transform the margins; by themselves they provide no evidence about whether an estimated treatment effect is equivalent.

When Are the Log Bounds Symmetric?

The log-scale bounds are symmetric around zero when the ratio limits are reciprocals: \(L=1/U\). In that case, \(\ln(L)=\ln(1/U)=-\ln(U)\). The equivalence region can then be written as \(-\Delta<\theta<\Delta\), where \(\Delta=\ln(U)>0\). This is a useful special case, not a general property of ratio margins.

If the ratio limits are not reciprocal, the transformed bounds need not be symmetric around zero. Their midpoint is \(\{\ln(L)+\ln(U)\}/2=\ln(\sqrt{LU})\). Consequently, the log-scale interval is centered at zero only when \(LU=1\). The geometric midpoint \(\sqrt{LU}\) describes the ratio-scale center of the bounds; it need not be one.

Worked Example: Asymmetric Ratio Limits

Consider hypothetical limits \(L=0.90\) and \(U=1.20\). Calculate both log bounds and check whether they are symmetric around zero.

Lower bound. \(\ln(0.90)=-0.10536\).

Upper bound. \(\ln(1.20)=0.18232\).

The transformed region is \(-0.10536<\theta<0.18232\). Because the magnitudes of the bounds differ, they are not symmetric about zero. Their midpoint is \((-0.10536+0.18232)/2=0.03848\), which also equals \(\ln(\sqrt{0.90\times1.20})=\ln(\sqrt{1.08})\), approximately \(0.03848\). The associated ratio midpoint is \(\sqrt{1.08}\approx1.039\), not 1. The TOST null consists of \(\theta\leq-0.10536\) or \(\theta\geq0.18232\); using equal positive and negative log margins here would change the specified equivalence region.

Orientation, Units, and a Second Translation

A ratio limit only has meaning relative to its numerator and denominator. If \(R\) is test/reference, then \(R=1.10\) means the test value is 10% higher than the reference value on that ratio scale. Reversing the ratio to reference/test changes \(R\) to \(1/R\), and changes the log-ratio from \(\theta\) to \(-\theta\). The corresponding limits must also be reversed and reciprocated: \(L\) and \(U\) become \(1/U\) and \(1/L\). Transforming the parameter but not the bounds would mismatch the hypothesis with the estimand.

Worked Example: Translating Limits and a Log-Ratio Estimate

Suppose a hypothetical study uses test/reference limits \(L=0.75\) and \(U=1.30\), and reports a log-ratio point estimate \(\hat{\theta}=\ln(1.10)\). Translate the margins, locate the estimate, and express the estimate as a ratio.

Lower bound. \(\ln(0.75)=-0.28768\). Upper bound. \(\ln(1.30)=0.26236\). Thus the log-scale equivalence region is \(-0.28768<\theta<0.26236\), and its null is \(\theta\leq-0.28768\) or \(\theta\geq0.26236\).

Estimate. \(\hat{\theta}=\ln(1.10)=0.09531\), which lies between the transformed bounds. Exponentiating gives \(\exp(\hat{\theta})=\exp(0.09531)=1.10\), the original ratio estimate. This location alone does not establish equivalence: a TOST conclusion depends on the estimate’s uncertainty as well as the bounds.

Assumptions, Diagnostics, and Alternatives

The conversion from ratio limits to log limits is an algebraic identity, but it is meaningful only when its inputs are defined consistently. Check that the ratio estimand is positive, that \(0<L<U\), that the limits were specified on the same ratio scale as the estimand, and that numerator and denominator orientation has not changed. The logarithm of zero or of a negative number is not a real-valued log-ratio.

Before using the transformed bounds, verify the arithmetic and the direction of the inequalities. Since the logarithm is increasing, the lower ratio limit must map to the lower log limit. A quick check is to exponentiate each transformed limit: \(\exp(\ln(L))=L\) and \(\exp(\ln(U))=U\). For reciprocal margins, another check is whether the two log bounds sum to zero.

Ratio-scale margins are suitable when acceptable differences are naturally proportional rather than absolute. For example, a fixed relative departure from a reference value can have a different absolute size at different baseline levels. A difference-scale margin instead specifies an acceptable absolute departure in the measurement units. These approaches answer different questions; taking logarithms does not make an absolute-difference margin into a ratio margin.

The log-scale representation is especially convenient when the statistical model estimates a contrast between log-transformed outcomes: the model contrast can then be compared directly with \(\ln(L)\) and \(\ln(U)\). The transformation does not justify a particular model, standard error, or degrees of freedom. Those must come from the analysis of the data. Nor does a symmetric log interval imply symmetric percentage departures on the original scale: exponentiating \(-\Delta\) and \(\Delta\) gives reciprocal ratio limits, not equal additive deviations from 1.

Common Mistakes

Logging the width instead of the boundaries. The expression \(\ln(U-L)\) is not the transformed equivalence interval. Transform \(L\) and \(U\) separately.
Assuming every pair of ratio limits becomes symmetric. Symmetry around zero requires \(LU=1\). Otherwise retain the two distinct log bounds.
Changing the ratio orientation without changing the margins. Reversing test/reference to reference/test negates the log-ratio and requires reciprocal, reversed limits.
Treating an estimate inside the interval as proof of equivalence. The point estimate’s location does not account for its uncertainty; the two one-sided tests must be evaluated.
Taking logs of percentages as written. Convert percentages to ratios first: for example, 80% is \(0.80\), not 80, when used as a ratio boundary.

Key takeaway: For positive ordered ratio limits \(L\) and \(U\), compare the log-ratio parameter with \(\ln(L)\) and \(\ln(U)\). The TOST null is that the parameter is at or beyond either transformed boundary; symmetry around zero occurs only for reciprocal ratio limits.

Check Your Understanding

Answer each question, then compare with the brief solutions.

Transform ratio limits \(0.80\) and \(1.25\). Are the log bounds symmetric around zero?
For limits \(0.85\) and \(1.10\), write the log-scale equivalence region and the union null in terms of \(\theta\).
If \(R\) is test/reference, what are the limits for reference/test when the original limits are \(0.80\) and \(1.25\)?
Why does a log-ratio estimate lying inside the transformed bounds not by itself establish equivalence?
What condition on \(L\) and \(U\) makes the transformed bounds symmetric around zero?

Solutions. (1) The bounds are \(\ln(0.80)=-0.22314\) and \(\ln(1.25)=0.22314\), so they are symmetric. (2) The equivalence region is \(\ln(0.85)<\theta<\ln(1.10)\), approximately \(-0.16252<\theta<0.09531\); the null is \(\theta\leq-0.16252\) or \(\theta\geq0.09531\). (3) The reversed limits are \(1/1.25=0.80\) and \(1/0.80=1.25\), in reversed order after taking reciprocals; these particular limits are reciprocal and remain the same pair. (4) The estimate alone does not account for sampling uncertainty, which is needed for the two one-sided tests. (5) The condition is \(L=1/U\), equivalently \(LU=1\).