Diagnosing Translation Differences from Transformations

Why Transformations Can Look Like Model Differences

When two software implementations report different parameter values, the difference may arise before any model calculation: one value may be on a log scale, another on the natural scale, or the same physical quantity may use different units. Random effects add another source of apparent mismatch. In an exponential model, an individual parameter is formed by exponentiating a random effect, so its distribution and interpretation differ from those of an additive model.

The aim of diagnosis is to establish what each number represents and then express both implementations in the same parameterization and units. A numerical difference is not, by itself, evidence that the model has changed. The key is to trace the definitions, transformations, and units through the model equations.

Definition: In an exponential random-effects parameterization, an individual parameter \(P_i\) is \(P_i=P_{\text{typ}}\exp(\eta_i)\), where \(P_{\text{typ}}\) is the typical population value, \(\eta_i\) is an individual random effect on the log scale, and \(\exp\) is the natural exponential function. If \(\eta_i\sim N(0,\omega^2)\), then \(P_i\) is lognormally distributed. Its median is \(P_{\text{typ}}\), while its arithmetic mean is \(P_{\text{typ}}\exp(\omega^2/2)\).

Identify the Scale Before Comparing Values

A parameter that is positive by definition, such as clearance or volume, is often represented with an exponential random effect. The typical value \(P_{\text{typ}}\) is on the natural scale and has the physical units of the parameter. The random effect \(\eta_i\) is dimensionless and acts on the log scale. The product \(P_{\text{typ}}\exp(\eta_i)\) is therefore on the natural scale and retains the units of \(P_{\text{typ}}\).

Software may also represent the typical value through a fixed effect on the log scale. If that fixed effect is \(\beta=\log(P_{\text{typ}})\), then \(P_{\text{typ}}=\exp(\beta)\). The log is the natural logarithm unless a different base is explicitly stated. A value of \(\beta\) is dimensionless; it is not a clearance or a volume that can be substituted directly where that physical parameter is required.

Check the model equation and the parameter definition, not only the label in a results table. For instance, “CL” might denote a typical clearance in one output and the fixed effect \(\log(\text{CL}_{\text{typ}})\) in another. A parameter-name match does not establish a scale match.

Worked Example: Converting a Log-Scale Typical Clearance

Suppose a hypothetical implementation reports a fixed effect \(\beta=1.5686\) for clearance, defined by \(\beta=\log(\text{CL}_{\text{typ}})\). A second implementation reports typical clearance directly. Convert the first value and compare the definitions.

Because the first value is explicitly defined as the natural log of typical clearance, invert the transformation:

\text{CL}_{\text{typ}}=\exp(\beta)=\exp(1.5686)=4.800\ \text{L/h}.

The directly reported value should therefore be 4.800 L/h, if it uses the same units and typical-value definition. Comparing 1.5686 directly with 4.800 would be an invalid comparison: the former is a dimensionless log-scale fixed effect, and the latter is a natural-scale clearance.

The definition matters even when the same numerical value appears in both files. If a parameter is defined as a direct typical clearance of 1.5686 L/h, exponentiating it would also be wrong. The model equation determines whether the transformation is needed.

Exponential Random Effects and Lognormal Summaries

With \(P_i=P_{\text{typ}}\exp(\eta_i)\) and \(\eta_i\sim N(0,\omega^2)\), zero is the mean of the random effect, not of the multiplicative factor \(\exp(\eta_i)\). Consequently, the typical value is the median of the individual parameter distribution, not its arithmetic mean, unless \(\omega^2=0\). This distinction can explain apparently inconsistent reported summaries even when the parameterization is equivalent.

The following derivation establishes the arithmetic mean. For a normal random effect with mean zero and variance \(\omega^2\), integrating its density gives the expectation of its exponential. Completing the square in the exponent yields:

\begin{aligned} E[\exp(\eta_i)] &=\int_{-\infty}^{\infty}\exp(x) \frac{1}{\sqrt{2\pi\omega^2}}\exp\left(-\frac{x^2}{2\omega^2}\right)\,dx\\ &=\exp\left(\frac{\omega^2}{2}\right) \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\omega^2}} \exp\left(-\frac{(x-\omega^2)^2}{2\omega^2}\right)\,dx\\ &=\exp\left(\frac{\omega^2}{2}\right). \end{aligned}

The remaining integral is one because its integrand is a normal density. Multiplying by \(P_{\text{typ}}\) gives \(E[P_i]=P_{\text{typ}}\exp(\omega^2/2)\). Also, \(\log(P_i)=\log(P_{\text{typ}})+\eta_i\) is normally distributed with median on the natural scale equal to \(P_{\text{typ}}\). These results assume a single normal random effect with mean zero and finite variance \(\omega^2\) in the stated exponential relationship.

The notation for variability must also be checked. If a report gives \(\omega\), it may give a standard deviation on the log scale; if it gives \(\omega^2\), it gives a variance. For example, \(\omega=0.30\) corresponds to \(\omega^2=0.09\), not 0.30. A variance on the log scale is not a variance in clearance units. For a single lognormal parameter, the natural-scale variance is \(P_{\text{typ}}^2\exp(\omega^2)\{\exp(\omega^2)-1\}\).

Worked Example: Clearance With an Exponential Random Effect

Suppose a hypothetical model has \(\text{CL}_{\text{typ}}=5.00\) L/h, \(\eta_{\text{CL},i}\sim N(0,0.09)\), and \(\text{CL}_i=\text{CL}_{\text{typ}}\exp(\eta_{\text{CL},i})\). Calculate the individual clearance for \(\eta_{\text{CL},i}=-0.20\), and distinguish the population median from the arithmetic mean.

For the stated individual:

\text{CL}_i=5.00\exp(-0.20)=5.00(0.8187)=4.094\ \text{L/h}.

The random effect is dimensionless, and a negative value multiplies the typical clearance by a factor below one. The median is 5.00 L/h. The arithmetic mean is larger because exponentiation makes the multiplicative distribution asymmetric:

E[\text{CL}_i]=5.00\exp(0.09/2) =5.00\exp(0.045)=5.230\ \text{L/h}.

Thus, a report of 5.00 L/h as the typical value and a report of about 5.230 L/h as an arithmetic mean can describe the same model. Treating them as competing estimates without checking the summary definition would mistake a difference in summaries for a difference in parameterization.

Unit Conversions Must Preserve the Physical Quantity

Clearance has dimensions of volume per time, such as L/h or mL/min. Volume has dimensions of volume, such as L or mL. A unit conversion changes the numerical value but not the physical quantity or model. Convert both the numerator and denominator of a derived quantity consistently; comparing numbers with incompatible units has no physical meaning.

For an exponential random-effects model, changing the units of \(P_{\text{typ}}\) does not change the dimensionless \(\eta_i\) or its variance. If the unit conversion multiplies the parameter by a positive constant \(c\), the natural-scale value becomes \(cP_{\text{typ}}\exp(\eta_i)\). On a log scale, the fixed effect shifts by \(\log(c)\), while the random-effect variance remains the same.

Worked Example: Converting Clearance and Volume Consistently

A hypothetical parameter table gives \(\text{CL}=6.00\) L/h and \(V=30.0\) L. Convert both to mL-based units, then calculate and compare the ratio \(\text{CL}/V\). Use \(1\) L \(=1000\) mL and \(1\) h \(=60\) min.

For clearance, convert the volume and time units:

6.00\ \text{L/h}\times\frac{1000\ \text{mL}}{1\ \text{L}} \times\frac{1\ \text{h}}{60\ \text{min}} =100.0\ \text{mL/min}.

For volume, only the volume unit changes: \(30.0\ \text{L}\times1000=30{,}000\ \text{mL}\). In the original units, \(\text{CL}/V=6.00\ \text{L/h}/30.0\ \text{L}=0.2000\ \text{h}^{-1}\). In the converted units, \(100.0\ \text{mL/min}/30{,}000\ \text{mL}=0.003333\ \text{min}^{-1}\), which equals \(0.2000\ \text{h}^{-1}\) because \(60\ \text{min}=1\ \text{h}\).

The agreement confirms the unit conversion. Dividing \(100.0\) mL/min by \(30.0\) L would mix units and produce a number that cannot be interpreted as a valid rate constant. The value 100.0 is not a different clearance from 6.00 L/h; it is the same clearance expressed in different units.

A Practical Diagnostic Sequence

A reliable comparison follows the definitions rather than trying to reconcile isolated output numbers. Keep the same hypothetical or fitted parameter values and translate the equations explicitly before judging whether the implementations differ.

Write each parameter definition.
Record whether the reported fixed effect is a natural-scale typical value, its logarithm, or another transformed quantity. Include the units.

Write the individual-level relationship.
For example, distinguish \(P_i=P_{\text{typ}}\exp(\eta_i)\) from \(P_i=P_{\text{typ}}+\eta_i\). These equations define different models and do not use random effects on the same scale.

Check the random-effect scale and summary.
Determine whether the reported quantity is \(\omega\) or \(\omega^2\), whether the random effect is centered at zero, and whether the reported parameter summary is a median, typical value, or arithmetic mean.

Convert units and test a known value.
Convert each parameter to shared units, then calculate the same transformed value from both definitions. A small hand calculation can reveal an omitted exponential, an extra exponential, or an inconsistent time conversion.

These checks also help separate common causes of disagreement. A log-scale fixed effect used as if it were a natural-scale parameter points to a scale mismatch. Correct typical values but different individual values point to the random-effect equation, sign, or value of \(\eta_i\). A consistent multiplicative discrepancy across values may point to a unit factor. If definitions, random effects, and units all agree but values still differ, the cause lies elsewhere and should not be attributed to transformations without further evidence.

Common Mistakes and Limits

Exponentiating every reported parameter. Exponentiation is appropriate only when the parameter is defined on a log scale. A directly reported typical clearance should not be exponentiated.
Calling the typical value the arithmetic mean. Under a mean-zero normal random effect in an exponential model, the typical value is the median; the arithmetic mean is multiplied by \(\exp(\omega^2/2)\).
Using the wrong variability quantity. A log-scale standard deviation and its variance are different values. Neither should be described as a natural-scale standard deviation without transforming the distribution.
Applying an additive random effect to an exponential model. \(P_{\text{typ}}+\eta_i\) is not algebraically equivalent to \(P_{\text{typ}}\exp(\eta_i)\). It also does not guarantee positive individual parameters.
Mixing units in a ratio. Convert clearance and volume to compatible volume units and convert the time unit in clearance before interpreting \(\text{CL}/V\).

The lognormal formulas above rely on the stated normal random-effect assumption and exponential relationship. They do not automatically apply if the random effect has a different distribution, is not centered at zero, or enters the parameter through another transformation. A transformation can explain a discrepancy only when the equations and assumptions actually match.

A useful comparison is therefore between explicit definitions, not software names or table labels. One can transform fixed effects and units directly, and derive the implied lognormal summaries under the stated assumptions. By contrast, a nonlinear transformation of a parameter does not generally permit its uncertainty to be transformed by applying the same function to a standard error alone. Keep the parameter definition and its variability description distinct.

Key takeaway: Before comparing parameter values, put both implementations on the same scale, use the same random-effects relationship and summary definition, and express each physical quantity in the same units. Only then is a remaining numerical difference evidence of a substantive mismatch.

Check Your Understanding

For each question, first identify the relevant definition or units. Solutions follow each problem.

A hypothetical model reports \(\beta=2.3026\), defined as \(\log(V_{\text{typ}})\). What is \(V_{\text{typ}}\)?
Solution: \(V_{\text{typ}}=\exp(2.3026)\approx10.00\) in the volume units specified by the model. The value 2.3026 itself is a log-scale fixed effect, not a volume.
If \(P_i=P_{\text{typ}}\exp(\eta_i)\) and \(\eta_i=0.10\), is \(P_i\) above or below \(P_{\text{typ}}\), and by what factor?
Solution: It is above the typical value by the factor \(\exp(0.10)=1.105\), so \(P_i\approx1.105P_{\text{typ}}\).
For a mean-zero normal random effect with \(\omega^2=0.16\), is the arithmetic mean of \(P_i\) equal to \(P_{\text{typ}}\)? Give the multiplier.
Solution: No. The arithmetic mean is \(P_{\text{typ}}\exp(0.16/2)=P_{\text{typ}}\exp(0.08)\approx1.083P_{\text{typ}}\). The typical value is the median.
Convert a hypothetical clearance of 3.00 L/h to mL/min.
Solution: \(3.00\times1000/60=50.0\) mL/min. The time conversion is required because clearance is volume per time.
Two implementations report the same typical clearance in different units, but one reports \(\omega=0.20\) and the other reports \(\omega^2=0.04\). Are these necessarily different random-effect variabilities?
Solution: No. Since \(0.20^2=0.04\), they may be reporting the same log-scale variability using standard deviation and variance, respectively. Confirm that both use the same random-effect definition and scale.