Why a Range Does Not Specify a Distribution
A PBPK input may be supported by only a few measurements, a literature range, or a statement that values are “typically between” two limits. Those summaries are useful, but a range alone does not tell us how values are distributed within it. Treating every reported lower and upper value as if it had the same statistical meaning can therefore create unjustified precision in an uncertainty description.
The first task is to identify what the reported numbers represent. They might be hard physical limits, the smallest and largest observations in a sample, selected plausible limits, or specified percentiles of a distribution. These interpretations imply different models. A distribution chosen from sparse evidence is best understood as an explicit representation of assumptions, not as a uniquely discovered fact.
For a positive PBPK parameter, a lognormal distribution may be defensible if multiplicative variation is plausible or if the available summaries are naturally expressed as percentile ratios. For a quantity with true finite bounds, a bounded distribution may be appropriate. A uniform distribution assigns equal density across a chosen interval; a triangular distribution adds a specified most plausible value. Each choice makes a different claim about values inside the interval.
Interpreting Reported Percentile Intervals
Suppose a positive parameter \(X\) is modeled as lognormal, meaning \(\ln X\sim N(\mu,\sigma^2)\). Here \(\mu\) and \(\sigma\) are the mean and standard deviation of the natural logarithm of \(X\); they are not generally the mean and standard deviation of \(X\). If the reported lower and upper endpoints are explicitly the 5th and 95th percentiles, the standard-normal quantiles are approximately \(-1.6449\) and \(1.6449\). The lognormal parameters can then be obtained by matching those two quantiles.
Adding the equations gives the midpoint on the log scale, while subtracting them gives the log-scale spread. Thus the implied median is the geometric midpoint of the endpoints, not their arithmetic midpoint.
This calculation is justified only if the endpoints really are the stated percentiles and a lognormal model is reasonable. A paper that reports a “range” without describing its statistical meaning does not provide enough information to use these equations. The moment formulas also depend on the lognormal assumption: \(E[X]=\exp(\mu+\sigma^2/2)\), whereas the median is \(\exp(\mu)\).
Worked Example: Recovering a Lognormal Distribution from a Central 90% Interval
Suppose a hypothetical source reports a central 90% interval of 1.5 to 6.0 L/h for a positive clearance parameter. Assume, specifically, that these are the 5th and 95th percentiles and that a lognormal distribution is suitable.
The log-scale centre is \(\mu=[\ln(1.5)+\ln(6.0)]/2=\ln(3)=1.0986\). The median clearance is therefore \(\exp(\mu)=3.00\) L/h.
The log-scale standard deviation is \(\sigma=[\ln(6.0)-\ln(1.5)]/[2(1.6449)]\). Substituting the ratio \(6.0/1.5=4\) gives \(\sigma=\ln(4)/3.2898=1.3863/3.2898\approx0.4214\).
The implied arithmetic mean is \(E[X]=\exp(\mu+\sigma^2/2)=\exp(1.0986+0.4214^2/2)\approx3.279\) L/h. Thus the median is 3.00 L/h, the mean is about 3.279 L/h, and the assumed central 90% interval is 1.5 to 6.0 L/h. The mean exceeds the median because the lognormal distribution is right-skewed.
Using Bounds When Distributional Information Is Sparse
If the endpoints are genuine physical limits but there is no evidence about where values concentrate, a uniform distribution is a simple, transparent assumption. It has constant density between its limits. This does not mean that nature is known to assign equal probability to equal-width subintervals; it means that equal density is being used as a modeling choice on the specified scale. Changing the scale changes what “equal” means, so the scale should be stated.
Worked Example: A Uniform Model from Physical Bounds
Suppose a hypothetical dimensionless partition parameter is known, for the model context under consideration, to lie between 0.50 and 2.00. Assume the limits are hard bounds and, because there is no basis for preferring one interior value to another on the linear scale, use a uniform distribution on that interval.
The mean is \((0.50+2.00)/2=1.25\). The variance is \((2.00-0.50)^2/12=1.50^2/12=0.1875\), so the standard deviation is \(\sqrt{0.1875}=0.4330\). The model therefore has mean 1.25 and standard deviation 0.4330, with all its probability between 0.50 and 2.00.
The calculation is internally consistent with the assumed hard bounds. It does not establish that the parameter is equally likely throughout the interval. If evidence suggests concentration near one end, a uniform model may be a poor representation even though it respects the same limits.
When a plausible minimum, most likely value, and plausible maximum are available, a triangular distribution can encode that additional information. It is bounded and has a linear increase in density from the minimum to the mode, followed by a linear decrease to the maximum. It is useful as a transparent approximation, but the stated mode is an assumption or elicited summary, not an observed fact unless supported by data.
Worked Example: A Triangular Model with an Explicit Mode
For a hypothetical positive clearance parameter, suppose a subject-matter summary gives a plausible minimum of 2 L/h, a most likely value of 3 L/h, and a plausible maximum of 8 L/h. Use a triangular distribution with \(a=2\), \(c=3\), and \(b=8\). These are elicited modeling inputs, not claims that the endpoints are exact physical laws.
The mean is \((2+3+8)/3=13/3\approx4.333\) L/h. The variance is \([2^2+3^2+8^2-(2)(3)-(2)(8)-(3)(8)]/18\). The numerator is \(4+9+64-6-16-24=31\), so the variance is \(31/18\approx1.722\) \((\text{L/h})^2\), and the standard deviation is \(\sqrt{31/18}\approx1.312\) L/h.
The mean exceeds the mode because the interval extends farther above the mode than below it. A symmetric interval or a uniform distribution would not express that same concentration near the lower end.
Distinguishing Range Types and Comparing Choices
An observed minimum and maximum should not automatically be treated as hard limits or percentile endpoints. A finite sample may not include values near the true extremes, and its maximum is not automatically the 95th percentile. With sparse observations, the range may be especially unstable: one additional measurement can change an endpoint substantially. Before fitting a distribution, record the sample size, units, measurement context, and exactly how the reported limits were obtained.
The same pair of endpoints can support different candidate models if the evidence does not distinguish them. For example, a uniform distribution on 2 to 9 L/h has mean \((2+9)/2=5.50\) L/h and standard deviation \(\sqrt{7^2/12}\approx2.021\) L/h. A triangular distribution on the same bounds with mode 3 L/h has mean \((2+3+9)/3=4.667\) L/h. The bounds alone do not select between these models or establish which mean is more credible.
| Available information | Possible representation | Key assumption to state |
|---|---|---|
| Explicit central percentiles for a positive parameter | Lognormal, if multiplicative spread is plausible | The endpoints are the stated percentiles and the lognormal shape is adequate |
| True finite bounds with little interior information | Uniform on the stated scale | Equal density on that scale is a deliberate simplification |
| Plausible minimum, mode, and maximum | Triangular | The mode and limits are meaningful summaries for the intended context |
| Only an observed sample range | No unique distribution follows from the range alone | Further data or explicit modeling assumptions are required |
A defensible choice should also respect the variable’s mathematical support. A positive clearance should not receive a normal model that places substantial probability below zero. A fraction must remain between zero and one; the earlier material on bounded distributions for fractions and probabilities addresses suitable bounded forms. For positive parameters, the earlier material on lognormal distributions explains the positive support and multiplicative interpretation. These support considerations rule out some candidates but do not, by themselves, prove that one remaining distribution is correct.
Diagnostics, Limitations, and Common Pitfalls
Check a proposed distribution against the information that motivated it. Do its support limits match genuine constraints? If percentile endpoints were used, does the fitted distribution reproduce those quantiles? Does its implied mean or mode conflict with the reported central estimate? For sparse measurements, a visually plausible fit is weak evidence: a small sample may not reveal skewness or tail behaviour reliably. Make these checks on the scale relevant to the parameter and report the assumptions alongside the distribution.
A common mistake is to read “range 1.5–6.0” as “central 90% interval,” then calculate a lognormal standard deviation from the endpoints. That reasoning is incorrect unless the source identifies those numbers as the 5th and 95th percentiles. If the values are merely observed extremes or rough plausible limits, the quantile calculation has no stated basis. The correction is to clarify the range definition, obtain further information if possible, or compare explicitly labeled alternatives rather than presenting one inferred distribution as uniquely determined.
Distribution choice is also limited by model purpose. A simple distribution can be useful when information is sparse, but its tails may matter for PBPK predictions even when the central region appears reasonable. A reported plausible interval may not define tail probabilities, and no calculation from its endpoints alone can recover them. The appropriate response is to document the uncertainty and consider how conclusions depend on defensible alternatives, rather than disguising an assumption as a measurement.
Check Your Understanding
Answer each question using the stated information only.
- A paper reports the smallest and largest values from six assays. Can these endpoints be used directly as the 5th and 95th percentiles? Explain.
- A positive parameter has a stated central 90% interval from 2 to 8 units. Under a lognormal assumption, what is the log-scale standard deviation? Use 1.6449 for the 95th-percentile normal quantile.
- For a uniform distribution on 1 to 5 units, calculate the mean and variance.
- A triangular distribution has minimum 1, mode 2, and maximum 7 units. Calculate its mean and variance.
Solutions. (1) No. They are observed sample extremes; the source has not identified them as population percentiles, and six assays cannot make that interpretation automatic. (2) \(\sigma=[\ln(8)-\ln(2)]/[2(1.6449)]=\ln(4)/3.2898\approx0.4214\). (3) The mean is \((1+5)/2=3\) units; the variance is \((5-1)^2/12=16/12\approx1.333\) units squared. (4) The mean is \((1+2+7)/3=10/3\approx3.333\) units. The variance is \([1^2+2^2+7^2-(1)(2)-(1)(7)-(2)(7)]/18=(1+4+49-2-7-14)/18=31/18\approx1.722\) units squared.