Why Examine Sensitivity to Maturation Assumptions?
A pediatric clearance prediction depends on the assumed course of maturation as well as on the mature clearance level. Two maturation curves can agree at one age yet differ substantially at younger or older ages. Likewise, changing a half-maturation age may have little effect in one part of the age range and a much larger effect in another. Sensitivity analysis makes these consequences visible: it asks how much the predicted clearance changes when a stated model assumption or parameter value changes, while other inputs are held fixed.
This analysis is most useful after candidate curves have been evaluated against observed data. It does not select a correct model by itself, establish that one parameter value is more probable than another, or replace evaluation of the underlying data. Its purpose here is narrower: to quantify the prediction consequences of alternative maturation assumptions.
Set Up Predictions So the Comparison Is Interpretable
To isolate maturation, consider a clearance model at a fixed reference body size:
Here \(CL(a)\) is clearance at age \(a\), \(CL_{\mathrm{mat}}\) is the clearance level after maturation is complete at that reference size, and \(M(a)\) is a maturation fraction between zero and one. Age and any age-related parameters must use consistent units. In the examples, age is measured in years and clearance in L/h. Body size is held fixed, so differences in predictions arise only from the maturation assumptions.
A common sigmoid Emax maturation function is
For \(a\geq 0\), \(TM_{50}>0\), and \(h>0\), this function rises from zero toward one. At \(a=TM_{50}\), its value is \(1/2\). The Hill coefficient \(h\) controls the steepness around that transition: larger values give a sharper transition. An alternative with the same limiting values is the exponential function
where \(k>0\) has units of inverse years when age is in years. Setting \(k=\ln(2)/TM_{50}\) makes the exponential function reach one-half at the same age as the sigmoid. Matching this single feature does not make the curves equivalent; their shapes on either side of that age can still differ.
The comparison assumes that the model forms being evaluated are meaningful over the stated age range, that the mature clearance level and reference size are held fixed, and that age is measured consistently. It also assumes the candidate parameter values are chosen deliberately and reported transparently. A sensitivity analysis cannot compensate for a maturation function that is biologically implausible or contradicted by the available observations.
Worked Example: Comparing Two Functions at the Same Half-Maturation Age
Suppose \(CL_{\mathrm{mat}}=4.000\ \mathrm{L/h}\), and compare a sigmoid with \(h=2\) and \(TM_{50}=1.00\) year against an exponential with \(k=\ln(2)/1.00=0.6931\ \mathrm{year}^{-1}\). The curves have the same half-maturation age. Calculate predictions at ages 0.50, 1.00, 2.00, and 8.00 years.
At 0.50 years, the sigmoid fraction is \(0.50^2/(1.00^2+0.50^2)=0.25/1.25=0.2000\), giving \(CL=4.000(0.2000)=0.8000\ \mathrm{L/h}\). The exponential fraction is \(1-\exp[-0.6931(0.50)]=1-0.7071=0.2929\), giving \(CL=1.172\ \mathrm{L/h}\). The exponential prediction is higher by \(1.172-0.8000=0.3716\ \mathrm{L/h}\).
At 1.00 year, both fractions equal 0.5000, so both predictions are \(4.000(0.5000)=2.000\ \mathrm{L/h}\).
At 2.00 years, the sigmoid fraction is \(2.00^2/(1.00^2+2.00^2)=4/5=0.8000\), giving \(3.200\ \mathrm{L/h}\). The exponential fraction is \(1-\exp[-0.6931(2.00)]=1-0.2500=0.7500\), giving \(3.000\ \mathrm{L/h}\). Here the sigmoid prediction is higher by \(0.2000\ \mathrm{L/h}\).
At 8.00 years, the sigmoid fraction is \(64/65=0.9846\), giving \(3.938\ \mathrm{L/h}\). The exponential fraction is \(1-\exp[-0.6931(8.00)]=1-0.003906=0.9961\), giving \(3.984\ \mathrm{L/h}\). The exponential prediction is higher by \(0.04591\ \mathrm{L/h}\). Thus, agreement at the half-maturation age does not guarantee agreement elsewhere. The greatest difference among these four evaluated ages occurs at 0.50 years.
Quantifying Parameter Sensitivity
A second comparison keeps the function form fixed and varies one parameter. For the sigmoid function, changing \(TM_{50}\) shifts the age at which the maturation fraction reaches one-half. A larger \(TM_{50}\) means slower maturation at any fixed positive age, so it produces a lower maturation fraction and, with all else fixed, lower clearance.
Worked Example: Changing the Half-Maturation Age
Use \(CL_{\mathrm{mat}}=2.000\ \mathrm{L/h}\), \(h=2\), and age \(a=2.00\) years. Compare \(TM_{50}=0.4000\) years with \(TM_{50}=0.6000\) years.
For \(TM_{50}=0.4000\), the prediction is
For \(TM_{50}=0.6000\), it is
The absolute prediction range between these two scenarios is \(1.9230769-1.8348624=0.0882145\ \mathrm{L/h}\), or \(0.08821\ \mathrm{L/h}\) to four significant figures. This is the change in predicted clearance across the specified parameter values at age 2.00 years; it is not an estimate of the probability that the true clearance lies between the two predictions.
Where Does Parameter Uncertainty Matter Most?
The age at which predictions are most sensitive depends on which parameter is varied, the range of values considered, the function form, and the scale used to quantify change. The following derivation identifies the age of greatest absolute spread for one specific setup; it is not a universal rule for all maturation models.
For \(x>0\), the derivative of the factor \(x/(x^2+0.52x+0.0576)\) has the sign of
It is positive when \(x<0.24\) and negative when \(x>0.24\), so the absolute spread is largest at \(x=0.24\), or \(a=\sqrt{0.24}=0.4899\) years. This finding is specific to the selected parameter range and model. A relative difference, such as the prediction ratio, can have a different pattern and should be assessed separately when clinically relevant.
Worked Example: Comparing the Age Pattern of Prediction Spread
For the same \(CL_{\mathrm{mat}}=2.000\ \mathrm{L/h}\), \(h=2\), and \(TM_{50}\) values of 0.4000 and 0.6000 years, calculate the absolute spread at four ages using the two scenario predictions.
At age 0.2000 years, the predictions are \(2.000(0.04/0.20)=0.4000\ \mathrm{L/h}\) and \(2.000(0.04/0.40)=0.2000\ \mathrm{L/h}\). The spread is \(0.2000\ \mathrm{L/h}\).
At age 0.5000 years, they are \(2.000(0.25/0.41)=1.2195\ \mathrm{L/h}\) and \(2.000(0.25/0.61)=0.8197\ \mathrm{L/h}\). The spread is \(1.2195-0.8197=0.3998\ \mathrm{L/h}\).
At age 2.00 years, the spread is \(0.08821\ \mathrm{L/h}\), as calculated above. At age 5.00 years, the predictions are \(50/25.16=1.9873\ \mathrm{L/h}\) and \(50/25.36=1.9716\ \mathrm{L/h}\), giving a spread of \(0.01567\ \mathrm{L/h}\). Among these ages, the largest spread is at 0.5000 years, consistent with the derived maximum near 0.4899 years. The calculation shows why an age-specific sensitivity table or plot can be more informative than reporting a single parameter change alone.
Diagnostics, Alternatives, and Limitations
A useful sensitivity analysis makes it possible to see when predictions change meaningfully, not merely to generate multiple curves. Check that the curves have the intended limits and parameter interpretation, that the age and parameter units are consistent, and that all predictions use the same mature clearance level and reference size. Tabulate predictions across the relevant age range or plot the curves with clearly identified parameter values. Look especially for unexpected curve crossings, predictions outside the intended range, or large differences concentrated where there are few observations. Such findings may indicate a coding error, an unsuitable function, or genuine extrapolation sensitivity; the calculation alone cannot distinguish among them.
The main alternatives answer related but different questions. Comparing function forms, as in the first example, tests structural sensitivity. Varying one parameter while holding the form fixed tests parameter sensitivity. Varying several assumptions together can show a combined scenario envelope, but it obscures which assumption drove the change unless additional comparisons isolate each contribution. A fitted statistical uncertainty interval instead requires an explicit probability model and uncertainty estimation; a hand-selected low-to-high scenario range does not provide that inferential interpretation.
Sensitivity results are conditional on the chosen mature clearance, body size, and candidate maturation assumptions. Holding body size constant is appropriate for isolating maturation, but it does not describe the full uncertainty of a prediction when size also varies. The values tested should be scientifically defensible and relevant to the intended population. Sparse observations at young ages, parameter correlation, misspecification, or extrapolation can all make apparently precise predictions misleading. Finally, a large difference between scenarios signals dependence on assumptions, not evidence that either scenario is correct.
Check Your Understanding
Answer each question, then compare with the worked solutions below.
- With \(CL_{\mathrm{mat}}=5.00\ \mathrm{L/h}\), \(h=2\), \(TM_{50}=2.00\) years, and age 1.00 year, what clearance does the sigmoid model predict?
- An exponential maturation function has a half-maturation age of 3.00 years. What value of \(k\) matches that half-maturation age, and what fraction is predicted at age 6.00 years?
- Why can two maturation functions that share the same half-maturation age still predict different clearances at other ages?
- With \(CL_{\mathrm{mat}}=2.00\ \mathrm{L/h}\), \(h=2\), and age 1.00 year, calculate the prediction spread for \(TM_{50}\) values of 0.4000 and 0.6000 years. Is that spread a probability interval?
Solutions. (1) The maturation fraction is \(1^2/(2^2+1^2)=1/5=0.2000\), so \(CL=5.00(0.2000)=1.00\ \mathrm{L/h}\). (2) \(k=\ln(2)/3.00=0.2310\ \mathrm{year}^{-1}\). At age 6.00 years, using the unrounded \(k=\ln(2)/3.00\), \(M=1-\exp[-(\ln(2)/3.00)(6.00)]=1-\exp[-2\ln(2)]=0.7500\). (3) Matching the half-maturation age makes both fractions equal to 0.5 at that age only; the functions have different shapes elsewhere. (4) The predictions are \(2.00/(1+0.16)=1.724\ \mathrm{L/h}\) and \(2.00/(1+0.36)=1.471\ \mathrm{L/h}\). Using the unrounded predictions, their spread is \(2.00/1.16-2.00/1.36=0.2535497\ \mathrm{L/h}\), or \(0.2535\ \mathrm{L/h}\) to four decimal places. It is a scenario range, not a probability interval.