Sigmoid Emax Models of Maturation

Why Use a Sigmoid Maturation Function?

A maturation effect may rise slowly at young ages, increase more rapidly over an intermediate age range, and then approach a plateau. A single straight line does not represent that bounded pattern well: it keeps increasing, even at ages where the process should be near its mature level. A sigmoid Emax function represents a gradual transition from a lower value toward a finite maximum.

Here $A$ denotes age measured from a specified origin, such as birth or another defined developmental reference. The earlier material on postmenstrual age versus postnatal age motivates choosing that origin carefully: $A$ must be the age scale intended by the model. The function itself does not decide which age definition is scientifically appropriate.

Definition: For $A\geq 0$, half-maturation age $T_{50}>0$, and Hill coefficient $\gamma>0$, define the normalized sigmoid maturation curve $$ q(A)=\frac{A^\gamma}{A^\gamma+T_{50}^\gamma}. $$ It has $q(0)=0$, approaches 1 as age increases, and equals $1/2$ at $A=T_{50}$. A response with baseline $E_0$ and maximum increase $E_{\max}$ is $$ E(A)=E_0+E_{\max}q(A). $$

For a mature fraction that starts at zero and reaches one, use $E_0=0$ and $E_{\max}=1$, so the mature fraction is simply $q(A)$. If the fraction has a nonzero baseline $E_0$, or its maximum is below one, the general response form keeps those features explicit. For a fraction, the baseline and maximum must be compatible with the interval from zero to one; in particular, $E_0\geq 0$ and $E_0+E_{\max}\leq 1$. In the usual increasing-maturation setting, $E_{\max}>0$.

What the Parameters Do

The parameter $E_{\max}$ is the greatest increase in the response above its baseline $E_0$. It is not necessarily the maximum total response: that is $E_0+E_{\max}$. When the outcome is a normalized mature fraction with zero baseline and maximum one, $E_{\max}=1$, so the curve’s upper limit is one.

The parameter $T_{50}$ has units of age and identifies the midpoint of the increase. Substitution gives $q(T_{50})=1/2$, and therefore $E(T_{50})=E_0+E_{\max}/2$. Thus, “half-maturation age” means the age at which half the maximum increase above baseline has occurred. It means half of the total response only when the baseline is zero.

The Hill coefficient $\gamma$ is dimensionless and controls the curve’s shape around its transition. Larger values produce a more concentrated, steeper rise near $T_{50}$; smaller positive values spread the transition over a wider range of ages. Changing $\gamma$ does not change the value at $T_{50}$, and changing $E_{\max}$ changes the response scale rather than the normalized midpoint.

The age and half-maturation age must use the same units. The ratio $A/T_{50}$ is dimensionless, which makes raising it to the power $\gamma$ interpretable without giving the exponent units. Equivalently, the normalized function can be written as

q(A)=\frac{(A/T_{50})^\gamma}{1+(A/T_{50})^\gamma}.

This form is useful for calculating values and checking that an age entered in months is not accidentally compared with a $T_{50}$ entered in years.

Deriving the Midpoint and Curve Behavior

The midpoint follows directly from the function rather than from a separate approximation. At $A=T_{50}$, the two terms in the denominator are equal:

q(T_{50}) = \frac{T_{50}^{\gamma}}{T_{50}^{\gamma}+T_{50}^{\gamma}} = \frac{1}{2}.

More generally, the normalized value determines age relative to $T_{50}$. For $0<q(A)<1$, rearranging the function gives

\frac{q(A)}{1-q(A)} = \left(\frac{A}{T_{50}}\right)^\gamma.

The left side is the ratio of the attained fraction to the fraction still remaining before the upper limit. This identity shows why $T_{50}$ is the midpoint and why $\gamma$ changes how rapidly the curve moves through intermediate fractions.

For $A>0$, differentiation gives

\frac{dq}{dA} = \frac{\gamma T_{50}^{\gamma}A^{\gamma-1}} {\left(T_{50}^{\gamma}+A^{\gamma}\right)^2}.

This derivative is positive for $A>0$, so the normalized curve increases with age. At the midpoint its slope is $\gamma/(4T_{50})$, which has units of inverse age. At the boundary $A=0$, the right derivative must be considered separately: it is unbounded for $0<\gamma<1$, equals $1/T_{50}$ for $\gamma=1$, and equals zero for $\gamma>1$. The function itself remains defined at zero for every $\gamma>0$, with $q(0)=0$.

At large ages, $q(A)$ approaches one, so $E(A)$ approaches $E_0+E_{\max}$. This is an asymptote: at any finite positive age, $q(A)<1$. A model may therefore predict a value very close to its maximum without reaching it exactly.

Worked Example: Calculating a Mature Fraction at Three Ages

Suppose a hypothetical normalized maturation fraction follows $q(A)=A^2/(A^2+T_{50}^2)$, with $T_{50}=2$ months. Calculate the fraction at ages 1, 2, and 4 months.

At 1 month, $$ q(1)=\frac{1^2}{1^2+2^2}=\frac{1}{5}=0.2000. $$ The normalized mature fraction is 0.2000, or 20.00%.

At 2 months, $$ q(2)=\frac{2^2}{2^2+2^2}=\frac{4}{8}=0.5000. $$ This confirms that the fraction is one-half at $T_{50}$.

At 4 months, $$ q(4)=\frac{4^2}{4^2+2^2}=\frac{16}{20}=0.8000. $$ The function gives 80.00% at four months. These are model values for the stated hypothetical curve, not observations from a study.

Including a Baseline and a Maximum Below One

A response does not always begin at zero or reach one. The general form separates the normalized curve from its response scale: first calculate $q(A)$, then multiply it by $E_{\max}$ and add $E_0$. This distinction prevents confusing a normalized mature fraction with a response measured on another scale.

Worked Example: Applying the Curve to a Bounded Response

Consider a hypothetical response with $E_0=0.10$, $E_{\max}=0.80$, $T_{50}=4$ years, and $\gamma=1$. Its maximum total response is $0.10+0.80=0.90$. Calculate the response at ages 2, 4, and 12 years.

At 2 years, the normalized curve is $q(2)=2/(2+4)=1/3$. Therefore, $$ E(2)=0.10+0.80\left(\frac{1}{3}\right) =0.10+0.2667 =0.3667. $$

At 4 years, $q(4)=4/(4+4)=1/2$, so $$ E(4)=0.10+0.80(0.50)=0.5000. $$ This is halfway between the baseline 0.10 and the maximum 0.90.

At 12 years, $q(12)=12/(12+4)=0.75$, giving $$ E(12)=0.10+0.80(0.75)=0.7000. $$ The response is below its asymptotic maximum of 0.90, as expected at a finite age.

The calculation also illustrates an important interpretation rule: at the half-maturation age, the response is $E_0+E_{\max}/2$, not automatically 0.50. Here it happens to equal 0.50 because the baseline and maximum increase have those particular values.

How the Hill Coefficient Changes the Transition

The Hill coefficient changes the distribution of the rise across ages while preserving the midpoint when $T_{50}$ is fixed. The next example compares two curves directly. Both are hypothetical normalized fractions, so their baseline is zero and their maximum is one.

Worked Example: Comparing Two Hill Coefficients

Let $T_{50}=6$ months. Compare $\gamma=1$ and $\gamma=3$ at ages 3 and 12 months.

At 3 months, $A/T_{50}=1/2$. With $\gamma=1$, $$ q(3)=\frac{(1/2)^1}{1+(1/2)^1} =\frac{1/2}{3/2} =\frac{1}{3} =0.3333. $$ With $\gamma=3$, $$ q(3)=\frac{(1/2)^3}{1+(1/2)^3} =\frac{1/8}{9/8} =\frac{1}{9} =0.1111. $$

At 12 months, $A/T_{50}=2$. With $\gamma=1$, $$ q(12)=\frac{2}{1+2}=\frac{2}{3}=0.6667. $$ With $\gamma=3$, $$ q(12)=\frac{2^3}{1+2^3}=\frac{8}{9}=0.8889. $$

Both curves equal 0.50 at 6 months. The curve with $\gamma=3$ is lower at the younger age and higher at the older age in this comparison; its rise is more concentrated around the same midpoint. The coefficient changes shape, not the specified half-maturation age or upper limit.

Assumptions, Diagnostics, and Alternatives

The function is a useful description only when its assumptions make sense for the application. It assumes a monotone, saturating age pattern with a single lower asymptote and upper asymptote. In the normalized version, the lower value is zero and the upper limit is one. The general response adds a baseline and scales the increase. The standard age domain is $A\geq 0$, with $T_{50}>0$ and $\gamma>0$; using a negative age with a non-integer exponent is generally not defined as a real-valued function.

A fitted curve should be checked against both the data’s age coverage and its observed pattern. Useful checks include plotting observed responses against age with the fitted curve, examining whether residuals show systematic age trends, and checking whether fitted values respect scientifically meaningful bounds. If the data cover only young ages and never approach a plateau, the upper asymptote may be weakly informed. If observations cluster on one side of $T_{50}$, the midpoint and curve shape may also be difficult to distinguish. These are support and identifiability concerns, not evidence that the formula is wrong.

Common signs of a problem include a predicted curve that decreases despite an intended increasing maturation effect, a midpoint that appears at the wrong age because age units were mixed, or a fit whose asymptote is driven mainly by extrapolation beyond observed ages. Also check whether a response’s baseline is being mistaken for zero: with $E_0>0$, half of the maximum increase is not half of the total response.

A linear age effect is simpler and may describe a narrow observed age range, but it does not have a finite upper asymptote. A power function can describe a nonlinear trend, but without an added saturation mechanism it need not level off. Age categories can make fewer shape assumptions, at the cost of a stepwise representation and no smooth interpolation between categories. The sigmoid Emax curve is appropriate when a smooth, monotone transition to a plateau is scientifically plausible and supported by the age range. It should not be chosen merely because maturation is expected in general.

In population pharmacokinetics, the maturation function may later be used as part of a covariate model for a pharmacokinetic parameter. This tutorial defines the function and its values; deciding which parameter it should modify, or how to combine it with body-size effects, requires a separate model specification.

Common Mistakes

Calling $T_{50}$ the age at which total response is 0.50. It marks half of the increase above $E_0$; the response there is $E_0+E_{\max}/2$.
Treating $E_{\max}$ as the maximum total response. The total upper limit is $E_0+E_{\max}$.
Changing $\gamma$ and assuming the midpoint also changes. The function remains at one-half of its normalized increase when $A=T_{50}$, for every $\gamma>0$.
Mixing age units. A value of 6 months cannot be substituted alongside $T_{50}=1$ year without converting one of them first.
Using the derivative formula at zero without checking its domain. The displayed derivative formula applies for $A>0$; the right derivative at zero depends on $\gamma$.

Key takeaway: The sigmoid Emax maturation function separates the response scale ($E_0$ and $E_{\max}$), the age of its midpoint ($T_{50}$), and the steepness of its transition ($\gamma$). Calculate the normalized curve first, then apply the baseline and maximum increase.

Check Your Understanding

Try each question, then compare your work with the solutions below.

For $T_{50}=5$ months and $\gamma=2$, calculate the normalized mature fraction at $A=5$ months.
If $E_0=0.20$, $E_{\max}=0.60$, and $T_{50}=3$ years, what is the response at $A=3$ years?
For a fixed $T_{50}$, what does increasing $\gamma$ change, and what does it leave unchanged?
For which positive values of $\gamma$ is the right derivative of $q(A)$ at $A=0$ unbounded, equal to $1/T_{50}$, or zero?

Solutions. (1) At $A=T_{50}$, $q=1/2=0.5000$. (2) At $A=T_{50}$, $E=E_0+E_{\max}/2=0.20+0.60/2=0.5000$. (3) Increasing $\gamma$ makes the transition more concentrated around $T_{50}$; it leaves the midpoint and the normalized upper limit unchanged. (4) The right derivative is unbounded for $0<\gamma<1$, equals $1/T_{50}$ for $\gamma=1$, and equals zero for $\gamma>1$.