Hill Coefficient as a Log-Concentration Slope

Why Put the Sigmoid Emax Fraction on a Logit Scale?

The sigmoid Emax model describes a normalized effect fraction that changes with concentration. Its Hill coefficient, $\gamma$, controls how sharply that fraction changes around the concentration producing half of the maximum effect. The coefficient can also be understood as a slope—but only after specifying the axes: it is the slope of the fraction's logit against the natural logarithm of concentration.

This connection is useful for interpreting the model's shape parameter and for checking the mathematical form of a proposed concentration-effect relationship. It does not mean that the untransformed fraction is linear in concentration or in log concentration. The straight-line relationship applies to the log odds of the fraction.

Definition: For positive concentration $C$, positive half-maximal concentration $EC_{50}$, and positive Hill coefficient $\gamma$, the sigmoid Emax fraction is $$ f(C)=\frac{C^\gamma}{EC_{50}^\gamma+C^\gamma} =\frac{1}{1+(EC_{50}/C)^\gamma}. $$ For $0<f<1$, its logit is $\operatorname{logit}(f)=\ln\!\left(\frac{f}{1-f}\right)$, the natural logarithm of the fraction's odds.

Here $f(C)$ is the fraction of the model's maximum effect, not necessarily the measured response itself. For a baseline-relative response, for example, one may write the modeled effect as the baseline plus a maximum effect change multiplied by $f(C)$. The logit result concerns $f(C)$, after that normalization.

Deriving the Logit-Slope Result

The derivation requires $C>0$, $EC_{50}>0$, and $\gamma>0$. These conditions make the concentration ratio well-defined and ensure that the model fraction lies strictly between zero and one, so its odds and logit exist. Also, $C$ and $EC_{50}$ must be expressed in the same concentration units; their ratio is dimensionless.

Write the fraction and its complement.
From the sigmoid Emax equation, $1-f(C)=\frac{EC_{50}^\gamma}{EC_{50}^\gamma+C^\gamma}$. Both quantities are positive under the stated conditions.

Form the odds.
Dividing the fraction by its complement cancels the shared denominator: $\frac{f(C)}{1-f(C)}=\frac{C^\gamma}{EC_{50}^\gamma}=(C/EC_{50})^\gamma$.

Take natural logarithms.
Applying $\ln$ to both sides gives $\operatorname{logit}(f(C))=\gamma\ln(C/EC_{50})$.

Using the logarithm rule $\ln(C/EC_{50})=\ln C-\ln EC_{50}$, the result can be written as a straight-line equation:

\operatorname{logit}(f(C))=\gamma\ln C-\gamma\ln EC_{50}.

Thus, on a graph with $\ln C$ on the horizontal axis and $\operatorname{logit}(f(C))$ on the vertical axis, the model is a line with slope $\gamma$ and intercept $-\gamma\ln EC_{50}$. Equivalently, the derivative of the logit with respect to natural log concentration is $\gamma$. At $C=EC_{50}$, the concentration ratio is one, so the logit is zero and $f(C)=0.5$.

Theorem: Under the sigmoid Emax fraction model, $\operatorname{logit}(f(C))$ is linear in $\ln C$, with slope $\gamma$. This slope is dimensionless: both axes use logarithmic transformations of dimensionless ratios or quantities represented relative to a fixed unit.

The slope depends on the log base used for the horizontal axis. If the horizontal coordinate is $\log_{10} C$, while the logit continues to use the natural logarithm, then $\ln C=(\ln 10)\log_{10} C$. The slope against $\log_{10} C$ is therefore $\gamma\ln 10$, not $\gamma$. The natural-log convention gives the direct slope-equals-$\gamma$ statement.

Worked Numerical Examples

Worked Example: Calculating Fractions and Their Logits

Suppose a hypothetical concentration-effect relationship has $EC_{50}=4$ mg/L and $\gamma=2$. Calculate the model fractions at $C=2$, $4$, and $8$ mg/L, and verify the logit-slope relationship.

At $C=2$ mg/L, the concentration ratio is $C/EC_{50}=0.5$, so

f(2)=\frac{1}{1+(4/2)^2}=\frac{1}{5}=0.2000.

Its odds are $0.2/0.8=0.25$, so $\operatorname{logit}(f(2))=\ln(0.25)=-1.3863$. At $C=4$ mg/L, $f(4)=1/(1+1)=0.5000$, with logit $\ln(1)=0$. At $C=8$ mg/L,

f(8)=\frac{1}{1+(4/8)^2}=\frac{1}{1.25}=0.8000, \qquad \operatorname{logit}(f(8))=\ln(0.8/0.2)=\ln(4)=1.3863.

The horizontal coordinates for $C=2$ and $C=8$ are $\ln 2$ and $\ln 8$. The logit change is $1.3863-(-1.3863)=2.7726$, and the log-concentration change is $\ln 8-\ln 2=\ln 4=1.3863$. Their ratio is $2.7726/1.3863=2.000$, which equals $\gamma$. All three points lie on the line described by the model.

Worked Example: Reading the Slope Across a Fourfold Concentration Change

For a hypothetical model with $EC_{50}=3$ mg/L and $\gamma=1.5$, compare concentrations of $1.5$ and $6$ mg/L. These are respectively one-half and twice $EC_{50}$.

The fractions are

f(1.5)=\frac{1}{1+2^{1.5}}=0.2612, \qquad f(6)=\frac{1}{1+(1/2)^{1.5}}=0.7388.

The corresponding logits can be calculated from the model's logit equation: $\operatorname{logit}(f(1.5))=1.5\ln(0.5)=-1.0397$, and $\operatorname{logit}(f(6))=1.5\ln(2)=1.0397$. The horizontal change is $\ln(6)-\ln(1.5)=\ln(4)=1.3863$, while the vertical change is $1.0397-(-1.0397)=2.0794$. Therefore the slope is $2.0794/1.3863=1.500$. The slope is not the change in fraction per mg/L; it is the change in logit per unit change in natural log concentration.

Worked Example: Recovering the Hill Coefficient From One Fraction

Suppose a hypothetical sigmoid Emax model has $EC_{50}=4$ mg/L, and the model fraction at $C=2$ mg/L is $f=0.2000$. Find $\gamma$.

The odds at that concentration are $f/(1-f)=0.2/0.8=0.25$. From the odds equation, $0.25=(2/4)^\gamma=(0.5)^\gamma$. Taking natural logarithms of both sides gives

\gamma=\frac{\ln(0.25)}{\ln(0.5)} =\frac{-1.3863}{-0.6931} =2.000.

The result is positive, as required for the stated increasing sigmoid Emax model. This calculation illustrates the slope identity in reverse: a known fraction determines a logit, and the logit divided by $\ln(C/EC_{50})$ gives the Hill coefficient when $C\ne EC_{50}$.

Interpretation, Diagnostics, and Alternatives

A larger positive $\gamma$ means a steeper logit-versus-log-concentration line. In the original fraction scale, it corresponds to a sharper transition in the neighborhood of $EC_{50}$. It does not, by itself, establish a biological mechanism such as receptor cooperativity. In a PK/PD model, $\gamma$ is a shape parameter unless further mechanistic evidence and assumptions justify a more specific interpretation.

The straight-line result can serve as a mathematical diagnostic. If exact fractions generated by the stated model are plotted as logits against $\ln C$, they must fall on a line with slope $\gamma$. Systematic curvature in a plot constructed from reliable, appropriately normalized fractions can indicate that the sigmoid Emax shape does not describe those values, or that the normalization or concentration scale is incorrect. With noisy observations, however, scatter is expected; a few departures from a line do not by themselves disprove the model.

A common pitfall is to take the logit of a measured response and treat the resulting points as exact model fractions. The logit is defined only for values strictly between zero and one, whereas noisy measurements may equal or exceed either boundary. Moreover, transformation changes the error behavior: even when measurement errors are additive on the original response scale, they are not thereby made additive with constant variance on the logit scale. The slope identity is an exact statement about the model fraction, not a guarantee that ordinary linear regression on transformed observations is an appropriate estimation procedure.

The main alternative for describing or fitting this concentration-effect relationship is to retain the original sigmoid Emax equation and work on the response scale. That approach directly represents the modeled fraction and can accommodate a response model and error assumptions chosen for the data. The logit-line form is especially helpful for interpretation and algebraic checks; it does not replace assessment of whether the response normalization, model assumptions, and measurement-error description are appropriate. Neither representation removes the need for informative observations across the concentration range.

The result also has limits. It concerns the normalized sigmoid Emax fraction under the stated direct-effect model, not every observed response or every exposure metric. It relies on positive concentration and a positive $EC_{50}$, and the logit cannot be evaluated at a fraction of exactly zero or one. Using $\log_{10}$ concentration without adjusting the slope, or using concentrations in inconsistent units inside a ratio, also leads to an incorrect interpretation.

Key takeaway: For the sigmoid Emax fraction, $\operatorname{logit}(f(C))=\gamma\ln(C/EC_{50})$. With natural log concentration on the horizontal axis, the slope is exactly the Hill coefficient; the statement applies to the model fraction, not automatically to noisy measured responses.

Check Your Understanding

Answer each question using the sigmoid Emax fraction and the natural-log definition of logit.

If $C=EC_{50}$, what is the model fraction and what is its logit?
A model has $\gamma=1.2$. What is the change in logit when concentration increases fourfold?
For $EC_{50}=5$ mg/L and $\gamma=1$, calculate the fraction at $C=10$ mg/L.
Why is the slope against $\log_{10} C$ not $\gamma$ when the logit uses natural logarithms?
Why can a measured response equal to 1 not be directly given a finite logit?

Solutions. (1) At $C=EC_{50}$, the concentration ratio is 1, so $f=1/(1+1)=0.5$, and its logit is $\ln(1)=0$. (2) The logit change is $\gamma\ln(4)=1.2(1.3863)=1.664$. (3) $f(10)=1/[1+(5/10)^1]=1/1.5=0.6667$. (4) Since $\ln C=(\ln 10)\log_{10}C$, the slope against $\log_{10}C$ is $\gamma\ln 10$. (5) Its odds would be $1/(1-1)$, which is undefined, so the logit is not finite.

Hill Coefficient as a Log-Concentration Slope

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