Scaling Dose for Logistic Toxicity Models

Why Scale Dose Before Modeling Toxicity?

Dose levels in a clinical trial are positive quantities, often separated by multiplicative rather than additive changes. Moving from 10 mg to 20 mg doubles the dose; moving from 20 mg to 40 mg doubles it again. A model based on raw dose treats both changes as an increase of 10 or 20 mg, even though their relative size differs. For many dose-escalation applications, a log-dose scale is therefore a useful way to express changes in dose proportionally.

The scale considered here starts by dividing each dose by a positive reference dose. Taking the logarithm of that ratio produces a dimensionless predictor. In a logistic toxicity model, this predictor allows a linear change in log odds to correspond to a multiplicative change in dose. This tutorial concerns the construction and interpretation of that dose scale, not the full Bayesian regression model or its estimation.

Definition: Let $d>0$ be a dose and let $d_{\mathrm{ref}}>0$ be a chosen reference dose, expressed in the same units. The reference-based natural log dose is $$ x(d)=\log\!\left(\frac{d}{d_{\mathrm{ref}}}\right). $$ The ratio $d/d_{\mathrm{ref}}$ is dimensionless. At the reference dose, $x(d_{\mathrm{ref}})=0$; doses above the reference have positive transformed values, and doses below it have negative transformed values.

The reference dose should be a positive dose relevant to the study, such as a dose in the planned range or one with particular clinical or prior information. It is a coordinate choice, not a claim that the reference dose is safe, effective, or the recommended dose. A toxicity model may use $x(d)$ as a predictor; in a generic logistic representation, its linear predictor can be written $a+b\,x(d)$, where $a$ and $b$ are the intercept and slope on this scale. The full two-parameter Bayesian model is a separate subject.

Calculating and Reading the Scale

The logarithm is the natural logarithm unless stated otherwise. Because $\log(1)=0$, the reference dose is the zero point. Equal multiplicative dose changes produce equal differences on the transformed scale: $$ x(cd)-x(d) =\log\!\left(\frac{cd}{d_{\mathrm{ref}}}\right) -\log\!\left(\frac{d}{d_{\mathrm{ref}}}\right) =\log(c), \qquad c>0. $$ Thus a doubling always adds $\log(2)$, approximately 0.6931, whether it occurs at a low or high dose. A change by a factor of $c$ adds $\log(c)$; a halving subtracts $\log(2)$.

Worked Example: Transforming a Planned Dose Range

Suppose a hypothetical study plans dose levels of 5, 10, 20, and 40 mg. Choose 10 mg as the reference dose and calculate each transformed value.

For 5 mg, $x(5)=\log(5/10)=\log(0.5)=-0.6931$. For 10 mg, $x(10)=\log(10/10)=0$. For 20 mg, $x(20)=\log(20/10)=\log(2)=0.6931$. For 40 mg, $x(40)=\log(40/10)=\log(4)=1.3863$.

The resulting values are $-0.6931,\ 0,\ 0.6931,\ 1.3863$. Each doubling advances by approximately 0.6931. The transformed values are not doses in milligrams; they are dimensionless log ratios. The calculation does not itself imply that any of these levels is appropriate for a patient.

For a generic logistic toxicity relationship, changing $x$ by $\Delta x$ changes the linear predictor by $b\Delta x$. Since the logistic function converts a linear predictor to a probability, this change corresponds to multiplying the toxicity odds by $\exp(b\Delta x)$. In particular, the odds ratio for a dose multiplied by $c$, at a fixed slope $b$, is $\exp\{b\log(c)\}=c^b$. This is an interpretation of the chosen scale, not a claim that the probability itself changes by a constant amount at every dose.

Worked Example: Interpreting a Doubling on the Log Scale

For illustration, suppose a generic logistic toxicity model has slope $b=0.80$ per unit of natural log dose. Suppose its toxicity probability at a reference dose of 10 mg is 0.20. Calculate the odds ratio for doubling the dose and the implied probability at 20 mg under this specified model.

A doubling changes the predictor by $\log(2)=0.6931$, so the log-odds change is $0.80(0.6931)=0.5545$. The corresponding odds ratio is $$ \exp\{0.80\log(2)\}=\exp(0.5545)=1.741. $$

At 10 mg the odds are $0.20/(1-0.20)=0.25$. At 20 mg, the odds under the stated model are $0.25(1.741)=0.4353$. Converting back to probability gives $$ \frac{0.4353}{1+0.4353}=0.3033. $$ Thus this hypothetical model implies a toxicity probability of about 0.303 at 20 mg. The 0.103 increase in probability is specific to the starting probability and slope; it is not a constant probability increment associated with every doubling.

Changing the Reference Dose

The reference dose determines where the transformed scale is zero. If the reference changes, the same dose receives a different transformed value. With an intercept that is adjusted consistently, however, the model can represent exactly the same dose-toxicity curve. The algebra makes clear both why the reference is a coordinate choice and why changing it carelessly can matter.

Result: Suppose $x(d)=\log(d/d_{\mathrm{ref}})$ and the linear predictor is $\eta(d)=a+b\,x(d)$. If the reference is changed to $d_0>0$, define $x_0(d)=\log(d/d_0)$. The equivalent intercept is $$ a_0=a+b\log\!\left(\frac{d_0}{d_{\mathrm{ref}}}\right), $$ so that $\eta(d)=a_0+b\,x_0(d)$ for every positive dose $d$.

To verify the result, note that $$ x(d)=\log\!\left(\frac{d}{d_0}\right)+\log\!\left(\frac{d_0}{d_{\mathrm{ref}}}\right) =x_0(d)+\log\!\left(\frac{d_0}{d_{\mathrm{ref}}}\right). $$ Substitution gives $a+b\,x(d)=a+b\,x_0(d)+b\log(d_0/d_{\mathrm{ref}})=a_0+b\,x_0(d)$, as claimed. The slope is unchanged, while the intercept now refers to the new zero point, the dose $d_0$.

Worked Example: Re-expressing a Curve at a New Reference

Suppose the reference is initially 10 mg and the illustrative linear predictor is $\eta(d)=-1.3863+0.80\log(d/10)$. Re-express it using 20 mg as the reference, then check the predictor at 40 mg.

Here $a=-1.3863$, $b=0.80$, $d_{\mathrm{ref}}=10$, and $d_0=20$, with doses in mg. The new intercept is $$ a_0=-1.3863+0.80\log(20/10) =-1.3863+0.80(0.6931) =-0.8318. $$

At 40 mg, the original predictor is $-1.3863+0.80\log(40/10)=-1.3863+1.1090=-0.2773$. Using the new reference gives $-0.8318+0.80\log(40/20)=-0.8318+0.5545=-0.2773$, the same value up to rounding. Therefore the implied probability from the logistic function is also unchanged.

This equivalence concerns the parameterization of the same curve. In Bayesian analysis, priors are also attached to parameters. If one changes the reference but keeps the same independent prior distributions for the intercept and slope, one has not necessarily preserved the same prior distribution over dose-toxicity curves: the transformed intercept depends on the slope. Reference choice can therefore affect prior meaning and computation even though the likelihood model can be re-expressed without changing its fitted curve. Prior calibration is a distinct task; the practical point here is to document the reference and avoid treating different parameterizations as interchangeable without checking the priors.

Choosing and Checking a Dose Scale

A useful reference is positive, stated in the same units as the administered doses, and located so that the transformed values are interpretable over the candidate range. Choosing a dose near the center of the range often keeps values from becoming needlessly large in magnitude. A reference with a clinical or evidential interpretation can also make the intercept easier to discuss. These are practical considerations rather than mathematical requirements for the transformation.

Before using the scale, verify the following:

Every modeled dose is positive. The logarithm is undefined at zero and for negative values.
The dose and reference use matching units. A ratio such as $20\text{ mg}/10\text{ mg}$ is meaningful; dividing a dose in mg by a reference value in micrograms without conversion is not.
The transformed values match the intended dose ratios. A larger dose must have a larger $x(d)$, and doubling must add $\log(2)$.
The reference is reported explicitly. A transformed value such as 0.6931 is uninterpretable without knowing the reference and logarithm convention.

A raw-dose linear predictor is a possible alternative: it makes equal absolute dose increments correspond to equal changes in the linear predictor. That may suit a context where additive changes are clinically meaningful, but it is sensitive to the chosen dose units and does not encode equal relative changes equally. A log-base-10 transformation is another option. It preserves the same ordering and dose ratios, but its numerical scale differs: $\log_{10}(d/d_{\mathrm{ref}})=\log(d/d_{\mathrm{ref}})/\log(10)$. The associated slope must be rescaled to preserve the same model predictions. Natural logarithms are conventional here because the slope’s odds-ratio expression is straightforward.

Common Mistakes and Limitations

Taking the logarithm of the dose without a reference. A logarithm of a dimensional quantity depends on the units used. Use the dimensionless ratio $d/d_{\mathrm{ref}}$; do not interpret $\log(d)$ as unit-invariant.
Putting zero dose into the same log-dose formula. Since $\log(0)$ is undefined, a placebo or zero-dose group cannot be transformed this way. It requires a separately specified modeling treatment, not an arbitrary small positive replacement presented as if it were the actual dose.
Thinking the reference dose is a target or recommended dose. It only fixes the scale’s zero point. Safety and dose recommendations require clinical criteria and inferential procedures beyond this transformation.
Interpreting the slope as a probability change per mg. On this scale, the slope applies per unit change in natural log dose. A dose ratio $c$ changes log odds by $b\log(c)$; the probability change depends on the probability at the starting dose.
Changing the reference but retaining parameter interpretations unchanged. The intercept moves to the new reference. In Bayesian work, unchanged numerical priors after reparameterization can also imply a different prior over curves.

Log scaling is not universally appropriate. It cannot resolve uncertainty about the true dose-toxicity shape, guarantee that toxicity increases with dose, or correct for an unsuitable dose range or poorly measured exposure. It is a covariate construction that makes proportional changes linear in the predictor; the adequacy of the model and its assumptions must still be assessed in the analysis.

Key takeaway: Express each positive dose as a ratio to a declared positive reference, then take the natural logarithm. Equal dose ratios then correspond to equal predictor differences, while the reference sets the intercept’s interpretation rather than determining a clinically preferred dose.

Check Your Understanding

Answer each question, then compare with the solutions below.

With a reference dose of 8 mg, calculate the transformed value for 16 mg. What transformed-scale increment corresponds to tripling any positive dose?
A modeled dose is 25 micrograms and the reference is 0.01 mg. May the values be used directly in the ratio? If not, calculate the transformed value after conversion.
If the slope is 0.50, what odds ratio corresponds to a fourfold dose increase? Give the expression and an approximate value.
Why can a zero-dose placebo not be entered into the reference-based log-dose formula, and what should not be done to hide this problem?

Solutions. (1) $x(16)=\log(16/8)=\log(2)=0.6931$. Tripling adds $\log(3)=1.099$. (2) Convert 0.01 mg to 10 micrograms. Then $x=\log(25/10)=\log(2.5)=0.9163$; the units cancel. (3) The log-odds change is $0.50\log(4)=0.6931$, so the odds ratio is $\exp(0.6931)=2.000$. (4) $\log(0)$ is undefined, so the formula does not apply. Do not substitute an arbitrary tiny positive dose and treat it as the actual placebo dose; a separate modeling treatment must be specified.