Why Recode Dose?
A logistic dose-toxicity model can be written using dose in milligrams, an ordered dose index, or a centered and scaled covariate. These choices can make the model easier to interpret and fit, but they are not interchangeable in every sense. An affine change of units preserves the model's fitted toxicity probabilities when its coefficients are transformed accordingly. It does not preserve coefficient meanings if those coefficients are held fixed, and it does not turn a misspecified dose-response shape into a suitable one.
This tutorial focuses on that distinction. Centering chooses where the intercept is interpreted; scaling chooses the dose increment represented by the slope. Both can also affect the numerical work needed to estimate the model. Neither operation, by itself, changes the observations or the set of fitted probability curves represented by a linear logistic model.
With ordered dose levels, using the level number as \(d_i\) assumes a constant change in log odds from one level to the next. Using the actual dose in milligrams assumes a constant change in log odds per milligram. Those assumptions can differ substantially when dose levels are not equally spaced. Centering and scaling are affine transformations of a chosen covariate; they do not alter its assumed shape.
Centered and Scaled Parameterizations
Choose a reference dose \(c\) and a positive dose increment \(s_d\), and define \(x_i=(d_i-c)/s_d\). The model becomes \(\operatorname{logit}(p_i)=\alpha+\beta x_i\). For example, \(c\) might be a clinically interpretable dose, while \(s_d\) might be a dose increment of practical interest. Setting \(s_d=1\) centers without scaling; setting \(c=0\) scales without centering.
The word “odds” matters: \(\exp(\beta)\) is not a risk ratio. A positive slope means the modeled DLT probability increases with dose; a negative slope means it decreases. When \(s_d\) is positive, changing the scale does not reverse the slope's sign. The intercept may be much easier to interpret after centering: with raw dose in milligrams, \(\gamma_0\) refers to zero milligrams, which may be outside the modeled dose range and clinically uninformative.
Worked Example: Centering at a Clinically Relevant Dose
Suppose a hypothetical analysis of a fixed-window DLT endpoint uses doses of 10, 20, and 30 mg. Assume the fitted raw-dose model is \(\operatorname{logit}(p)=-2.00+0.0800d\), with dose \(d\) measured in mg. Re-express it using \(c=20\) mg and \(s_d=10\) mg.
The coefficient conversion gives \(\alpha=\gamma_0+\gamma_1c=-2.00+(0.0800)(20)=-0.400\), and \(\beta=\gamma_1s_d=(0.0800)(10)=0.800\). The centered covariate is \(x=(d-20)/10\), so the equivalent model is \(\operatorname{logit}(p)=-0.400+0.800x\).
At 10 mg, \(x=-1\), giving a logit of \(-0.400+0.800(-1)=-1.200\), and a DLT probability of \(1/(1+\exp(1.200))=0.2315\). At 20 mg, \(x=0\), so the probability is \(1/(1+\exp(0.400))=0.4013\). At 30 mg, \(x=1\), giving \(1/(1+\exp(-0.400))=0.5987\). Substitution into the raw-dose model gives the same three logits: \(-1.200\), \(-0.400\), and \(0.400\).
The transformed intercept says that the modeled DLT probability at 20 mg is 0.4013. The transformed slope says that a 10 mg increase multiplies the modeled odds by \(\exp(0.800)=2.226\). It does not say that the probability increases by 0.800 or by a fixed amount for each 10 mg.
Why Fitted Risks Are Preserved
The conversion is an algebraic identity, not an approximation. It also explains which aspects of the likelihood remain unchanged. For independent Bernoulli outcomes, the log-likelihood (apart from terms that do not involve the coefficients) is \(\ell=\sum_i[y_i\eta_i-\log(1+\exp(\eta_i))]\), where \(\eta_i\) is the linear predictor. Replacing the raw-dose coefficients by the transformed coefficients preserves each \(\eta_i\), so it preserves every fitted probability and the likelihood at the corresponding parameter values.
Consequently, if both parameterizations are fit by maximum likelihood, they represent the same fitted curves and have corresponding maximum-likelihood estimates, provided the optimization reaches the same finite optimum. Uncertainty measures also need conversion: for example, the standard error for \(\gamma_1\) is the standard error for \(\beta\) divided by \(s_d\). More generally, covariance matrices transform with the same linear coefficient conversion. Reporting transformed point estimates alongside untransformed standard errors would be inconsistent.
This equivalence concerns the likelihood and the represented curves. If a Bayesian analysis assigns priors to the coefficients, the prior distributions must also be transformed to preserve the same prior over curves. Reusing the same numerical prior for \(\alpha,\beta\) after changing the dose coding generally changes the prior implications. Centering and scaling do not make such priors equivalent merely because the likelihood parameterizations are equivalent.
Centering, Scaling, and Numerical Conditioning
Logistic regression is commonly estimated by numerical optimization. Very large or very small covariate values can make computations less well conditioned, while an intercept and slope can be strongly correlated when the intercept refers to a dose far from the observed doses. Centering near the doses with substantial information can reduce that correlation. Scaling to a meaningful increment can keep the coefficient and changes in the linear predictor on manageable numerical scales.
A local way to see the role of centering is through the information matrix. For grouped binomial observations at doses \(d_j\), let \(n_j\) be the number treated at dose \(d_j\), and let \(p_j\) be the fitted DLT probability. The logistic model's local information for an intercept and a slope on the covariate \(d_j-c\) has off-diagonal element \(\sum_j w_j(d_j-c)\), where \(w_j=n_jp_j(1-p_j)\). This term measures local coupling between the intercept and slope.
At a particular fitted model, choosing \(c=\sum_j w_jd_j/\sum_jw_j\) makes this off-diagonal element zero, because \(\sum_jw_j(d_j-c)=\sum_jw_jd_j-c\sum_jw_j=0\). This is the information-weighted mean dose at that fit. It is not a universal best reference: the weights depend on fitted probabilities, and a different reference may be preferable for interpretation or may be more stable across fits. Scaling changes the units of the slope but does not, by itself, eliminate this coupling.
Worked Example: Centering an Ordered Dose Index
Consider a hypothetical escalation design with four ordered levels, coded \(d=1,2,3,4\). Suppose an illustrative fitted model on this index scale has \(\gamma_0=-2.40\) and \(\gamma_1=0.70\). Recode using the midpoint \(c=2.5\) and \(s_d=1\), so \(x=d-2.5\).
The transformed intercept is \(\alpha=-2.40+(0.70)(2.5)=-0.650\), and the transformed slope remains \(\beta=0.70\). The raw-dose model predicts logits \(-1.70,-1.00,-0.300,0.400\) at levels 1 through 4. The centered model gives \(x=-1.5,-0.5,0.5,1.5\), and its logits are \(-0.650+0.70x=-1.70,-1.00,-0.300,0.400\), respectively.
The centered intercept is the log odds halfway between levels 2 and 3. That midpoint is a reference for the linear predictor; it need not be an administered dose. The slope still represents the change in log odds per one dose level, and the odds ratio per level is \(\exp(0.70)=2.014\). Centering improves the intercept's reference point here, but the dose-index model still assumes the same log-odds increment between every adjacent pair.
For a one-predictor model, choosing a centered reference near the observed doses can make estimation more stable, but it cannot guarantee convergence. In a model with additional covariates, scaling those predictors and checking their joint information may also matter. Numerical improvements are practical, not a change in the statistical information in the data: if a fit is unstable because the data provide little information about the slope, recoding cannot create that information.
Choosing Among Dose Codings
Dose coding should follow the scientific structure and the intended interpretation. The main options below impose different assumptions; only centering and positive scaling of the same covariate preserve the same linear-logit curve family.
| Coding | What the slope represents | Main implication |
|---|---|---|
| Actual dose \(d\) | Change in log odds per one physical dose unit | Assumes a linear logit in the physical dose scale |
| Centered, scaled actual dose \((d-c)/s_d\) | Change in log odds per \(s_d\) physical dose units | Same curve family as actual dose; reference and units change |
| Ordered level index | Change in log odds per one ordered level | Assumes equal log-odds increments across adjacent levels |
| Log dose or another prespecified transformation | Change in log odds per unit of the transformed dose | Allows a different dose-response shape; requires a suitable transformation |
| Categorical dose indicators | Separate log-odds contrasts against a reference level | Avoids a linear trend assumption but uses more parameters |
Worked Example: Unequally Spaced Dose Levels
Suppose a hypothetical trial uses four doses of 5, 10, 20, and 40 mg. An ordered-level model assigns indices 1, 2, 3, and 4. If its fitted slope is \(0.60\) per index step, each adjacent step has an odds ratio of \(\exp(0.60)=1.822\). This is the same multiplier from 5 to 10 mg as from 20 to 40 mg, even though the physical dose increments are 5 mg and 20 mg.
If instead the predictor is actual dose in mg, a slope of \(0.020\) per mg implies an odds ratio of \(\exp((0.020)(5))=1.105\) for a 5 mg increase and \(\exp((0.020)(20))=1.492\) for a 20 mg increase. These are different models, not just different numerical representations of one model. Centering and scaling actual dose—for example, using \((d-10)/10\)—would retain the actual-dose model's assumed shape while expressing its slope per 10 mg.
The coding decision therefore cannot be made solely by asking which coefficient is easiest to fit. It must reflect which change in log odds is assumed constant across the doses. A categorical model avoids imposing either of these linear trends, but its additional parameters may be weakly identified when few participants are treated at some levels.
Diagnostics, Limitations, and Common Mistakes
Check that the recoding has done what was intended. Confirm the reference dose and increment with units; calculate a few linear predictors using both codings; and verify that the resulting fitted probabilities agree. Review convergence warnings, unusually large coefficients or standard errors, and whether the fitted probabilities are effectively zero or one at observed dose levels. These checks can reveal implementation errors or weak information, though a numerically converged fit does not establish that the assumed dose-response shape is appropriate.
A common mistake is to center dose and then interpret the new intercept as the toxicity probability at zero dose. With \(x=(d-c)/s_d\), \(x=0\) means \(d=c\), so the intercept is the log odds at \(c\); the corresponding probability is \(1/(1+\exp(-\alpha))\). Another mistake is to scale the dose but keep the old slope unchanged. If the dose is divided by \(s_d\), the equivalent slope is multiplied by \(s_d\), not left unchanged.
Centering and scaling also do not remedy complete or quasi-complete separation. If the observed outcomes allow a coefficient to grow without bound while improving the likelihood, an ordinary finite maximum-likelihood estimate may not exist. Nor do they cure a poor linear-logit assumption, account for partial follow-up, or address dependence between outcomes. Those are modeling or data problems, not choices of units. Finally, the one-slope model imposes a monotone dose-toxicity relationship when the slope is nonzero; it cannot represent a nonmonotone pattern without changing the model.
Check Your Understanding
Use the model definitions and distinctions above to answer the questions. Solutions are included for checking your reasoning.
- A raw-dose model has \(\gamma_0=-1.5\) and \(\gamma_1=0.05\) per mg. What are \(\alpha\) and \(\beta\) when \(c=20\) mg and \(s_d=10\) mg?
- In the recoded model, what does \(\exp(\beta)\) represent? Is it a risk ratio?
- Why can centering reduce local intercept-slope coupling, and why is centering not guaranteed to solve every convergence problem?
- For doses 5, 10, 20, and 40 mg, what assumption is made by coding them 1, 2, 3, and 4 in a one-slope logistic model?
Solutions. (1) \(\alpha=-1.5+(0.05)(20)=-0.500\), and \(\beta=(0.05)(10)=0.500\). (2) \(\exp(\beta)=\exp(0.500)=1.649\) is the modeled odds ratio for a 10 mg increase; it is not a risk ratio. (3) Centering at the information-weighted mean makes the local off-diagonal information term \(\sum_jw_j(d_j-c)\) equal zero at that fit. It cannot add information or fix separation, misspecification, or other sources of instability. (4) It assumes a constant change in log odds for each adjacent level, including the 5-to-10 mg and 20-to-40 mg increases.