From a Skeleton to a Family of Toxicity Curves
In the one-parameter power CRM, the skeleton is a set of reference toxicity probabilities, one for each dose. The model parameter transforms the entire skeleton into a candidate toxicity curve. Understanding this transformation is important because a single parameter value determines probabilities at all doses: it does not adjust each dose independently.
Let the trial have ordered dose levels \(d_1,\ldots,d_J\), and let \(p_{0j}\) denote the skeleton probability at dose \(d_j\). The dose index \(j\) identifies a location on this discrete grid. The one-parameter power model uses one real-valued parameter \(\beta\), shared across every dose. Its toxicity probability at dose \(d_j\) is denoted \(p_j(\beta)\).
This is a probability model for dose-limiting toxicity (DLT) at each dose, not a statement that the probability can be read directly from an observed proportion. In the usual Bernoulli formulation, each participant’s DLT outcome is binary, and, conditional on the assigned dose and \(\beta\), the DLT probability is \(p_j(\beta)\). The power transformation describes the probability curve; the CRM’s likelihood and posterior calculations use participant outcomes to learn about \(\beta\).
The exponent \(\exp(\beta)\) is always positive. That fact will ensure that probabilities remain between zero and one and, when the skeleton is ordered, that the transformed curve remains ordered. The parameter itself is not a toxicity probability and need not lie between zero and one.
How the Parameter Changes the Curve
At \(\beta=0\), the exponent is one, so the model reproduces the skeleton exactly. If \(\beta>0\), the exponent exceeds one. Raising a number strictly between zero and one to a power greater than one makes it smaller, so every model probability is below its corresponding skeleton value. If \(\beta<0\), the exponent is between zero and one, and the transformed probabilities are above the skeleton values.
The last claim can also be checked by differentiation. Rewriting the model as \(p_j(\beta)=\exp\{\exp(\beta)\log(p_{0j})\}\), its derivative is
because \(p_j(\beta)>0\), \(\exp(\beta)>0\), and \(\log(p_{0j})<0\). Thus each dose-specific probability decreases strictly as \(\beta\) increases. This derivative describes parameter sensitivity; it does not say that the probability decreases as dose increases.
Worked Example: Comparing Curves Across the Dose Grid
Suppose a hypothetical five-dose trial uses skeleton probabilities \(0.05, 0.10, 0.20, 0.35, 0.50\). Calculate the model probabilities for \(\beta=-0.5,0,\) and \(0.5\), and interpret the comparison.
For each dose, use \(p_j(\beta)=\exp\{\exp(\beta)\log(p_{0j})\}\). The exponents for the two nonzero parameter values are \(\exp(-0.5)\approx0.6065\) and \(\exp(0.5)\approx1.6487\). For example, at the first dose with \(\beta=-0.5\), $$ p_1(-0.5)=0.05^{0.6065} =\exp\{0.6065\log(0.05)\} \approx\exp(-1.8170) \approx0.1625. $$
Applying the same calculation at each dose gives:
| Dose index | Skeleton \(p_{0j}\) | \(p_j(-0.5)\) | \(p_j(0)\) | \(p_j(0.5)\) |
|---|---|---|---|---|
| 1 | 0.0500 | 0.1625 | 0.0500 | 0.007161 |
| 2 | 0.1000 | 0.2474 | 0.1000 | 0.02245 |
| 3 | 0.2000 | 0.3767 | 0.2000 | 0.07040 |
| 4 | 0.3500 | 0.5289 | 0.3500 | 0.1771 |
| 5 | 0.5000 | 0.6567 | 0.5000 | 0.3189 |
At \(\beta=0\), the curve equals the skeleton. The negative parameter raises all five probabilities, while the positive parameter lowers all five. Within each column the probabilities still increase with dose. These are model-implied DLT probabilities, not observed DLT rates from a trial.
Mapping a Probability at One Dose to \(\beta\)
The power model can be inverted when a skeleton probability and a desired model probability at one dose are specified. Suppose the skeleton probability at dose \(d_k\) is \(p_{0k}\), and the desired model probability there is \(q\), with both values strictly between zero and one. Solving \(q=p_{0k}^{\exp(\beta)}\) gives
Both logarithms in the ratio are negative, so the ratio is positive and its logarithm is defined. The solution is unique: the mapping from \(\beta\) to \(p_k(\beta)\) is strictly decreasing, as shown by the derivative above. A desired value \(q\) above the skeleton value yields \(\beta<0\); one below it yields \(\beta>0\). Retain unrounded values in intermediate calculations, since rounding the log ratio before taking its logarithm can change the final reported parameter.
Worked Example: Matching a Probability at a Reference Dose
Suppose a hypothetical skeleton assigns \(p_{0k}=0.20\) to a reference dose, and a specified model curve should have probability \(q=0.30\) at that dose. Find \(\beta\), then map the full skeleton \(0.05,0.10,0.20,0.35,0.50\).
Using the inversion formula, $$ \beta=\log\!\left(\frac{\log(0.30)}{\log(0.20)}\right) =\log\!\left(\frac{-1.2039728}{-1.6094379}\right) \approx\log(0.748071) \approx-0.2903. $$
The exponent is \(\exp(\beta)\approx0.748071\). Applying it to the full skeleton gives $$ \begin{aligned} 0.05^{0.748071}&\approx0.1063, & 0.10^{0.748071}&\approx0.1787, & 0.20^{0.748071}&=0.3000,\\ 0.35^{0.748071}&\approx0.4560, & 0.50^{0.748071}&\approx0.5955. \end{aligned} $$
The reference probability is above its skeleton value, so the negative parameter is consistent with the model’s direction of change. The resulting curve is increasing across the ordered skeleton. The reported \(\beta\) uses the unrounded log ratio; taking a logarithm of a ratio already rounded to \(0.7481\) can produce a slightly different last digit.
Why the Power Transformation Preserves Dose Monotonicity
The model does not need a separate dose-order constraint on \(\beta\). Instead, dose monotonicity follows directly from the construction, provided the skeleton is ordered and the same parameter is applied at every dose. Write \(c=\exp(\beta)\), so \(c>0\). For any two dose indices \(j<\ell\) with \(p_{0j}\leq p_{0\ell}\), the function \(x\mapsto x^c\) is increasing for positive \(x\), because its derivative is \(cx^{c-1}>0\). Therefore:
This is a structural property of the model, not evidence from observed outcomes that toxicity truly increases with dose. If two skeleton values are equal, their transformed probabilities remain equal; strict increase follows only when the skeleton is strictly increasing. If the skeleton is unordered, the power transformation does not repair its ordering.
Worked Example: Checking Monotonicity and a Model Boundary
For a hypothetical three-dose skeleton \(0.10,0.25,0.40\), take \(\beta=\log(2)\). Then \(\exp(\beta)=2\), so the model probabilities are $$ p_1(\beta)=0.10^2=0.0100,\qquad p_2(\beta)=0.25^2=0.0625,\qquad p_3(\beta)=0.40^2=0.1600. $$
They are ordered \(0.0100<0.0625<0.1600\), as predicted by the monotonicity argument. For contrast, if a skeleton were \(0.10,0.08,0.40\), the same transformation would give \(0.0100,0.0064,0.1600\), which is not increasing. The model preserves the given ordering; it cannot turn a wrongly ordered skeleton into an ordered one.
Limits, Diagnostics, and Alternatives
The extremes of the parameter clarify the range of curves the model can represent. As \(\beta\to\infty\), \(\exp(\beta)\to\infty\), and \(p_{0j}^{\exp(\beta)}\to0\) for each \(p_{0j}\in(0,1)\). As \(\beta\to-\infty\), \(\exp(\beta)\to0\), and \(p_{0j}^{\exp(\beta)}\to1\). Thus every finite parameter value gives probabilities strictly between zero and one, while very large positive or negative values can place the entire curve close to a boundary.
The shared parameter also limits the shapes available. It shifts all probabilities in the same direction relative to the skeleton and preserves the skeleton’s ordering, but it cannot independently raise one dose probability while lowering another. In particular, flexibility across doses comes from the skeleton’s shape; the single parameter scales the transformation globally. A fixed skeleton corresponds to setting \(\beta=0\). At the other end, assigning an independent probability to every dose is more flexible but uses more dose-specific quantities and does not by itself guarantee monotonicity. The power CRM is a structured compromise: it retains a prespecified ordered shape while allowing a common transformation.
When checking a calculated curve, verify that every skeleton value lies strictly between zero and one, that the dose ordering is correct, and that the same \(\beta\) was used at every dose. Check the direction: a positive \(\beta\) must lower all probabilities relative to the skeleton, and a negative one must raise them. If a computed probability falls outside \((0,1)\), or if an ordered skeleton produces a reversed curve, there is an arithmetic or implementation error. Also distinguish model limitations from calculation mistakes: an ordered output does not establish that the real dose-toxicity relationship is monotone, and an implausible curve may indicate that the chosen skeleton cannot represent the pattern of interest.
Common mistake. A frequent error is to interpret \(\beta>0\) as increasing toxicity because “positive means more.” In this parameterization, that reasoning reverses the direction: since \(0<p_{0j}<1\), a power greater than one makes \(p_{0j}^{\exp(\beta)}\) smaller. The correction is to inspect the exponent and the base, or substitute a simple value such as \(p_{0j}=0.20\): with \(\beta=\log(2)>0\), the model probability is \(0.20^2=0.04\), below the skeleton value.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If \(p_{0j}=0.30\) and \(\beta=0\), what is \(p_j(\beta)\)?
- For \(p_{0j}=0.20\), determine whether \(\beta=0.4\) makes the model probability larger or smaller than the skeleton value. Give the reason.
- A reference skeleton probability is \(0.25\), and the desired model probability there is \(0.40\). Is the corresponding \(\beta\) positive or negative? Write the inversion formula.
- Why does an ordered skeleton remain ordered after the power transformation?
- Can the one-parameter power model independently increase the probability at one dose and decrease it at another relative to their skeleton values? Explain.
Solutions. (1) \(p_j(0)=0.30^{\exp(0)}=0.30^1=0.30\). (2) Since \(\exp(0.4)>1\), \(0.20^{\exp(0.4)}<0.20\); the model probability is smaller. (3) The desired value is above the skeleton value, so \(\beta<0\). Specifically, \(\beta=\log\{\log(0.40)/\log(0.25)\}\). (4) The exponent \(c=\exp(\beta)\) is positive, and \(x^c\) is increasing for \(x>0\), so \(p_{0j}\leq p_{0\ell}\) implies \(p_{0j}^c\leq p_{0\ell}^c\). (5) No. A single shared \(\beta\) makes all probabilities lower than their skeleton values when positive, or higher when negative; it cannot change directions dose by dose.