Why the Two Parameters Need Separate Interpretations
In a Bayesian logistic regression model for dose-limiting toxicity (DLT), the intercept and slope play different roles. The intercept anchors the fitted toxicity curve at a chosen reference dose; the slope describes how the log odds of DLT change as dose moves away from that reference on the transformed dose scale. Understanding both interpretations helps translate model parameters into clinically meaningful comparisons between doses.
The focus here is interpretation, not estimation or dose selection. The numerical parameter values in the examples are hypothetical. In an analysis, parameter uncertainty matters: a fitted curve based on one parameter pair is not the same thing as the full posterior distribution of possible curves.
The BLRM and Its Parameter Meanings
Let \(Y_i=1\) indicate that participant \(i\) experiences a DLT during the specified assessment window, and let \(Y_i=0\) otherwise. Let \(d_i>0\) be that participant’s dose, and choose a positive reference dose \(d_{\mathrm{ref}}\). The two-parameter logistic regression model is
Here, the dose is represented by its natural logarithm relative to the reference dose. At \(d_i=d_{\mathrm{ref}}\), the transformed dose is zero. The fitted probability is obtained by applying the inverse-logit function, \(p(d)=\exp(\eta(d))/\{1+\exp(\eta(d))\}\), where \(\eta(d)=\alpha+\beta\log(d/d_{\mathrm{ref}})\).
The usual model assumptions include a common dose-toxicity curve for the participants being modeled, a binary DLT outcome defined over a consistent assessment window, and conditionally independent outcomes given doses and model parameters. The logit of the DLT probability is assumed linear in log dose over the dose range of interest. A positive \(\beta\) gives a strictly increasing fitted probability as dose increases, as established in the earlier material on monotonicity. A zero slope gives a constant fitted probability; a negative slope would imply decreasing toxicity with dose and is inconsistent with the usual monotone-increasing dose-toxicity assumption.
Interpreting the Intercept at the Reference Dose
The intercept is not itself a probability. It is a log odds value, so its meaning becomes clearer by converting it first to odds, then to probability. At the reference dose, \(\log(d/d_{\mathrm{ref}})=0\), giving \(\operatorname{logit}\{p(d_{\mathrm{ref}})\}=\alpha\). Therefore the fitted odds are \(\exp(\alpha)\), and the fitted probability is \(\exp(\alpha)/\{1+\exp(\alpha)\}\).
Worked Example: Converting the Reference-Dose Intercept
Suppose a hypothetical BLRM uses a reference dose of 80 mg and has intercept \(\alpha=-0.5546\). Interpret the fitted toxicity probability at 80 mg.
At 80 mg the dose ratio is \(80/80=1\), so its logarithm is zero. The fitted log odds are therefore \(-0.5546\), and the fitted odds are \(\exp(-0.5546)\approx 0.5743\).
Convert the odds to a probability: \[ p(80)=\frac{0.5743}{1+0.5743} =\frac{0.5743}{1.5743} \approx 0.3648. \]
Thus, under this hypothetical parameter pair, the fitted DLT probability at 80 mg is about 36.48%. The intercept is the log odds value \(-0.5546\), not a 36.48% probability. That probability is obtained only after applying the inverse-logit transformation.
Holding \(\beta\) fixed, increasing \(\alpha\) raises the fitted log odds by the same amount at every dose. It therefore raises the fitted probability at every dose, although the probability increase is not constant across doses. For example, the same increment in log odds produces different probability increments depending on the starting probability. Intercept changes are vertical shifts on the log-odds scale, not fixed percentage-point shifts on the probability scale.
Interpreting the Slope Across Dose Ratios
The slope is tied to a change in transformed dose, so it should not be described simply as “the change in probability per mg.” For two positive doses \(d_1\) and \(d_2\), subtract the model’s two logit values. The intercept cancels, leaving the change in log odds as \(\beta\log(d_2/d_1)\). Exponentiating gives the corresponding odds ratio:
This makes the dose-ratio interpretation direct. Multiplying dose by \(k\) multiplies fitted odds by \(k^\beta\); multiplying dose by \(e\) multiplies fitted odds by \(e^\beta\). The slope is not itself an odds ratio unless the dose ratio is \(e\). Nor is the odds ratio a probability ratio: probabilities depend on the starting probability as well as the odds multiplier.
Worked Example: Comparing Two Doses Using the Slope
Suppose a hypothetical model has reference dose 100 mg, fitted DLT odds of 0.25 at that dose, and slope \(\beta=1.2\). Find the fitted odds and probability at 150 mg.
The dose ratio is \(150/100=1.5\). By the model, the odds at 150 mg equal the reference-dose odds multiplied by \(1.5^{1.2}\): \[ \operatorname{odds}\{p(150)\} =0.25(1.5^{1.2}) \approx 0.25(1.6267) \approx 0.4067. \]
Convert these odds to a probability: \[ p(150)=\frac{0.4067}{1+0.4067} \approx 0.2891. \]
The fitted DLT probability is about 28.91% at 150 mg, compared with \(0.25/(1+0.25)=0.2000\), or 20.00%, at 100 mg. The dose increase multiplies the odds by approximately 1.6267; it does not multiply the probability by 1.6267.
A larger positive slope makes the fitted curve change more rapidly with log dose. One justification comes from differentiating the fitted probability. Since \(p=\exp(\eta)/(1+\exp(\eta))\) and \(d\eta/d\log(d/d_{\mathrm{ref}})=\beta\), the chain rule gives
For \(0<p(d)<1\), the probability increases with log dose when \(\beta>0\). The derivative also shows why the slope is not a constant probability change: the factor \(p(d)\{1-p(d)\}\) varies along the curve. On the log-odds scale, however, the slope has a constant interpretation.
Worked Example: How a Steeper Slope Changes the Curve
Consider two hypothetical curves with the same fitted odds of 0.25 at a reference dose of 100 mg, but slopes \(\beta=0.6\) and \(\beta=1.2\). Compare their fitted probabilities at 50 mg, below the reference dose.
The dose ratio is \(50/100=0.5\). For \(\beta=0.6\), the odds are \[ 0.25(0.5^{0.6})\approx 0.1649, \] so the probability is \(0.1649/(1+0.1649)\approx 0.1416\).
For \(\beta=1.2\), the odds are \[ 0.25(0.5^{1.2})\approx 0.1088, \] so the probability is \(0.1088/(1+0.1088)\approx 0.0981\).
Both curves have the same probability, 20.00%, at 100 mg. Below that reference dose, the steeper curve has the lower fitted probability. Above the reference dose, it has the higher fitted probability. Thus, with a common reference-dose intercept and positive slopes, increasing the slope makes the curve depart more sharply from its reference-dose value in either direction.
Changing the Reference Dose
The choice of reference dose determines where the intercept is read, but it does not change the fitted curve when the parameters are re-expressed correctly. Suppose the reference dose changes from \(d_0\) to \(d_1\). Since \(\log(d/d_0)=\log(d/d_1)+\log(d_1/d_0)\), substitution into the original linear predictor gives
The new intercept is therefore \(\alpha_{\mathrm{new}}=\alpha+\beta\log(d_1/d_0)\), while the slope remains \(\beta\). The new intercept represents the fitted log odds at \(d_1\). The change is a reparameterization, not a change in the underlying fitted toxicity probabilities.
Worked Example: Re-expressing an Intercept at a New Reference Dose
Suppose a hypothetical model has reference dose 100 mg, odds 0.25 at that dose, and slope \(\beta=1.2\). Re-express the intercept using 200 mg as the reference dose, and calculate the fitted probability there.
The original intercept is \(\alpha=\log(0.25)\approx -1.3863\). The new intercept is \[ \alpha_{\mathrm{new}} =-1.3863+1.2\log(200/100) =-1.3863+1.2\log(2) \approx -0.5545. \]
At the new reference dose, the odds are \(\exp(-0.5545)\approx 0.5743\), giving probability \(0.5743/(1+0.5743)\approx 0.3648\). Equivalently, starting from the old reference-dose odds, the odds at 200 mg are \(0.25(2^{1.2})\approx 0.5743\). Both calculations give the same fitted probability, about 36.48%.
Diagnostics, Alternatives, and Limitations
A parameter interpretation is only as useful as the model’s fit and the relevance of its assumptions. Inspect whether the assumed monotone dose-toxicity pattern is plausible, whether the logit-linear dose relationship is credible over the dose range, and whether the observed outcomes are compatible with the fitted curve. Sparse observations at some doses can leave the slope weakly informed; a very steep or nearly flat fitted curve should prompt careful review rather than automatic acceptance. With Bayesian analyses, uncertainty in both parameters should be kept in view instead of interpreting only one selected parameter pair.
The main alternative parameterization is to describe the reference-dose probability \(p_{\mathrm{ref}}\) directly rather than use its logit \(\alpha\). The two forms are equivalent when \(\alpha=\log\{p_{\mathrm{ref}}/(1-p_{\mathrm{ref}})\}\). Probability-scale language may be more intuitive clinically, while the logit intercept and slope make the logistic model algebraically convenient. The choice of reference dose also affects the numerical intercept, so intercepts from models using different reference doses should not be compared without re-expression.
The BLRM imposes a specific shape: linearity of log odds in log dose. It does not guarantee that this shape describes every biological dose-response relationship, and a good interpretation of \(\alpha\) or \(\beta\) does not establish good model fit. These parameters also do not, by themselves, identify an optimal dose or define a clinical recommendation.
- Mistake: calling \(\alpha\) the DLT probability. The intercept is the log odds at the reference dose. Apply the inverse logit to obtain the probability.
- Mistake: saying \(\beta\) is the probability increase per mg. The slope is a change in log odds per unit increase in log dose. Use \((d_2/d_1)^\beta\) for the odds ratio between two doses.
- Mistake: interpreting an odds multiplier as a probability multiplier. Convert the resulting odds back to probability; the probability change depends on its starting value.
- Mistake: assuming a different reference dose changes the fitted curve. The intercept changes by \(\beta\log(d_1/d_0)\); the slope and fitted probabilities remain unchanged when the reparameterization is exact.
Check Your Understanding
Answer each question, then check the solutions below.
- A model has \(\alpha=\log(0.25)\) at its reference dose. What are the fitted odds and probability at that dose?
- If \(\beta=0.8\), by what factor do the fitted odds change when dose is doubled?
- A model has reference dose 100 mg, \(\alpha=\log(0.25)\), and \(\beta=1.2\). What is the new intercept if the reference dose is changed to 200 mg?
- Why does a slope of 1.2 not mean that a fixed 1.2-unit increase in dose raises DLT probability by a fixed amount?
Solutions. (1) The odds are \(\exp\{\log(0.25)\}=0.25\), and the probability is \(0.25/(1+0.25)=0.2000\), or 20.00%. (2) The dose ratio is 2, so the odds multiplier is \(2^{0.8}\approx 1.741\). (3) The new intercept is \(\log(0.25)+1.2\log(2)\approx -0.5545\). (4) The slope describes change in log odds per unit change in log dose, not probability per mg. Probability changes nonlinearly with log odds, as shown by \(dp/d\log(d/d_{\mathrm{ref}})=\beta p(1-p)\), which depends on the current fitted probability.