Deriving the BLRM Overdose Boundary

From a Toxicity Limit to a Parameter Boundary

A BLRM overdose boundary answers a pointwise question: for a given candidate dose, which combinations of model parameters imply a dose-limiting toxicity (DLT) probability above the specified upper toxicity limit? The answer is a line in intercept-slope space. It is a property of the model and the chosen dose and limit; by itself, it is not a probability statement about uncertain parameters.

The earlier material on Defining Overdose in a BLRM establishes the toxicity limit used to classify a dose. Here, the task is to turn that classification into an explicit parameter-space boundary. We use the BLRM log-odds parameterization and a positive dose slope, as in the earlier material on the Two-Parameter Bayesian Logistic Regression Model and Why a Positive Dose Slope Enforces Monotonicity.

Definition: Let $d>0$ be a candidate dose and $d_{\mathrm{ref}}>0$ a reference dose. Define the scaled dose $x(d)=\log(d/d_{\mathrm{ref}})$. In the two-parameter BLRM, let $\alpha>0$ denote the odds of a DLT at the reference dose and let $\beta>0$ be the dose slope. Then $$ \operatorname{logit} p(d)=\log\!\left(\frac{p(d)}{1-p(d)}\right)=\log(\alpha)+\beta x(d), $$ where $p(d)$ is the modelled probability of a DLT at dose $d$. Write $a=\log(\alpha)$ for the intercept in log-odds space. For an upper toxicity limit $\phi_U$, with $0<\phi_U<1$, the dose is an overdose at a parameter point $(a,\beta)$ when $p(d)>\phi_U$.

The ratio $d/d_{\mathrm{ref}}$ is dimensionless, so changing dose units consistently does not alter $x(d)$. The reference dose has $x(d_{\mathrm{ref}})=0$, and therefore $p(d_{\mathrm{ref}})=\alpha/(1+\alpha)$. Because $\alpha$ is an odds, it is not itself a probability. The intercept coordinate for the boundary is $a=\log(\alpha)$, not $\alpha$.

Deriving the Boundary

The logistic function is strictly increasing: if one log-odds value is greater than another, its corresponding probability is also greater. Therefore the overdose condition can be transformed by applying the logit to both sides. Define $\operatorname{logit}(\phi_U)=\log\{\phi_U/(1-\phi_U)\}$. Then

\begin{aligned} p(d)>\phi_U &\iff \operatorname{logit}p(d)>\operatorname{logit}(\phi_U)\\ &\iff a+\beta x(d)>\operatorname{logit}(\phi_U)\\ &\iff a>\operatorname{logit}(\phi_U)-\beta x(d). \end{aligned}

Equality gives the boundary itself, where the modelled toxicity probability is exactly the upper limit. Under the strict definition $p(d)>\phi_U$, equality is not classified as overdose. For a fixed candidate dose, the line is

a_{\mathrm{boundary}}(\beta;d)=\operatorname{logit}(\phi_U)-\beta x(d).

The overdose region is above this line in $(a,\beta)$ coordinates: points with larger $a$ have larger log odds, and thus larger toxicity probabilities, at the fixed dose. In the original $(\alpha,\beta)$ parameterization, the same boundary is

\alpha_{\mathrm{boundary}}(\beta;d) =\exp\!\left\{\operatorname{logit}(\phi_U)-\beta x(d)\right\}, \qquad p(d)>\phi_U \iff \alpha>\alpha_{\mathrm{boundary}}(\beta;d).

In log-intercept-slope space, the line’s slope with respect to $\beta$ is $-x(d)$. A dose above the reference dose has $x(d)>0$, so the boundary slopes downward as $\beta$ increases. A dose below the reference dose has $x(d)<0$, so it slopes upward. At the reference dose, $x(d)=0$, and the boundary does not depend on $\beta$.

Worked Example: At the Reference Dose

Suppose the reference dose is $10$ mg, the candidate dose is also $10$ mg, and the upper toxicity limit is $\phi_U=0.30$. Find the boundary and classify a parameter point with $\alpha=0.50$.

Because the candidate equals the reference dose, $x(d)=\log(10/10)=0$. The boundary in log-intercept space is

a_{\mathrm{boundary}} =\operatorname{logit}(0.30)-\beta(0) =\log\!\left(\frac{0.30}{0.70}\right) =-0.8473.

Equivalently, the boundary in $\alpha$ is $\exp(-0.8473)=0.4286$, the odds corresponding to a probability of $0.30$. The selected parameter point has $a=\log(0.50)=-0.6931$, which is greater than $-0.8473$. Directly, its reference-dose toxicity probability is

p(10)=\frac{0.50}{1+0.50}=0.3333.

Thus $p(10)=0.3333>0.30$, so this point is in the overdose region. The result does not depend on the value of $\beta$, since the scaled dose is zero.

Worked Example: A Dose Above the Reference Dose

Let $d_{\mathrm{ref}}=10$ mg, $d=20$ mg, and $\phi_U=0.25$. Derive the boundary, then classify the two parameter points $(\alpha,\beta)=(0.20,1.20)$ and $(0.10,1.20)$.

The scaled dose is $x(d)=\log(20/10)=\log(2)=0.6931$. The limit’s logit is $\operatorname{logit}(0.25)=\log(0.25/0.75)=-1.0986$. Hence

a_{\mathrm{boundary}}=-1.0986-(0.6931)\beta.

At $\beta=1.20$, this is $a_{\mathrm{boundary}}=-1.0986-0.8318=-1.9304$, or $\alpha_{\mathrm{boundary}}=\exp(-1.9304)=0.1451$. For $\alpha=0.20$, $a=\log(0.20)=-1.6094$, which is above the boundary. Checking the probability directly gives

\operatorname{logit}p(20) =-1.6094+(1.20)(0.6931) =-0.7777, \qquad p(20)=\frac{1}{1+\exp(0.7777)}=0.3148.

Since $0.3148>0.25$, this point implies overdose at $20$ mg. For $\alpha=0.10$, $a=\log(0.10)=-2.3026$, below the boundary. Its log odds are $-2.3026+0.8318=-1.4708$, giving $p(20)=1/\{1+\exp(1.4708)\}=0.1868$. This point is below the limit.

Worked Example: A Dose Below the Reference Dose

Let $d_{\mathrm{ref}}=10$ mg, $d=5$ mg, and $\phi_U=0.30$. Find the boundary at $\beta=1.50$, then check $\alpha=1.50$ and $\alpha=1.00$.

Here $x(d)=\log(5/10)=-0.6931$, so

a_{\mathrm{boundary}} =-0.8473-\beta(-0.6931) =-0.8473+0.6931\beta.

At $\beta=1.50$, the boundary is $-0.8473+1.0397=0.1924$, corresponding to $\alpha_{\mathrm{boundary}}=\exp(0.1924)=1.212$. For $\alpha=1.50$, $a=\log(1.50)=0.4055$, above the boundary. The model gives log odds approximately $-0.6343$, and therefore

p(5)=\frac{1}{1+\exp(0.6343)}=0.3465>0.30.

This parameter point is in the overdose region. For $\alpha=1.00$, $a=0$, below the boundary. The log odds are $0+(1.50)(-0.6931)=-1.0397$, giving $p(5)=1/\{1+\exp(1.0397)\}=0.2612$, which is below the limit. The upward slope of the boundary as $\beta$ increases is consistent with this dose being below the reference: a steeper positive dose slope lowers toxicity at the lower dose when the reference-dose intercept is held fixed.

What the Boundary Does—and Does Not—Say

The boundary is useful because it converts a probability threshold at a dose into a simple comparison of model parameters. It also makes the role of dose scaling visible: the same upper limit gives different lines at different candidate doses, while the line at the reference dose depends only on the intercept. The boundary can be evaluated for each candidate dose without changing the model parameterization.

The derivation assumes the specified logistic model, positive doses, a fixed reference dose, and an upper limit strictly between zero and one. With $\beta>0$, toxicity probability increases with dose, as required by the monotone BLRM specification. The algebraic boundary formula still describes the condition at a particular dose if a slope is allowed to be negative, but such a slope no longer enforces increasing toxicity with dose. Thus the formula alone does not validate the model’s monotonicity assumption.

A parameter-space boundary also does not state how much posterior support lies on either side. It classifies each fixed parameter pair. Quantifying posterior probability of overdose requires combining this classification with the posterior distribution of the parameters; that is a different calculation from deriving the boundary.

Approach	Question answered	What it cannot establish alone
Parameter-space boundary	For a fixed $(a,\beta)$, is $p(d)$ above the specified limit?	How probable that parameter region is under a posterior distribution
Direct probability calculation	For a fixed parameter pair, what value does the model assign to $p(d)$?	Whether other parameter pairs in a posterior distribution also exceed the limit
Posterior overdose probability	How much posterior probability corresponds to $p(d)>\phi_U$?	It is not obtained merely by checking one point estimate or drawing the boundary

The boundary’s straight-line form relies on plotting $a=\log(\alpha)$ against $\beta$. In $(\alpha,\beta)$ coordinates it is exponential, not a straight line. A common error is to plot $\alpha$ as though it were the log-odds intercept, or to use $\alpha$ itself in $a+\beta x(d)$. The correction is to calculate $a=\log(\alpha)$ before using the linear log-odds equation. Another frequent error is to assign the same direction to the boundary at every dose: its slope is $-x(d)$, so it changes sign around the reference dose.

Scale the candidate dose.
Compute $x(d)=\log(d/d_{\mathrm{ref}})$; verify that a dose above the reference gives a positive value and one below gives a negative value.

Transform the limit.
Compute $\operatorname{logit}(\phi_U)=\log\{\phi_U/(1-\phi_U)\}$, keeping the chosen limit and its complement in the correct order.

Write and check the boundary.
Use $a_{\mathrm{boundary}}=\operatorname{logit}(\phi_U)-\beta x(d)$. The overdose side has $a>a_{\mathrm{boundary}}$; substitute a point into the model if the direction is uncertain.

Key takeaway: At a fixed candidate dose, the BLRM overdose boundary is $a+\beta x(d)=\operatorname{logit}(\phi_U)$, with overdose strictly above it in intercept-slope space. Its orientation is determined by the signed scaled dose $x(d)$.

Check Your Understanding

Answer each question using the stated strict overdose definition. Short solutions are included for checking your work.

With $d_{\mathrm{ref}}=10$ mg, $d=40$ mg, $\phi_U=0.20$, and $\beta=0.50$, calculate the boundary value of $a$. Is a point on the boundary an overdose?
For a candidate dose below the reference dose, what is the sign of the boundary’s slope when $a$ is plotted against $\beta$? Explain using the boundary equation.
At the reference dose, why does changing $\beta$ leave the boundary unchanged?
A candidate dose has $x(d)=0.50$, the limit is $0.25$, and $\beta=1$. For $\alpha=0.10$, determine whether the parameter point is in the overdose region.
Does a boundary plot alone give the posterior probability of overdose? State what it does provide.

Solutions. (1) $x(d)=\log(4)=1.3863$ and $\operatorname{logit}(0.20)=-1.3863$, so $a_{\mathrm{boundary}}=-1.3863-(0.50)(1.3863)=-2.0794$. A point exactly on the boundary has $p(d)=0.20$, so it is not an overdose under the strict definition. (2) Since $x(d)<0$, the slope $-x(d)$ is positive. (3) At the reference dose, $x(d)=0$, making $a_{\mathrm{boundary}}=\operatorname{logit}(\phi_U)$, independent of $\beta$. (4) The boundary is $-1.0986-(1)(0.50)=-1.5986$. Since $a=\log(0.10)=-2.3026$ is below it, this point is not an overdose. (5) No. The boundary classifies fixed parameter values; posterior overdose probability additionally requires the posterior distribution of those values.

Approach	Question answered	What it cannot establish alone
Parameter-space boundary	For a fixed \((a,\beta)\), is \(p(d)\) above the specified limit?	How probable that parameter region is under a posterior distribution
Direct probability calculation	For a fixed parameter pair, what value does the model assign to \(p(d)\)?	Whether other parameter pairs in a posterior distribution also exceed the limit
Posterior overdose probability	How much posterior probability corresponds to \(p(d)>\phi_U\)?	It is not obtained merely by checking one point estimate or drawing the boundary

Deriving the BLRM Overdose Boundary

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