Optimal Benchmark for Evaluating Drug-Combination Dose-Finding Clinical Trials

Abstract

Numerous dose-finding methods have been proposed for drug-combination trials. A head-to-head comparison of the performance of these designs is difficult and often not very meaningful because different designs use different models and decision rules that often require judicious calibration to obtain good small-sample performance. It is desirable to have a general benchmark that can be used to evaluate the absolute performance of combination dose-finding designs. In this article, we propose an optimal nonparametric benchmark for evaluating drug-combination dose-finding methods, which provides an upper bound of accuracy beyond which further improvements are generally not achievable without making parametric assumptions of the dose-toxicity relationship. Our method is based on a new concept called critical information, which provides an upper bound on the information that we could possibly learn from patients while explicitly accounting for the partial order of the dose combinations, a fundamental feature of drug-combination trials. Our numerical study shows that the proposed benchmark provides a sharp upper bound that is useful for evaluating the performance of combination dose-finding designs.

Appendix

1.1 Proof of Proposition 2

For a given patient, let $d(c_1,c_2)$ denote the critical dose combination. For any combination d(j, k), let y(j, k) and z(j, k) denote the toxicity outcomes from the complete information and critical information with unknown outcomes imputed, respectively. By the partial order of the doses, we know that for $j\le j'$ and $k\le k'$, $y(j,k)\le y(j',k')$, denoted by $d(j,k)\preceq d(j',k')$. Here we demonstrate that $z(j,k)\le z(j',k')$ for $j\le j'$ and $k\le k'$. Note that $y(j,k)=z(j,k)$ for the critical dose and any dose that is ordered in relation to the critical dose. Based on whether $y(j,k)=z(j,k)$ and $y(j',k')=z(j',k')$, there are 3 possible situations.

• Neither is true, which implies that both d(j, k) and $d(j',k')$ have an unknown order in relation to $d(c_1,c_2)$. Since the true probabilities satisfy $\pi (j,k)\le \pi (j',k')$, $z(j,k)=\text{ Pr }(\hat{Y}(j,k)\ge \hat{Y}(c_1,c_2))\le \text{ Pr }(\hat{Y}(j',k')\ge \hat{Y}(c_1,c_2))=z(j',k')$. So the order is preserved.

• Only one of them is true, which implies one combination is ordered in relation to $d(c_1,c_2)$ and the other has an unknown order in relation to $d(c_1,c_2)$.

  • If $y(j,k)=y(j',k')=0$, then neither combination is the critical combination. Since $d(j,k)\preceq d(j',k')$, it must be true that $d(j,k)\preceq d(c_1,c_2)$ and $d(j',k')$ has an unknown order in relation to $d(c_1,c_2)$. So $z(j,k)=y(j,k)=0$ and $z(j',k')=\text{ Pr }(\hat{Y}(j',k')\ge \hat{Y}(c_1,c_2))$, which is a value between 0 and 1. So $z(j,k)\le z(j',k')$ holds.

  • If $y(j,k)=0$ and $y(j',k')=1$, then $z(j,k)\le z(j',k')$ holds, because no matter which one is replaced with the predictive probability, the value is between 0 and 1.

  • If $y(j,k)=y(j',k')=1$, then neither combination is the critical combination, because if one of them is the critical combination, then the other has an unknown order in relation to the critical dose, but here $d(j,k)\preceq d(j',k')$. In this case, $d(c_1,c_2)\preceq d(j',k')$ and d(j, k) has an unknown order in relation to $d(c_1,c_2)$. So $z(j',k')=1$ and z(j, k) is between 0 and 1. The order is satisfied.

• Both are true. The order is naturally preserved.

1.2 A Brief Sketch of the Proof for Proposition 3

For any patient with critical dose $d(c_1,c_2)$, the values of Y are identical in the complete information and in the critical information for doses in $\mathcal{H}(c_1, c_2)$ and $\mathcal{L}(c_1, c_2)$. For any other dose d(j, k) not in these two sets, let $\hat{Y}(j,k)$ and $\hat{Y}(c_1,c_2)$ be the observed toxicity rates at d(j, k) and $d(c_1,c_2)$, which follow independent binomial distributions $\text{ Binomial }(n/2,\pi (j,k))$ and $\text{ Binomial }(n/2,\pi (c_1,c_2))$, respectively, up to a constant. As the sample size n goes to infinity, we can use a normal approximation for the asymptotic distributions.

$$\begin{aligned} \hat{Y}(j,k)\cong & {} N\left( \pi (j,k),\frac{\pi (j,k)(1-\pi (j,k))}{n/2}\right) \\ \hat{Y}(c_1,c_2)\cong & {} N\left( \pi (c_1,c_2),\frac{\pi (c_1,c_2)(1-\pi (c_1,c_2))}{n/2}\right) . \end{aligned}$$

Since $\hat{Y}(j,k)$ and $\hat{Y}(c_1,c_2)$ are independent, the difference $\hat{Y}(j,k)-\hat{Y}(c_1,c_2)$ asymptotically follows a normal distribution with the mean and variance given below,

$$\begin{aligned} \hat{Y}(j,k)-\hat{Y}(c_1,c_2)\cong & {} N\Big (\pi (j,k)-\pi (c_1,c_2),\\&\times \frac{\pi (j,k)(1-\pi (j,k))+\pi (c_1,c_2)(1-\pi (c_1,c_2))}{n/2}\Big ). \end{aligned}$$

The variance $V(\hat{Y}(j,k)-\hat{Y}(c_1,c_2))$ converges to 0 when n goes to infinity. So when n is sufficiently large, $P(\hat{Y}(j,k)-\hat{Y}(c_1,c_2)\ge 0)$ converges to 1 when $\pi (j,k)>\pi (c_1,c_2)$ and converges to 0 when $\pi (j,k)<\pi (c_1,c_2)$.

In other words, $\lambda$ in equation (1) converges to 1 if d(j, k) is more toxic than the critical dose $d(c_1,c_2)$, and converges to 0 if d(j, k) is less toxic than the critical dose $d(c_1,c_2)$. As a result, the critical information converges to the complete information as the sample size n goes to infinity. Since complete information yields consistent estimators of the true toxicity probabilities, $\hat{\pi }(j, k)$ is a consistent estimator for $\pi (j, k)$.