Dependence Between Efficacy and Toxicity Predictions

Why Dependence Changes a Joint Dose Decision

A dose may have a high probability of meeting an efficacy target and a high probability of staying below a toxicity limit, yet the probability of meeting both targets depends on how efficacy and toxicity predictions vary together. If favorable efficacy predictions tend to occur alongside unfavorable toxicity predictions, the joint success probability can be lower than an independence calculation suggests. If favorable efficacy tends to accompany acceptable toxicity, the joint probability can be higher.

The relevant dependence is defined for a particular dose, target definition, and uncertainty distribution. It can arise because efficacy and toxicity share uncertain parameters, because the same simulated patient characteristics affect both predictions, or because the model links the endpoints. For model-informed decisions, the key calculation is usually based on paired draws: each draw preserves the efficacy and toxicity predictions that belong together under the same simulated uncertainty realization.

Definition: At dose $d$, let $E_d$ denote the efficacy prediction, where larger values are better, and let $T_d$ denote the toxicity burden prediction, where smaller values are better. For efficacy threshold $e^*$ and toxicity limit $t^*$, the joint target-attainment event is $E_d \ge e^*$ and $T_d \le t^*$. Its probability is $$ P(E_d \ge e^*,\,T_d \le t^*). $$ The marginal attainment probabilities are $P(E_d \ge e^*)$ and $P(T_d \le t^*)$. Their product equals the joint probability only when the two threshold events are independent.

These quantities must be tied to a clearly stated uncertainty distribution. For example, a probability may describe uncertainty in model parameters conditional on the available data, uncertainty in a future patient's covariates, or a predictive distribution that also includes outcome variability. The simulation must represent whichever uncertainty the decision is intended to address. Parameter uncertainty alone does not automatically include residual or individual-level outcome variability.

Earlier material on joint efficacy and safety under parameter uncertainty introduced the joint decision perspective. The specific issue here is dependence: marginal probabilities alone do not identify the joint probability, and a correlation coefficient alone will not generally identify it either.

Joint Probability, Bounds, and Independence

Let $A$ be the event that efficacy meets its threshold and $B$ the event that toxicity stays within its limit. Then the joint probability can be expressed as

P(A \cap B)=P(A)P(B)+\operatorname{Cov}(I_A,I_B),

where $I_A$ and $I_B$ are indicator variables equal to 1 when their events occur and 0 otherwise. This follows directly from $E[I_A I_B]=P(A\cap B)$, $E[I_A]=P(A)$, and the definition of covariance. Thus, it is the dependence between the attainment events that determines the difference from the independence product.

Correlation between the continuous predictions $E_d$ and $T_d$ can help describe dependence, but it is not itself the covariance of the two attainment indicators. Thresholds, distribution shapes, and nonlinear relationships matter. In particular, a Pearson correlation between predictions does not by itself determine the joint target probability unless a sufficiently specific joint distributional model is also assumed.

Even without specifying dependence, the marginals imply useful bounds:

\max\{0,\,P(A)+P(B)-1\} \le P(A \cap B) \le \min\{P(A),P(B)\}.

The upper bound holds because $A\cap B$ is contained in each event. For the lower bound, $P(A\cup B)\le 1$, while $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Rearranging gives $P(A\cap B)\ge P(A)+P(B)-1$, and probabilities are also at least zero. These bounds can be wide; multiplying the marginals selects one particular value within them, justified only under independence.

Worked Example: Same Marginals, Different Joint Success

Suppose 10 hypothetical paired uncertainty draws at one dose produce the following counts. “Efficacy met” means $E_d\ge e^*$, and “safe” means $T_d\le t^*$.

	Safe	Not safe	Total
Efficacy met	5	1	6
Efficacy not met	1	3	4
Total	6	4	10

The marginal probabilities are $P(A)=6/10=0.60$ and $P(B)=6/10=0.60$. The joint simulation estimate is the proportion of draws in the upper-left cell:

\widehat{P(A\cap B)}=\frac{5}{10}=0.50.

An independence calculation would instead give $0.60\times0.60=0.36$, underestimating the paired estimate by $0.14$. To see the dependence directly, the empirical covariance of the two indicators is $0.50-(0.60)(0.60)=0.14$. The empirical correlation is $0.14/[0.60(1-0.60)]=0.5833$, because both indicators have variance $0.24$.

The same marginals could have a different pairing: 2 draws could meet both targets, 4 could meet efficacy only, 4 could be safe only, and none could meet neither. The marginals would still be $0.60$ each, but the joint estimate would be $2/10=0.20$. The product remains $0.36$, but neither paired result equals it. It is the pairing—not the marginal counts—that carries the joint information.

When a Distributional Assumption Makes Correlation Informative

A joint distribution can make correlation more directly useful. For illustration, suppose standardized efficacy $U$ and toxicity $V$ follow a bivariate normal distribution with zero means, unit variances, and correlation $\rho$. Suppose also that both thresholds are at their respective means, so success requires $U\ge0$ and $V\le0$. Under these specific assumptions,

P(U\ge0,\,V\le0)=\frac14-\frac{\arcsin(\rho)}{2\pi}.

A short geometric justification makes the direction clear. Write $U=A$ and $V=\rho A+\sqrt{1-\rho^2}B$, where $A$ and $B$ are independent standard normal variables. The pair $(A,B)$ has a rotationally symmetric density, so its angle is uniform around the origin. The conditions $A\ge0$ and $\rho A+\sqrt{1-\rho^2}B\le0$ select a wedge of angle $\pi/2-\arcsin(\rho)$. Dividing that angle by $2\pi$ gives the stated probability. The formula is not a general correlation-to-joint-probability conversion; it relies on the centered bivariate normal assumptions and these particular thresholds.

Worked Example: Correlation Can Help or Hurt Joint Attainment

Consider hypothetical standardized predictions with the bivariate normal assumptions above. With $\rho=0$, independence gives $P(U\ge0,V\le0)=0.25$.

If efficacy and toxicity are negatively correlated with $\rho=-0.60$, substitute into the formula:

\frac14-\frac{\arcsin(-0.60)}{2\pi} =0.25+\frac{0.6435}{6.2832} \approx0.3524.

Here high efficacy predictions tend to accompany lower toxicity predictions, so the chance of meeting both centered thresholds is higher than under independence. If instead $\rho=+0.60$, then

\frac14-\frac{\arcsin(0.60)}{2\pi} =0.25-\frac{0.6435}{6.2832} \approx0.1476.

In this case, high efficacy tends to accompany high toxicity, and joint attainment is less frequent than under independence. These are probabilities under the stated hypothetical distribution, not universal consequences of the correlation values. With other thresholds or non-normal distributions, the joint probabilities can differ.

Estimating Joint Attainment by Paired Simulation

In a model-based analysis, generate $M$ draws from the uncertainty distribution relevant to the decision. For each draw $m$, calculate both predictions at the same dose using that draw's parameter values and any jointly simulated covariates or other uncertain inputs. Define $I_{A,m}=1$ if efficacy meets its threshold and 0 otherwise; define $I_{B,m}$ similarly for safety. The joint estimate is

\widehat{P(A\cap B)} =\frac{1}{M}\sum_{m=1}^{M}I_{A,m}I_{B,m}.

This is the Monte Carlo estimate of the probability under the simulated distribution. Its approximate Monte Carlo standard error, when draws are independent, is $\sqrt{\hat q(1-\hat q)/M}$, where $\hat q$ is the estimated joint probability. This quantifies simulation error, not uncertainty about whether the model or its assumptions are correct. With correlated simulation output, such as a Markov chain, the independent-draw formula can understate simulation error; effective sample size or a suitable chain-based calculation is needed.

The independence comparison is $\hat p_E\hat p_S$, where each marginal is estimated from the same simulation output. This comparison asks what the joint probability would be if the attainment events were independent while retaining their estimated marginal probabilities. It is useful as a diagnostic benchmark, but should not replace the paired estimate unless independence is justified.

Specify the estimand.
State the dose, efficacy and toxicity quantities, thresholds, and which uncertainties the probability includes.

Generate coherent paired draws.
For each simulation draw, use one jointly generated parameter and covariate realization to calculate both predictions.

Record both threshold events.
For every draw, classify efficacy attainment and safety attainment separately, and record whether both occurred.

Compare and check.
Report the joint fraction, the two marginal fractions, and their product as an independence benchmark. Check pairing and Monte Carlo precision before drawing a decision conclusion.

Worked Example: Dependence Changes a Chance-Constrained Choice

Suppose three hypothetical candidate doses were each evaluated using $M=1{,}000$ paired draws. A decision rule requires at least 0.75 probability of meeting both targets; among doses satisfying that constraint, the decision maker prefers the one with the highest efficacy-attainment probability. The simulated counts are:

Dose	Efficacy met	Safe	Both	Paired joint estimate	Marginal product
A	900	900	820	0.820	0.810
B	950	850	720	0.720	0.8075
C	800	950	770	0.770	0.760

For dose B, for example, the paired estimate is $720/1{,}000=0.720$. The marginal probabilities are $950/1{,}000=0.950$ and $850/1{,}000=0.850$, whose product is $0.950\times0.850=0.8075$. Under the paired simulation, B does not meet the 0.75 constraint. If the product were mistakenly treated as the joint probability, B would appear to meet it and would be preferred for its higher efficacy-attainment probability.

The paired estimates show that A and C satisfy the stated constraint, with C's estimate $0.770$ and A's $0.820$. Under the stated preference rule, A is selected because it has the higher efficacy-attainment probability among the doses that pass. This example illustrates why marginal products can change a decision, not just a reported probability. For B, the approximate Monte Carlo standard error of the joint estimate is $\sqrt{0.720(1-0.720)/1{,}000}\approx0.0142$; its proximity to the decision threshold should therefore prompt consideration of simulation precision and decision uncertainty rather than false precision.

Diagnostics, Alternatives, and Limitations

A first diagnostic is to verify that efficacy and toxicity predictions remain paired throughout the computation. Sorting one vector independently, simulating the endpoints separately, or joining output files by the wrong draw identifier can destroy the dependence while leaving marginal summaries unchanged. Compare the sample correlation of continuous predictions and the association of the two attainment indicators, but do not treat either summary as a substitute for checking the paired joint fraction.

Also inspect whether the specified joint uncertainty model is appropriate. Parameter draws should preserve estimated parameter dependence; covariates shared by both predictions should be sampled consistently; and endpoint-specific residual variability should be included only if it belongs to the target probability. Very few joint successes, a probability near a decision boundary, or sensitivity to plausible parameter or covariate assumptions signals that the estimate may be unstable or model-dependent. Increasing $M$ reduces Monte Carlo error but does not repair misspecified dependence, biased parameter estimates, or unrepresented uncertainty.

The main alternatives have different roles:

Independence product: $\hat p_E\hat p_S$ is simple and appropriate when the attainment events are independent under the stated distribution. Otherwise, it is a benchmark, not the joint estimate.
Marginal bounds: the lower and upper bounds show what can be concluded from marginal probabilities alone when dependence is unspecified. They may be too broad to support a dose decision.
Parametric joint distribution: a joint model can summarize dependence and sometimes yield analytic probabilities. Its result depends on the distributional form, thresholds, and tail behavior being credible.
Paired joint simulation: this directly estimates the target probability under a specified model and uncertainty distribution, but inherits their limitations and requires correct pairing and adequate simulation.

Correlation is especially limited as a stand-alone decision summary. It measures linear association between continuous predictions, not the probability of crossing two thresholds together. Nonlinear dependence, skewness, tail behavior, and threshold placement can matter even if the correlation is modest. Conversely, two endpoints can have a noticeable correlation while their threshold events show a different degree of association. Report the joint probability and the marginal probabilities in context, and describe the uncertainty distribution that generated them.

Common Mistake

Common pitfall: “Efficacy attainment is 90% and safety attainment is 90%, so the probability of both is $0.90\times0.90=0.81$.” This multiplication assumes independence of the attainment events. If both predictions come from paired model draws, estimate the joint probability by counting draws meeting both thresholds. Use the product only when independence is justified, and remember that a correlation value alone does not generally determine the joint probability.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

At one dose, $P(A)=0.70$ and $P(B)=0.80$. What range of joint probabilities is possible without a dependence assumption? Is the independence product guaranteed to lie in that range?
In 500 paired draws, 340 meet the efficacy threshold, 400 meet the safety threshold, and 300 meet both. Calculate the joint estimate and the independence benchmark.
Under the centered bivariate normal example, what is the joint probability for $\rho=0$, and why does the formula not apply automatically to arbitrary thresholds?
Give one reason that independently simulating efficacy and toxicity predictions can give an incorrect joint attainment estimate even when both marginal estimates are accurate.

Solutions. (1) The lower bound is $\max(0,0.70+0.80-1)=0.50$, and the upper bound is $\min(0.70,0.80)=0.70$. The independence product is $0.70\times0.80=0.56$, which is within the bounds but is not guaranteed to be the true joint probability. (2) The paired joint estimate is $300/500=0.600$. The marginal estimates are $340/500=0.680$ and $400/500=0.800$, so the independence benchmark is $0.680\times0.800=0.544$. (3) At $\rho=0$, the predictions are independent under the assumed bivariate normal model, giving $0.25$ for these centered thresholds. The formula was derived for centered thresholds and a bivariate normal distribution; moving either threshold changes the relevant probability. (4) Separate simulation can erase shared parameter or covariate dependence. It may preserve each marginal distribution while changing which efficacy and toxicity values occur together, and therefore change the joint threshold probability.