Why Joint Success Is a Separate Question
A trial with co-primary endpoints is successful only if every prespecified endpoint criterion is met. For example, a treatment might need to produce both a sufficiently large reduction in symptom severity and a sufficiently large improvement in physical function. Knowing the posterior probability of meeting each criterion separately does not, by itself, give the probability of meeting both.
The missing ingredient is the joint posterior distribution of the endpoint effects. If favorable effects on the endpoints tend to occur together in posterior draws, the chance of meeting both criteria may be relatively high. If one endpoint tends to be favorable when the other is not, the joint chance may be lower. The calculation must therefore preserve the dependence represented by the model.
The direction and scale of each effect must be stated carefully. If larger values are favorable, a criterion might be \(\theta_j\geq \delta_j\), where \(\delta_j\) is a prespecified clinically meaningful threshold. If smaller values are favorable, the effect can be defined with its sign reversed or the success set can be written in the appropriate direction. Criteria may also be interval-based. The joint event is the intersection of the endpoint-specific success events, not a new average of the endpoint effects.
The probability \(P(A\mid D)\) describes uncertainty about the joint event conditional on the observed data, model, and prior. It is not the probability that the model is true, nor does it establish that treatment caused an effect beyond the assumptions of the trial and analysis. Its clinical interpretation depends on the endpoints, scales, thresholds, and model being appropriate.
Calculating the Probability From Joint Posterior Draws
Suppose posterior computation supplies \(M\) draws \((\theta_1^{(m)},\ldots,\theta_J^{(m)})\), each retaining the endpoint effects from the same draw \(m\). For each draw, check whether every criterion is satisfied. The fraction of draws satisfying all criteria estimates the joint posterior probability.
Here \(\mathbf{1}\{\cdot\}\) equals 1 when the condition inside the braces is true and 0 otherwise. This estimator follows directly by writing the posterior probability as the posterior expectation of an indicator:
The draws must be joint draws. Pairing \(\theta_1^{(m)}\) with \(\theta_2^{(m)}\) preserves their posterior association. Independently shuffling one endpoint’s draws, or sampling each endpoint separately and pairing arbitrary values, can destroy that association and produce the wrong joint probability. For correlated MCMC output, the estimate is still the indicator average; its Monte Carlo precision depends on the effective information in the indicator sequence, not just the raw number of saved iterations.
Worked Example: Counting Draws That Meet Both Criteria
Hypothetical example. Consider a trial with two co-primary endpoints. Let \(\theta_1\) be the reduction in symptom score, in points, and \(\theta_2\) the improvement in a function score, also in points. The prespecified criteria are \(\theta_1\geq 3\) points and \(\theta_2\geq 4\) points. The following ten paired values are selected posterior draws, shown only to illustrate the calculation.
| Draw | \(\theta_1\) (symptom points) | \(\theta_2\) (function points) | Both criteria met? |
|---|---|---|---|
| 1 | 4.2 | 5.1 | Yes |
| 2 | 3.5 | 3.8 | No |
| 3 | 2.8 | 6.0 | No |
| 4 | 5.0 | 4.4 | Yes |
| 5 | 3.1 | 4.2 | Yes |
| 6 | 4.0 | 2.9 | No |
| 7 | 3.8 | 4.7 | Yes |
| 8 | 2.5 | 3.2 | No |
| 9 | 4.6 | 4.0 | Yes |
| 10 | 3.0 | 4.5 | Yes |
There are six draws in which both values meet their thresholds. Therefore,
In these same draws, eight of ten symptom effects meet the first criterion and seven of ten function effects meet the second. The estimated marginal probabilities are therefore 0.80 and 0.70. Their product is \(0.80\times0.70=0.56\), not the joint estimate of 0.60. With only ten illustrative draws, this difference should not be treated as precise evidence of a particular dependence pattern; it shows why the joint event should be counted directly.
What Dependence Changes
For two endpoints, let \(A_1\) and \(A_2\) denote their respective success events. The probability rule \(P(A_1\cap A_2\mid D)=P(A_1\mid D)P(A_2\mid A_1,D)\) shows exactly when multiplying marginal probabilities works: it works if the success events are posterior independent, so that \(P(A_2\mid A_1,D)=P(A_2\mid D)\). Independence of the endpoint effects is sufficient for this condition, but independence should not be assumed just because the endpoints are analyzed separately.
With marginal success probabilities \(p_1=P(A_1\mid D)\) and \(p_2=P(A_2\mid D)\), the joint probability must satisfy \(\max(0,p_1+p_2-1)\leq P(A_1\cap A_2\mid D)\leq\min(p_1,p_2)\). The upper bound holds because meeting both criteria implies meeting each one. For the lower bound, the probability of the union cannot exceed 1, while \(P(A_1\cup A_2\mid D)=p_1+p_2-P(A_1\cap A_2\mid D)\). Rearranging gives the stated lower bound, along with the fact that probabilities cannot be negative. These bounds are useful checks, but they do not determine the joint probability: the joint posterior distribution supplies the additional information.
Worked Example: Same Marginals, Different Joint Probabilities
Hypothetical example. Suppose 1,000 paired posterior draws give a marginal success rate of 0.80 for endpoint 1 and 0.70 for endpoint 2. In one possible joint draw pattern, 650 draws meet both criteria, 150 meet only endpoint 1, 50 meet only endpoint 2, and 150 meet neither. The marginal counts are \(650+150=800\) for endpoint 1 and \(650+50=700\) for endpoint 2. Thus the joint estimate is \(650/1000=0.65\).
For comparison, if the two success events were posterior independent, the joint probability would be
A different dependence pattern can give a joint probability of 0.50 while keeping the same marginals: 500 draws meet both, 300 meet only endpoint 1, 200 meet only endpoint 2, and none meet neither. The endpoint-specific probabilities remain 0.80 and 0.70, but now \(P(A_1\cap A_2\mid D)=500/1000=0.50\). Both examples respect the bounds \(0.50\leq P(A_1\cap A_2\mid D)\leq0.70\). Marginal probabilities alone cannot distinguish these patterns.
A Normal-Posterior Illustration
Dependence is often summarized by posterior correlation between continuous endpoint effects, but correlation is not itself the probability of joint success. The effect of correlation also depends on where the success thresholds lie relative to each marginal posterior distribution. A simple special case makes the relationship visible.
Worked Example: Joint Success at Centered Thresholds
Hypothetical example. Suppose the standardized posterior effects \(X\) and \(Y\) have a bivariate normal distribution with mean zero, unit marginal variances, and correlation \(\rho=0.60\). Let success require \(X\geq0\) and \(Y\geq0\). Each marginal success probability is 0.50.
For this centered bivariate normal case, the joint probability is \(P(X\geq0,Y\geq0)=\frac14+\frac{\arcsin(\rho)}{2\pi}\). One justification is to differentiate the bivariate-normal quadrant probability with respect to \(\rho\), obtaining \(1/(2\pi\sqrt{1-\rho^2})\), and then integrate from \(\rho=0\), where symmetry and independence give probability \(1/4\).
Under posterior independence, the result would instead be \(0.50\times0.50=0.25\). In this centered example, positive correlation increases the probability that both standardized effects are on the favorable side of their thresholds. If \(\rho=-0.60\), the same formula gives approximately \(0.1476\). These are probabilities for the specified centered, standardized normal case; they are not general formulas for arbitrary thresholds or posterior distributions.
Assumptions, Diagnostics, and Common Pitfalls
The calculation is only as meaningful as the event and posterior distribution used to compute it. Before interpreting a joint probability, confirm that each endpoint, effect direction, scale, threshold, and combination rule was specified appropriately for the clinical question. Confirm also that the posterior model represents the endpoint data jointly, including relevant within-patient association when patients contribute measurements to multiple endpoints. A model that imposes independence may be unsuitable if the endpoint data support meaningful dependence.
Practical diagnostics follow from the calculation. Check that each paired draw uses effects from the same posterior iteration and that every endpoint criterion is applied with the correct inequality and units. Compare the joint estimate with the marginal probabilities and the bounds above; a result outside those bounds signals a calculation or coding error. If MCMC is used, inspect sampling and mixing for the joint quantities and for the indicator of joint success. Few effective draws near the success boundary can make the estimated probability imprecise even when each marginal probability appears stable.
The estimated fraction is subject to Monte Carlo error. For \(M\) independent posterior draws and estimated probability \(\hat{q}\), its approximate Monte Carlo standard error is \(\sqrt{\hat{q}(1-\hat{q})/M}\). MCMC draws are generally dependent, so substituting the raw number of iterations for \(M\) can understate this error; precision should reflect the effective sample size of the indicator sequence.
- Wrong: “The marginal probabilities are 0.80 and 0.70, so the probability of both is 0.56.” Correction: That product requires posterior independence of the success events. Otherwise, estimate the joint probability from paired draws or the specified joint posterior.
- Wrong: “The endpoints are positively correlated, so the joint success probability must be high.” Correction: Correlation alone is not enough; the marginal locations and thresholds matter too. Calculate the probability of the actual joint event.
- Wrong: “I can pair any draw for endpoint 1 with any draw for endpoint 2.” Correction: Arbitrary pairing removes the dependence in the joint posterior and can change the answer.
A marginal-probability calculation answers how likely each criterion is to be met separately. A product calculation adds an independence assumption. A joint posterior calculation answers the all-criteria question while retaining the dependence in the model. This rule concerns the conjunction of co-primary criteria; it is distinct from a rule in which success requires only one of several endpoint criteria.
The result remains conditional on the chosen model, prior, data, and thresholds. It does not remove the need to justify clinical thresholds, assess model adequacy, or consider whether the chosen endpoints capture the intended benefit. When those inputs are uncertain or poorly specified, a precise-looking joint probability can still be misleading.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Two endpoint success events have posterior probabilities 0.75 and 0.60. What joint-probability bounds must hold? What is the joint probability under posterior independence?
- Out of 500 paired posterior draws, 210 satisfy both endpoint criteria. Estimate the joint success probability.
- A student independently shuffles the posterior draws for endpoint 2 before counting how often both endpoints meet their criteria. What is the problem with this procedure?
- Two standardized effects have a centered bivariate normal posterior with correlation \(-0.40\), and success requires both to be at least zero. What is the joint probability, approximately?
Solutions. 1. The bounds are \(\max(0,0.75+0.60-1)=0.35\) and \(\min(0.75,0.60)=0.60\). Under posterior independence, the joint probability is \(0.75\times0.60=0.45\), which lies within the bounds. 2. The estimate is \(210/500=0.420\), or 42.0%. 3. Shuffling breaks the pairing and therefore removes the dependence represented by the joint posterior draws. The resulting count need not estimate the original joint probability; it corresponds to an artificial pairing. 4. Using the centered bivariate-normal formula gives \(\frac14+\frac{\arcsin(-0.40)}{2\pi}\approx0.1845\). The negative correlation lowers the probability below the independent value of 0.25 in this centered case.