Why Dependence Matters
In a multi-arm trial, disjunctive power is the probability that at least one prespecified treatment comparison meets its success rule. Pairwise powers describe the chance of success for each comparison separately. But the chance of at least one success also depends on how often those successes occur together. If two comparisons tend to succeed on the same trial replicates, their successes overlap; if they tend to succeed on different replicates, either one may provide the only success. The marginal powers alone do not record that overlap.
This tutorial focuses on that dependence. It does not require a particular test statistic or multiplicity procedure: the success rule for each comparison must simply be specified first. For example, success might mean that a prespecified adjusted test rejects its null hypothesis. Whatever the rule, calculate power under a stated design and treatment-effect scenario.
Dependence may arise because comparisons use shared data, because their test statistics are correlated, or because their success rules use the same multiplicity procedure. Marginal powers remain meaningful in those settings, but treating the events as independent can give the wrong disjunctive power.
Two Comparisons: Exact Calculation and Bounds
For two comparisons, write \(A\) and \(B\) for their success events, \(p_A=P(A)\), \(p_B=P(B)\), and \(q=P(A\cap B)\) for the probability that both succeed. The exact disjunctive power follows by counting successes in \(A\) or \(B\), while subtracting the overlap counted twice.
If \(q\) is known from a joint probability model or a joint simulation, this formula gives the disjunctive power directly. If only the two marginal powers are known, \(q\) is unknown, but it cannot take just any value. The four possible outcomes—both succeed, only \(A\) succeeds, only \(B\) succeeds, or neither succeeds—must all have nonnegative probabilities.
Each expression is at least zero. In particular, \(q\leq p_A\), \(q\leq p_B\), and \(q\geq p_A+p_B-1\); also \(q\geq0\). Combining these restrictions gives the possible range for the joint success probability.
Substituting the smallest and largest possible values of \(q\) into the union formula gives bounds that require no independence assumption.
Worked Example: Calculating Power From a Joint Probability
Suppose a hypothetical trial has two prespecified comparisons, with marginal powers \(p_A=0.70\) and \(p_B=0.60\). Under a joint model for their outcomes, the probability that both succeed is \(q=0.45\). Calculate the disjunctive power and compare it with the independence calculation.
The exact calculation uses the joint success probability: $$ P(A\cup B)=p_A+p_B-q=0.70+0.60-0.45=0.85. $$ Thus, under this joint model and scenario, the probability of at least one successful comparison is 0.85, or 85%.
If independence were assumed, the probability that neither succeeds would be \((1-p_A)(1-p_B)\). The resulting disjunctive power would be $$ 1-(1-0.70)(1-0.60)=1-(0.30)(0.40)=0.88. $$ That is 3 percentage points higher than the value calculated from the specified joint probability. The independence result is not a second estimate justified by the same information; it is the answer under an additional assumption.
The bounds from the marginals alone are \( \max(0.70,0.60)=0.70 \) and \( \min(1,0.70+0.60)=1.00 \). The computed value of 0.85 lies within these limits, as it must.
Why Marginal Powers Do Not Pin Down the Answer
The bounds are not merely formal possibilities: different joint patterns can have identical marginal powers. In the next example, each pattern has the same chance of success for comparison \(A\) and the same chance for comparison \(B\), but the overlap between their successes differs. Therefore the probability of at least one success differs as well.
Worked Example: Same Marginal Powers, Different Disjunctive Powers
Consider two hypothetical joint outcome patterns for comparisons \(A\) and \(B\). Each row gives the probability of one outcome. In both patterns, \(P(A)=0.80\) and \(P(B)=0.50\).
| Outcome | Pattern 1 | Pattern 2 |
|---|---|---|
| Both succeed | 0.30 | 0.50 |
| Only \(A\) succeeds | 0.50 | 0.30 |
| Only \(B\) succeeds | 0.20 | 0.00 |
| Neither succeeds | 0.00 | 0.20 |
Each column sums to 1.00, so each is a valid joint distribution. In Pattern 1, \(P(A)=0.30+0.50=0.80\) and \(P(B)=0.30+0.20=0.50\). Its disjunctive power is \(0.30+0.50+0.20=1.00\).
In Pattern 2, the marginal powers are again \(P(A)=0.50+0.30=0.80\) and \(P(B)=0.50+0.00=0.50\). But its disjunctive power is \(0.50+0.30+0.00=0.80\). The difference arises because every success for \(B\) in Pattern 2 occurs alongside a success for \(A\), whereas Pattern 1 includes outcomes where only \(B\) succeeds.
The independence calculation from these marginal powers would be \(1-(0.20)(0.50)=0.90\). It matches neither pattern. Without information about the joint outcomes, the marginal powers cannot select between them.
Estimation by Joint Simulation
When an analytic joint probability is difficult to obtain, a design simulation can estimate disjunctive power. Each simulated replicate must generate outcomes for all comparisons together under the same trial replicate. Counting successes separately and then combining marginal estimates as if independent discards the dependence the simulation is intended to capture.
Worked Example: Estimating Disjunctive Power From Replicates
Suppose a hypothetical simulation generates 10,000 complete trial replicates under one specified scenario. Comparison \(A\) succeeds in 7,200 replicates, comparison \(B\) succeeds in 5,800, and both succeed in 4,600. Estimate the disjunctive power.
The number of replicates with at least one success is the number with \(A\), plus the number with \(B\), minus the number with both: $$ 7{,}200+5{,}800-4{,}600=8{,}400. $$ Thus the estimated disjunctive power is \(8{,}400/10{,}000=0.84\), or 84%. The estimated marginal powers are 0.72 and 0.58, and the estimated joint success probability is 0.46. Substitution gives the same result: \(0.72+0.58-0.46=0.84\).
For a rough measure of simulation error, the Monte Carlo standard error of this estimated probability is $$ \sqrt{\frac{0.84(1-0.84)}{10{,}000}}=\sqrt{0.00001344}\approx0.00367. $$ This describes simulation variability from using a finite number of replicates; it is not uncertainty about the treatment effect in a future trial. The independence calculation would give \(1-(0.28)(0.42)=0.8824\), which differs because the simulated joint outcomes need not be independent.
Choosing a Calculation and Checking Its Assumptions
A defensible calculation begins with the success events, not with an independence formula. Each event must correspond to the prespecified decision rule being evaluated, and the scenario must state the assumed treatment effects and design conditions. If a joint probability model is used, its assumptions about the dependence of the comparison outcomes must also be appropriate. If joint simulation is used, each replicate must preserve the trial’s shared data structure and apply the success rules to all comparisons.
State exactly what counts as success for each comparison, including any adjustment or decision rule in the design.
Use a suitable joint probability model or complete simulated replicates to estimate how often successes occur together. If this information is unavailable, report bounds rather than silently assuming independence.
For two comparisons, combine the marginal and joint probabilities, then check that the result is between the larger marginal power and \(\min(1,p_A+p_B)\).
Useful diagnostics include checking that all simulated outcome frequencies sum to one, that estimated joint success is no greater than either marginal power, and that the calculated union probability lies within its bounds. For simulation, also report the number of replicates and consider whether the Monte Carlo standard error is small enough for the intended precision. A result outside the bounds indicates an arithmetic, coding, or event-definition error.
The main alternative to a joint calculation is the independence formula. It is appropriate only when the success events are independent under the specified scenario; that assumption should be justified by the design or model, not inferred from separate marginal power calculations. If dependence is unknown but the marginals are credible, the distribution-free bounds remain valid, though they may be wide. A joint model or simulation can be more informative, but it relies on correctly representing how the comparisons vary together.
A common mistake is to calculate \(1-(1-p_A)(1-p_B)\) automatically because each marginal power is available. That expression multiplies the two failure probabilities, which is valid only if the failure events are independent. The correction is to obtain or estimate the joint probability of both successes (or equivalently the joint probability of neither success) and use it in the calculation. If neither joint quantity is available, give a bound or state the independence assumption explicitly.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- Two comparisons have marginal powers 0.65 and 0.55, and their joint success probability is 0.40. What is their disjunctive power?
- For marginal powers 0.65 and 0.55, give the distribution-free lower and upper bounds on disjunctive power.
- Why is the independence calculation not justified merely because two marginal powers have been calculated separately?
- In 2,000 joint simulation replicates, comparison \(A\) succeeds 1,300 times, \(B\) succeeds 900 times, and both succeed 600 times. Estimate disjunctive power.
Solutions. (1) \(0.65+0.55-0.40=0.80\), so disjunctive power is 80%. (2) The bounds are \(\max(0.65,0.55)=0.65\) and \(\min(1,0.65+0.55)=1.00\). (3) Separate marginal probabilities do not establish independence; the success events may overlap more or less than independence predicts. (4) There are \(1{,}300+900-600=1{,}600\) replicates with at least one success, so estimated disjunctive power is \(1{,}600/2{,}000=0.80\), or 80%.