Why Marginal Power Is Not a Joint Probability
A multi-arm trial can have a well-defined power for each treatment comparison and still have no uniquely determined probability of overall success. The reason is that “at least one comparison succeeds” and “every comparison succeeds” concern joint events. Their probabilities depend not only on how often each comparison succeeds individually, but also on how successes co-occur.
Fix a treatment-effect configuration and the trial’s testing procedure, including its success criterion. Let \(A_i\) be the event that comparison \(i\) meets that criterion, and let \(p_i=P(A_i)\) be its marginal power under this configuration. The disjunctive success event is \(\bigcup_{i=1}^m A_i\), while the conjunctive success event is \(\bigcap_{i=1}^m A_i\). These definitions are the same for any dependence structure; what can change is the probability assigned to those events.
The probabilities are understood under the same specified trial design, analysis rule, and treatment-effect configuration. If those change, the events or their probabilities may change as well. In what follows, hold them fixed and vary only the dependence among the success events. This isolates the issue: identical marginal powers can coexist with different joint powers.
Two Comparisons: The Missing Joint Probability
For two comparisons, write \(q=P(A_1\cap A_2)\). Inclusion–exclusion gives the disjunctive power, and the definition gives the conjunctive power:
Thus, once the two marginal powers are known, both joint criteria still depend on \(q\). The event probabilities must be feasible: the overlap cannot exceed either marginal probability, and the union cannot have probability greater than 1. These facts imply the sharp bounds
The lower bound follows because \(P(A_1\cup A_2)=p_1+p_2-q\leq1\), and because a probability cannot be negative. The upper bound follows from \(A_1\cap A_2\subseteq A_1\) and \(A_1\cap A_2\subseteq A_2\). Each endpoint can be attained by a valid joint distribution, so marginal powers alone cannot generally narrow the range further.
Worked Example: Same Marginal Powers, Different Two-Arm Results
Suppose two comparisons each have marginal power \(0.70\) under a fixed alternative. Consider two hypothetical joint distributions. Each row records the probability that the indicated comparisons succeed.
| Dependence pattern | Both succeed | Only comparison 1 | Only comparison 2 | Neither succeeds |
|---|---|---|---|---|
| High overlap | 0.70 | 0.00 | 0.00 | 0.30 |
| Lower overlap | 0.40 | 0.30 | 0.30 | 0.00 |
Both rows are valid probability distributions: their entries are nonnegative and sum to 1. In each row, comparison 1 has marginal power equal to the probability of “both” plus “only comparison 1,” and likewise for comparison 2. The marginals are therefore \(0.70\) in both cases.
For high overlap, the conjunctive power is \(0.70\). The disjunctive power is \(0.70+0.00+0.00=0.70\). For lower overlap, the conjunctive power is \(0.40\), while the disjunctive power is \(0.40+0.30+0.30=1.00\). Equivalently, inclusion–exclusion gives \(0.70+0.70-0.40=1.00\). The marginal powers have not changed, but the joint powers have.
For these marginals, the overlap bounds are \(\max(0,0.70+0.70-1)=0.40\) and \(\min(0.70,0.70)=0.70\). The two examples attain the lower and upper bounds. The lower-overlap distribution does not make the success events mutually exclusive: both comparisons succeed with probability \(0.40\). Rather, the events overlap as little as is possible given the two marginal probabilities.
Bounds for Several Comparisons
For \(m\) comparisons, let \(I_i\) equal 1 when \(A_i\) occurs and 0 otherwise. Let \(U\) indicate that at least one comparison succeeds, and let \(J\) indicate that all comparisons succeed. Pointwise, \(I_i\leq U\) for every \(i\), and \(U\leq\sum_i I_i\). Taking expectations yields bounds for disjunctive power. Also, \(J\leq I_i\) for every \(i\), and \(\sum_i I_i\leq (m-1)+J\). Taking expectations yields bounds for conjunctive power.
For the conjunctive lower bound, the pointwise inequality gives \(\sum_i p_i\leq(m-1)+P(\bigcap_i A_i)\), and hence \(P(\bigcap_i A_i)\geq\sum_i p_i-(m-1)\); the probability is also at least zero. These bounds use only the marginal powers. Without further information about the joint distribution, one should not report a single joint power as though the marginals uniquely determine it.
Worked Example: Three Comparisons Under Three Dependence Structures
Suppose three treatment comparisons each have marginal power \(0.60\). The following are hypothetical joint distributions for their success indicators, written as the probability of each success pattern. A 1 denotes success and a 0 denotes no success.
| Pattern | Independent outcomes | All-or-none outcomes | Exactly two or none |
|---|---|---|---|
| 111 | 0.216 | 0.600 | 0.000 |
| 110 | 0.144 | 0.000 | 0.300 |
| 101 | 0.144 | 0.000 | 0.300 |
| 011 | 0.144 | 0.000 | 0.300 |
| 100 | 0.096 | 0.000 | 0.000 |
| 010 | 0.096 | 0.000 | 0.000 |
| 001 | 0.096 | 0.000 | 0.000 |
| 000 | 0.064 | 0.400 | 0.100 |
Each column sums to 1. In the independent column, the probability of a pattern with \(k\) successes is \(0.60^k0.40^{3-k}\), so each comparison has marginal power \(0.60\). The disjunctive power is \(1-P(000)=1-0.064=0.936\), and the conjunctive power is \(P(111)=0.216\).
In the all-or-none column, the three comparisons always succeed together or fail together. Each has marginal power \(0.600\), but both the disjunctive and conjunctive powers are \(0.600\). In the final column, each comparison succeeds in two of the three patterns with probability \(0.300\), giving marginal power \(0.300+0.300=0.600\). At least one succeeds with probability \(0.300+0.300+0.300=0.900\), while all three succeed with probability \(0\). Thus the same three marginal powers are compatible with disjunctive powers \(0.600\), \(0.900\), and \(0.936\), and conjunctive powers \(0.600\), \(0\), and \(0.216\), respectively.
Independence is an assumption in the first column, not a consequence of the three marginal powers. The other columns are also legitimate joint distributions with the same marginals. The example does not claim that all such patterns arise from a particular trial design; it shows mathematically why the marginal values do not identify the joint answer.
What Can Be Reported When the Joint Distribution Is Unavailable?
When the joint distribution is known or can be calculated under a justified model, use it to obtain the probability of the specified success event. For two comparisons, this may mean obtaining \(P(A_1\cap A_2)\) and then applying inclusion–exclusion. For several comparisons, it means evaluating the probability of the union or intersection under the joint distribution. If only marginal powers are available, report the applicable bounds or state explicitly that the joint power is undetermined from the available information.
A useful contrast is the independence calculation. Under mutual independence, the probability that all comparisons succeed is \(\prod_i p_i\), and the probability that at least one succeeds is \(1-\prod_i(1-p_i)\). Those formulas are correct only when the success events are mutually independent. They cannot be selected just because independence makes the calculation convenient. Conversely, a full joint model can provide the needed probability, but its dependence assumptions need justification. Marginal calculations and joint calculations answer different questions; neither replaces the other.
The bounds are a third option. They require no particular dependence model, but can be wide. For example, in the three-comparison example, the general bounds give disjunctive power between \(0.60\) and 1, and conjunctive power between 0 and \(0.60\). The three displayed distributions give values within those bounds. Bounds communicate what is identified by the marginals without implying a dependence structure that has not been specified.
Worked Example: Applying the Two-Event Bounds to Unequal Powers
Suppose two hypothetical comparisons have marginal powers \(p_1=0.80\) and \(p_2=0.65\). Their overlap must satisfy
If \(q=0.45\), a valid joint distribution has probabilities 0.45 for both succeeding, 0.35 for only comparison 1 succeeding, 0.20 for only comparison 2 succeeding, and 0 for neither. The marginals are \(0.45+0.35=0.80\) and \(0.45+0.20=0.65\). Its conjunctive power is \(0.45\), and its disjunctive power is \(0.45+0.35+0.20=1.00\).
If \(q=0.65\), a valid distribution has probabilities 0.65 for both, 0.15 for only comparison 1, 0 for only comparison 2, and 0.20 for neither. Again the marginals are 0.80 and 0.65. The conjunctive power is now 0.65, and the disjunctive power is \(0.65+0.15=0.80\). Therefore the marginal powers permit conjunctive power from 0.45 to 0.65 and disjunctive power from 0.80 to 1.00.
In both cases the entries are nonnegative, sum to 1, and reproduce the stated marginals. This checks that the examples are valid rather than merely algebraic endpoint values.
Diagnostics, Limitations, and a Common Mistake
A calculation is suspect if it reports disjunctive or conjunctive power from marginal powers alone without stating a dependence assumption or providing joint information. A practical diagnostic is to ask what joint success patterns the calculation allows: how often do all comparisons succeed, how often does exactly one succeed, and how often do none succeed? If the analysis uses simulation, record success patterns jointly within each simulated trial replicate. Simulating each comparison separately and combining marginal estimates discards the co-occurrence information needed for joint power.
Even pairwise dependence summaries may be insufficient for a joint event involving three or more comparisons. In general, knowing marginal probabilities and pairwise associations does not specify every higher-order joint probability. The complete joint distribution, or a model that determines the event probability, may be required. This tutorial does not identify the dependence structure for any particular multi-arm design; that requires information about the design and its statistical model.
Common mistake: “Each of two comparisons has power 0.70, so the chance both succeed is \(0.70^2=0.49\), and the chance at least one succeeds is \(1-(1-0.70)^2=0.91\).” Both calculations silently assume independence. With the same marginal powers, the examples above show that conjunctive power can be 0.40 or 0.70, and disjunctive power can be 1.00 or 0.70. The correction is to use the independence formulas only when independence is justified; otherwise obtain joint information or report bounds.
Check Your Understanding
Try each question, then check the concise solutions below.
- Two comparisons each have marginal power 0.60. What are the possible bounds for the probability that both succeed?
- For the same two marginal powers, what are the corresponding bounds for disjunctive power?
- Under independence, calculate the disjunctive and conjunctive power for three comparisons, each with marginal power 0.50.
- Explain why multiplying two marginal powers is not generally a valid calculation of conjunctive power.
- For three comparisons with marginal powers 0.80, 0.70, and 0.60, give the general bounds for conjunctive power.
Solutions. (1) The overlap is between \(\max(0,0.60+0.60-1)=0.20\) and \(\min(0.60,0.60)=0.60\). (2) Since disjunctive power is \(0.60+0.60-q\), it ranges from \(0.60\) (when \(q=0.60\)) to \(1.00\) (when \(q=0.20\)). (3) Under mutual independence, conjunctive power is \(0.50^3=0.125\), and disjunctive power is \(1-(1-0.50)^3=0.875\). (4) Multiplication gives the joint probability only under independence; marginal probabilities alone do not specify co-occurrence. (5) The lower bound is \(\max(0,0.80+0.70+0.60-2)=0.10\), and the upper bound is \(\min(0.80,0.70,0.60)=0.60\).