The Complement Rule for Disjunctive Power

Why the Complement Rule Is Useful

In a multi-arm trial, a disjunctive objective is met if at least one comparison in a prespecified target set succeeds. This can be calculated by adding probabilities of success patterns, but the number of patterns grows quickly with the number of comparisons. A simpler route is to identify the one event that means the objective was not met: every target comparison fails. The complement rule then gives disjunctive power directly.

The calculation depends on defining success consistently with the trial’s decision procedure. For example, success might mean rejecting a particular treatment-versus-control null hypothesis using a prespecified multiplicity procedure. The probability must then be evaluated under the specified treatment-effect configuration. Changing the target set or success rule changes the event whose probability is being calculated.

Definition: Let \(J\) be a fixed, nonempty set of target comparisons. For each \(j\in J\), let \(S_j\) be the event that comparison \(j\) meets its prespecified success criterion, and let \(F_j=S_j^c\) be its failure event. Disjunctive success is \(D=\bigcup_{j\in J}S_j\), meaning that at least one target comparison succeeds. Under a specified parameter configuration \(\theta\), disjunctive power is \(P_\theta(D)\).

The failure of the disjunctive objective is not merely that one comparison fails. It is that all target comparisons fail. In event notation, this is \(D^c=\bigcap_{j\in J}F_j\). Thus the complement rule gives a direct expression for power.

Theorem: For a fixed target set and success rule, disjunctive power is

P_\theta(D) =1-P_\theta\!\left(\bigcap_{j\in J}F_j\right).

The probability subtracted from one is the joint probability that every target comparison fails, evaluated under the same parameter configuration \(\theta\).

Derivation: Complementing the All-Fail Event

The identity follows from two basic event relationships. First, \(D\) is the union of the success events. By De Morgan’s law, the complement of that union is the intersection of their complements:

D^c =\left(\bigcup_{j\in J}S_j\right)^c =\bigcap_{j\in J}S_j^c =\bigcap_{j\in J}F_j.

An event and its complement partition the sample space, so their probabilities sum to one. Therefore,

P_\theta(D)+P_\theta(D^c)=1 \quad\Longrightarrow\quad P_\theta(D)=1-P_\theta\!\left(\bigcap_{j\in J}F_j\right).

This is an event identity, not an approximation. It requires the target set to be fixed for the calculation and the success events to be defined. It does not require the comparisons to be independent. However, it does require the joint probability of all failures, or enough information to calculate that probability.

Worked Example: Power From a Joint Failure Probability

Suppose a hypothetical trial has three prespecified treatment-versus-control comparisons. Success for each comparison is defined as rejection under the trial’s prespecified multiplicity procedure. Under a specified alternative configuration, the probability that all three comparisons fail is \(0.28\). Calculate disjunctive power.

Let \(F_1,F_2,F_3\) denote failure of the three target comparisons. By the complement formula,

P_\theta(D) =1-P_\theta(F_1\cap F_2\cap F_3) =1-0.28 =0.72.

The hypothetical disjunctive power is \(0.72\), or 72%. Under the specified alternative configuration and decision rule, the probability of at least one successful target comparison is 72%. The calculation does not say that each individual comparison has 72% power; it concerns the trial-level event that at least one of the three succeeds.

Calculating Joint Failure From Outcome Patterns

Sometimes the joint all-fail probability is not supplied as a single number. It can be obtained from a complete distribution of success and failure patterns. Each pattern records which target comparisons succeed. The all-fail pattern is the one in which none succeeds. If the listed patterns are mutually exclusive and exhaustive, the probability of that pattern is the joint failure probability needed by the complement rule.

The pattern probabilities must refer to the same procedure, target set, and parameter configuration. In particular, “success” must not be interpreted as an unadjusted rejection in one row and a multiplicity-adjusted rejection in another.

Worked Example: Reading the All-Fail Pattern From a Joint Distribution

Consider a hypothetical three-comparison trial. Under a specified alternative, suppose its joint distribution of successful comparisons is as follows. For example, “A and C” means those two comparisons succeed and B fails.

Successful comparisons	Probability
None	0.31
A only	0.18
B only	0.12
C only	0.09
A and B only	0.11
A and C only	0.08
B and C only	0.06
A, B, and C	0.05

The patterns cover all possible outcomes and their probabilities sum to \(0.31+0.18+0.12+0.09+0.11+0.08+0.06+0.05=1.00\). The all-fail event is exactly the “None” pattern, with probability \(0.31\). Hence,

P_\theta(D)=1-P_\theta(\text{none succeed}) =1-0.31 =0.69.

The disjunctive power is 0.69, or 69%. The other seven rows collectively describe outcomes in which at least one comparison succeeds; their probabilities sum to \(0.69\), agreeing with the complement calculation.

When Power Is Averaged Across Scenarios

A design may be assessed over a prespecified collection of plausible parameter scenarios rather than at only one fixed configuration. The complement rule still applies within each scenario. If scenario \(k\) has weight \(w_k\), with nonnegative weights summing to one, an average power can be formed by weighting the scenario-specific powers. This is an average across the specified scenarios, not the power at a single parameter value.

The scenario weights and the meaning of the average should be stated. For example, weights might describe a design criterion chosen to summarize performance across a range of effects. They should not be treated as probabilities about which treatment effect will occur unless that interpretation is specifically justified.

Worked Example: Combining Scenario-Specific All-Fail Probabilities

A hypothetical design is evaluated over two prespecified effect scenarios. Scenario 1 has weight \(0.40\), and its probability that all target comparisons fail is \(0.20\). Scenario 2 has weight \(0.60\), and its all-fail probability is \(0.45\). Calculate the weighted average disjunctive power.

First, the scenario-specific powers are \(1-0.20=0.80\) and \(1-0.45=0.55\). Weighting these powers gives

\bar{P}(D) =0.40(0.80)+0.60(0.55) =0.32+0.33 =0.65.

Equivalently, average the all-fail probabilities first: \(0.40(0.20)+0.60(0.45)=0.08+0.27=0.35\), then take the complement: \(1-0.35=0.65\). The weighted average power is 0.65, or 65%, for these specified scenario weights. It is not a claim that power equals 65% at every effect configuration.

Choosing a Calculation Route and Checking It

There are two natural ways to calculate the same quantity. One is to calculate the probability of at least one success directly, by considering success outcomes. The other is to calculate the probability of joint failure and subtract it from one. The complement route is especially concise when the all-fail probability is available, while direct enumeration can be convenient when the success-pattern probabilities are already tabulated. Both routes must use the same event definitions.

A sound calculation can be checked in several ways:

Confirm that the target comparisons and their success criteria were fixed before evaluating the probability.
Check that the all-fail event means every target comparison fails, not merely that one comparison fails.
Verify that any outcome-pattern probabilities are mutually exclusive, exhaustive, and sum to one, up to rounding.
Check that the resulting power lies between 0 and 1 and moves in the expected direction: a larger all-fail probability means lower disjunctive power.
Make sure the joint all-fail probability corresponds to the specified effect configuration and analysis procedure.

If only the separate failure probabilities are known, the joint all-fail probability is generally not determined by them alone. Substituting a product of marginal probabilities is not justified unless the needed independence condition has been established. The complement identity remains valid, but it does not supply missing joint information.

Common Mistakes and Limitations

Complementing the wrong event. “At least one failure” is not the complement of “at least one success.” The complement is that all target comparisons fail.
Using a different target set. Including comparisons that are not part of the stated objective changes the union of success events and therefore changes power.
Mixing success definitions. If the planned procedure requires an adjusted rejection, a nominal unadjusted rejection cannot be substituted as the success event without changing the quantity being calculated.
Inferring joint failure from marginals without justification. The complement rule needs the probability of simultaneous failure. Separate probabilities for individual failures do not, by themselves, provide it.
Over-interpreting an average. A weighted average across scenarios summarizes performance under those weights; it does not describe power at each scenario separately.

The complement rule is a compact way to express disjunctive power, not a method for generating the joint failure probability from incomplete information. That probability may require a suitable joint model or a complete joint outcome distribution. The rule also does not determine whether the chosen multiplicity procedure or target set is appropriate; those are design decisions that must be specified separately.

Key takeaway: Define the target success events first. Disjunctive power is exactly one minus the probability that every comparison in that target set fails.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

A hypothetical trial has two target comparisons, and the probability that both fail is \(0.37\). What is its disjunctive power?
For three target comparisons, what event is the complement of “at least one comparison succeeds”?
A complete outcome distribution assigns probability \(0.24\) to the pattern in which none of four target comparisons succeeds. Calculate disjunctive power.
Two prespecified scenarios have weights \(0.25\) and \(0.75\), and all-fail probabilities \(0.10\) and \(0.30\), respectively. Calculate the weighted average disjunctive power.
Why is it generally incorrect to calculate the joint probability that all comparisons fail by multiplying their separate failure probabilities?

Solutions. (1) \(1-0.37=0.63\), so disjunctive power is 63%. (2) All three comparisons fail; the complement is the intersection of their failure events. (3) \(1-0.24=0.76\), or 76%. (4) The scenario-specific powers are \(0.90\) and \(0.70\), so the weighted average is \(0.25(0.90)+0.75(0.70)=0.225+0.525=0.75\), or 75%. (5) Multiplying marginal failure probabilities requires an appropriate independence condition. Without it, the joint all-fail probability is not determined by the marginal probabilities alone.