Inclusion–Exclusion for Three-Arm Disjunctive Power

Why Three Success Events Need a Joint Calculation

In a three-treatment trial, each treatment may be compared with a control, producing three prespecified comparison success events. Disjunctive power is the probability that at least one of those comparisons is successful under a specified treatment-effect scenario and analysis procedure. If the comparison events are dependent, their individual powers alone are not enough to calculate this probability. Inclusion–exclusion gives an exact calculation when the pairwise and triple intersection probabilities are available.

The calculation concerns a fixed set of three events, not a choice made after seeing results. For example, the trial protocol might define success as rejecting the relevant null hypothesis for each prespecified treatment-control comparison under a stated testing procedure. The exact success criterion matters: changing the test or decision rule changes the events and therefore their probabilities.

Definition: Let \(A_1,A_2,A_3\) denote the success events for the three prespecified treatment comparisons under one fixed trial design, testing procedure, and parameter scenario. The three-arm disjunctive power is \(P(A_1\cup A_2\cup A_3)\), the probability that one or more comparisons succeed. Write \(\pi_i=P(A_i)\), \(\pi_{ij}=P(A_i\cap A_j)\), and \(\pi_{123}=P(A_1\cap A_2\cap A_3)\).

These probabilities are taken under the scenario for which power is being evaluated, not under the global null unless that is the scenario of interest. Each \(\pi_i\) is a marginal power, while \(\pi_{ij}\) and \(\pi_{123}\) describe joint success. The events need not be independent. The method requires that the events be precisely defined and that the probabilities refer to the same design, procedure, and scenario.

The Three-Event Formula

Adding the three marginal probabilities counts outcomes with multiple successes too many times. An outcome in \(A_1\cap A_2\), for instance, contributes once to \(\pi_1\) and once to \(\pi_2\). Subtracting the three pairwise intersections corrects this double counting, but outcomes in all three events have then been subtracted too often. Adding the triple intersection once completes the correction.

Theorem: For any three events \(A_1,A_2,A_3\), with no independence assumption required,

P(A_1\cup A_2\cup A_3) =\pi_1+\pi_2+\pi_3 -\pi_{12}-\pi_{13}-\pi_{23} +\pi_{123}.

A short derivation makes the signs in the formula clear. Let \(I_i\) equal 1 when \(A_i\) occurs and 0 otherwise. For every outcome, the indicator of at least one success satisfies

I(A_1\cup A_2\cup A_3) =I_1+I_2+I_3-I_1I_2-I_1I_3-I_2I_3+I_1I_2I_3.

If no event occurs, the right side is 0. If exactly one occurs, it is 1; if exactly two occur, it is \(2-1=1\); and if all three occur, it is \(3-3+1=1\). Thus the identity holds outcome by outcome. Taking expectations gives the theorem, because the expectation of an event indicator is its probability and the expectation of a product of indicators is the corresponding intersection probability.

This identity explains why the triple intersection must be added, even when the events are strongly dependent. The formula is not an approximation based on pairwise information: the triple intersection is one of its required inputs.

Worked Example: Calculating Disjunctive Power From Joint Probabilities

Suppose a hypothetical design evaluation gives marginal success probabilities \(0.60\), \(0.55\), and \(0.50\) for the three comparisons. The pairwise intersection probabilities are \(0.30\), \(0.28\), and \(0.27\), and the probability of all three comparisons succeeding is \(0.15\). Calculate disjunctive power.

Here \(\pi_1=0.60\), \(\pi_2=0.55\), \(\pi_3=0.50\), \(\pi_{12}=0.30\), \(\pi_{13}=0.28\), \(\pi_{23}=0.27\), and \(\pi_{123}=0.15\). Substitution gives

\begin{aligned} P(A_1\cup A_2\cup A_3) &=0.60+0.55+0.50-0.30-0.28-0.27+0.15\\ &=1.65-0.85+0.15\\ &=0.95. \end{aligned}

The disjunctive power is \(0.95\), or 95% under this hypothetical scenario and procedure. This is the probability of at least one successful comparison; it is not the probability that all three comparisons succeed.

Checking the Joint Probability Inputs

Because the formula combines several probabilities, plausible-looking inputs can still be mutually inconsistent. A useful check is to calculate the probabilities of the eight possible success patterns: no events, each of the three single-event-only patterns, each of the three exactly-two-event patterns, and the all-three pattern. Every pattern probability must be at least 0, and their sum must be 1. In particular, a negative pattern probability signals that the supplied marginal and intersection probabilities cannot all describe the same joint distribution.

Worked Example: Verifying a Three-Event Probability Specification

For a second hypothetical evaluation, let \(\pi_1=0.72\), \(\pi_2=0.65\), and \(\pi_3=0.58\). Let \(\pi_{12}=0.48\), \(\pi_{13}=0.41\), \(\pi_{23}=0.37\), and \(\pi_{123}=0.30\). Calculate the probability of at least one success and check the pattern probabilities.

Applying inclusion–exclusion,

\begin{aligned} P(A_1\cup A_2\cup A_3) &=0.72+0.65+0.58-0.48-0.41-0.37+0.30\\ &=1.95-1.26+0.30\\ &=0.99. \end{aligned}

The single-event-only probabilities are found by subtracting the relevant pairwise intersections and adding back the triple intersection. For example,

P(A_2\text{ only})=0.65-0.48-0.37+0.30=0.10.

The other single-event-only probabilities are \(P(A_1\text{ only})=0.72-0.48-0.41+0.30=0.13\), and \(P(A_3\text{ only})=0.58-0.41-0.37+0.30=0.10\). The exactly-two-event probabilities are the pairwise intersections minus the triple intersection: \(0.48-0.30=0.18\) for \(A_1,A_2\) only, \(0.41-0.30=0.11\) for \(A_1,A_3\) only, and \(0.37-0.30=0.07\) for \(A_2,A_3\) only. The all-three probability is \(0.30\).

Summing the seven nonzero-success patterns gives \(0.13+0.10+0.10+0.18+0.11+0.07+0.30=0.99\). The remaining probability, for no successes, is \(1-0.99=0.01\). All eight pattern probabilities are nonnegative and sum to 1, so this specification passes this consistency check.

What the Triple Intersection Contributes

The triple intersection is not a small technical detail that can be dropped by default. If it is omitted, the calculation subtracts outcomes with all three successes three times but adds them back zero times. The resulting value is too low by exactly \(\pi_{123}\). Conversely, adding \(\pi_{123}\) more than once also gives the wrong answer.

Worked Example: Diagnosing an Omitted Triple Intersection

Use the probabilities in the preceding example. Suppose an analyst adds the marginal probabilities and subtracts the pairwise intersections but forgets the triple intersection. The resulting calculation is

$$ 0.72+0.65+0.58-0.48-0.41-0.37=0.69. $$

This is not the disjunctive power. It is \(0.30\) below the correct value of \(0.99\), precisely the omitted triple intersection probability. The error is especially visible from the success-pattern calculation: outcomes in all three events were subtracted three times in the pairwise correction, so the final \(+0.30\) is necessary to count them once. A separate check is that the omitted-intersection value \(0.69\) does not equal the sum of the seven nonzero-success pattern probabilities.

How This Relates to Other Calculations

Inclusion–exclusion and the complement rule are two ways to calculate the same target. The complement rule writes the probability of at least one success as one minus the probability of no successes. If the eight success-pattern probabilities are available, this can be convenient. Inclusion–exclusion is often useful when the marginal, pairwise, and triple intersection probabilities are the available inputs.

A product calculation for the probability of no successes is justified only when the three events are mutually independent. Under that assumption, the disjunctive power is \(1-(1-\pi_1)(1-\pi_2)(1-\pi_3)\). When tests share data or otherwise have dependent outcomes, that product generally does not apply. Inclusion–exclusion remains valid under either independence or dependence, provided its intersection probabilities describe the joint distribution in the scenario being evaluated.

The key practical limitation is that the marginal powers alone do not supply the pairwise or triple intersections. Those quantities must come from a suitable joint model, a justified calculation, or another source that represents dependence under the same design and scenario. Inserting guessed intersection probabilities can produce a numerical answer without a defensible power calculation.

Diagnostics and Common Mistakes

Leaving out the triple intersection. The pairwise subtractions remove triple-success outcomes repeatedly; add \(\pi_{123}\) once.
Assuming independence without justification. Separate comparisons are not automatically independent. Use a product formula only when mutual independence is justified; otherwise use appropriate joint probabilities.
Mixing probability scenarios. Marginals and intersections must all correspond to the same fixed treatment-effect scenario and testing procedure.
Accepting impossible inputs. Check that the success-pattern probabilities are nonnegative and sum to 1. A result outside \([0,1]\) is an immediate warning, though a result inside that interval alone does not establish that every input is coherent.
Misstating the target. Disjunctive power is the probability of at least one success, not the probability that every comparison succeeds.

Key takeaway: For three prespecified comparison success events, disjunctive power equals the sum of the three marginal probabilities, minus the three pairwise intersections, plus the triple intersection. The identity is exact without independence, but the inputs must form a coherent joint probability specification.

Check Your Understanding

Answer each question, then compare with the solutions below.

If \(\pi_1=0.50\), \(\pi_2=0.40\), \(\pi_3=0.30\), \(\pi_{12}=0.20\), \(\pi_{13}=0.15\), \(\pi_{23}=0.10\), and \(\pi_{123}=0.05\), what is the disjunctive power?
Why must the triple intersection be added in the three-event formula?
Can the independence product formula be used just because the three comparisons are all treatment-versus-control tests? Explain.
For the inputs in question 1, calculate the probability that only \(A_1\) occurs.

Solutions. 1. Inclusion–exclusion gives \(0.50+0.40+0.30-0.20-0.15-0.10+0.05=0.80\). 2. Outcomes in all three events are counted three times in the marginal sum and subtracted three times in the pairwise correction; adding the triple intersection once leaves them counted once. 3. No. The fact that comparisons share a control does not establish independence. A product calculation needs mutual independence, while inclusion–exclusion uses the actual intersection probabilities. 4. \(P(A_1\text{ only})=\pi_1-\pi_{12}-\pi_{13}+\pi_{123}=0.50-0.20-0.15+0.05=0.20\).