Why Pairwise Assignment Probabilities Matter
In a sequential trial, a participant’s probability of receiving treatment may depend on the assignments and responses observed earlier. Marginal inclusion probabilities summarize each participant’s chance of treatment, but they do not say whether two participants tend to receive treatment together. That dependence is part of the assignment design: ignoring it can misrepresent the uncertainty of estimates based on the randomization.
The key calculation is to trace the assignment histories that can lead to both participants receiving treatment. The calculation must respect the order of assignment and the probability used at each step. When the rule adapts to observed responses, those responses belong in the history too.
The joint probability is not generally the product of the marginal probabilities. That product would equal \(\pi_{ij}\) if the two assignment indicators were independent, but sequential adaptation can make them dependent. For example, an early treatment assignment may change the allocation probability for later participants.
Calculating Probabilities From Assignment Histories
Suppose there are \(N\) participants assigned in a fixed order. Let \(H_{t-1}\) denote the information available to the assignment algorithm just before participant \(t\) is assigned. It can include earlier assignments, observed responses, and other information used by the prespecified rule. Write \(q_t(H_{t-1})=P(Z_t=1\mid H_{t-1})\). For a binary assignment, the conditional probability of control is \(1-q_t(H_{t-1})\).
An assignment path \(z=(z_1,\ldots,z_N)\) has a probability found by multiplying the appropriate conditional probabilities along that path. If responses enter the rule, the history and response observed on each path must be consistent with the assignments on that path. For randomization-based calculations with fixed potential outcomes, this consistency is determined by the assignment and the potential outcomes; the algorithm is then applied along each possible path.
To obtain \(\pi_{ij}\), sum the probabilities of all complete paths on which \(z_i=z_j=1\). Equivalently, sum over histories up to the later of the two assignments and include the conditional probability of the later treatment assignment. If the participants are not consecutive, paths for the intervening assignments must be included or integrated out. Multiplying two marginal probabilities skips this shared-history calculation.
Worked Calculations
Worked Example: All Pairwise Probabilities in a Three-Person Design
Consider a hypothetical trial with three participants assigned in order. Participant 1 has treatment probability \(0.5\). Participant 2 has treatment probability \(0.8\) if participant 1 received treatment and \(0.2\) otherwise. Participant 3 has treatment probability \(0.75\) if participant 1 received treatment and \(0.25\) otherwise. These specified probabilities define the assignment rule; participant 3’s probability does not depend on participant 2’s assignment.
The marginal probabilities are \(\pi_1=0.5\) and \[ \pi_2=(0.5)(0.8)+(0.5)(0.2)=0.5,\qquad \pi_3=(0.5)(0.75)+(0.5)(0.25)=0.5. \] Each marginal averages the relevant conditional probability over the possible earlier assignment histories.
For participants 1 and 2, both are treated only on paths where participant 1 is treated and participant 2 then receives treatment: \[ \pi_{12}=P(Z_1=1)P(Z_2=1\mid Z_1=1)=(0.5)(0.8)=0.4. \] For participants 1 and 3, sum over the possible assignment of participant 2. Since participant 3’s probability depends on participant 1 but not participant 2, the intervening probabilities sum to one: \[ \pi_{13}=(0.5)(0.8+0.2)(0.75)=0.375. \]
For participants 2 and 3, both possible histories for participant 1 contribute: \[ \pi_{23} =(0.5)(0.8)(0.75)+(0.5)(0.2)(0.25) =0.300+0.025=0.325. \] Thus \(\pi_{12}=0.4\), \(\pi_{13}=0.375\), and \(\pi_{23}=0.325\). For instance, \(\pi_1\pi_2=0.25\), not \(\pi_{12}=0.4\); the two assignments are positively dependent under this design.
Worked Example: Summing Over an Intervening Assignment
In a second hypothetical design, participant 1 has treatment probability \(0.4\). Participant 2’s treatment probability is \(0.7\) if participant 1 is treated and \(0.3\) otherwise. Participant 3’s treatment probability is \(0.8\) if participant 2 is treated and \(0.2\) otherwise. Calculate \(\pi_{13}\), allowing for the assignment of participant 2.
For both participants 1 and 3 to receive treatment, participant 1 must first be treated. Conditional on that, participant 2 can be treated or assigned control. The two compatible histories contribute: \[ \begin{aligned} \pi_{13} &=P(Z_1=1)\bigl[ P(Z_2=1\mid Z_1=1)P(Z_3=1\mid Z_2=1)\\ &\qquad\qquad+ P(Z_2=0\mid Z_1=1)P(Z_3=1\mid Z_2=0) \bigr]\\ &=(0.4)\bigl[(0.7)(0.8)+(0.3)(0.2)\bigr]\\ &=(0.4)(0.56+0.06)=0.248. \end{aligned} \]
For comparison, the marginal probability of treatment for participant 3 is \[ \pi_3=(0.4)\bigl[(0.7)(0.8)+(0.3)(0.2)\bigr] +(0.6)\bigl[(0.3)(0.8)+(0.7)(0.2)\bigr] =0.248+0.228=0.476. \] Since \(\pi_1=0.4\), multiplying marginals would give \(0.4(0.476)=0.1904\), not the joint probability \(0.248\). That product is not a substitute for summing the shared histories.
Connection to Design-Based Uncertainty
The role of a joint inclusion probability can be seen in a Horvitz–Thompson estimator. Suppose \(Y_i(1)\) is a fixed potential outcome under treatment for participant \(i\), and the target is the finite-population treatment total \(\sum_i Y_i(1)\). With positive treatment inclusion probabilities, the estimator is \(\widehat{T}_1=\sum_i Z_iY_i(1)/\pi_i\). Its randomization covariance between the contributions from participants \(i\) and \(j\) depends on \(\pi_{ij}\):
This identity follows because \(E(Z_iZ_j)=\pi_{ij}\) and \(E(Z_i)E(Z_j)=\pi_i\pi_j\). Therefore \(\operatorname{Cov}(Z_i,Z_j)=\pi_{ij}-\pi_i\pi_j\); scaling each indicator by its inclusion probability and its fixed potential outcome gives the displayed expression. The estimator’s overall design variance includes such pairwise covariance contributions as well as the individual contributions. The next calculation shows why using only the marginals can discard a nonzero covariance.
Worked Example: A Pairwise Covariance Contribution
Consider a hypothetical two-person design. Participant 1 has treatment probability \(0.5\); participant 2 has treatment probability \(0.8\) after participant 1 is treated and \(0.2\) after participant 1 is assigned control. Thus \(\pi_1=\pi_2=0.5\) and \(\pi_{12}=(0.5)(0.8)=0.4\). Suppose the fixed treatment potential outcomes are \(Y_1(1)=10\) and \(Y_2(1)=20\), measured in outcome units.
The covariance between the two Horvitz–Thompson contributions is \[ 10(20)\frac{0.4-(0.5)(0.5)}{(0.5)(0.5)} =200\frac{0.15}{0.25}=120 \] squared outcome units. The positive value reflects that treatment assignments tend to occur together relative to what independence with these marginals would imply. Assuming independence would set this covariance to zero and would therefore omit this pair’s contribution. This number is one covariance contribution, not by itself the variance of the full estimator.
Checks, Alternatives, and Limitations
Before using calculated joint probabilities, check that they describe a coherent assignment design. For distinct participants, probabilities must satisfy \[ \max(0,\pi_i+\pi_j-1)\leq \pi_{ij}\leq \min(\pi_i,\pi_j), \] because the probability of both events cannot exceed either event, and the probability of their intersection cannot be smaller than \(P(Z_i=1)+P(Z_j=1)-1\). Also check symmetry, \(\pi_{ij}=\pi_{ji}\), and the diagonal identity \(\pi_{ii}=\pi_i\). These checks can reveal arithmetic errors, but passing them alone does not prove that all probabilities match the assignment algorithm.
For a small design, explicit enumeration is often clearest. For a large design, exact enumeration may be impractical; Monte Carlo simulation from the fully specified algorithm can approximate joint probabilities by counting the simulated paths on which both participants are treated. The estimate then has simulation error, so report simulation precision and use enough draws for the required accuracy. In either approach, the probabilities must be generated under the actual sequential rule, not a simplified rule that ignores adaptation.
A common mistake is to calculate \(\pi_{ij}\) as \(\pi_i\pi_j\) because the two participants have known marginal probabilities. This is correct only when their assignment indicators are independent under the design. In an adaptive design, they can share earlier assignments and responses, so the correction is to condition on histories, multiply the conditional probabilities along each compatible path, and sum those path probabilities.
Joint inclusion probabilities do not solve every inferential problem. They rely on a correctly specified assignment mechanism and, when response-dependent rules are involved, correct reconstruction of the histories that feed that rule. If some pair has \(\pi_{ij}=0\), that pair can never both be treated, which can limit variance estimation for estimators requiring that pair’s joint inclusion probability. Very small probabilities can also make inverse-probability contributions unstable. These are properties of the design, not defects repaired by multiplying marginal probabilities.
Write each conditional treatment probability as a function of the history available at that assignment.
Identify the histories for which both target participants receive treatment, including intervening assignments and any response-dependent branches.
Calculate each path probability from its conditional probabilities, then add the probabilities of the compatible paths.
Verify basic probability bounds and explain how the result differs from, or equals, the product of marginals under this design.
Check Your Understanding
Answer each question using the assignment mechanism described. Solutions follow the questions.
- If \(\pi_i=0.6\) and \(\pi_j=0.5\), what are the general lower and upper bounds for \(\pi_{ij}\)?
- Participant 1 is treated with probability \(0.5\). Participant 2’s treatment probability is \(0.7\) after participant 1 is treated and \(0.3\) otherwise. Calculate \(\pi_{12}\) and compare it with \(\pi_1\pi_2\).
- Why must the assignment of an intervening participant be included when calculating a joint probability for participants 1 and 3?
- In one sentence, explain how \(\pi_{ij}\) enters the covariance of two Horvitz–Thompson treatment contributions.
Solutions. (1) The bounds are \(\max(0,0.6+0.5-1)=0.1\) and \(\min(0.6,0.5)=0.5\), so \(0.1\leq\pi_{ij}\leq0.5\). (2) \(\pi_{12}=(0.5)(0.7)=0.35\). The marginal probability for participant 2 is \((0.5)(0.7)+(0.5)(0.3)=0.5\), so \(\pi_1\pi_2=(0.5)(0.5)=0.25\), which differs from the joint probability. (3) The intervening assignment may affect the later participant’s conditional treatment probability, so its possible histories must be weighted and summed. (4) The covariance is proportional to \(\pi_{ij}-\pi_i\pi_j\), the covariance of the two assignment indicators, with scaling by their inclusion probabilities and fixed potential outcomes.