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Randomization-Based Inference · Tutorial 465 of 501

Marginal Inclusion Probabilities In Sequential Designs

Learn to average sequential assignment probabilities over possible histories, calculate marginal inclusion probabilities, and distinguish them from probabilities conditional on the history that actually occurred.

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What You'll Learn

  • Define marginal inclusion probabilities for participants in sequentially randomized designs.
  • Derive marginal probabilities by weighting conditional assignment probabilities by history probabilities.
  • Calculate history probabilities from sequential randomization rules.
  • Distinguish design-based averaging over assignment paths from averaging under a response model.
  • Diagnose errors caused by equal-weighting histories or using a realized conditional probability as a marginal one.
  • Recognize what marginal probabilities can and cannot summarize about an adaptive design.

Why Average Over Histories?

In a sequentially randomized trial, the probability of assigning a participant to treatment may depend on what happened earlier. For example, a response-adaptive rule may change the probability of treatment after observing earlier participants’ outcomes. The probability specified by the rule for the next assignment is therefore conditional on the history available at that moment.

A different question is the probability that a particular participant would be assigned to treatment across repeated runs of the design, before that participant’s prior history is known. This probability averages the conditional assignment probabilities over all possible histories, giving more weight to histories that are more likely to occur. It is useful to distinguish the two quantities: the realized-history probability describes the next randomization given what has occurred; the marginal inclusion probability describes assignment probability averaged over the design’s possible prior histories.

Definition: Let \(A_i\) denote the treatment assigned to participant \(i\), and let \(H_{i-1}\) denote the information available immediately before that assignment, including earlier assignments and any outcomes used by the allocation rule. The conditional assignment probability for treatment \(a\) is \(e_i(a\mid h)=P(A_i=a\mid H_{i-1}=h)\). The marginal inclusion probability for participant \(i\) and treatment \(a\) is \(\pi_i(a)=P(A_i=a)\), averaging over the possible histories \(H_{i-1}\) under the specified design and probability model.

The marginal probability is obtained by the law of total probability:

$$ \pi_i(a)=\sum_h P(H_{i-1}=h)e_i(a\mid h). $$

The sum is over histories that can occur before participant \(i\) is assigned. A history’s weight is its probability under the design, not an equal weight assigned simply because the history is listed. When there are continuous histories, the corresponding expression is an expectation, \(\pi_i(a)=E[e_i(a\mid H_{i-1})]\).

Deriving History Weights

To average correctly, first determine the probability of each relevant history. For an assignment path \(a_1,\ldots,a_{i-1}\), its probability is the product of the sequential assignment probabilities along that path, provided the path is compatible with the design. Earlier outcomes may also be part of the history. Under a design-based calculation with fixed potential outcomes, those outcomes along a path are determined by the assignments on that path. If instead responses are treated as random under a specified outcome model, their probabilities must also be included.

$$ P(A_1=a_1,\ldots,A_{i-1}=a_{i-1}) = \prod_{j=1}^{i-1} e_j(a_j\mid h_{j-1}). $$

This product gives the probability of a particular assignment path. If more than one path leads to the same history relevant to participant \(i\), add their probabilities to obtain \(P(H_{i-1}=h)\). Then multiply each history probability by the conditional probability of treatment \(a\) at that history, and sum.

Justification: The histories \(H_{i-1}=h\) partition the possible events before participant \(i\)'s assignment. For each history, the probability of assigning treatment \(a\) together with that history is \(P(H_{i-1}=h)e_i(a\mid h)\). Summing these disjoint-event probabilities gives \(P(A_i=a)\), which is the marginal inclusion probability.

A design-based marginal probability may be conditional on fixed baseline covariates and fixed potential outcomes, particularly when the allocation rule uses responses. In that setting, the assignments are random, but the potential outcomes are treated as fixed. The design’s assignment rule alone is enough to calculate the average if the relevant potential outcomes are specified. In a superpopulation calculation that treats responses as random, an outcome model is needed to determine how often response-dependent histories arise. These are different probability calculations and should not be silently substituted for one another.

Worked Example: Averaging Over a Two-Participant History

Suppose participant 1 is assigned to treatment \(T\) or control \(C\) with probability \(0.5\) each. For participant 2, the rule assigns \(T\) with probability \(0.8\) if participant 1 received \(T\), and with probability \(0.2\) if participant 1 received \(C\). Find participant 2’s marginal probability of treatment.

The two histories before participant 2 are \(A_1=T\) and \(A_1=C\). Each has probability \(0.5\), so

$$ \pi_2(T) =P(A_1=T)(0.8)+P(A_1=C)(0.2) =(0.5)(0.8)+(0.5)(0.2) =0.40+0.10 =0.50. $$

If participant 1 actually received \(T\), the conditional assignment probability for participant 2 is \(0.8\), not \(0.5\). The value \(0.5\) is the marginal probability across repeated runs before the first assignment is known. The two quantities answer different questions.

Response-Adaptive Histories

When allocation depends on responses, histories can differ in both previous assignments and observed outcomes. A practical calculation must account for how likely each response-dependent history is. The next example explicitly uses a stochastic response model; therefore, its answer averages over both the first assignment and the first response. It is not a purely design-based calculation conditional on fixed potential outcomes.

Worked Example: Averaging Over an Uncertain Response

In a hypothetical two-participant trial, participant 1 is assigned \(T\) or \(C\) with probability \(0.5\) each. Under a stipulated response model, the probability of a response is \(0.8\) following \(T\) and \(0.4\) following \(C\). Participant 2 is assigned \(T\) with probability \(0.75\) if participant 1 responds, and with probability \(0.25\) otherwise. Find participant 2’s marginal probability of \(T\).

First find the probability of a response for participant 1 by averaging over that participant’s randomized treatment:

$$ P(R_1=1) =(0.5)(0.8)+(0.5)(0.4) =0.40+0.20 =0.60. $$

Thus \(P(R_1=0)=1-0.60=0.40\). Averaging participant 2’s conditional assignment probability over these response histories gives

$$ \pi_2(T) =(0.60)(0.75)+(0.40)(0.25) =0.45+0.10 =0.55. $$

Under the stated response model, participant 2 has a marginal probability of \(0.55\) of receiving \(T\). If participant 1’s response is already observed to be 1, the conditional probability is \(0.75\); if it is observed to be 0, the conditional probability is \(0.25\). Neither realized conditional probability is the marginal value \(0.55\).

The response probabilities in this example are necessary because the calculation averages over possible responses. In a design-based analysis with fixed potential outcomes, one would instead enumerate the response that each possible assignment path produces from those fixed outcomes. If the potential outcomes needed to determine the histories are not specified, a response-adaptive marginal probability cannot generally be obtained from the allocation rule alone.

More Than Two Treatment Arms

The same averaging principle applies with any number of treatment arms. For a given participant, the marginal probabilities across all arms must sum to 1, because the participant receives exactly one arm. This provides a useful check on calculations, but it does not mean the probabilities are equal.

Worked Example: Marginal Probabilities for Three Arms

Suppose participant 1 is assigned to arms \(A\), \(B\), and \(C\) with probabilities \(0.5\), \(0.3\), and \(0.2\). For participant 2, the conditional probabilities of arms \(A,B,C\) are \((0.2,0.5,0.3)\) after history \(A_1=A\), \((0.6,0.1,0.3)\) after \(A_1=B\), and \((0.4,0.4,0.2)\) after \(A_1=C\). Find participant 2’s marginal inclusion probabilities.

For arm \(A\), average its conditional probability across the three histories:

$$ \pi_2(A)=(0.5)(0.2)+(0.3)(0.6)+(0.2)(0.4) =0.10+0.18+0.08=0.36. $$

For arm \(B\) and arm \(C\), the corresponding calculations are

$$ \begin{aligned} \pi_2(B)&=(0.5)(0.5)+(0.3)(0.1)+(0.2)(0.4) =0.25+0.03+0.08=0.36,\\ \pi_2(C)&=(0.5)(0.3)+(0.3)(0.3)+(0.2)(0.2) =0.15+0.09+0.04=0.28. \end{aligned} $$

The check is \(0.36+0.36+0.28=1.00\). These are participant 2’s marginal probabilities across the three arms, averaged over participant 1’s possible assignments.

What to Check, and What Marginal Probabilities Do Not Tell You

A calculation is more credible when its history enumeration is explicit and its probabilities pass basic checks. For each stage, verify that the conditional probabilities across available arms sum to 1 for every history. Verify that the probabilities of all mutually exclusive histories being averaged also sum to 1. Finally, verify that the marginal probabilities across treatment arms sum to 1 for each participant. If any check fails, a history may have been omitted, counted twice, or assigned the wrong probability.

A common mistake is to average conditional probabilities equally over the listed histories. In the three-arm example, this would give an incorrect value for arm \(A\): the histories \(A_1=A,B,C\) do not have equal probabilities \(0.5,0.3,0.2\). The correct calculation weights each conditional probability by its history probability. Another mistake is to report the conditional probability from the history actually observed as though it were the marginal probability. That answers a question conditional on the realized past, not one averaged over possible pasts.

Marginal inclusion probabilities can summarize each participant’s chance of receiving each arm and can be relevant to weighting procedures. However, these probabilities alone do not describe the dependence among participants’ assignments. For example, knowing each participant’s marginal probability does not by itself determine how often two participants are assigned to the same arm. Analyses that require the joint assignment structure need information beyond individual marginals.

The main alternatives are not competing calculations of the same quantity. A conditional assignment probability is appropriate when describing or simulating the next randomization given the history currently observed. A marginal inclusion probability is appropriate when the question averages over prior histories under a specified design and, when needed, outcome model or fixed potential-outcome schedule. A fixed allocation probability applies when the assignment probability does not vary with history; in that special case, the conditional and marginal probabilities coincide.

The method has an important practical limitation: the number of possible histories can grow rapidly with the number of participants and the complexity of the adaptation rule. Exact enumeration may then be cumbersome. Any approximation must still represent histories with their correct probabilities, rather than treating them as equally likely. Simulation can approximate the average by repeatedly generating histories under the specified model and design, but the result is a numerical approximation, not an exact enumeration.

Key takeaway: The marginal inclusion probability for treatment \(a\) is the probability-weighted average of the conditional assignment probability over prior histories. State what is being averaged over, calculate the history weights under the relevant design and outcome assumptions, and keep the result distinct from the conditional probability for the history that occurred.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. Participant 1 is assigned to \(T\) with probability \(0.4\). Participant 2 receives \(T\) with probability \(0.7\) if participant 1 received \(T\), and \(0.3\) otherwise. What is participant 2’s marginal probability of \(T\)?
  2. Why is an unweighted average of conditional probabilities generally not the marginal inclusion probability?
  3. In a response-adaptive design, what extra information may be needed to average over histories under a superpopulation calculation?
  4. A participant has marginal probabilities \(0.35\) for arm \(A\), \(0.40\) for arm \(B\), and \(0.25\) for arm \(C\). What check can be made, and does it establish that different participants’ assignments are independent?
  5. If the conditional probability of treatment is \(0.6\) for every possible prior history, what is the marginal probability of treatment?

Solutions. (1) The histories have probabilities \(0.4\) and \(0.6\), so \(\pi_2(T)=(0.4)(0.7)+(0.6)(0.3)=0.28+0.18=0.46\). (2) Histories generally have different probabilities; the marginal calculation weights each conditional probability by its history probability. (3) If responses are treated as random, their distribution under a specified outcome model may be needed to determine the probabilities of response-dependent histories. (4) The probabilities sum to \(0.35+0.40+0.25=1.00\), as required for one assignment among three arms. This check does not establish independence across participants. (5) The marginal probability is \(0.6\), since averaging a constant conditional probability over histories leaves it unchanged.

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