Why the Assignment Direction Matters
In a pattern-mixture analysis, the distribution of outcomes can differ across response patterns. When an outcome is missing, its distribution within that pattern is not directly observed; an identifying restriction specifies which observed pattern’s distribution will stand in for it. In a three-visit study, two assignments can use nearby patterns yet supply a missing outcome from different sources. The resulting assumptions, and therefore the completed pattern distributions, need not agree.
Consider monotone dropout: participants either complete all three visits, drop out after visit 2, or drop out after visit 1. Write \(Y_1,Y_2,Y_3\) for the visit outcomes and \(R\) for the observed pattern. Let \(R=111\) denote three observed outcomes, \(R=110\) two observed outcomes, and \(R=100\) only the first observed outcome. The pattern labels are in visit order.
These are distinct restriction choices, not consequences of the pattern labels alone. To state them precisely, let \(f_r\) denote a conditional probability distribution within pattern \(r\). For continuous outcomes, read \(f\) as a density. The forward assignment is
The backward assignment uses
The first assignment uses two donor patterns for \(R=100\): \(R=110\) supplies the distribution of \(Y_2\), and \(R=111\) supplies the conditional distribution of \(Y_3\). The second takes the entire missing block from \(R=111\). In its reverse-order factorization, it uses \(f_{111}(y_3\mid y_1)\) followed by \(f_{111}(y_2\mid y_1,y_3)\). Naming the direction is useful only when the rule is stated this explicitly; the name by itself does not determine a donor.
Assumptions and Donor Map
The assignments require monotone observation patterns as defined above, a well-defined visit outcome, and comparable histories across the pattern distributions. The displayed equations condition on \(Y_1\), and where relevant on \(Y_2\). If baseline covariates are included, the same conditioning variables must be used consistently on both sides of each restriction. The distributions used as donors must have support for the histories at which they are applied. For example, estimating \(f_{111}(y_3\mid y_1,y_2)\) for a history requires that history to occur in \(R=111\).
The restrictions are assumptions about unobserved outcomes; observed data alone do not establish that a pattern’s missing outcomes follow its donor’s distribution. A donor map makes the assumptions auditable:
| Target pattern and missing outcome | Forward donor | Backward donor |
|---|---|---|
| \(R=110\): \(Y_3\) | \(R=111\), conditional on \(Y_1,Y_2\) | \(R=111\), conditional on \(Y_1,Y_2\) |
| \(R=100\): \(Y_2\) | \(R=110\), conditional on \(Y_1\) | \(R=111\), as part of the joint distribution given \(Y_1\) |
| \(R=100\): \(Y_3\) | \(R=111\), conditional on \(Y_1,Y_2\) | \(R=111\), as part of the joint distribution given \(Y_1\) |
The third row needs care. Under the forward rule, \(Y_3\) is not assigned an unconditional distribution: its distribution depends on the assigned \(Y_2\). Under the backward rule, the joint donor distribution retains the dependence between \(Y_2\) and \(Y_3\) seen in \(R=111\).
Worked Example: Tracing the Donors in a Binary-Outcome Study
Suppose \(Y_1\) is fixed within a stratum and \(Y_2,Y_3\) are binary. The following hypothetical conditional probabilities describe \(R=111\):
Suppose \(P(Y_2=1\mid Y_1,R=110)=0.80\). First obtain the conditional distributions from the \(R=111\) table: \(P(Y_3=1\mid Y_2=0,Y_1,R=111)=0.10/0.40=0.25\), and \(P(Y_3=1\mid Y_2=1,Y_1,R=111)=0.42/0.60=0.70\).
For \(R=110\), both assignments give \(Y_3\) from those conditional probabilities. Mixing over the observed \(Y_2\) distribution in \(R=110\) gives \(P(Y_3=1\mid Y_1,R=110\text{ with assigned }Y_3)=0.20(0.25)+0.80(0.70)=0.61\).
For \(R=100\), the forward rule takes \(P(Y_2=1)=0.80\) from \(R=110\), then uses the \(R=111\) conditional distribution of \(Y_3\). The four assigned joint probabilities, in order \((Y_2,Y_3)=(0,0),(0,1),(1,0),(1,1)\), are \(0.20(0.75)=0.15\), \(0.20(0.25)=0.05\), \(0.80(0.30)=0.24\), and \(0.80(0.70)=0.56\). They sum to \(1.00\).
The backward rule instead borrows the joint distribution directly from \(R=111\), giving the four probabilities \(0.30,0.10,0.18,0.42\). Thus the assigned probability of \(Y_2=1\) is \(0.80\) forward but \(0.60\) backward. The assigned probability of \(Y_3=1\) is \(0.05+0.56=0.61\) forward but \(0.10+0.42=0.52\) backward. The donor choice changes both the joint distribution and, here, its marginal summaries.
Why the Two Assignments Can Differ
The distinction follows from the chain rule for joint distributions. Under the backward assignment, factor the \(R=111\) joint distribution in forward order:
The forward assignment has the same second factor but replaces the first with \(f_{110}(y_2\mid y_1)\). Therefore the two assigned distributions agree precisely when their \(Y_2\) factors agree, on the histories with positive probability.
Conversely, if the two joint distributions are equal, summing or integrating both sides over \(y_3\) gives \(f_{110}(y_2\mid y_1)=f_{111}(y_2\mid y_1)\). This establishes the agreement condition; equality is not guaranteed merely because both procedures use the same complete-case distribution for the conditional \(Y_3\) step.
Worked Example: Checking the Agreement Condition
Suppose, hypothetically, that within a fixed \(Y_1\) stratum the \(R=111\) distribution has \(P(Y_2=1)=0.60\), while the \(R=110\) distribution has \(P(Y_2=1)=0.60\) as well. Let \(P(Y_3=1\mid Y_2=0,R=111)=0.25\) and \(P(Y_3=1\mid Y_2=1,R=111)=0.70\).
The forward probability of \((Y_2,Y_3)=(1,1)\) in \(R=100\) is \(0.60(0.70)=0.42\); the backward probability is also \(0.60(0.70)=0.42\). For \((0,1)\), both give \(0.40(0.25)=0.10\). The other two cells are \(0.40(0.75)=0.30\) and \(0.60(0.30)=0.18\) under either assignment. All four cells therefore agree. This agreement results from equality of the donor distributions for \(Y_2\), not from an algebraic identity that always makes the procedures equivalent.
Diagnostics, Alternatives, and Limitations
A useful diagnostic is to lay out each target pattern, missing outcome, conditioning history, and donor pattern before calculating anything. Then check whether each conditional distribution is estimable in its proposed donor. Sparse histories can make a donor distribution unstable; a history absent from the donor pattern makes the specified conditional distribution unavailable without an additional modeling or pooling assumption. A second diagnostic is to compare observed distributions that are being substituted, such as \(Y_2\) in \(R=110\) and \(R=111\). A difference signals that forward and backward assignments may have different implications. Similar observed distributions do not prove that either assignment is correct for outcomes that are missing.
These are identifying restrictions, not testable facts about the unseen outcomes. Complete-case restrictions and available-case restrictions are alternative ways to specify relationships between pattern distributions; neighboring assignments instead make the donor choice locally and outcome by outcome. None is automatically justified by having a convenient observed pattern. The earlier material on neighboring-case restrictions motivates borrowing across adjacent patterns; the present comparison shows why the direction and conditioning order must also be stated.
Common mistake: “Both assignments use neighboring patterns, so they must give the same completed distribution.” That reasoning overlooks that the forward assignment takes \(Y_2\) from \(R=110\), whereas the backward assignment takes its joint \((Y_2,Y_3)\) distribution from \(R=111\). The correction is to compare the actual conditional factors and donor patterns, not just the word “neighboring.”
Neither assignment removes uncertainty about the missing outcomes, and neither should be interpreted as a fact learned from the observed data. Their value is to make the identifying choice explicit and to reveal which comparisons drive the implied distribution. When the choices yield materially different results, that difference is a reason to report the sensitivity of conclusions to the assignment, not to select one solely because it is easier to compute.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Under the forward assignment, which pattern supplies the distribution of \(Y_2\) missing in \(R=100\)?
- In a fixed \(Y_1\) stratum, suppose \(P(Y_2=1\mid R=110)=0.70\), \(P(Y_2=1\mid R=111)=0.50\), and \(P(Y_3=1\mid Y_2=0,R=111)=0.20\), \(P(Y_3=1\mid Y_2=1,R=111)=0.60\). Find the forward assigned probability of \(Y_3=1\) in \(R=100\).
- Using the values in question 2, what is the backward assigned probability of \(Y_3=1\) in \(R=100\), and why can it differ?
- State the distributional condition under which the forward and backward assignments agree.
- Why does an observed difference between \(Y_2\) distributions in \(R=110\) and \(R=111\) not establish which assignment is correct for \(R=100\)?
Solutions. (1) \(R=110\), conditional on \(Y_1\). (2) The forward mixture is \(0.30(0.20)+0.70(0.60)=0.06+0.42=0.48\). (3) The backward rule borrows the joint distribution from \(R=111\), so its \(Y_3\) marginal is \(0.50(0.20)+0.50(0.60)=0.10+0.30=0.40\). It differs because backward uses the \(R=111\) distribution of \(Y_2\), while forward uses the \(R=110\) distribution. (4) They agree when \(f_{110}(y_2\mid y_1)=f_{111}(y_2\mid y_1)\) on the relevant support. (5) Both are assumptions about outcomes not observed in \(R=100\); observed pattern differences can diagnose sensitivity but cannot reveal the missing-outcome distribution or decide which restriction is true.