From a neighboring-case assumption to a mean equation
A neighboring-case restriction becomes useful for calculation when it is written as an equation for the mean of an outcome that is missing in one response pattern. The equation does not make that mean observable: it specifies how to identify it by borrowing a conditional mean from a neighboring pattern. The central calculation is then to average the borrowed conditional means over the histories found in the pattern whose outcomes are missing.
Consider a longitudinal study with outcomes \(Y_1,Y_2,\ldots,Y_T\). Let \(R_t=1\) if \(Y_t\) is observed and \(R_t=0\) otherwise. A response pattern is the vector \(r=(R_1,\ldots,R_T)\). For example, under monotone dropout, patterns \(111\) and \(110\) mean that outcomes are observed at all three visits and at the first two visits, respectively. Monotone dropout means that once an outcome is missing, later outcomes are also missing.
For a two-visit study in which everyone has \(Y_1\), suppose the patterns are \(11\) and \(10\). A conditional neighboring-case restriction for the missing \(Y_2\) values in pattern \(10\) can be written \(E(Y_2\mid Y_1=y,R=10)=E(Y_2\mid Y_1=y,R=11)\). This is an assumption about the conditional mean, not a claim that the two patterns have the same distribution of \(Y_1\), or that their marginal means of \(Y_2\) must be equal.
To use this equation, the conditional mean on the right must be estimable from the neighboring pattern, and the histories represented in pattern \(10\) must have support in pattern \(11\). The outcomes must have finite means, and the chosen history must be observed in both patterns. Pattern proportions are observable from response indicators. The restriction itself, which concerns outcomes missing in pattern \(10\), is not verified merely by observing the data.
Why marginalization matters
A conditional equality does not generally imply an equality of pattern-specific marginal means. To obtain the mean in the pattern with missing outcomes, average the donor conditional mean over the history distribution in the recipient pattern. If \(H\) is the shared observed history and the restriction is \(E(Y_t\mid H,R=r)=m_t(H)\), then
The second equality is the law of iterated expectation. The final expectation uses the distribution of \(H\) among pattern-\(r\) individuals, not the distribution among the donor cases. This is why the same conditional donor means can produce different marginal means in two patterns.
If history is grouped into categories \(h_1,\ldots,h_K\), the calculation becomes a weighted average:
where \(s\) is the neighboring donor pattern and the restriction equates the conditional means in patterns \(r\) and \(s\). If the target is the population mean at visit \(t\), partition the population by response pattern. With \(\pi_r=P(R=r)\),
This last calculation requires a mean for every pattern with positive weight, whether that mean is observed directly or supplied by an identifying restriction.
Worked Example: Stratified Borrowing for a Two-Visit Mean
Suppose a hypothetical study has 150 participants, all observed at visit 1. There are 100 participants in pattern \(11\) and 50 in pattern \(10\). Divide the observed history \(Y_1\) into low and high strata. In pattern \(11\), the mean \(Y_2\) is 8.0 units in the low stratum and 12.0 units in the high stratum. Of the 50 participants in pattern \(10\), 35 are in the low stratum and 15 are in the high stratum. Assume the conditional neighboring-case mean restriction in each stratum.
Assumption check. The history \(Y_1\) is observed for both patterns, and both strata occur in pattern \(11\), so these donor conditional means are available for the histories in pattern \(10\). The identifying assumption is that the conditional mean \(Y_2\) in pattern \(10\) equals the corresponding pattern-\(11\) mean within each stratum.
The pattern-\(10\) stratum weights are \(35/50=0.70\) and \(15/50=0.30\). Therefore the missing-outcome mean implied for this pattern is
The pattern-\(11\) mean is based on 50 low-stratum and 50 high-stratum participants: \(\widehat{\mu}_{2,11}=(50/100)(8.0)+(50/100)(12.0)=10.0\) units. The full-sample mean under the restriction uses pattern proportions \(100/150\) and \(50/150\):
The pattern-specific means differ because the two patterns have different proportions in the low and high strata. The restriction equated conditional means, not marginal means.
Calculations with several visits and quantitative history
For a later visit, the relevant shared history can include several earlier outcomes. Under monotone dropout, a restriction for \(Y_3\) in pattern \(110\) might equate its conditional mean to the mean among pattern \(111\), given \(Y_1\) and \(Y_2\). The more detailed history can make the borrowing assumption more targeted, but it also creates a practical requirement: donor cases must provide information for the histories in the recipient pattern.
Worked Example: A Three-Visit Pattern-Specific Mean
In a hypothetical three-visit study, consider pattern \(111\), with 80 participants, and pattern \(110\), with 40 participants. For simplicity, classify the shared history \((Y_1,Y_2)\) into two cells, A and B. In pattern \(111\), the mean \(Y_3\) is 20 units in cell A and 30 units in cell B. Of the 40 participants in pattern \(110\), 28 have a history in cell A and 12 in cell B. Assume that, conditional on these history cells, the mean \(Y_3\) in pattern \(110\) equals the corresponding mean in pattern \(111\).
Assumption check. Both history components are observed in patterns \(111\) and \(110\). Each recipient history cell has donor cases, and the restriction is stated separately by cell. It is a conditional mean assumption about the unobserved \(Y_3\) values in pattern \(110\).
The pattern-\(110\) history weights are \(28/40=0.70\) and \(12/40=0.30\), so
Suppose the 80 completers are evenly divided between cells A and B. Their observed mean is \(\widehat{\mu}_{3,111}=(0.50)(20)+(0.50)(30)=25\) units. The implied mean among all 120 participants is
This calculation uses the observed pattern proportions and the assumed pattern-\(110\) mean. It does not claim that all participants with the same history have the same individual outcome; the restriction concerns conditional means.
A quantitative history need not be divided into categories. If the donor conditional mean is represented by a specified function \(m(h)\), the recipient-pattern mean is the average of that function over recipient histories. In practice, a fitted model may be used to estimate \(m(h)\); its adequacy and the identifying restriction are separate issues. The following calculation treats the function as given to make the averaging step explicit.
Worked Example: Averaging a Conditional Mean Function
Suppose, hypothetically, that a neighboring pattern supports the conditional mean function \(E(Y_2\mid Y_1=y,R=11)=4+0.5y\) units. In pattern \(10\), the observed \(Y_1\) values are 4, 8, and 12 for three equally weighted participants. Assume the same conditional mean function applies to pattern \(10\).
Assumption check. \(Y_1\) is observed for each pattern-\(10\) participant, and the assumed donor function is defined at all three observed values. The calculation assumes conditional mean equality across patterns at these histories.
Evaluate the function at each recipient history: \(4+0.5(4)=6\), \(4+0.5(8)=8\), and \(4+0.5(12)=10\) units. Average these predicted conditional means using the pattern-\(10\) history distribution:
Equivalently, because the function is linear and the three histories have mean \((4+8+12)/3=8\), its average is \(4+0.5(8)=8.0\) units. For a nonlinear function, averaging the function values is generally not the same as evaluating the function at the average history.
Derivation and practical checks
The key calculation follows from conditioning on the history \(H\). For a recipient pattern \(r\), start with the definition of its pattern-specific mean and apply iterated expectation:
\(\mu_{t,r}=E(Y_t\mid R=r)=E\{E(Y_t\mid H,R=r)\mid R=r\}\), provided the conditional and marginal means exist.
Replace \(E(Y_t\mid H,R=r)\) by the assumed donor conditional mean \(E(Y_t\mid H,R=s)\), where \(Y_t\) is observed in neighboring pattern \(s\).
Take the expectation of the donor conditional mean using the distribution of \(H\) given \(R=r\). This yields the implied pattern-specific mean.
Multiply each pattern-specific mean by its pattern proportion and sum over the patterns in the target population.
This argument justifies the equation; it does not establish that the neighboring-case assumption is true. It also shows why replacing recipient history weights with donor history weights changes the calculation: that replacement computes an average over a different conditional distribution.
Before reporting a result, check that the response patterns have been defined consistently, the outcome units match across patterns, and the conditioning history is observed where required. For categorical histories, inspect the cell counts in both patterns. For quantitative histories, check whether recipient values fall within the donor range and whether a conditional mean model is being extrapolated. Very small donor counts, sparse history cells, and extreme fitted predictions can make the calculation unstable even when the equation is written correctly.
How this restriction differs from alternatives
A complete-case analysis uses observed outcomes from pattern \(111\) and does not, by itself, supply the missing-pattern mean. A neighboring-case mean restriction supplies that mean through conditional borrowing from an adjacent pattern, then averages over the recipient pattern's histories. An unconditional marginal equality instead sets the two pattern means equal directly; that is a different, generally stronger or simply different assumption, and it need not follow from conditional equality when the history distributions differ. An available-case calculation summarizes outcomes among those observed at a visit; it does not automatically identify the mean for the full target population.
These methods answer different questions because they impose different restrictions on unobserved outcomes. A calculated value is therefore conditional on the chosen assumption. It is not a data-only discovery about outcomes that were never observed, and the arithmetic cannot compensate for a poorly supported donor comparison.
Common mistakes and limitations
- Using donor-pattern weights. The missing-pattern mean must be averaged over recipient histories. Donor weights answer a different averaging question.
- Confusing conditional and marginal equality. Equality within history strata does not generally imply equal pattern-specific means if the patterns have different history distributions.
- Ignoring lack of support. If a recipient history has no comparable donor cases, its conditional mean cannot be obtained from observed donor data without additional modeling or assumptions.
- Treating the restriction as empirically confirmed. The missing outcomes prevent direct verification of the assumed equality. Observed-data checks can reveal sparse support or implausible extrapolation, but cannot confirm the unobserved conditional means.
- Interpreting an imposed mean equality as equality of distributions. A mean restriction says nothing by itself about equality of variances, shapes, or individual outcomes.
The method's appeal is its transparent link between an identifying assumption and a numerical target. Its limitation is the same: another defensible conditional mean restriction can imply another target. When history is high-dimensional or patterns are sparse, estimation of donor conditional means may also be imprecise or dependent on modeling choices.
Check Your Understanding
Answer each question using the conditional-mean and pattern-weighting steps above. Solutions follow the questions.
- A recipient pattern has 30% of its histories in a low stratum and 70% in a high stratum. The donor conditional means are 5 and 9 units. Under the neighboring-case mean restriction, what is the recipient-pattern mean?
- Why is it generally incorrect to use the donor pattern's history proportions when calculating a recipient-pattern mean?
- A study has patterns with proportions 0.75 and 0.25. The first pattern's visit mean is 10 units and the restriction implies a mean of 8 units in the second pattern. Calculate the full-population mean.
- What should be checked if a recipient history value lies outside the range observed in the donor pattern?
Solutions. (1) The recipient mean is \((0.30)(5)+(0.70)(9)=1.5+6.3=7.8\) units. (2) The target pattern-specific mean is an expectation conditional on membership in the recipient pattern, so its weights must describe the recipient histories; donor weights average over a different distribution. (3) The full mean is \((0.75)(10)+(0.25)(8)=7.5+2.0=9.5\) units. (4) Check donor support and determine whether the conditional mean calculation requires extrapolation. If no donor information supports that history, the restriction alone does not provide an empirically informed donor mean there; further modeling or assumptions would be needed.