Pattern Aggregation and Restriction-Induced Borrowing

Why Pattern Aggregation Changes a Restriction

A pattern-mixture restriction assigns a distribution to an outcome that is missing for one response pattern by relating it to outcomes observed in other patterns. When more than one observed pattern is used as a donor, the restriction borrows from a mixture of their distributions. That mixture is not generally the distribution in any single donor pattern, and its weights are determined by how the donor patterns are combined.

Consider a three-visit study with monotone dropout patterns. Write $G=111$ for participants observed at all three visits, $G=110$ for those observed at visits 1 and 2, and $G=100$ for those observed only at visit 1. The second outcome $Y_2$ is observed in patterns 111 and 110, but missing in pattern 100. A restriction for $Y_2$ in pattern 100 can borrow from either observed pattern separately or from their aggregate. Those choices imply different distributions whenever the donor distributions differ or their mixture weights change.

Definition: Let $Y_j$ be the outcome at visit $j$, and let $G$ denote a response pattern. For an outcome missing in a target pattern $t$, let $A$ be a set of donor patterns in which that outcome is observed. An aggregated donor restriction assigns the target pattern the outcome distribution of the pooled donor group: $$ F_{Y_j\mid G=t,H=h}(y) = F_{Y_j\mid G\in A,H=h}(y), $$ where $H=h$ denotes the history or covariate stratum on which the restriction is defined. The right-hand distribution is a weighted mixture of the donor-pattern distributions within that stratum.

The equality in this definition is an identifying restriction, not a consequence of the observed data. The donor outcome is observed in patterns in $A$, while it is missing in the target pattern. To use the restriction, specify the target pattern, the donor set, the outcome, and any history or covariates being conditioned on. Donor patterns must actually observe that outcome. The target is not included in its own donor set.

Deriving the Mixture Weights

Within a fixed history $h$, define the donor-pattern weight as the proportion of the pooled donor group in pattern $g$: $$ w_g(h)=P(G=g\mid G\in A,H=h) = \frac{P(G=g\mid H=h)} {\sum_{k\in A}P(G=k\mid H=h)},\qquad g\in A. $$ These weights are nonnegative and sum to one. By partitioning the pooled donor group into its patterns, its observed outcome distribution is $$ F_{Y_j\mid G\in A,H=h}(y) = \sum_{g\in A}w_g(h)F_{Y_j\mid G=g,H=h}(y). $$ This follows because the event $Y_j\le y$ within the pooled donor group can occur in any one of the mutually exclusive donor patterns. The probability of that event is the sum, over patterns, of the probability of being in that pattern times the conditional probability of $Y_j\le y$ there.

In data, the weights are commonly estimated from pattern counts within the relevant stratum: $\hat w_g(h)=n_{gh}/\sum_{k\in A}n_{kh}$, where $n_{gh}$ is the number of observed participants in pattern $g$ with history $h$. For a mean restriction, the implied target mean is the same weighted average of donor means: $$ E(Y_j\mid G=t,H=h) = \sum_{g\in A}w_g(h)E(Y_j\mid G=g,H=h). $$ For a distribution restriction, however, the mixture formula—not just the mean—is needed. For example, a mixture of normal distributions with different means is generally not itself normal.

If donor-pattern means and variances exist, the pooled variance also reflects differences between pattern means: $$ \operatorname{Var}(Y_j\mid G\in A,H=h) = \sum_{g\in A}w_g(h) \left[ \operatorname{Var}(Y_j\mid G=g,H=h) + \{\mu_g(h)-\mu_A(h)\}^2 \right], $$ where $\mu_g(h)=E(Y_j\mid G=g,H=h)$ and $\mu_A(h)=\sum_g w_g(h)\mu_g(h)$. The second term inside the brackets accounts for between-pattern variation. Omitting it understates the variance unless all donor means are equal.

Worked Example: Pooling Two Donor Patterns for Visit 2

Suppose, hypothetically, that within one history stratum there are 40 participants in pattern 111 and 60 in pattern 110. The observed $Y_2$ distribution in pattern 111 assigns probabilities 0.25, 0.50, and 0.25 to values 8, 10, and 12 points. In pattern 110 it assigns probabilities 0.50 and 0.50 to values 6 and 10 points. For the 50 participants in pattern 100, impose the restriction that their missing $Y_2$ distribution equals the distribution pooled across patterns 111 and 110.

The donor weights are $40/(40+60)=0.40$ for pattern 111 and $60/(40+60)=0.60$ for pattern 110. The pooled probability at each possible value is the weighted sum of the donor probabilities:

\begin{aligned} P(Y_2=6\mid G\in A)&=0.40(0)+0.60(0.50)=0.30,\\ P(Y_2=8\mid G\in A)&=0.40(0.25)+0.60(0)=0.10,\\ P(Y_2=10\mid G\in A)&=0.40(0.50)+0.60(0.50)=0.50,\\ P(Y_2=12\mid G\in A)&=0.40(0.25)+0.60(0)=0.10. \end{aligned}

The probabilities sum to $0.30+0.10+0.50+0.10=1.00$. Thus the restriction assigns this distribution to the missing $Y_2$ values in pattern 100. Its mean is $$ 0.30(6)+0.10(8)+0.50(10)+0.10(12)=8.80\text{ points}. $$ For comparison, borrowing from pattern 111 alone would assign mean $0.25(8)+0.50(10)+0.25(12)=10.0$ points; borrowing from pattern 110 alone would assign mean $0.50(6)+0.50(10)=8.00$ points. The pooled restriction gives 8.80 points because it combines the two observed distributions in the proportions 0.40 and 0.60. These are different assumptions about the missing outcomes, not three estimates that the observed data alone can distinguish.

What the Aggregated Distribution Implies

The mixture weights affect more than the mean. They determine the probability assigned to every part of the missing-outcome distribution, and therefore can change tail probabilities, quantiles, and variance. The next example shows why a pooled distribution should not be replaced automatically by a normal distribution with its pooled mean and variance.

Worked Example: A Continuous Mixture Is Not Necessarily Normal

Suppose, hypothetically, two observed donor patterns have $Y_2$ distributions $N(8,4)$ and $N(12,9)$, where the second parameter is the variance. There are 75 participants in the first pattern and 25 in the second. A target pattern has missing $Y_2$, and the chosen restriction borrows from the pooled donor group.

The donor weights are $0.75$ and $0.25$, so the assigned distribution is the mixture $0.75N(8,4)+0.25N(12,9)$. Its mean is $$ \mu_A=0.75(8)+0.25(12)=9.00. $$ Its variance includes both within-pattern variances and the differences between each pattern mean and the mixture mean: $$ \begin{aligned} \sigma_A^2 &=0.75\{4+(8-9)^2\}+0.25\{9+(12-9)^2\}\\ &=0.75(5)+0.25(18)=8.25. \end{aligned} $$

For example, the mixture probability of $Y_2\le 10$ is $$ 0.75\Phi\left(\frac{10-8}{2}\right) +0.25\Phi\left(\frac{10-12}{3}\right) =0.75\Phi(1)+0.25\Phi(-0.667) \approx 0.694. $$ Here $\Phi$ is the standard normal cumulative distribution function. The assigned missing-outcome distribution therefore has a mean of 9.00 outcome units, variance 8.25 squared outcome units, and about 0.694 probability of being at most 10 units. It is the stated two-component mixture; it is not generally $N(9,8.25)$. Replacing it with a single normal distribution would be an additional modeling choice.

History-Specific Aggregation

Weights should correspond to the population to which the restriction refers. If the restriction is defined within histories, calculate the donor mixture separately within each history. Pooling patterns across histories first can produce different weights and a different missing-outcome distribution. This is especially relevant when histories are distributed differently across response patterns.

Worked Example: Aggregation Within Two Histories

Suppose, hypothetically, patterns A and B both observe $Y_2$, while target pattern T does not. There are two histories, $H=0$ and $H=1$. Within $H=0$, pattern A has 30 observations all equal to 5 points and pattern B has 70 observations all equal to 7 points. Within $H=1$, pattern A has 80 observations all equal to 10 points and pattern B has 20 observations all equal to 14 points. Target pattern T contains 40 participants with $H=0$ and 60 with $H=1$. Impose a restriction that matches T to the A/B pooled donor distribution within each history.

For $H=0$, the donor weights are $30/100=0.30$ for A and $70/100=0.70$ for B. The assigned conditional distribution puts probability 0.30 at 5 and 0.70 at 7, with mean $0.30(5)+0.70(7)=6.40$ points. For $H=1$, the weights are $80/100=0.80$ for A and $20/100=0.20$ for B. The assigned conditional distribution puts probability 0.80 at 10 and 0.20 at 14, with mean $0.80(10)+0.20(14)=10.8$ points.

Across the target pattern's histories, the implied missing-outcome distribution puts probabilities $0.40(0.30)=0.12$ at 5, $0.40(0.70)=0.28$ at 7, $0.60(0.80)=0.48$ at 10, and $0.60(0.20)=0.12$ at 14. These probabilities sum to 1.00. Its mean is $$ 0.12(5)+0.28(7)+0.48(10)+0.12(14)=9.04\text{ points}. $$

For contrast, pooling all 200 donor observations without conditioning on history gives probabilities 0.15 at 5, 0.35 at 7, 0.40 at 10, and 0.10 at 14, with mean $0.15(5)+0.35(7)+0.40(10)+0.10(14)=8.60$ points. The difference, 0.44 points, arises because the donor pool's history composition differs from the target pattern's history composition. The marginal pooled restriction and the history-specific restriction are distinct choices; the latter yields 9.04 points for this target composition.

Assumptions, Checks, and Alternatives

An aggregated restriction is meaningful only after its assumptions and target are stated. It requires that the selected donor patterns observe the outcome, that the pooling set and weights are defined, and that the chosen distributional equality is accepted as an identifying assumption for the target pattern. If conditioning on history, the donor data must support the relevant histories; a donor pattern with no observations in a target history cannot supply an empirical within-history distribution without further modeling. Estimated weights also require nonzero donor counts in the strata where they are used.

Several checks help catch implementation errors. Confirm that the donor weights are nonnegative and sum to one, and that every pooled probability distribution sums to one. Verify that the pooled mean lies between the smallest and largest donor means. If the donor distributions are identical, changing their weights should not change the pooled distribution. If they differ, changing weights generally changes the result. For history-specific restrictions, compare the history distribution in the donor pool with that in the target; a large discrepancy warns that marginal pooling may not represent the intended conditional restriction. Sparse or empty pattern-by-history cells signal unstable or unavailable empirical borrowing, not evidence that the missing distribution equals another one.

Pattern-specific borrowing assigns the target distribution from one donor pattern and avoids choosing a mixture across patterns, but its result depends on which donor is selected. Aggregated borrowing uses multiple donors and makes that choice explicit through weights; it can be less dependent on one pattern, but the pooled distribution may conceal meaningful differences between donors. An available-case restriction uses a broader observed group, while a neighboring-case restriction selects a donor according to pattern proximity. These are alternative identifying restrictions, not interchangeable calculations. The introductory material on pattern-mixture models establishes the role of restrictions in identifying distributions for missing outcomes; aggregation specifies the observed mixture used by one such restriction.

A common mistake is to average donor means equally even when donor patterns have unequal sizes, or to treat the mean of a pooled distribution as though it fully describes that distribution. In the first worked example, an equal-weight mean would be $(10+8)/2=9.00$ points, not the count-weighted 8.80 points. The correction is to use the weights implied by the stated pooling rule. Another mistake is to pool across histories and then claim a history-conditional restriction; calculate the mixture within each history when that is the assumption being made. Finally, no observed-data comparison can reveal whether the target pattern truly follows the assigned distribution, because its outcome is missing. The restriction must be reported as an assumption, and conclusions should be understood in light of its sensitivity.

Key takeaway: Aggregation changes the donor distribution through its pattern weights. State the donor set and conditioning variables, calculate the mixture distribution with the corresponding weights, and distinguish the resulting identifying assumption from what the observed data establish.

Check Your Understanding

Answer each question, then compare with the solutions below.

Two donor patterns have means 4 and 10 units, with 25 and 75 observations. What mean does their count-weighted aggregate assign to a target pattern?
Two donor distributions are identical within a history. What happens to their pooled distribution if the donor weights change?
A target pattern has two histories, but one donor pattern has no observations in the second history. Can its empirical distribution directly define the donor mixture in that history?
Why can the variance of a pooled donor distribution exceed the weighted average of the donor variances?

Solutions. (1) The weights are $25/100=0.25$ and $75/100=0.75$, so the assigned mean is $0.25(4)+0.75(10)=8.50$ units. (2) It does not change: a weighted mixture of identical distributions equals that common distribution for any nonnegative weights summing to one. (3) No. There is no empirical within-history distribution from that pattern; another donor, a different restriction, or an explicitly stated model would be needed. (4) The pooled variance includes between-pattern variation in donor means as well as within-pattern variation. The additional term is $\sum_g w_g(\mu_g-\mu_A)^2$, which is nonnegative and is positive when donor means differ.