Why Cluster Size Changes the Question
In nested data, clusters may contain different numbers of individuals. That fact alone does not mean that cluster size is informative. It does, however, affect what a summary means: a mean over all individuals gives larger clusters more influence, while a mean of cluster means gives every cluster equal influence. These summaries answer different questions, and can differ substantially when cluster size and cluster outcomes are associated.
In multiple imputation, the distinction matters twice. The imputation model must represent the relevant clustering and any observed association with cluster size, while the analysis of each completed data set must estimate the intended target. Imputing values does not choose the estimand for you. Weighting an analysis also does not, by itself, repair an imputation model that fails to represent the missing-data structure.
“Informativeness” is a relationship, not a property that can be declared without naming the outcome and target. For example, if the expected cluster mean changes with \(n_j\), then size is informative for the cluster-level outcome mean. A different feature, such as within-cluster variability, might depend on size even when the expected cluster mean does not. An observed association between size and outcome can suggest informativeness, but in a finite sample it can also arise by chance.
Two Means, Two Targets
The individual-weighted mean gives each person equal weight. The cluster-weighted mean gives each cluster equal weight, regardless of its size.
The first describes the average individual in the population represented by the sampled individuals. The second describes the average cluster mean in the population represented by the sampled clusters. Neither is universally preferable. The scientific question determines which target is appropriate. If the aim is to describe the average patient, for example, a large clinic should generally contribute more patients than a small clinic. If the aim is to describe the average clinic, each clinic should contribute equally.
To verify the identity, write \(n_j=\bar{n}+(n_j-\bar{n})\) and \(\bar{Y}_j=\bar{Y}_{\mathrm{cluster}}+(\bar{Y}_j-\bar{Y}_{\mathrm{cluster}})\). On summing their products, the terms involving a single centered quantity vanish because each set of deviations sums to zero. Therefore \(\sum_j n_j\bar{Y}_j=J\bar{n}\bar{Y}_{\mathrm{cluster}}+J\operatorname{Cov}_J(n,\bar{Y})\). Since \(N=J\bar{n}\), division by \(N\) gives the stated result. Positive covariance means larger clusters tend to have higher means, so the individual-weighted mean exceeds the cluster-weighted mean; negative covariance reverses the direction.
Worked Example: Unequal Sizes Without an Expected Size Effect
Consider a hypothetical set of three clusters with sizes 2, 4, and 8. Suppose the outcome-generating process has the same expected cluster mean, 50 points, at each of these sizes. If the observed cluster means in this example are also 50, both summaries equal 50 points: \[ \bar{Y}_{\mathrm{cluster}}=(50+50+50)/3=50 \] and \[ \bar{Y}_{\mathrm{individual}}=(2(50)+4(50)+8(50))/(2+4+8)=700/14=50. \] The sizes are unequal, but there is no size-outcome association in these illustrative means. Unequal size alone is therefore not evidence that size is informative. In real data, observed means will usually vary; assessing informativeness requires considering whether that variation is systematically related to size, not simply whether the sizes differ.
Worked Example: Positive Size-Outcome Association Changes the Mean
Now consider hypothetical clusters of sizes 2, 4, and 8 with cluster means 10, 20, and 30 points, respectively. The equal-cluster target is \[ \bar{Y}_{\mathrm{cluster}}=(10+20+30)/3=20\text{ points}. \] The individual-weighted target is \[ \bar{Y}_{\mathrm{individual}} =\frac{2(10)+4(20)+8(30)}{2+4+8} =\frac{20+80+240}{14} =24.29\text{ points}. \] Larger clusters have higher means, so the individual-weighted mean is higher. The difference is 4.286 points. This difference is not a computational error: the two calculations answer different questions.
Weights and the Imputation Target
For an equal-cluster target, a convenient way to weight individual records is to give each person in cluster \(j\) weight \(1/n_j\). The total weight contributed by that cluster is \(n_j(1/n_j)=1\); after normalizing by the total weight \(J\), its contribution is exactly its cluster mean divided by \(J\). For an individual-weighted target, each person receives equal weight, so the unweighted mean over records already has the desired weighting.
This algebra explains the role of weights in a mean calculation; it is not a general instruction to insert inverse-size weights into every imputation model. The imputation model and the analysis weights have different jobs. Imputation should preserve the clustered data structure and include observed information needed to make the conditional distributions credible. Analysis weights should define the desired target. In particular, weighting only the final analysis cannot undo biased or incompatible imputations.
Worked Example: Equal-Cluster Weighting by Hand
Use the hypothetical clusters with sizes 2, 4, and 8 and means 10, 20, and 30 points. Under person weights, the total weight is \(2+4+8=14\), and the weighted sum is \(2(10)+4(20)+8(30)=340\), giving \(340/14=24.29\) points. Under inverse-size weights, each person in the first cluster has weight \(1/2\), each in the second has weight \(1/4\), and each in the third has weight \(1/8\). Each cluster’s total weight is one, so the total weight is \(3\). The weighted sum is \[ 2(1/2)(10)+4(1/4)(20)+8(1/8)(30)=10+20+30=60. \] The normalized result is \(60/3=20\) points, the equal-cluster mean. The weight changes the target from an average person to an average cluster; it does not make the two targets interchangeable.
Multiple Imputation and Pooling the Chosen Summary
Suppose some \(Y_{ij}\) values are missing and multiple imputation produces \(m\) completed data sets. For each completed data set, calculate the same target-specific estimate \(Q_k\) and its complete-data variance estimate \(U_k\), for \(k=1,\ldots,m\). For a scalar estimate, Rubin’s rules combine these quantities as follows:
Here \(\bar{Q}\) is the pooled estimate, \(\bar{U}\) is the average within-imputation variance, \(B\) is the between-imputation variance, and \(T\) is the total variance estimate. The standard error is \(\sqrt{T}\). For a scalar estimate with \(B>0\), a common large-sample degrees-of-freedom approximation is \((m-1)\{1+\bar{U}/[(1+1/m)B]\}^2\); the resulting interval uses a \(t\) critical value with those degrees of freedom. The assumptions behind this calculation include proper imputations and a valid within-imputation variance estimate for the chosen estimand. The imputation procedure must also be defensible for the missingness mechanism and conditional data structure. A pooled estimate cannot correct a mismatch between the scientific target and the summary calculated in each completed data set.
Worked Example: Pooling an Equal-Cluster Estimate
Suppose five hypothetical imputations produce equal-cluster mean estimates \(Q_k=18,19,20,21,22\) points, each with within-imputation variance \(U_k=4\) points squared. The pooled estimate is \[ \bar{Q}=(18+19+20+21+22)/5=20\text{ points}. \] The average within-imputation variance is \(\bar{U}=4\). The between-imputation variance is \[ B=\{(18-20)^2+(19-20)^2+(20-20)^2+(21-20)^2+(22-20)^2\}/4 =(4+1+0+1+4)/4=2.5. \] Thus \[ T=4+(1+1/5)(2.5)=4+3=7\text{ points squared}, \qquad SE=\sqrt{7}=2.646\text{ points}. \] The approximate degrees of freedom are \(4\{1+4/[(1.2)(2.5)]\}^2=21.78\). A two-sided 95% \(t\) interval is approximately \(20\pm2.075(2.646)\), or 14.51 to 25.49 points. Under the stated imputation and variance assumptions, this interval quantifies uncertainty about the equal-cluster mean. It is not an interval for the individual-weighted mean.
Assumptions, Diagnostics, and Limits
A useful analysis starts by fixing the target and clarifying what \(n_j\) counts. Cluster size should refer to the relevant eligible or sampled individuals, not merely the number with observed outcomes; otherwise missing outcomes can change the apparent size and complicate interpretation. If sizes are known and are part of the observed information, the imputation model can condition on size where that is appropriate. For example, a random-intercept imputation model may include cluster size as a predictor when the conditional outcome distribution varies with size. Such inclusion is not a universal cure: a model that assumes cluster effects independent of size can remain inadequate if that assumption fails.
The usual missing-at-random justification requires that, conditional on observed information included in the imputation model, missingness does not depend on the unobserved values. It also requires a sufficiently well-specified conditional model, adequate overlap in the observed information, and imputations that represent uncertainty rather than merely filling in conditional means. If cluster size is related to missingness or outcomes in ways omitted from the model, these conditions may be implausible. An association between size and outcome can be modeled as observed information, but it does not prove that missingness is ignorable or resolve all dependence between cluster formation and outcomes.
Practical diagnostics should examine both the data and the completed imputations. Plot or tabulate cluster means, missingness rates, and relevant predictors against cluster size; compare the equal-cluster and individual-weighted summaries in every completed data set; and check that imputed values reproduce plausible within- and between-cluster variation across the range of sizes. Large shifts across sizes, sparse representation of very large clusters, or imputed distributions that erase observed size patterns are warnings. A difference between weighted and unweighted estimates is informative about the target and the data, but is not by itself proof that an imputation model failed.
| Choice | What receives equal influence? | Typical target |
|---|---|---|
| Equal person weights | Each individual | Mean for the represented individual population |
| Inverse cluster-size weights | Each cluster | Mean of cluster-specific means |
| Different target-specific weighting or modeling | Defined by the analysis | A population or estimand explicitly specified by the study |
The principal alternative to choosing one of these summaries is to report both when both questions matter, making their interpretations explicit. More elaborate selection or weighting approaches may be appropriate when cluster inclusion or size is tied to the sampling process, but they require assumptions beyond the arithmetic distinction described here. In particular, neither the cluster-weighted mean nor a size covariate automatically addresses outcome-dependent cluster formation. When conclusions depend strongly on assumptions about that process, a sensitivity analysis is preferable to presenting one weighting choice as universally correct.
Common Pitfall
A frequent mistake is to say, “The cluster sizes differ, so the data have informative cluster size; therefore I should weight every record by \(1/n_j\).” This reasoning makes two unsupported jumps. Unequal sizes do not establish an outcome-size relationship, and inverse-size weighting changes the target to an equal-cluster mean whether or not size is informative. First state whether the scientific question concerns an average individual or an average cluster. Then examine how size relates to the outcome and choose a model and analysis that support that target. If both questions are scientifically relevant, report both summaries rather than treating one as a correction for the other.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- Clusters have sizes 3 and 9. Does this alone prove informative cluster size? What additional relationship would be relevant for a cluster-mean target?
- Two clusters have sizes 2 and 6 and means 5 and 11 points. Calculate the equal-cluster and individual-weighted means.
- Why does assigning each person weight \(1/n_j\) give equal total weight to each cluster?
- In multiple imputation, should the equal-cluster mean or the individual-weighted mean be calculated in each completed data set? What determines the choice?
- What does a difference between the two pooled means establish, and what does it not establish?
Solutions. (1) No. One would examine whether the outcome distribution relevant to the target, such as the expected cluster mean, varies with size. (2) The equal-cluster mean is \((5+11)/2=8\) points. The individual-weighted mean is \(\{2(5)+6(11)\}/8=76/8=9.5\) points. (3) Cluster \(j\) has \(n_j\) people, each weighted \(1/n_j\), so its total is \(n_j/n_j=1\). (4) Calculate the summary matching the scientific estimand in every completed data set; pool those estimates and their variances. (5) It establishes that the summaries differ for these data and targets. It does not by itself prove informative size in the population or show that either imputation model is misspecified.