Why Marginal Covariance Depends on Dropout Patterns
A population covariance summarizes how outcomes vary together across everyone in the target population. In a dropout setting, however, individuals belong to different patterns, and their outcome distributions can differ. The covariance in the full population therefore reflects two sources of variation: variation among individuals within the same pattern, and variation between patterns whose mean outcome vectors differ.
This distinction matters because averaging the covariance matrices within patterns does not generally give the population covariance matrix. That average omits the variation attributable to different pattern means. Conversely, a large population covariance need not mean that individuals have strong within-pattern association: it can arise partly, or even mainly, because patterns have different means.
Let \(G\) denote the dropout pattern, taking values \(1,\ldots,K\), and let \(\mathbf{Y}\) be a common \(q\)-dimensional outcome vector, such as measurements at two visits. The vector is defined for every individual in the target population, including outcomes that are unobserved after dropout. For pattern \(g\), define its weight \(\pi_g=P(G=g)\), mean vector \(\boldsymbol{\mu}_g=E(\mathbf{Y}\mid G=g)\), and covariance matrix \(\boldsymbol{\Sigma}_g=\operatorname{Cov}(\mathbf{Y}\mid G=g)\). The population mean is \(\boldsymbol{\mu}=E(\mathbf{Y})\), and the population covariance matrix is \(\boldsymbol{\Sigma}=\operatorname{Cov}(\mathbf{Y})\).
The weights are nonnegative and sum to one, and the patterns form an exhaustive, mutually exclusive partition of the population. The outcome vector must have finite second moments for its covariance to exist. No normality assumption is needed for this decomposition. In a dropout application, the essential additional requirement is that each pattern-specific distribution be defined for the same full outcome vector; this includes assumptions or sensitivity specifications for outcomes that are unobserved in a pattern.
Deriving the Decomposition
The population mean is the weighted average of the pattern-specific means:
To decompose covariance, write the centered outcome as the sum of a within-pattern deviation and a pattern-mean deviation:
Here \(\boldsymbol{\mu}_G\) means the mean vector for the pattern an individual belongs to. Take the outer product of this equality and then its expectation. The two cross-product terms have expectation zero: conditional on \(G=g\), \(E(\mathbf{Y}-\boldsymbol{\mu}_g\mid G=g)=\mathbf{0}\), while \(\boldsymbol{\mu}_g-\boldsymbol{\mu}\) is fixed. The remaining terms yield the decomposition.
This is the multivariate law of total covariance. Each diagonal entry decomposes a population variance into within-pattern and between-pattern variance. Each off-diagonal entry decomposes a population covariance in the same way. The between-pattern matrix is positive semidefinite because it is a weighted sum of outer products; its off-diagonal entries, however, can be positive, negative, or zero.
Worked Example: Two Patterns and Two Visits
Suppose a hypothetical population has two dropout patterns: pattern A, with weight \(\pi_A=0.60\), and pattern B, with weight \(\pi_B=0.40\). Consider the full two-visit outcome vector \(\mathbf{Y}=(Y_1,Y_2)'\). For this calculation, assume the pattern-specific distributions, including the unobserved visit-2 outcome in pattern B, have the following means and covariance matrices:
First calculate the population mean:
The within-pattern component is the weighted average of the two covariance matrices:
The mean deviations are \(\boldsymbol{\mu}_A-\boldsymbol{\mu}=(-0.8,-1.2)'\) and \(\boldsymbol{\mu}_B-\boldsymbol{\mu}=(1.2,1.8)'\). Thus the between-pattern component is
Adding the two components gives the population covariance matrix:
For example, the population covariance between visits is \(5.64\) outcome-units squared. Of this, \(4.2\) comes from within-pattern covariance and \(1.44\) from the fact that pattern B has higher means at both visits. The calculation is algebraically valid given the stated pattern-specific distributions. In an actual dropout analysis, the visit-2 mean and covariance in pattern B are not directly observed for those who dropped out; their specification requires assumptions beyond the observed outcomes.
What the Components Mean
The within-pattern term answers a conditional question: after fixing dropout pattern, how do outcomes vary and covary? The between-pattern term answers a different question: how much population variation is associated with the pattern groups having different mean vectors? These are components of a population covariance, not necessarily separate causal effects of dropout. Pattern membership can be associated with observed or unobserved characteristics, and this decomposition alone does not explain why the means differ.
For any pair of visits \(j\) and \(k\), the marginal covariance is
If patterns with above-average means at one visit also have above-average means at the other, the between-pattern covariance tends to be positive. If one visit's pattern means are high where the other's are low, it can be negative. Consequently, the marginal covariance can differ substantially from every pattern-specific covariance.
Worked Example: Positive Within-Pattern but Negative Marginal Covariance
Consider a hypothetical two-pattern population with equal weights. At both visits, let each pattern have covariance matrix \(\begin{pmatrix}4&2\\2&4\end{pmatrix}\). Let the pattern means be \((0,10)'\) and \((10,0)'\), respectively. Both within-pattern covariances between visits are \(2\), so the within-pattern component is
The population mean is \((5,5)'\). The deviations of the two pattern means from it are \((-5,5)'\) and \((5,-5)'\). Therefore the between-pattern component is
The marginal covariance matrix is consequently \(\begin{pmatrix}29&-23\\-23&29\end{pmatrix}\). Its correlation is \(-23/\sqrt{29(29)}=-23/29\), or approximately \(-0.793\). Thus positive within-pattern covariance does not guarantee positive marginal covariance: here the opposing pattern means produce a larger negative between-pattern component.
Assumptions, Diagnostics, and Limitations
The decomposition itself is an identity under the stated partition and finite-moment conditions. The challenging assumptions in dropout applications concern the inputs, not the identity. Pattern weights can often be estimated from observed pattern membership. By contrast, pattern-specific means and covariances for the full outcome vector can involve unobserved outcomes. The observed-data distribution alone may not determine those quantities; assumptions, restrictions, or sensitivity choices are needed.
Useful checks include verifying that weights are nonnegative and sum to one; that each covariance matrix is symmetric with nonnegative variances; and that the calculated between-pattern and total covariance matrices are symmetric and positive semidefinite. A negative diagonal variance or a markedly asymmetric result signals an arithmetic or implementation error. An off-diagonal covariance can be negative without being an error. Also check that every pattern uses the same outcome ordering and units, and that the full vector's definition is consistent across patterns.
When pattern-specific means are nearly equal, the between-pattern component should be small. If the means are exactly equal, it is the zero matrix, although within-pattern covariance matrices can still differ. When there is only one pattern, the between-pattern component is zero and the population covariance equals that pattern's covariance.
A common mistake is to report \(\sum_g\pi_g\boldsymbol{\Sigma}_g\) as the marginal covariance. This is only the within-pattern component; it equals the marginal covariance only when all pattern means are equal. Another mistake is to calculate a covariance for a dropout pattern using only individuals with both visits observed and treat it as \(\boldsymbol{\Sigma}_g\). That calculation describes an observed subset, not necessarily the full pattern-specific distribution. The correction is to state which distribution the covariance represents and how unobserved outcomes are handled.
The decomposition is useful for interpreting a marginal covariance, but it does not select a pattern-mixture model, identify missing outcomes, or establish a causal explanation for differences between patterns. It also does not replace comparisons with other modeling approaches. A selection model represents the joint distribution through a different factorization; a pattern-mixture model makes the pattern-specific outcome distributions explicit. Regardless of representation, a marginal covariance derived for the same full population must account for both within-group variation and differences in group means.
Worked Example: Equal Pattern Means Remove the Between Component
Suppose two hypothetical patterns have weights \(0.75\) and \(0.25\), and both have mean vector \((5,7)'\). Their covariance matrices are \(\begin{pmatrix}1&0.2\\0.2&4\end{pmatrix}\) and \(\begin{pmatrix}9&-1\\-1&1\end{pmatrix}\). Since the pattern means equal the population mean, each mean deviation is zero, so the between-pattern component is the zero matrix. The within-pattern component is
Thus this weighted average is also the marginal covariance matrix. The result does not require equal pattern covariance matrices; it follows because the pattern means are identical.
Check Your Understanding
Answer each question using the decomposition and show the key calculation.
- Two patterns have weights \(0.25\) and \(0.75\), with scalar means \(2\) and \(6\), and within-pattern variances \(1\) and \(3\). Find the population mean, within component, between component, and population variance.
- For a multivariate outcome, what does a zero between-pattern component imply about the pattern means when all pattern weights are positive?
- Can the between-pattern covariance between two visits be negative even if each pattern-specific covariance is positive? Explain.
- Why might the pattern-specific covariance matrix for a dropout group require assumptions about unobserved outcomes?
Solutions. (1) The mean is \(0.25(2)+0.75(6)=5\). The within component is \(0.25(1)+0.75(3)=2.5\). The between component is \(0.25(2-5)^2+0.75(6-5)^2=2.25+0.75=3\). The population variance is \(2.5+3=5.5\). (2) Each diagonal entry of the between component is a weighted sum of squared deviations. If the whole matrix is zero and every weight is positive, those deviations must all be zero, so every pattern mean equals the population mean. (3) Yes. If patterns have opposite-signed deviations at the two visits, their products in the between-pattern covariance are negative; their weighted sum can outweigh positive within-pattern covariance. (4) The full vector includes outcomes after dropout, which are not observed for that group. The observed records alone therefore do not supply all moments of that full pattern-specific vector.