Why Estimate Within Each Pattern?
In a pattern-mixture analysis, the distribution of an outcome can differ across dropout patterns. Estimating a mean or a covariate effect within one pattern therefore uses the information contributed by subjects in that pattern, rather than imposing a common mean or regression across patterns. This matters both for interpretation and for the calculation: a subject contributes only the outcome components observed for that subject. An unobserved outcome is not turned into an observation merely because it has a place in a model.
Weighted estimating equations make this principle explicit. They specify a set of equations whose solutions define the estimates, without requiring a full probability model for every outcome. The weights can account for a sampling or analysis scheme, while the pattern-specific model describes the conditional mean being estimated. These are distinct roles.
For a scalar outcome \(Y_i\) observed for subjects in pattern \(r\), let \(a_i\) be a specified nonnegative analysis weight. The simplest equation estimates the pattern-specific mean \(\mu_r\):
Here \(I(R_i=r)\) equals 1 for a subject in pattern \(r\) and 0 otherwise. If the included weights sum to a positive value, solving gives
When all included \(a_i=1\), this is the ordinary mean of the observed outcomes in pattern \(r\). With unequal weights, it is a weighted mean; its target depends on what those weights represent. For example, prespecified survey expansion weights target a survey-weighted mean, whereas equal weights target the unweighted mean among the observed pattern members.
Covariate Effects and the General Equation
Suppose the conditional mean in pattern \(r\) is modeled as \(E(Y_i\mid X_i,R_i=r)=X_i^\mathsf{T}\beta_r\), where \(X_i\) is a vector of covariates (including a constant if an intercept is wanted). The weighted estimating equation is
If the weighted design matrix has full column rank, the solution is the weighted least-squares estimate:
Each pattern has its own parameter vector \(\beta_r\). A coefficient therefore describes the fitted mean relationship within that pattern, under the chosen model and weights; it is not automatically a common covariate effect across patterns. For binary or other non-normal outcomes, the mean can instead be linked to a linear predictor, and the equation uses the corresponding mean-model derivative and working covariance. The essential restriction remains: include only observed outcome components for the pattern.
For repeated outcomes, write \(Y_{ir}\) for the vector of outcomes observed for subject \(i\) in pattern \(r\), and \(X_{ir}\) for the corresponding design matrix. A general weighted estimating equation is
Here \(m_{ir}(\beta_r)\) is the modeled mean vector for the observed occasions, and \(V_{ir}\) is a positive-definite working covariance matrix for those components. The scalar \(a_i\) is an analysis weight; \(V_{ir}^{-1}\) weights residual directions according to the working covariance. A covariance model can improve efficiency when it is reasonable, but the mean model and the observed components still determine what is being estimated.
Assumptions and a Short Justification
For the scalar equations, a clear set of conditions is: the pattern labels and observed outcomes are correctly recorded; the conditional mean model is correct for the target within pattern; weights are specified and finite, with positive total weight; and, for regression, the weighted design matrix has full rank. For an equation to have mean zero at the target parameter, the weighted residual must have conditional expectation zero. A sufficient condition is that, within pattern and covariate values, the conditional mean is correctly specified and the analysis weights do not depend on the residual in a way that shifts its weighted expectation.
The mean formula follows directly from the estimating equation. Let \(W_r=\sum_i I(R_i=r)a_i\) and \(T_r=\sum_i I(R_i=r)a_iY_i\). Then
provided \(W_r>0\). For regression, expanding \(U_r(\beta_r)=0\) gives the weighted normal equations
The full-rank condition makes the coefficient solution unique. If the matrix is singular, the data and model do not identify all requested coefficients within that pattern; changing the numerical algorithm does not create the missing information.
Worked Example: A Weighted Mean in One Pattern
Suppose three hypothetical subjects in pattern \(r\) have observed outcome values \(4\), \(6\), and \(10\), with prespecified analysis weights \(1\), \(2\), and \(1\). Assume the target is the weighted mean for this pattern and the recorded values are the outcomes being modeled. The equation is
The total weighted outcome is \(4+12+10=26\), and the total weight is \(1+2+1=4\). Thus
The estimate is \(6.5\) outcome units for the weighted pattern-specific mean. With equal weights, the ordinary mean would be \((4+6+10)/3=6.667\) outcome units. That difference is not an arithmetic discrepancy: the two calculations use different weighting targets.
Worked Example: A Weighted Covariate Regression
In a second hypothetical pattern, three subjects have observed pairs \((x_i,Y_i)=(0,2),(1,4),(2,8)\) and weights \(1,2,1\). Fit the pattern-specific linear mean \(E(Y_i\mid x_i,R_i=r)=\beta_{0r}+\beta_{1r}x_i\). The weighted sums are
The weighted slope and intercept are
The fitted mean is \(1.5+3x\): the estimated slope is \(3\) outcome units per one-unit increase in \(x\), within this pattern. As a check, the residuals are \(0.5\), \(-0.5\), and \(0.5\). Their weighted sum is \(0.5+2(-0.5)+0.5=0\), and their weighted covariate sum is \(0(0.5)+2(1)(-0.5)+2(0.5)=0\). Both regression estimating equations are satisfied.
Worked Example: Using Only the Observed Visits
Consider a hypothetical pattern in which subjects are observed at visits 1 and 2 but not at visit 3. Three subjects have observed vectors \((5,7)\), \((7,9)\), and \((6,8)\), with weights \(1,2,1\). Estimate a separate mean at each observed visit in this pattern. For visit 1, the weighted equation gives
For visit 2,
These estimates use the observed components in the pattern. No equation for the visit-3 mean can be formed from these data alone, because none of these subjects has an observed visit-3 outcome. Estimating that unobserved component would require additional modeling assumptions or information; it is not a consequence of the weighted mean equations above.
Choosing Weights, Comparing Approaches, and Checking Fit
A weight must have a stated purpose. Survey or sampling weights can make a weighted equation target a population represented by the sampling design. Precision weights may be used when the variance structure justifies them. In contrast, the inverse of a dropout probability is not automatically appropriate just because dropout occurred: using such a weight changes the estimating target and requires assumptions about the dropout process. Weighted estimating equations do not, by themselves, establish those assumptions.
Nor should the analysis weight \(a_i\) be confused with the proportion of the sample in pattern \(r\). The equation above estimates a mean or regression within pattern \(r\); it does not combine pattern-specific estimates into a whole-population summary. Likewise, a covariance weight \(V_{ir}^{-1}\) handles residual dependence or scale in a repeated-outcome equation, whereas \(a_i\) represents the subject-level analysis weighting scheme.
Compared with a full likelihood for a pattern-specific distribution, estimating equations can focus on the mean and avoid specifying a complete outcome distribution. This can be useful when a mean model is credible but distributional details are uncertain. A likelihood uses a fuller probability model and may be more efficient if correctly specified, but its conclusions depend on that model. Neither approach makes an unobserved outcome directly observed, and neither removes the need to state the target and assumptions.
Useful checks include verifying that each pattern has enough subjects, that all weights are finite and have a positive sum, and that covariates vary enough within each pattern to estimate the requested coefficients. Examine whether a small number of large weights dominate the weighted sums. Compare fitted means with observed values across covariate ranges and check whether the model form is plausible. Sparse patterns, near-collinear covariates, extreme weights, or systematic residual patterns can produce unstable or misleading estimates. Standard errors require additional variance calculations; the point estimate alone does not quantify its uncertainty.
Check Your Understanding
Answer each question, then compare your work with the solutions below.
- In a pattern, two observed outcomes are \(3\) and \(9\), with weights \(2\) and \(1\). What is the weighted mean?
- What does the slope in a pattern-specific linear mean model describe?
- A pattern has no observed outcomes at visit 4. Can its visit-4 mean be estimated by the observed-outcome mean equation?
- Name one condition required for a unique weighted linear regression estimate within a pattern.
Solutions. (1) The weighted mean is \(\{2(3)+1(9)\}/(2+1)=15/3=5\) outcome units. (2) It describes the fitted change in the conditional mean, in outcome units per one-unit increase in the covariate, within that pattern and under the specified model. (3) No. The equation has no observed visit-4 outcomes to contribute; additional assumptions or information are needed. (4) The weighted design matrix must have full column rank (and the included weights must have positive total weight).