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Rubin's Rules and Pooling · Tutorial 52 of 497

Estimands And Notation For Multiple-Imputation Pooling

Learn to define a common target across completed datasets and interpret each component of Rubin’s pooling notation without confusing estimates, variances, or scales.

Intermediate 11 min read

What You'll Learn

  • Define a common estimand for analyses across multiple imputed datasets.
  • Distinguish imputation-specific estimates \(Q_i\) from their estimated variances \(U_i\).
  • Interpret \(M\), \(\bar Q\), \(\bar U\), \(B\), and \(T\), including their units.
  • Explain why the between-imputation variance uses \(M-1\) in its denominator.
  • Diagnose scale mismatches and other conditions that make pooling meaningless.
  • Compare multiple-imputation pooling notation with single-imputation summaries.

Why Pooling Needs a Shared Target

Multiple imputation produces \(M\) completed datasets from the same incomplete data. An analysis is then applied to each completed dataset. The results can be combined only if the analyses address the same estimand: the same precisely defined population quantity, for the same population, under the same model and on the same measurement scale. Otherwise, the resulting numbers do not represent repeated estimates of one target.

For example, a treatment-effect coefficient adjusted for age is not automatically the same target as an unadjusted treatment-effect coefficient. Nor are a regression coefficient on the log-odds scale and its exponentiated odds ratio interchangeable as pooling inputs. A common estimand is not just a common variable name; it includes the quantity being estimated and the scale on which it is expressed.

Definition: In scalar Rubin’s rules, \(Q\) denotes the common scalar estimand, and \(Q_i\) denotes its estimate from the \(i\)th imputed dataset, for \(i=1,\ldots,M\). \(U_i\) is the estimated sampling variance of \(Q_i\) from the analysis of that completed dataset. The \(M\) estimates and their variances must refer to the same estimand and scale.

The \(Q_i\) values will generally differ because the imputed values differ. The \(U_i\) values can also differ because the completed datasets differ. A variance is not a standard error: \(U_i\) is in squared units of \(Q_i\), whereas its square root is in the units of \(Q_i\). For instance, if \(Q_i\) is a mean difference in mmHg, \(U_i\) is in \(\text{mmHg}^2\).

The Pooling Notation

The symbols below label distinct summaries of the imputation-specific results. Their definitions are straightforward, but keeping their roles and units separate is essential.

Definition: \(M\) is the number of completed datasets analyzed. The pooled point estimate is \(\bar Q\), the arithmetic mean of the \(Q_i\). The average within-imputation variance is \(\bar U\), the arithmetic mean of the \(U_i\). The between-imputation variance is \(B\), the sample variance of the \(Q_i\). Rubin’s total variance is denoted by \(T\); in the standard scalar rules it combines \(\bar U\) and \(B\) as \(T=\bar U+(1+1/M)B\).
$$ \bar Q=\frac{1}{M}\sum_{i=1}^{M}Q_i,\qquad \bar U=\frac{1}{M}\sum_{i=1}^{M}U_i $$
$$ B=\frac{1}{M-1}\sum_{i=1}^{M}(Q_i-\bar Q)^2,\qquad T=\bar U+\left(1+\frac{1}{M}\right)B $$

These are definitions and the standard scalar total-variance expression, not a claim that every inferential detail has been settled by calculating \(T\). The derivation and small-sample adjustments to inference require additional results. Here, the key is to identify what each quantity summarizes: \(\bar U\) averages the estimated variances within completed datasets, while \(B\) summarizes how much the estimates vary across imputations. Both are measured in squared units of \(Q\); \(T\) is also in squared units, and \(\sqrt{T}\) is in the units of \(Q\).

Why the Between-Imputation Variance Has \(M-1\)

The denominator in \(B\) is the sample-variance denominator. Its role can be justified by treating the imputation-specific estimates, under ideal independent draws from the relevant imputation distribution, as draws with a common variance. The sample mean is estimated from those same draws, which costs one degree of freedom in the deviations around the mean.

$$ \sum_{i=1}^{M}(Q_i-\bar Q)=0 $$

This equality gives the deviations one linear constraint, so they have \(M-1\) degrees of freedom. It does not make the deviations statistically independent. For example, with \(M=2\), the two deviations are exact negatives of each other. Under independent draws with finite variance, dividing the sum of squared deviations by \(M-1\) makes the sample variance an unbiased estimate of the variance of a draw. This explains the conventional form of \(B\); it does not mean the imputations themselves are always independent in practice. Chained-equation draws may retain dependence if chains have not mixed well, so convergence and adequate Monte Carlo sampling matter.

The \(U_i\) values have a different role. Each is obtained from the complete-data analysis applied to one completed dataset, using that analysis’s variance estimator. These values represent estimated sampling variance conditional on the completed data; they are not the spread of the \(Q_i\) across imputations. The distinction prevents treating uncertainty within a completed-data analysis and variation between imputations as if they were the same source.

Worked Example: Labeling the Components in a Mean-Difference Analysis

Suppose an analyst estimates a treatment-group mean difference in systolic blood pressure, in mmHg, in \(M=4\) hypothetical imputed datasets. The estimates are \(Q_1=2\), \(Q_2=3\), \(Q_3=2\), and \(Q_4=5\) mmHg. The corresponding estimated variances are \(U_1=0.64\), \(U_2=0.81\), \(U_3=1.00\), and \(U_4=0.49\) \(\text{mmHg}^2\). Each analysis uses the same definition of the group contrast and the same model.

First, average the estimates:

$$ \bar Q=\frac{2+3+2+5}{4}=\frac{12}{4}=3.00\ \text{mmHg}. $$

Next, average the estimated variances:

$$ \bar U=\frac{0.64+0.81+1.00+0.49}{4} =\frac{2.94}{4}=0.735\ \text{mmHg}^2. $$

For \(B\), the deviations from \(3\) mmHg are \(-1,0,-1,2\) mmHg. Thus:

$$ B=\frac{(-1)^2+0^2+(-1)^2+2^2}{4-1} =\frac{6}{3}=2.00\ \text{mmHg}^2. $$

The total-variance expression gives \(T=0.735+(1+1/4)(2.00)=3.235\ \text{mmHg}^2\), and \(\sqrt{T}\approx1.799\) mmHg. This example identifies the notation and its units. It does not, by itself, provide a confidence interval or a complete test; those require the relevant degrees of freedom and inferential procedure.

Check the Estimand and Scale Before Calculating

The formulas do not repair incompatible analysis results. Before pooling, verify that each completed-data analysis estimates the same target, uses the same parameterization, and returns the same kind of variance estimate. A useful check is dimensional: every \(Q_i\) must have the same units, and every \(U_i\) must have the corresponding squared units.

Worked Example: Pooling a Coefficient on a Consistent Scale

In three hypothetical imputations, a logistic regression analysis reports the same treatment coefficient on the log-odds scale. The estimates are \(0.20\), \(0.35\), and \(0.25\), and their estimated variances are \(0.0400\), \(0.0900\), and \(0.0625\), respectively. Here \(M=3\); each \(Q_i\) is a log-odds coefficient, and each \(U_i\) is in squared log-odds units.

The average estimate is \(\bar Q=(0.20+0.35+0.25)/3=0.80/3\approx0.2667\). The average within-imputation variance is \(\bar U=(0.0400+0.0900+0.0625)/3=0.1925/3\approx0.06417\).

The deviations from \(\bar Q\) are approximately \(-0.0667\), \(0.0833\), and \(-0.0167\). Therefore:

$$ B=\frac{(-0.0667)^2+(0.0833)^2+(-0.0167)^2}{3-1} \approx\frac{0.01167}{2}=0.005833. $$

The total variance is \(T=0.06417+(1+1/3)(0.005833)\approx0.07194\) squared log-odds units. The result is internally consistent because all three estimates use the same coefficient scale and all three variances correspond to that scale. Exponentiating an estimate to report an odds ratio can be useful for interpretation, but it does not change the fact that the quantities used here were defined on the log-odds scale.

Worked Example: Detecting a Target Mismatch

Suppose two hypothetical completed-data analyses report treatment coefficients in mmHg. One is an unadjusted group difference of \(1.8\) mmHg; the other is an age-adjusted coefficient of \(2.6\) mmHg. Their arithmetic average is \((1.8+2.6)/2=2.2\) mmHg. This calculation is numerically correct, but it is not a valid \(\bar Q\) for Rubin’s rules if the two analyses target different quantities.

The problem is the estimand, not the units: both numbers are in mmHg, but one is an unadjusted contrast and the other is a conditional, age-adjusted model coefficient. The analyst must choose and apply a common analysis model to all completed datasets, then define each \(Q_i\) as the estimate of the same parameter from that model. Only then does the arithmetic mean of those \(Q_i\) values have a coherent interpretation as \(\bar Q\).

Assumptions, Diagnostics, and Alternatives

The notation is simple; the validity of the resulting inference depends on the analyses and imputations behind it. At minimum, the missing-data and imputation models must be defensible for the intended target, and the completed-data analysis must estimate that target appropriately. Rubin’s usual variance interpretation also relies on proper representation of imputation uncertainty and sufficiently well-behaved Monte Carlo draws. The notation alone cannot guarantee these conditions.

Practical diagnostics include checking that chains have mixed and stabilized, inspecting whether imputed values are plausible, and verifying that each completed dataset produces the same model parameter and variance definition. Implausibly erratic \(Q_i\) values can reflect an unstable analysis, sparse data, or poor chain behavior. A very small \(B\) is not proof that missing-data uncertainty is negligible: it may also occur when the imputation model or sampling process fails to reflect relevant uncertainty.

Multiple-imputation pooling differs from analyzing just one completed dataset. A single-imputation analysis supplies one \(Q\) and one \(U\), but it has no across-imputation \(B\) to represent differences among plausible completed datasets. Complete-case analysis is another alternative, but it uses a different subset of observations and may target a different quantity or require different assumptions. Neither alternative can be made equivalent to pooling merely by relabeling its standard error.

The scalar notation is suitable when the target is one number, such as a mean difference or regression coefficient. For a vector of parameters, the estimates are vectors and the variance quantities are covariance matrices; joint inference must account for their covariances. Applying scalar formulas independently does not, by itself, provide a valid joint test.

Common Mistakes

  • Confusing \(U_i\) and \(B\). \(U_i\) is an estimated variance within completed dataset \(i\); \(B\) is the sample variance of the \(Q_i\) values across imputations.
  • Averaging standard errors as if they were variances. The definition of \(\bar U\) averages \(U_i\), not \(\sqrt{U_i}\). The squared units make the distinction visible.
  • Pooling estimates of different targets. A common unit is not enough; the estimand and model parameter must also match.
  • Calling the deviations independent because they have \(M-1\) degrees of freedom. The deviations have one linear constraint, not statistical independence.
  • Assuming \(T\) alone completes inference. Standard errors, degrees of freedom, intervals, and tests require their corresponding inferential rules.
Key takeaway: Define the common estimand first. Then \(Q_i\) is its estimate in completed dataset \(i\), \(U_i\) is that estimate’s within-dataset variance estimate, \(\bar Q\) and \(\bar U\) are their averages, \(B\) is the across-imputation sample variance, and \(T\) is the total-variance quantity used in Rubin’s rules.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. In one sentence, distinguish \(U_i\) from \(B\).
  2. If \(Q_i\) is measured in kilograms, what are the units of \(U_i\), \(B\), and \(\sqrt{T}\)?
  3. For \(M=3\) estimates \(4, 5, 9\), calculate \(\bar Q\) and \(B\).
  4. Why do estimates on the log-odds scale and odds-ratio scale not form a consistent set of \(Q_i\) values for direct pooling?
  5. What does the zero-sum constraint on \(Q_i-\bar Q\) imply, and what does it not imply?

Solutions. (1) \(U_i\) estimates the sampling variance of the analysis estimate within completed dataset \(i\); \(B\) measures the sample variance of estimates across imputations. (2) \(U_i\) and \(B\) are in kilograms squared, and \(\sqrt{T}\) is in kilograms. (3) \(\bar Q=(4+5+9)/3=6\). The deviations are \(-2,-1,3\), so \(B=(4+1+9)/(3-1)=14/2=7\) squared estimate-units. (4) The coefficient and odds ratio are different parameter scales; the \(Q_i\) must all estimate the same quantity on the same scale, with variance estimates corresponding to that scale. (5) The deviations sum to zero, giving one linear constraint and \(M-1\) degrees of freedom. It does not imply that the deviations are statistically independent.

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