Why Total Variance Has Two Sources
Multiple imputation produces several completed data sets, and the analysis is performed separately in each one. The resulting estimates differ for two reasons: each completed data set has its own sampling uncertainty, and the imputed values vary across data sets because the missing values are uncertain. Rubin’s total variance combines both sources. It also accounts for the fact that a finite number of imputations only approximates the result that would be obtained by averaging over infinitely many imputations.
The formula is for repeated estimates of the same scalar estimand—the same population quantity defined by the analysis model—in every completed data set. It is not a formula for averaging arbitrary statistics from different models or for pooling estimates with different meanings.
Here, \(U_m\) and \(T\) are variances, so they have the square of the estimand’s units. The standard error associated with the pooled estimate is \(\sqrt{T}\), in the estimand’s original units. The \(U_m\) values need not be equal: \(\bar{U}\) is their arithmetic mean. The variance \(B\) measures how much the estimates vary among imputations; its denominator is \(M-1\), as for a sample variance.
Deriving the Finite-Imputation Correction
The derivation separates the uncertainty that would remain with infinitely many imputations from the additional simulation error caused by using only \(M\). It relies on proper imputations: the imputation procedure must represent uncertainty about missing values and relevant model parameters, rather than inserting fixed predictions. It also assumes that each completed-data analysis estimates the same quantity, that its variance estimate \(U_m\) is appropriate, and that the imputation and analysis models support valid repeated-imputation inference. In practice, this includes suitable missingness and model assumptions; the variance formula cannot repair an unsuitable imputation model.
To see the variance decomposition, imagine first that the imputations could be averaged over indefinitely. Under the regularity conditions for Rubin’s rules, the variance of that ideal pooled estimate is approximated by \(\bar{U}+B\): \(\bar{U}\) represents complete-data sampling uncertainty, while \(B\) represents uncertainty associated with the missing values. With only \(M\) imputations, the calculated \(\bar{Q}\) also has Monte Carlo error. Conditional on the observed incomplete data, the variance of the mean of \(M\) independent imputation-specific estimates is approximately \(B/M\). Adding that finite-simulation contribution gives \(\bar{U}+B+B/M\).
Each completed-data analysis estimates the target with variance \(U_m\). Averaging these estimates gives the pooled contribution \(\bar{U}\).
Different plausible completed data sets yield different \(Q_m\). The sample variance \(B\) estimates the variance across imputations that contributes to the ideal, infinite-imputation variance.
The mean of \(M\) imputation-specific estimates has Monte Carlo variance approximately \(B/M\), in addition to the uncertainty represented by \(B\).
\(T=\bar{U}+B+B/M=\bar{U}+(1+1/M)B\).
The factor \(1+1/M\) therefore consists of two parts, not one: \(1\) multiplies the between-imputation contribution itself, and \(1/M\) allows for the error in approximating the infinite-imputation average with a finite set. This derivation uses \(B\) as an estimate of the relevant between-imputation variability. With a finite number of imputations, \(B\) is itself estimated imprecisely, so the formula is an approximation under the stated conditions, not an exact identity for every possible imputation procedure.
Worked Example: Combining Two Sources of Uncertainty
Suppose five imputations give hypothetical estimates of an adjusted systolic blood pressure difference, measured in mmHg: \(2.1,\ 1.8,\ 2.4,\ 2.0,\ 2.3\). Suppose their estimated within-imputation variances, in \(\text{mmHg}^2\), are \(0.16,\ 0.25,\ 0.09,\ 0.36,\ 0.14\). Calculate the pooled estimate and total variance.
First, the pooled estimate is \( \bar{Q}=(2.1+1.8+2.4+2.0+2.3)/5=10.6/5=2.12\) mmHg.
The pooled within-imputation variance is \( \bar{U}=(0.16+0.25+0.09+0.36+0.14)/5=1.00/5=0.2000\ \text{mmHg}^2\).
For the between-imputation variance, the deviations from \(2.12\) are \(-0.02,\ -0.32,\ 0.28,\ -0.12,\ 0.18\) mmHg. Thus \( B=[0.0004+0.1024+0.0784+0.0144+0.0324]/(5-1)=0.2280/4=0.05700\ \text{mmHg}^2\).
With \(M=5\), the total variance is \( T=0.2000+(1+1/5)(0.05700)=0.2000+1.2(0.05700)=0.2684\ \text{mmHg}^2\). The corresponding standard error is \(\sqrt{0.2684}=0.5181\) mmHg. This is the standard error for the pooled estimate; the variance formula alone does not determine a confidence interval or its degrees of freedom.
Worked Calculations and Interpretation
The arithmetic makes clear why neither source should be dropped. The following examples use hypothetical data and emphasize different features of the formula.
Worked Example: Unequal Within-Imputation Variances
In four hypothetical completed data sets, a regression coefficient estimating the change in an outcome per one-unit increase in a predictor is \(10,\ 12,\ 9,\ 11\) outcome units per predictor unit. The estimated coefficient variances are \(4,\ 9,\ 4,\ 9\) squared coefficient units. Find \(T\).
The pooled coefficient is \( \bar{Q}=(10+12+9+11)/4=42/4=10.5\) outcome units per predictor unit.
The average within-imputation variance is \( \bar{U}=(4+9+4+9)/4=26/4=6.5\) squared coefficient units. The deviations from \(10.5\) are \(-0.5,\ 1.5,\ -1.5,\ 0.5\), so \( B=[0.25+2.25+2.25+0.25]/(4-1)=5/3=1.6667\) squared coefficient units.
Therefore, \( T=6.5+(1+1/4)(1.6667)=6.5+2.0834=8.5834\) squared coefficient units (rounding the final value gives \(8.583\)). The standard error is \(\sqrt{8.5834}=2.930\) coefficient units. Averaging the four standard errors would not give this result: the pooling formula averages the variance estimates \(U_m\), then adds between-imputation uncertainty.
Worked Example: Seeing the \(1/M\) Term
Suppose two hypothetical imputations yield estimates of a mean difference of \(5\) and \(7\) minutes, and each has within-imputation variance \(1.5\ \text{minutes}^2\). Calculate the pooled total variance and identify the part added by the finite-\(M\) correction.
The pooled estimate is \(\bar{Q}=(5+7)/2=6\) minutes, and \(\bar{U}=1.5\ \text{minutes}^2\). The two deviations from \(6\) are \(-1\) and \(1\), so \( B=[(-1)^2+1^2]/(2-1)=2\ \text{minutes}^2\).
The total variance is \( T=1.5+(1+1/2)(2)=1.5+3.0=4.5\ \text{minutes}^2\). The \(B\) contribution is \(2.0\ \text{minutes}^2\); the finite-\(M\) addition \(B/M\) is \(2/2=1.0\ \text{minutes}^2\). If \(B\) stayed at \(2.0\ \text{minutes}^2\) while \(M\) increased, the \(B/M\) term would shrink; for example, at \(M=20\), it would be \(2/20=0.1000\ \text{minutes}^2\). The total standard error for the two-imputation calculation is \(\sqrt{4.5}=2.121\) minutes.
Assumptions, Diagnostics, and Alternatives
Rubin’s formula is a combining rule, not a test of whether the imputation model was sound. Its interpretation depends on a consistent estimand and appropriate imputation and analysis models. The imputation procedure should reflect uncertainty in the missing values and in model parameters; deterministic fills generally do not provide the between-imputation variation required by the rule. Inference may also be poor when the analysis is highly non-normal or near a boundary. The formula does not by itself establish that missingness assumptions hold or that the imputation model is correctly specified.
Before trusting a pooled variance, check that the imputation process has stabilized and that the estimates \(Q_m\) do not show unexplained trends across iterations. Compare the \(U_m\) values for unusual outliers, confirm they refer to the same estimand and scale, and inspect whether between-imputation variation is plausible given the missingness and imputation procedure. A value of \(B=0\) can occur if all \(Q_m\) are identical, but it can also be a warning that the imputation process is not reflecting uncertainty. With small \(M\), \(B\) can be particularly noisy.
| Approach | What it represents | Key limitation for variance |
|---|---|---|
| Rubin’s total variance | Within-imputation variance, between-imputation variance, and the finite-\(M\) correction | Requires suitable imputations, compatible analyses, and regularity conditions |
| Average within-imputation variance only | \(\bar{U}\) from completed-data analyses | Omits uncertainty reflected by variation among imputations |
| Single imputation | One completed data set and its analysis | Does not estimate between-imputation variability, so its reported variance does not reflect that source |
An important pitfall is to reason, “Each completed data set has a standard error, so the pooled standard error is just their average.” That calculation ignores variation in the estimates across imputations and averages standard errors rather than variances. The correction is to calculate \(\bar{U}\) from the \(U_m\) values, calculate \(B\) from the \(Q_m\) values, and then use \(T=\bar{U}+(1+1/M)B\). Another pitfall is to include \(B/M\) but omit \(B\): the correction is added to, not substituted for, the between-imputation contribution.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- In three imputations, the estimates are \(4,\ 5,\ 6\). Calculate \(\bar{Q}\) and \(B\).
- If \(\bar{U}=3.0\), \(B=1.2\), and \(M=5\), calculate \(T\) and the pooled standard error.
- Why does the total variance contain both \(B\) and \(B/M\)?
- A researcher averages the standard errors from each completed-data analysis and reports that as the pooled standard error. What is missing from that reasoning?
Solutions. (1) \(\bar{Q}=(4+5+6)/3=5\). The squared deviations are \(1,0,1\), so \(B=(1+0+1)/(3-1)=1\), in squared estimate units. (2) \(T=3.0+(1+1/5)(1.2)=3.0+1.44=4.44\); the standard error is \(\sqrt{4.44}=2.107\), in the estimate’s units. (3) \(B\) represents missing-data uncertainty in the ideal pooled estimate, while \(B/M\) represents additional Monte Carlo variance from estimating that pooled quantity with only \(M\) imputations. (4) The calculation neither averages the within-imputation variances as \(\bar{U}\) nor incorporates variation among the \(Q_m\) estimates through \(B\); the pooled standard error is \(\sqrt{\bar{U}+(1+1/M)B}\).