Why Compare the Added Variance With Within-Imputation Variance?
Multiple imputation produces several completed data sets, and the analysis of each one gives an estimate and an estimate of its variance. Those variances describe uncertainty within the completed data sets. The estimates can also differ across imputations, reflecting uncertainty associated with the unobserved values. A useful question is not only how much variance is added, but how large that addition is relative to the within-imputation uncertainty.
The relative increase in variance, \(r\), answers this comparison for a scalar estimate. It is a dimensionless ratio: \(r=0.20\), for example, means that the estimated added variance is 20% of the average within-imputation variance. It does not mean that 20% of observations are missing. Its value depends on the estimate, its within-imputation precision, the between-imputation variation, and the number of imputations.
Definition and Conditions for Use
Suppose the same scalar quantity is estimated in each of \(M\) imputed data sets. Let \(Q_m\) be its estimate in imputation \(m\), and let \(U_m\) be the estimated variance of \(Q_m\) from the analysis of that completed data set. The average within-imputation variance is \(\bar{U}\), and the between-imputation variance is \(B\).
The formula applies to a scalar estimate when all \(M\) analyses target the same estimand, use compatible variance estimates, and use imputations that adequately represent uncertainty about missing values. Rubin’s total variance is \(T=\bar{U}+(1+1/M)B\); this tutorial focuses on expressing its added component relative to \(\bar{U}\). In practice, \(U_m\) should be a valid variance estimate for \(Q_m\), and the imputation procedure should be proper for the inferential task. If the imputation model omits important uncertainty or the imputations have not adequately explored their distributions, \(B\) may not capture the uncertainty it is intended to reflect. The ratio cannot correct those problems.
The denominator must be positive for \(r\) to be defined. The numerator and denominator are both variances, so they have the same squared units and \(r\) has no units. The estimates themselves must be on the same analysis scale across imputations; a variance in one scale cannot sensibly be compared with a variance in another.
Why the Finite-Imputation Correction Appears
The between-imputation variance \(B\) is calculated from a finite set of \(M\) estimates. In Rubin’s total-variance formula, the contribution associated with between-imputation variation is therefore \((1+1/M)B\), rather than just \(B\). The factor approaches 1 as \(M\) grows, but is larger than 1 for any finite \(M\). The relative increase is the ratio of that corrected contribution to the average within-imputation variance.
Write the pooled variance as \(T=\bar{U}+(1+1/M)B\).
The part beyond the within-imputation variance is \(T-\bar{U}=(1+1/M)B\).
Divide the added part by \(\bar{U}\), giving \(r=(T-\bar{U})/\bar{U}=(1+1/M)B/\bar{U}\).
Since \(T=\bar{U}+(1+1/M)B\), division by \(\bar{U}\) gives \(T/\bar{U}=1+r\).
Thus \(r=0.25\) means the added variance is one quarter of \(\bar{U}\), and the total variance is \(1.25\bar{U}\). This is a statement about variances, not standard errors: the standard error is the square root of a variance, so its relative change is not \(r\).
Worked Example: Calculating the Ratio From Four Imputations
Consider hypothetical estimates of a mean change in symptom score, measured in points, from \(M=4\) imputed data sets. Suppose the estimates are \(12.1,\ 12.8,\ 11.9,\) and \(13.2\) points, and their estimated variances are \(4.00,\ 4.41,\ 3.61,\) and \(4.84\) points squared.
First calculate the average estimate:
Next calculate the within-imputation variance average:
The deviations from \(12.5\) are \(-0.4,\ 0.3,\ -0.6,\) and \(0.7\) points. Their squared deviations sum to \(0.16+0.09+0.36+0.49=1.10\) points squared. Therefore,
The relative increase is
The added variance is about 10.9% of the average within-imputation variance. For context, the total variance is \(4.215+0.4583=4.673\) points squared, and \(4.673/4.215\) is approximately \(1.109=1+r\). The ratio itself has no units. It does not say that 10.9% of patients had missing scores.
Interpreting the Size of the Ratio
A larger \(r\) indicates that between-imputation uncertainty makes a larger contribution relative to the within-imputation variance. A small \(r\) indicates that this contribution is comparatively small. Neither size, on its own, diagnoses whether the imputation model is appropriate. For example, a large ratio can be a plausible result when missing values leave substantial uncertainty; a small ratio can arise because missingness contributes little uncertainty, but it can also reflect a poorly specified model or inadequate exploration of the imputation distributions.
The same value of \(B\) can imply different relative increases for different analyses. If \(\bar{U}\) is small, a given added variance is large relative to it; if \(\bar{U}\) is large, that same addition is a smaller proportion. The ratio is therefore specific to the scalar estimate and its analysis, not a general summary of the data set’s missingness.
Worked Example: Applying the Finite-Imputation Correction
Suppose a hypothetical analysis of a treatment coefficient has \(M=5\), \(\bar{U}=6.00\) coefficient-units squared, and \(B=2.40\) coefficient-units squared. Calculate \(r\), and compare it with the ratio that would result from omitting the finite-imputation correction.
The correction factor is \(1+1/5=1.2\). The corrected added variance is
Thus,
The added variance is 48.0% of the average within-imputation variance, and the total variance is \(6.00+2.88=8.88\) coefficient-units squared. If the correction were omitted, the ratio would be \(B/\bar{U}=2.40/6.00=0.400\), or 40.0%. That is not the specified relative increase for these five imputations: it understates the corrected ratio by 0.080, or 8.0 percentage points. The correction would be smaller with a larger \(M\), but should not be dropped merely because it is convenient.
A Ratio Can Be Read Alongside the Variance Components
Reporting \(r\) without \(\bar{U}\), \(B\), and \(M\) can obscure what drives its value. In particular, identical ratios can result from very different absolute variances. It is often useful to retain the component values in calculations and reports, especially when comparing analyses of the same estimand on the same scale. The following example shows how the ratio and total variance can be calculated together.
Worked Example: Interpreting a Ratio for a Regression Slope
Suppose a hypothetical regression estimates the association between weekly exercise and systolic blood pressure. The scalar estimate is a slope in millimetres of mercury per exercise hour per week. Across \(M=5\) imputations, the estimates are \(2.10,\ 2.00,\ 2.40,\ 1.90,\) and \(2.30\) mmHg per hour. The corresponding within-imputation variances are \(0.09,\ 0.10,\ 0.08,\ 0.11,\) and \(0.09\) (mmHg per hour) squared.
The mean estimate is \((2.10+2.00+2.40+1.90+2.30)/5=2.14\) mmHg per hour. The average within-imputation variance is
The deviations from \(2.14\) are \(-0.04,\ -0.14,\ 0.26,\ -0.24,\) and \(0.16\) mmHg per hour. Squaring and summing gives \(0.0016+0.0196+0.0676+0.0576+0.0256=0.1720\) (mmHg per hour) squared. Therefore,
Then \(r=0.05160/0.0940=0.5489\). The added variance is about 54.9% of the within-imputation variance. The total variance is \(0.0940+0.0516=0.1456\) (mmHg per hour) squared; its square root is approximately \(0.3816\) mmHg per hour. This standard error is in the slope’s units, whereas \(r\) is a unitless variance ratio. The slope is an association in this regression; this calculation does not establish that exercise causes a change in blood pressure.
Diagnostics, Alternatives, and Limitations
Check the components before treating \(r\) as informative. Confirm that the same estimand and scale were used in every completed-data analysis, that all \(U_m\) values are nonnegative and plausible, and that \(\bar{U}>0\). Recheck the calculation if \(B\) is unexpectedly large, the estimates differ sharply across imputations, or a single estimate dominates the squared deviations. If \(B\) is close to zero, verify that the imputations represent uncertainty and that the computation used distinct, adequately explored imputations. A zero or near-zero estimate of \(B\) is not proof that missingness creates no uncertainty.
The formula includes a finite-\(M\) adjustment, but it does not remove Monte Carlo error from estimating \(B\). With few imputations, \(B\) can be unstable, and hence \(r\) can also be unstable. Increasing \(M\) can improve the numerical stability of pooled calculations; it cannot repair an unsuitable imputation model, inconsistent analysis models, or imputations that fail to capture relevant uncertainty. Likewise, if successive saved imputations are highly dependent, the nominal number \(M\) may overstate the amount of independent information in the set. These are reasons to assess the imputation and computation, not to reinterpret the formula.
A common arithmetic alternative is the uncorrected ratio \(B/\bar{U}\). It is smaller than \(r\) by the factor \(1+1/M\), and is not the finite-imputation-adjusted quantity defined here. Reporting \(B\) and \(\bar{U}\) separately can help explain the result, but neither alone gives the relative increase. A complete-case analysis avoids combining estimates across imputations but targets the same population quantity only under additional conditions; a single-imputation analysis does not represent between-imputation uncertainty. Those approaches are not substitutes for correctly calculating \(r\) from a multiple-imputation analysis.
Finally, \(r\) is not the proportion of records with missing data, nor should it be read as a percentage increase in the standard error. If \(r=0.48\), the variance is multiplied by \(1.48\) relative to \(\bar{U}\); the corresponding standard error is multiplied by \(\sqrt{1.48}\), not by \(1.48\). The ratio is a component comparison, not a complete assessment of model quality or a standalone inferential conclusion.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For \(M=9\), \(B=1.50\), and \(\bar{U}=5.00\), calculate \(r\). Interpret it in context as a variance ratio.
- If \(r=0.64\), by what factor is total variance larger than \(\bar{U}\)? What is the corresponding factor for the standard error?
- Why is \(B/\bar{U}\) not the same as the defined relative increase when \(M\) is finite?
- Does \(r=0.05\) establish that only 5% of records were missing? Explain.
Solutions. (1) The correction is \(1+1/9=1.1111\), so the added variance is \(1.1111(1.50)=1.6667\). Thus \(r=1.6667/5.00=0.3333\): the added variance is about 33.3% of the within-imputation variance. (2) Since \(T/\bar{U}=1+r\), total variance is \(1.64\) times \(\bar{U}\). The standard-error factor is \(\sqrt{1.64}=1.281\), approximately. (3) The defined ratio includes the factor \(1+1/M\); for finite \(M\), this factor exceeds 1, so \(B/\bar{U}\) omits part of the specified correction. (4) No. \(r\) compares two variance quantities and is not a missing-record proportion. It depends on the estimate, its within-imputation precision, the between-imputation variation, and \(M\).