Why the Number of Imputations Affects Precision
Multiple imputation represents uncertainty about missing values by creating several completed data sets. Rubin’s pooling rules combine the estimates and their variances across those data sets. With a finite number \(M\) of imputations, the pooled variance includes a finite-\(M\) correction; as \(M\) increases, that correction shrinks. The practical question is how large \(M\) must be for this additional uncertainty to have little effect on precision.
Relative efficiency gives an approximate answer. It compares the variance from using \(M\) imputations with the variance that would be obtained, under the same imputation and analysis procedures, if the number of imputations were effectively infinite. It is a planning aid, not a test of whether an imputation model is valid.
Deriving the Approximation
For imputation \(m\), let \(q_m\) be the estimate of the same scalar estimand and let \(U_m\) be its estimated variance within that completed data set. The pooled estimate is \(\bar{q}\), the average of the \(q_m\). Write \(\bar{U}\) for the average of the \(U_m\), and \(B\) for the between-imputation variance of the \(q_m\). Rubin’s total variance is
The term \(B/M\) is the finite-number correction. Under a valid imputation and analysis procedure, and with enough imputations for the variance components to be stable, \(\bar{U}\) and \(B\) approximate their limiting values as \(M\) grows. Denote those limits by \(U\) and \(B_\infty\). Then the infinite-imputation variance is \(T_\infty=U+B_\infty\).
The limiting fraction of missing information can be written \(\gamma_\infty=B_\infty/T_\infty\). Substituting the limiting variance components into the finite-\(M\) expression gives the approximation:
Since relative efficiency is the reciprocal variance ratio, this gives \(RE \approx (1+\gamma_\infty/M)^{-1}\). In practice, the estimated fraction of missing information from the scalar pooling calculation is used as a plug-in approximation for \(\gamma_\infty\). It is therefore an approximation, not an exact identity for every finite data set and finite \(M\).
Worked Example: Relative Efficiency at Several Values of M
Suppose a hypothetical scalar estimate has estimated fraction of missing information \(\gamma=0.40\). Compare \(M=5\), \(M=10\), and \(M=20\) imputations using the usual approximation.
For \(M=5\),
For \(M=10\),
For \(M=20\),
The approximation suggests that \(M=20\) yields about 98.0% of the infinite-imputation efficiency for this estimate, compared with about 92.6% at \(M=5\). These are variance-based comparisons. For example, the \(M=5\) variance is approximately \(1/0.9259=1.080\) times the infinite-imputation variance, or about 8.0% larger.
Planning M for a Target Efficiency
The approximation can be rearranged to find a minimum number of imputations for a desired relative efficiency \(R\), where \(0<R<1\). Starting from \(1/(1+\gamma/M)\ge R\), and using positive quantities, the inequality is equivalent to \(1+\gamma/M\le 1/R\). Therefore:
The required \(M\) must be rounded up to a whole number. This calculation uses a chosen target and an estimated \(\gamma\); it does not establish that the chosen imputation model is adequate.
Worked Example: Choosing M for at Least 99% Efficiency
Using the hypothetical \(\gamma=0.40\) above, find the smallest integer \(M\) that gives estimated relative efficiency of at least 99%.
Set \(R=0.99\) and substitute:
Because \(M\) must be an integer, the smallest qualifying choice is \(M=40\). Checking directly,
Thus the approximation gives relative efficiency slightly above 99% at \(M=40\). This is a planning calculation: if the fraction of missing information is uncertain or varies substantially across the estimates that matter, choosing \(M\) based on a single value may not provide the intended efficiency for every estimate.
Variance Efficiency and Standard-Error Change
A relative efficiency value is a variance ratio in reciprocal form; it is not a percentage change in standard error. If \(T_M/T_\infty\approx 1+\gamma/M\), then the corresponding standard-error ratio is approximately \(\sqrt{1+\gamma/M}\). This distinction matters when describing the practical size of the finite-\(M\) effect.
Worked Example: Translating a Variance Difference to Standard Error
Suppose the infinite-imputation variance is \(5.00\) squared units, and a hypothetical calculation gives \(\gamma=0.20\). At \(M=10\), the finite-\(M\) correction is \(B/M\). For illustration, let \(\bar{U}=4.00\) and \(B=1.00\) squared units, so that \(T_\infty=4.00+1.00=5.00\).
The finite-\(M\) total variance is
The resulting relative efficiency is \(T_\infty/T_{10}=5.00/5.10=0.9804\). Equivalently, the approximate formula gives \(1/(1+0.20/10)=1/1.02=0.9804\). The standard error based on \(M=10\), relative to the infinite-imputation standard error, is larger by
That is about a 0.995% increase in standard error. The variance is 2.0% larger than its infinite-imputation value, but the standard error increases by about 1.0%, because standard error is the square root of variance. The numerical values are hypothetical and are stated in the units of the estimate and its variance.
Diminishing Returns, Diagnostics, and Limits
The formula shows diminishing returns: increasing \(M\) reduces the term \(\gamma/M\), but each additional imputation has a smaller effect when \(M\) is already large. A high fraction of missing information calls for more imputations to achieve the same approximate efficiency. For example, doubling \(M\) halves \(\gamma/M\), not the variance itself and not necessarily the standard-error difference.
Before relying on an efficiency calculation, check that it applies to the estimate of interest and that the ingredients are credible. In particular:
- Use the fraction of missing information for the scalar estimate being considered. Different coefficients or estimands can have different values.
- Check that the imputation procedure has run adequately and that estimates and variance components are stable across imputations. Poorly mixing chained equations or unstable imputation-specific estimates can make a plug-in \(\gamma\) unreliable.
- Recognize that a very small \(M\) can make the estimated between-imputation variance itself noisy. The approximation does not remove Monte Carlo variability in \(\bar{q}\), \(\bar{U}\), or \(B\).
- Keep validity separate from efficiency. A high relative efficiency does not show that the missing-data assumptions, imputation models, or analysis model are appropriate, and it does not eliminate bias from model misspecification.
The main practical alternative to increasing \(M\) is not a different interpretation of the same formula: it is to assess whether the analysis is stable and whether the model is suitable for the scientific estimand. If computing resources are limited, a pilot run can help reveal the scale of \(\gamma\) and the stability of pooled estimates; a final run can then use a larger \(M\) if needed. Single imputation is not an equivalent way to obtain the same precision, because it fails to represent imputation uncertainty.
Relative efficiency is also not the only reason to choose \(M\). A larger number of imputations can reduce Monte Carlo noise in pooled estimates and support more stable downstream calculations. Conversely, when \(\gamma\) is near zero, the approximation predicts little variance penalty from a moderate \(M\), but it still does not verify model fit or convergence. The calculation is best treated as one targeted precision diagnostic, alongside checks of the imputation procedure and substantive analysis.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For \(\gamma=0.30\) and \(M=15\), calculate approximate relative efficiency.
- For \(\gamma=0.60\), what is the smallest integer \(M\) that gives approximate relative efficiency of at least 95%?
- If relative efficiency is 0.96, what is the approximate ratio of finite-\(M\) variance to infinite-imputation variance?
- In one sentence, explain why high relative efficiency does not establish that the imputation model is valid.
Solutions. 1. \(RE\approx1/(1+0.30/15)=1/1.02=0.9804\), or about 98.0%. 2. \(M\ge0.60/(1/0.95-1)=0.60/0.05263\ldots=11.4\), so round up to \(M=12\); checking gives \(1/(1+0.60/12)=1/1.05=0.9524\), which exceeds 95%. 3. The variance ratio is approximately \(1/0.96=1.0417\), so the finite-\(M\) variance is about 4.17% larger than the infinite-imputation variance. 4. Relative efficiency compares finite and infinite numbers of imputations under the same procedure; it measures the finite-\(M\) variance cost, not whether the procedure’s assumptions or models are correct.