Why Regression Estimates Must Be Pooled
Multiple imputation represents uncertainty about missing values by creating several completed datasets, rather than filling each missing entry with one fixed guess. The same regression model is then fitted to every completed dataset. This produces a set of estimates for each coefficient and a set of standard errors. The estimates will usually differ because the completed datasets differ.
The goal is not to choose one completed dataset or to average its standard error with the others. Pooling must reflect two sources of uncertainty: the usual sampling uncertainty within each completed dataset and the extra uncertainty indicated by variation in estimates across completed datasets. Rubin’s rules provide a standard way to combine these quantities for a coefficient.
For this tutorial, \(SE_j^2\) is the estimated variance of \(\hat{\beta}_j\) within completed dataset \(j\). It is called the within-imputation variance. Variation among the \(\hat{\beta}_j\) values measures how much the coefficient depends on which plausible completion was made; this is the between-imputation variance.
Rubin’s Rules for One Coefficient
Pool one regression coefficient at a time. If the model contains several coefficients, repeat the calculation for each coefficient using its estimate and variance from each completed dataset.
Here, \(\bar{\beta}\) is the pooled point estimate, \(\bar{U}\) is the average within-imputation variance, \(B\) is the between-imputation variance, and \(T\) is the total variance. The pooled standard error is the square root of that total variance. The factor \(1+1/m\) matters: simply adding \(\bar{U}\) and \(B\), or omitting \(B\), does not implement Rubin’s rules.
The variances have the square of the coefficient’s units. If a coefficient is measured in mmHg per year, then \(SE_j\) is also in mmHg per year, while \(SE_j^2\), \(\bar{U}\), \(B\), and \(T\) are in \((\text{mmHg per year})^2\), equivalently mmHg²/year². Taking the square root returns the pooled standard error to the coefficient’s original units.
Worked Calculation and Interpretation
Worked Example: Pooling an Exercise Slope
Suppose five hypothetical completed datasets are each analysed using the same regression of systolic blood pressure (mmHg) on weekly exercise (hours). The coefficient of interest is the exercise slope, measured in mmHg per additional hour of weekly exercise. The results are:
| Completed dataset | Slope estimate (mmHg/hour) | Standard error (mmHg/hour) |
|---|---|---|
| 1 | −1.8 | 0.60 |
| 2 | −2.2 | 0.55 |
| 3 | −1.5 | 0.65 |
| 4 | −2.0 | 0.58 |
| 5 | −2.5 | 0.62 |
Pooled coefficient. The sum of the five estimates is \(-10.0\), so
Within-imputation variance. Square each standard error before averaging:
Between-imputation variance. The deviations from \(-2.00\) are \(0.20\), \(-0.20\), \(0.50\), \(0.00\), and \(-0.50\) mmHg/hour. Thus,
Total variance and pooled standard error. With \(m=5\), the multiplier is \(1+1/5=1.2\):
The pooled slope is \(-2.00\) mmHg per additional hour of weekly exercise, with pooled standard error \(0.732\) mmHg/hour. In the fitted regression, an additional weekly exercise hour is associated with an estimated 2.00 mmHg lower systolic blood pressure, on average, according to the model. This is an association, not by itself evidence that exercise caused the difference.
Why the Total Variance Has Two Contributions
The total variance combines the uncertainty that would remain within a completed dataset with uncertainty arising because missing values could plausibly have been completed in different ways. A short variance decomposition explains the roles of the two terms.
The estimate has an estimated sampling variance \(U_j=SE_j^2\). Averaging these across the \(m\) fits gives \(\bar{U}\).
The estimates vary from one completed dataset to another. Their sample variance \(B\) measures this between-imputation component of uncertainty.
The observed between-imputation variation is itself estimated from only \(m\) fits. The standard multiple-imputation variance combination adds \(B\) and the finite-imputation adjustment \(B/m\) to \(\bar{U}\), giving \(T=\bar{U}+B+B/m\).
Combining the last expression gives \(T=\bar{U}+(1+1/m)B\). This justification also clarifies why the two terms are not interchangeable: \(\bar{U}\) comes from the regression standard errors within completed datasets, whereas \(B\) comes from the spread of their coefficient estimates.
What the Amount of Between-Imputation Variation Tells You
A relatively small \(B\) means the coefficient estimates are similar across the completed datasets. A larger \(B\), relative to \(\bar{U}\), means the estimates are more sensitive to the uncertain missing values represented by the imputations. The size of \(B\) should therefore be interpreted relative to the within-imputation variance, not in isolation. Both are expressed in squared coefficient units.
The following second example shows why \(B\) is not a substitute for ordinary regression uncertainty.
Worked Example: Similar Estimates Do Not Mean Zero Uncertainty
Suppose four hypothetical completed datasets give the same coefficient estimate, \(4.0\) score points, with a standard error of \(0.50\) score points in each fit. The pooled estimate is
Every estimate equals the pooled estimate, so all deviations are zero and \(B=0\). The average within-imputation variance is
Therefore, \(T=0.25+(1+1/4)(0)=0.25\text{ points}^2\), and \(SE_{\mathrm{pooled}}=\sqrt{0.25}=0.50\) points. The imputations introduce no observed between-imputation variation in this example, but the regression estimates still have sampling uncertainty within each completed dataset.
Pool Each Coefficient Separately
The same principle applies in a multiple regression. For every coefficient, use that coefficient’s estimates and standard errors from all \(m\) fits. Do not combine the estimate for one predictor with the standard error for another. The next example pools a treatment coefficient from a hypothetical regression; the unit of the coefficient is score points because treatment is an indicator in a model whose outcome is measured in score points.
Worked Example: Pooling a Treatment Coefficient
Four hypothetical completed datasets produce treatment coefficient estimates of \(3.1\), \(2.8\), \(3.4\), and \(2.7\) score points, with standard errors \(0.9\), \(0.8\), \(1.0\), and \(0.9\) score points. The same treatment coding and regression specification are used in all four fits.
First, the pooled estimate is \((3.1+2.8+3.4+2.7)/4=12.0/4=3.00\) points. The average within-imputation variance is
The deviations from \(3.00\) are \(0.10\), \(-0.20\), \(0.40\), and \(-0.30\) points, so
Using \(m=4\), the total variance is \(T=0.815+(1+1/4)(0.100)=0.815+0.125=0.940\text{ points}^2\). Thus, \(SE_{\mathrm{pooled}}=\sqrt{0.940}=0.9695\) points. The treatment coefficient is pooled as \(3.00\) score points with standard error \(0.970\) points. The coefficient’s substantive meaning depends on the regression model and treatment coding; pooling does not change that meaning.
The arithmetic in this example also shows why coefficient-specific pooling is essential: each coefficient has its own set of estimates, standard errors, units, and between-imputation variation. A standard error from one term cannot be used to pool a different term.
Common Mistakes
- Averaging the standard errors. The rule requires averaging the estimated variances \(SE_j^2\), then combining them with \(B\), and finally taking a square root. Averaging standard errors skips both the variance-scale calculation and the between-imputation contribution.
- Ignoring \(B\). This treats the completed datasets as if they gave the same answer, even when their coefficient estimates differ. The total variance must include \((1+1/m)B\).
- Using the wrong denominator for \(B\). The between-imputation variance is a sample variance across estimates, so its denominator is \(m-1\), not \(m\).
- Pooling unlike quantities. Estimates must refer to the same model coefficient with consistent coding and units. Otherwise their average does not have a clear common interpretation.
- Confusing variance units with standard-error units. A slope in mmHg/year has variance in \((\text{mmHg/year})^2\), not merely mmHg²/year. The pooled standard error returns to mmHg/year after taking the square root.
- Interpreting an association as a cause. A pooled regression coefficient remains an association conditional on the model; imputation and pooling alone do not establish causation.
Check Your Understanding
Answer each question using the pooling definitions and rules.
- In Rubin’s rules, what information is represented by \(\bar{U}\), and what information is represented by \(B\)?
- Four imputations give a coefficient estimate of \(1.2\) in every dataset. What is \(B\), and does that imply the pooled standard error must be zero? Explain.
- Why are the individual standard errors squared before calculating \(\bar{U}\)?
- A regression slope is measured in kilograms per month. State the units of its within-imputation variance, between-imputation variance, and pooled standard error.
- Why must the coefficient have the same coding and interpretation in all completed datasets before its estimates are pooled?