Worked Rubin Pooling For A Linear Regression Slope

Why Pooling a Slope Requires More Than Averaging

With multiple imputation, the same regression is fitted to each completed data set. Each fit gives a slope estimate and its estimated variance. Rubin pooling combines both sources of uncertainty: uncertainty within each completed-data analysis and variation in estimates across imputations. This tutorial works through that calculation for a linear-regression slope.

The pooled result is meaningful only when every completed-data analysis estimates the same slope on the same scale, with compatible model specification and variance estimates. The imputation procedure must also be appropriate for the analysis and yield imputations that represent uncertainty about missing values. These conditions do not guarantee that a regression model is correct, but pooling cannot repair a changed estimand, a misspecified model, or unsuitable imputations.

Definition: For \(m\) imputations, let \(Q_i\) be the slope estimate and \(U_i\) its estimated variance in imputation \(i\). The pooled estimate is \(\bar{Q}=m^{-1}\sum_{i=1}^m Q_i\). The average within-imputation variance is \(\bar{U}=m^{-1}\sum_{i=1}^m U_i\), the between-imputation variance is \(B=(m-1)^{-1}\sum_{i=1}^m(Q_i-\bar{Q})^2\), and the total variance is \(T=\bar{U}+(1+1/m)B\). The pooled standard error is \(\sqrt{T}\).

For a regression slope, \(U_i\) is the estimated variance of that slope, not its standard error. If a fitted model reports a standard error \(SE_i\), then \(U_i=SE_i^2\). The factor \(1+1/m\) in \(T\) accounts for the fact that the missing-data uncertainty is estimated using a finite number of imputations.

Worked Pooling Calculation

Worked Example: Pooling Five Slopes for an Exercise Association

In a hypothetical study, the outcome is systolic blood pressure in mmHg and the predictor is weekly exercise in hours. A linear regression with the same covariates is fitted to each of five imputed data sets. Suppose the slope estimates and their standard errors are as follows.

Imputation \(i\)	Slope \(Q_i\) (mmHg per exercise hour)	Standard error \(SE_i\)
1	0.42	0.10
2	0.50	0.11
3	0.46	0.09
4	0.54	0.10
5	0.48	0.10

The point estimate is the arithmetic mean of the five slopes:

\bar{Q}=\frac{0.42+0.50+0.46+0.54+0.48}{5} =\frac{2.40}{5}=0.4800\ \text{mmHg per exercise hour}.

Square each reported standard error to obtain its variance. Their sum is \(0.0100+0.0121+0.0081+0.0100+0.0100=0.0502\), so:

\bar{U}=\frac{0.0502}{5}=0.01004\ (\text{mmHg per exercise hour})^2.

The deviations from the pooled slope are \(-0.06, 0.02, -0.02, 0.06,\) and \(0.00\). Thus the sum of squared deviations is \(0.0036+0.0004+0.0004+0.0036+0=0.0080\), giving:

B=\frac{0.0080}{5-1}=0.002000\ (\text{mmHg per exercise hour})^2.

The total variance and pooled standard error are:

T=0.01004+\left(1+\frac{1}{5}\right)(0.002000) =0.01244,\qquad \sqrt{T}=0.1115\ \text{mmHg per exercise hour}.

The modest increase from \(\sqrt{\bar{U}}\) reflects the additional between-imputation component. The next step is to account for finite-sample uncertainty when choosing the \(t\) reference distribution.

Adjusted Degrees of Freedom and Interval

A common small-sample procedure is the Barnard–Rubin degrees-of-freedom adjustment. Let \(\nu_{\mathrm{com}}\) be the complete-data degrees of freedom for the slope in the corresponding regression. Define the relative increase in variance due to nonresponse as \(r=(1+1/m)B/\bar{U}\), and the estimated fraction of missing information as \(\lambda=(1+1/m)B/T\). The adjustment combines the degrees of freedom associated with the finite number of imputations and the information available in the observed data.

Result: The old degrees of freedom are \(\nu_{\mathrm{old}}=(m-1)(1+1/r)^2\). The observed-data degrees of freedom are \(\nu_{\mathrm{obs}}=((\nu_{\mathrm{com}}+1)/(\nu_{\mathrm{com}}+3))\nu_{\mathrm{com}}(1-\lambda)\). The adjusted degrees of freedom are \(\nu=(\nu_{\mathrm{old}}\nu_{\mathrm{obs}})/(\nu_{\mathrm{old}}+\nu_{\mathrm{obs}})\).

The final expression is the harmonic combination of the two degrees-of-freedom limits: it is smaller than either finite component. When between-imputation variation is zero, the old degrees of freedom tend to infinity; the calculation should then be interpreted by its limiting value rather than by dividing by zero.

Worked Example: Calculating the Adjusted Degrees of Freedom

Continue the hypothetical exercise analysis, and suppose the complete-data regression has \(\nu_{\mathrm{com}}=44\). Using the quantities already calculated, the relative increase and missing-information fraction are:

r=\frac{1.2(0.002000)}{0.01004}=0.2390,\qquad \lambda=\frac{1.2(0.002000)}{0.01244}=0.1929.

Then:

\nu_{\mathrm{old}}=4\left(1+\frac{1}{0.2390}\right)^2\approx107.5, \qquad \nu_{\mathrm{obs}}=\frac{45}{47}(44)(1-0.1929)\approx34.00.

Combining them gives \(\nu=(107.5)(34.00)/(107.5+34.00)\approx25.83\). For a 95% confidence interval, the critical value is approximately \(t_{0.975,25.83}=2.056\). Therefore:

\bar{Q}\pm t_{0.975,\nu}\sqrt{T} =0.4800\pm(2.056)(0.1115) \approx0.4800\pm0.2292 =[0.2508,\ 0.7092]\ \text{mmHg per exercise hour}.

Under the model and pooling assumptions, this procedure gives a 95% confidence interval for the population regression slope of about 0.251 to 0.709 mmHg per additional weekly exercise hour, conditional on the other covariates in the model. The interval does not mean there is a 95% probability that this fixed slope lies inside these particular limits.

The variance formula has a useful interpretation. The average within-imputation variance, \(\bar{U}\), captures the sampling uncertainty estimated by the regressions. Estimates differ across imputations because the missing values differ; \(B\) measures that variation. If infinitely many imputations could be used, the missing-data component would be represented by \(B\); with only \(m\) imputations, its contribution to the variance of the pooled estimate is inflated to \((1+1/m)B\). This explains both the total-variance formula and why simply averaging the standard errors omits relevant uncertainty.

Testing a Specified Slope

For a two-sided test of a specified slope \(\beta_0\), use \(H_0:\beta=\beta_0\) against \(H_a:\beta\ne\beta_0\). The pooled test statistic is \(t=(\bar{Q}-\beta_0)/\sqrt{T}\), compared with a \(t\) distribution using the adjusted degrees of freedom. This test and the corresponding two-sided confidence interval agree: a value is rejected at level \(\alpha\) exactly when it lies outside the matching \(100(1-\alpha)\%\) interval, subject to rounding.

Worked Example: Testing a Slope of 0.40

Using the pooled exercise result, test \(H_0:\beta=0.40\) against \(H_a:\beta\ne0.40\) at \(\alpha=0.05\). The estimates are in mmHg per exercise hour, so the null value has the same units.

t=\frac{0.4800-0.4000}{0.1115}\approx0.7175,\qquad df\approx25.83.

The two-sided \(p\)-value from a \(t\) distribution with 25.83 degrees of freedom is approximately 0.480. Since \(0.480>0.05\), the data do not provide sufficient evidence at the 5% level that the population slope differs from 0.40 mmHg per exercise hour. This is not evidence that the slope is exactly 0.40. Consistently, 0.40 lies inside the 95% interval calculated above.

Assumptions, Diagnostics, and Alternatives

Before interpreting the pooled answer, check that the regression in every imputed data set targets the same conditional slope, uses the same coding and units, and has an appropriate variance estimate. The complete-data inference should be adequately represented by the regression model and its degrees of freedom; in particular, the residual-based standard errors need to be defensible for the sample and model. Rubin pooling also relies on imputations that appropriately represent uncertainty, rather than treating filled-in values as known. The missing-data assumptions and imputation-model construction belong to the broader MICE framework; they are not established merely by doing the pooling arithmetic.

Useful diagnostics include comparing the five \(Q_i\) values and \(SE_i\) values, checking for unusual outlying estimates, and verifying that the variance components are on the correct squared scale. Large \(B\) relative to \(\bar{U}\) means estimates vary appreciably across imputations and can produce a larger pooled standard error and fewer effective degrees of freedom. Strong drift or unstable estimates across imputations may signal a poorly behaved imputation or analysis, not a reason to suppress \(B\). With only five imputations, Monte Carlo variability can also be noticeable; adding imputations may stabilize the numerical summaries but cannot fix incompatibility or model misspecification.

Complete-case regression is a different analysis: it discards records with missing values and may lose precision or target a different population under some missingness patterns. Single imputation treats one completed data set as if it were the observed data and generally fails to represent missing-data uncertainty. Rubin pooling retains the variation across multiple completed-data analyses. It is not equivalent to averaging the imputation-specific standard errors or \(p\)-values, because neither operation combines within- and between-imputation variance according to the pooling rules.

Worked Example: A Different Slope and a Less Precise Test

For a second hypothetical regression, five imputations give slopes \(1.8, 1.9, 2.0, 1.9,\) and \(2.1\), each with standard error \(0.20\) units of outcome per predictor unit. Suppose again that \(\nu_{\mathrm{com}}=44\). The mean slope is \(1.94\), the average within variance is \(0.20^2=0.0400\), and the squared deviations sum to \(0.0520\), so \(B=0.0520/4=0.0130\).

T=0.0400+1.2(0.0130)=0.05560,\qquad \sqrt{T}=0.2358.

Here \(r=0.0156/0.0400=0.3900\) and \(\lambda=0.0156/0.05560=0.2806\). The degrees of freedom are \(\nu_{\mathrm{old}}=4(1+1/0.3900)^2\approx50.81\), \(\nu_{\mathrm{obs}}=(45/47)(44)(1-0.2806)\approx30.31\), and \(\nu\approx18.98\). To test a specified slope of 1.50, the statistic is \(t=(1.94-1.50)/0.2358\approx1.866\). The two-sided \(p\)-value is about 0.078, so at \(\alpha=0.05\) the test does not reject \(H_0\). The relatively large between-imputation variation contributes to the standard error and reduces the adjusted degrees of freedom; ignoring either would overstate the precision of this test.

Common Mistakes and Limits

Averaging standard errors. The pooled variance uses the average of \(SE_i^2\), not the average of \(SE_i\); it then adds the between-imputation component.
Omitting the finite-imputation factor. Use \((1+1/m)B\), not just \(B\), in \(T\).
Using the wrong degrees of freedom. The number of imputations alone does not determine the test degrees of freedom. The adjustment also depends on the within- and between-imputation variances and complete-data degrees of freedom.
Pooling incompatible slopes. A change in outcome scale, predictor coding, covariates, or target population across imputations means the estimates may not represent the same quantity.
Reading a nonsignificant test as proof of no association. A large \(p\)-value says the data do not provide sufficient evidence against the specified null at the chosen level; it does not establish that the null slope is true.

Rubin pooling is a principled summary under its assumptions, not a safeguard against all inferential problems. If the number of imputations is small, degrees of freedom are especially limited, or imputations are poorly behaved, the nominal interval and test may not have their intended performance. Report the pooled slope, its units, standard error or interval, and the relevant inferential method so readers can see what was estimated and how uncertainty was represented.

Check Your Understanding

Answer each question, then compare with the brief solutions.

If five imputation-specific slope standard errors are all 0.12, what is each corresponding within-imputation variance?
Suppose \(m=5\), \(\bar{U}=0.020\), and \(B=0.005\). Calculate \(T\) and \(\sqrt{T}\).
For a pooled slope of 0.30, pooled standard error 0.10, and adjusted \(df=20\), calculate the test statistic for \(H_0:\beta=0.20\).
Why is it incorrect to average the five standard errors and call that the pooled standard error?

Solutions. (1) \(U_i=SE_i^2=0.12^2=0.0144\), in squared slope units. (2) \(T=0.020+1.2(0.005)=0.0260\), so \(\sqrt{T}\approx0.1612\). (3) \(t=(0.30-0.20)/0.10=1.00\). (4) Averaging standard errors omits the between-imputation variance and does not implement Rubin’s total-variance formula; pool the variances and include \((1+1/m)B\) instead.