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Reference-Based Imputation for Clinical Trials · Tutorial 292 of 499

Rubin-Rule Inference After Reference-Based Imputation

Combine treatment-effect estimates across reference-based imputations and use the pooled uncertainty for valid confidence intervals and tests.

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What You'll Learn

  • Define the pooled treatment-effect estimate and its within- and between-imputation variance components.
  • Calculate Rubin’s total variance and pooled standard error from completed-data analyses.
  • Use the appropriate degrees of freedom and a t distribution for inference.
  • Interpret pooled confidence intervals and hypothesis tests in the trial’s measurement units.
  • Recognize boundary cases and diagnose mistakes in pooling.

From Reference-Based Imputations to an Inferential Result

Reference-based imputation supplies plausible values for missing outcomes under an explicit assumption about post-dropout response. After imputing \(M\) completed data sets, the trial analysis is performed separately in each one. Rubin’s rules combine those analyses so that uncertainty reflects both the usual sampling variability within completed data sets and the variation in estimates across imputations.

The inferential target must be the same in every completed data set. For example, suppose the target is the adjusted difference in mean endpoint change between randomized treatment and reference groups, measured in pain-score points. Each completed-data analysis gives an estimate of that same contrast and an estimated variance for it. Pooling is not a way to remove the assumptions of the reference-based imputation model: it quantifies uncertainty conditional on the imputation and analysis procedures used.

Definition: For imputation \(m=1,\ldots,M\), let \(Q_m\) be the completed-data estimate of the treatment effect and let \(U_m\) be its estimated sampling variance. The pooled estimate is \(\bar{Q}=M^{-1}\sum_{m=1}^{M}Q_m\). The within-imputation variance is \(\bar{U}=M^{-1}\sum_{m=1}^{M}U_m\), and the between-imputation variance is \(B=(M-1)^{-1}\sum_{m=1}^{M}(Q_m-\bar{Q})^2\). Rubin’s total variance is \(T=\bar{U}+(1+1/M)B\), and the pooled standard error is \(SE_{\mathrm{MI}}=\sqrt{T}\).

The factor \(1+1/M\) accounts for the fact that the imputation process is estimated from a finite number of imputations. The term \(\bar{U}\) represents the average variance that would remain if the imputed data sets were complete and analyzed separately. The term \(B\) measures how much the treatment-effect estimate changes across imputations; its contribution does not disappear merely because each completed data set has a small standard error.

Degrees of Freedom and Inference

With a large-sample completed-data analysis, the pooled statistic is compared with a \(t\) distribution rather than automatically using a normal distribution. For the convention used here, define the relative between-imputation variance \(r=B/\bar{U}\). When \(M>1\), \(\bar{U}>0\), and \(B>0\), the infinite-complete-data-degrees-of-freedom approximation is

$$ \nu=(M-1)\left(1+\frac{1}{r}\right)^2. $$

The degrees of freedom \(\nu\) need not be an integer. A two-sided test of \(H_0:\Delta=\Delta_0\) uses \(t=(\bar{Q}-\Delta_0)/SE_{\mathrm{MI}}\) and the \(t\) distribution with \(\nu\) degrees of freedom. A \(100(1-\alpha)\%\) confidence interval is \(\bar{Q}\pm t_{1-\alpha/2,\nu}SE_{\mathrm{MI}}\). These expressions assume that the completed-data estimate is approximately normal with its estimated variance, and that imputations are proper and generated under the specified reference-based procedure. If the completed-data analysis has limited-sample degrees of freedom, an adjustment may be needed; the approximation above treats those degrees of freedom as infinite.

When \(B=0\), the displayed formula for \(\nu\) is undefined because \(r=0\). Under this infinite-complete-data-df approximation, its limiting value as \(B\) approaches zero is infinite. In that boundary case, use the normal distribution for inference. This convention does not mean that the imputation model is known to be correct; it only describes the limiting degrees of freedom implied by this approximation.

Key result: Rubin’s total variance is not the average of the completed-data variances alone. It adds the between-imputation contribution \((1+1/M)B\). The pooled estimate, pooled standard error, degrees of freedom, and reference distribution must be used together when forming an interval or test.

Worked Pooling and Inference

Worked Example: A Pooled Treatment Difference and 95% Interval

Consider a hypothetical randomized trial with a pain-score endpoint, where higher change scores indicate greater improvement. In four proper reference-based imputations, the estimated treatment-minus-reference differences are \(2.0,\ 2.5,\ 3.5,\) and \(4.0\) points. Each completed-data analysis reports \(SE_m=1.2629\) points, so \(U_m=1.2629^2\approx1.595\) points squared. Assume the same treatment-effect contrast and analysis are used in every imputation.

The pooled estimate is

$$ \bar{Q}=\frac{2.0+2.5+3.5+4.0}{4}=3.0\text{ points}. $$

The deviations from \(3.0\) are \(-1.0,-0.5,0.5,1.0\) points. Thus

$$ B=\frac{(-1.0)^2+(-0.5)^2+(0.5)^2+(1.0)^2}{4-1} =\frac{2.50}{3}=0.8333\text{ points}^2, \qquad \bar{U}=1.595\text{ points}^2. $$

With \(M=4\), the total variance and pooled standard error are

$$ T=1.595+\left(1+\frac{1}{4}\right)(0.8333) =1.595+1.0416\approx2.637\text{ points}^2, \qquad SE_{\mathrm{MI}}=\sqrt{2.637}\approx1.624\text{ points}. $$

For the degrees-of-freedom calculation, \(r=B/\bar{U}=0.8333/1.595\approx0.5225\). Therefore

$$ \nu=3\left(1+\frac{1}{0.5225}\right)^2 \approx3(2.914)^2\approx25.47. $$

For \(H_0:\Delta=0\), the statistic is \(t=3.0/1.624\approx1.847\). With about \(25.47\) degrees of freedom, the two-sided \(p\)-value is approximately \(0.076\). At \(\alpha=0.05\), this is not sufficient evidence against a zero treatment difference under the stated imputation and analysis assumptions. The approximate 95% interval uses \(t_{0.975,25.47}\approx2.059\):

$$ 3.0\pm2.059(1.624) =3.0\pm3.344 \approx(-0.344,\ 6.344)\text{ points}. $$

The interval includes zero. In context, the data and analysis are compatible with effects ranging from a small disadvantage of about \(0.34\) points to an improvement of about \(6.34\) points. It is not a statement that there is a 95% probability the fixed treatment effect lies in this particular interval.

Why the Variance Components Matter

The total-variance formula has a short justification. If the imputations were repeated, \(\bar{Q}\) would vary because of sampling uncertainty within a completed data set and because the imputed values—and hence the completed-data estimate—vary across imputations. The within-imputation contribution is estimated by \(\bar{U}\). The finite-\(M\) estimate of the additional variation due to missing values is \((1+1/M)B\). Adding these contributions gives \(T\). This is why averaging the completed-data standard errors, or averaging only the \(U_m\), is not a valid substitute for Rubin’s total variance.

1
Check the inputs.
Confirm that every \(Q_m\) estimates the same treatment contrast and that each \(U_m\) is a variance, not a standard error.
2
Pool the estimate and variances.
Calculate \(\bar{Q}\), \(\bar{U}\), and \(B\), retaining squared outcome units for \(U_m\), \(\bar{U}\), \(B\), and \(T\).
3
Make the inferential calculation.
Calculate \(T\), \(SE_{\mathrm{MI}}\), and the degrees of freedom, then use the corresponding \(t\) or boundary-case normal reference distribution.

A useful diagnostic is to compare \(B\) with \(\bar{U}\). A large \(B\) relative to \(\bar{U}\) signals substantial imputation-related variability and generally fewer effective degrees of freedom. Also inspect the individual \(Q_m\) values: a result driven by one unusual imputation may indicate unstable imputations or a computational problem. These checks do not establish that the reference-based assumption is scientifically appropriate; they help identify whether the pooling calculation is behaving as expected.

Worked Example: Unequal Completed-Data Standard Errors

Suppose a hypothetical study has three imputations with treatment-effect estimates \(0.8,\ 1.0,\) and \(1.6\) points. The corresponding completed-data standard errors are \(0.4,\ 0.5,\) and \(0.6\) points. All analyses estimate the same contrast. First convert each standard error to a variance: \(U_1=0.16,\ U_2=0.25,\ U_3=0.36\) points squared.

The pooled estimate is \(\bar{Q}=(0.8+1.0+1.6)/3=1.1333\) points. The average within-imputation variance is

$$ \bar{U}=\frac{0.16+0.25+0.36}{3}=0.2567\text{ points}^2. $$

The deviations from \(1.1333\) are approximately \(-0.3333,-0.1333,\) and \(0.4667\). Hence

$$ B=\frac{(-0.3333)^2+(-0.1333)^2+(0.4667)^2}{2} \approx0.1733\text{ points}^2. $$

Thus \(T=0.2567+(1+1/3)(0.1733)\approx0.488\) points squared and \(SE_{\mathrm{MI}}=\sqrt{0.488}\approx0.6986\) points. Here \(r=0.1733/0.2567\approx0.6753\), so \(\nu=2(1+1/0.6753)^2\approx12.31\). The main lesson is operational: average the variances \(U_m\), not the standard errors \(0.4,\ 0.5,\ 0.6\), and then add the between-imputation contribution.

Worked Example: The Boundary Case of Identical Estimates

Suppose three hypothetical imputations all give an estimated difference of \(4.0\) points, and each has \(U_m=0.25\) points squared. Then \(\bar{Q}=4.0\), \(\bar{U}=0.25\), and \(B=0\), because every estimate equals its mean. Therefore \(T=0.25+(1+1/3)(0)=0.25\), and \(SE_{\mathrm{MI}}=0.5\) points.

The formula involving \(1/r\) cannot be evaluated at \(r=0\). Under the stated limiting convention, use the normal reference distribution. A 95% interval is \(4.0\pm1.96(0.5)=(3.02,\ 4.98)\) points. Identical estimates do not prove that missing outcomes are known without uncertainty; they mean only that this finite set of imputations shows no between-imputation variation in the estimate.

Alternatives, Limitations, and Common Pitfalls

Rubin’s pooling differs from several tempting alternatives. A single imputation treats the filled-in outcomes as if they were observed, omitting uncertainty attributable to imputation. Averaging completed-data estimates but using only \(\bar{U}\) also omits the between-imputation contribution. Conversely, averaging standard errors and then treating that average as the pooled standard error does not implement the variance formula. Rubin’s rules combine estimates and variances, not estimates and standard errors.

  • Check scale and units. \(Q_m\) and \(SE_{\mathrm{MI}}\) have the outcome’s units; \(U_m\), \(B\), and \(T\) have squared units. Adding a standard error directly to a variance is a units error.
  • Check consistency of the analysis. The same estimand, contrast direction, and analysis method must be used in every imputation. If a contrast is treatment minus reference in one data set and reference minus treatment in another, the pooled result is meaningless.
  • Check \(M\) and the boundary. The between-imputation variance requires \(M>1\). If \(B=0\), do not calculate \(1/r\); apply the stated limiting normal convention.
  • Keep assumptions in view. Pooling correctly quantifies uncertainty under the imputations supplied. It does not validate the reference-based missing-data assumption or guarantee that the completed-data model is suitable.
Common mistake: “The four completed-data standard errors are all about 1.26 points, so the pooled standard error is 1.26 points.” This ignores variation among the \(Q_m\). The correction is to square each standard error to obtain \(U_m\), calculate \(B\) from the estimates, and use \(SE_{\mathrm{MI}}=\sqrt{\bar{U}+(1+1/M)B}\).

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. In two imputations, the estimates are \(2\) and \(4\) units. What are \(\bar{Q}\) and \(B\)?
  2. Why is it incorrect to pool completed-data standard errors by averaging them?
  3. If \(B=0\) and \(M>1\), should the formula for \(\nu\) be evaluated by substituting \(r=0\)? What is the stated boundary convention?
  4. Suppose \(M=4\), \(\bar{U}=2\), and \(B=1\), in squared outcome units. Calculate \(T\) and \(SE_{\mathrm{MI}}\).

Solutions. (1) \(\bar{Q}=(2+4)/2=3\). The deviations are \(-1\) and \(1\), so \(B=[(-1)^2+1^2]/(2-1)=2\) squared units. (2) Standard errors are not variances, and averaging them omits the between-imputation contribution. Convert each to \(U_m\), average those variances, calculate \(B\), and use Rubin’s total-variance formula. (3) No. \(r=0\) makes \(1/r\) undefined; the limiting degrees of freedom are infinite, so use the normal distribution under this approximation. (4) \(T=2+(1+1/4)(1)=3.25\) squared units, so \(SE_{\mathrm{MI}}=\sqrt{3.25}\approx1.803\) outcome units.

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