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Rubin's Rules and Pooling · Tutorial 65 of 497

Barnard–Rubin Small-Sample Degrees Of Freedom

Calculate Barnard–Rubin adjusted degrees of freedom and understand why they account for both missing information and finite complete-data information.

Intermediate 11 min read

What You'll Learn

  • Identify the inputs needed for the Barnard–Rubin small-sample degrees-of-freedom adjustment.
  • Calculate Rubin’s large-sample degrees of freedom and the adjusted observed-data degrees of freedom.
  • Combine the two degrees of freedom using the Barnard–Rubin formula.
  • Check how the adjustment behaves when complete-data degrees of freedom are limited.
  • Recognize input and interpretation errors that can make the adjustment misleading.

Why Adjust the Degrees of Freedom?

Rubin’s large-sample degrees of freedom reflect uncertainty due to missing information and the finite number of imputations. But they do not, on their own, account for limited degrees of freedom in the complete-data analysis. When that analysis has a small sample or a model with many estimated parameters, treating its variance estimate as if it came from unlimited data can make inference too optimistic.

The Barnard–Rubin adjustment combines the large-sample degrees of freedom with a second quantity that reflects the complete-data degrees of freedom. This tutorial focuses on calculating and interpreting that adjustment for a scalar estimate. It does not change the pooled point estimate or the pooled variance; it adjusts the degrees of freedom used to describe uncertainty.

Definition: Let \(m\) be the number of imputations, \(\bar U\) the average within-imputation variance, and \(B\) the between-imputation variance for a scalar estimate. Define the relative increase in variance as \(r=(1+1/m)B/\bar U\). Rubin’s large-sample degrees of freedom are \(\nu_{\text{old}}=(m-1)(1+1/r)^2\). Let \(\nu_{\text{com}}\) denote the complete-data degrees of freedom for the analysis. The Barnard–Rubin adjustment also uses \(\gamma=(r+2/(\nu_{\text{old}}+3))/(r+1)\) and \(\nu_{\text{obs}}=((\nu_{\text{com}}+1)/(\nu_{\text{com}}+3))\nu_{\text{com}}(1-\gamma)\), and sets \(\nu_{\text{BR}}=\nu_{\text{old}}\nu_{\text{obs}}/(\nu_{\text{old}}+\nu_{\text{obs}})\).

Here, \(\gamma\) is an adjusted estimate of the fraction of missing information for the scalar estimate. The term \(\nu_{\text{obs}}\) is the observed-data degrees-of-freedom component: it accounts for the fact that the complete-data analysis itself has finite degrees of freedom. The final value \(\nu_{\text{BR}}\) combines that component with \(\nu_{\text{old}}\).

The adjustment is used when the scalar estimate and its variance are appropriately pooled across imputations, and the complete-data analysis provides a suitable \(\nu_{\text{com}}\). The imputed data sets should arise from an adequate imputation procedure, and the estimate should have a distribution for which the usual approximate \(t\)-based degrees-of-freedom calculation is reasonable. The same scalar estimand must be analyzed in every imputation. If the complete-data degrees of freedom are not well defined or do not represent the analysis at hand, the adjustment cannot repair that problem.

How the Combination Works

The combination is easiest to understand by writing the formula for \(\nu_{\text{BR}}\) in reciprocal form:

$$ \frac{1}{\nu_{\text{BR}}} = \frac{1}{\nu_{\text{old}}} + \frac{1}{\nu_{\text{obs}}}. $$

This identity follows directly by taking the reciprocal of the product-over-sum formula. Since both degrees of freedom are positive, their reciprocal sum is larger than either reciprocal individually. Therefore, \(\nu_{\text{BR}}\) is smaller than either input. The adjustment does not simply choose whichever input is smaller; it combines both.

The observed-data component includes two corrections. The factor \((\nu_{\text{com}}+1)/(\nu_{\text{com}}+3)\) modifies the finite complete-data degrees of freedom. The factor \(1-\gamma\) reduces that quantity according to the estimated missing information. If complete-data degrees of freedom increase without bound while the other quantities remain fixed, \(\nu_{\text{obs}}\) increases without bound, and the harmonic combination approaches \(\nu_{\text{old}}\). Thus, in the large complete-data limit, the adjustment returns to Rubin’s large-sample degrees of freedom. For finite \(\nu_{\text{com}}\), it generally yields a smaller value.

A practical calculation follows a fixed order. First find \(r\) and \(\nu_{\text{old}}\). Next use \(\nu_{\text{old}}\) to calculate \(\gamma\), then calculate \(\nu_{\text{obs}}\) using \(\nu_{\text{com}}\), and finally combine the two degrees of freedom. Retaining unrounded intermediate values avoids avoidable discrepancies.

1
Find the relative increase and large-sample degrees of freedom.
Calculate \(r=(1+1/m)B/\bar U\), then \(\nu_{\text{old}}=(m-1)(1+1/r)^2\).
2
Calculate adjusted missing information.
Use \(\gamma=(r+2/(\nu_{\text{old}}+3))/(r+1)\).
3
Calculate observed-data degrees of freedom.
Use the complete-data input: \(\nu_{\text{obs}}=((\nu_{\text{com}}+1)/(\nu_{\text{com}}+3))\nu_{\text{com}}(1-\gamma)\).
4
Combine the two components.
Calculate \(\nu_{\text{BR}}=\nu_{\text{old}}\nu_{\text{obs}}/(\nu_{\text{old}}+\nu_{\text{obs}})\).

Worked Example: Many Large-Sample Degrees of Freedom, Limited Complete-Data Degrees of Freedom

Suppose \(m=10\) imputations give \(r=0.10\), and the complete-data analysis has \(\nu_{\text{com}}=30\) degrees of freedom. Calculate the Barnard–Rubin adjustment. These are hypothetical inputs.

First, the large-sample degrees of freedom are

$$ \nu_{\text{old}}=(10-1)\left(1+\frac{1}{0.10}\right)^2 =9(11)^2 =1089. $$

Next,

$$ \gamma = \frac{0.10+2/(1089+3)}{1.10} = \frac{0.10+2/1092}{1.10} \approx 0.09257. $$

The observed-data degrees of freedom are

$$ \nu_{\text{obs}} = \left(\frac{30+1}{30+3}\right)30(1-0.0925741) \approx 25.5729 \approx 25.57. $$

Using unrounded intermediate values for the final calculation,

$$ \nu_{\text{BR}} = \frac{1089(25.572912\ldots)}{1089+25.572912\ldots} \approx 24.986 \approx 24.99. $$

Although \(\nu_{\text{old}}\) is very large, the complete-data information limits the adjusted value to about 24.99. The adjusted degrees of freedom are below both inputs, as the reciprocal-sum identity requires.

Worked Calculations Under Different Conditions

Worked Example: A Small Complete-Data Analysis

Consider hypothetical results with \(m=5\), \(r=0.40\), and \(\nu_{\text{com}}=12\). Calculate \(\nu_{\text{BR}}\).

The large-sample degrees of freedom are

$$ \nu_{\text{old}} = (5-1)\left(1+\frac{1}{0.40}\right)^2 = 4(3.5)^2 = 49. $$

Then

$$ \gamma = \frac{0.40+2/(49+3)}{1.40} = \frac{0.40+2/52}{1.40} \approx 0.3132. $$

The complete-data adjustment gives

$$ \nu_{\text{obs}} = \left(\frac{12+1}{12+3}\right)12(1-0.3131868) = \frac{13}{15}(12)(0.6868132) \approx 7.1429. $$

Finally,

$$ \nu_{\text{BR}} = \frac{49(7.142857\ldots)}{49+7.142857\ldots} \approx 6.234. $$

The result is substantially below both 49 and 12 because the analysis has limited complete-data degrees of freedom and a nonzero estimated fraction of missing information. Reporting 49 as the adjusted degrees of freedom would ignore the finite complete-data contribution.

Worked Example: Small Missing Information but a Tight Complete-Data Limit

Suppose a hypothetical analysis has \(m=20\), \(r=0.02\), and only \(\nu_{\text{com}}=8\) complete-data degrees of freedom. Calculate the adjustment and identify which component is most limiting.

First,

$$ \nu_{\text{old}} = (20-1)\left(1+\frac{1}{0.02}\right)^2 = 19(51)^2 = 49{,}419. $$

The adjusted fraction of missing information is

$$ \gamma = \frac{0.02+2/(49{,}419+3)}{1.02} \approx 0.01965. $$

Therefore,

$$ \nu_{\text{obs}} = \left(\frac{8+1}{8+3}\right)8(1-0.0196475) \approx 6.4169. $$

The final combination is

$$ \nu_{\text{BR}} = \frac{49{,}419(6.4169\ldots)}{49{,}419+6.4169\ldots} \approx 6.4161. $$

Here \(\nu_{\text{old}}\) is much larger than \(\nu_{\text{obs}}\), so the harmonic combination is close to the smaller component. The result reflects the limited complete-data degrees of freedom, even though the estimated missing-information fraction is small.

Checks, Alternatives, and Limitations

A useful diagnostic is to check the direction and scale of the result. For positive finite inputs, \(\nu_{\text{BR}}\) should be positive and below both \(\nu_{\text{old}}\) and \(\nu_{\text{obs}}\). When one input is much smaller than the other, the result should be close to the smaller input, but slightly below it. If a calculation violates either check, inspect the arithmetic and the inputs.

The most direct alternative is to use \(\nu_{\text{old}}\) alone. That is the large-sample degrees-of-freedom approximation discussed in the earlier material on large-sample degrees of freedom. It can be adequate when complete-data degrees of freedom are effectively large, but it may overstate the degrees of freedom in a small complete-data analysis. Using \(\nu_{\text{com}}\) alone also omits the contribution of missing information and the finite number of imputations. Barnard–Rubin pooling incorporates both components, but it remains an approximation: it does not correct a misspecified imputation model, an inappropriate analysis model, or a poorly chosen complete-data degrees-of-freedom value.

The adjustment also depends on stable estimates of the inputs. The number of imputations \(m\) must be at least two for \(\nu_{\text{old}}\) to be defined. The calculation of \(r\) requires a meaningful positive \(\bar U\). When quantities are extremely close to zero or very large, direct computation can be numerically delicate; keep adequate precision and verify the result against the reciprocal form. Problems at a variance boundary require separate care and should not be handled by mechanically applying a formula outside its assumptions.

Common Mistakes

  • Using only \(\nu_{\text{old}}\). A large \(\nu_{\text{old}}\) does not imply that the complete-data analysis has abundant degrees of freedom. Include \(\nu_{\text{com}}\) through \(\nu_{\text{obs}}\).
  • Using \(\nu_{\text{com}}\) as the final answer. The complete-data value is an input, not the adjusted pooled degrees of freedom.
  • Omitting the finite correction in \(\gamma\). The term \(2/(\nu_{\text{old}}+3)\) is part of the stated adjustment; do not replace \(\gamma\) with \(r/(r+1)\).
  • Rounding too early. Carry unrounded values through \(\gamma\), \(\nu_{\text{obs}}\), and the final combination, then round the reported result.
  • Assuming the adjustment repairs poor modeling. It combines degrees-of-freedom information; it does not make an inadequate imputation or analysis model adequate.
Key takeaway: Calculate \(\nu_{\text{old}}\), \(\gamma\), and \(\nu_{\text{obs}}\) in that order, then combine the two degrees of freedom using the Barnard–Rubin formula. The adjusted value is below both inputs and accounts for both missing information and finite complete-data information.

Check Your Understanding

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

  1. What role does \(\nu_{\text{com}}\) play in the Barnard–Rubin adjustment?
  2. If \(\nu_{\text{old}}=40\) and \(\nu_{\text{obs}}=10\), calculate \(\nu_{\text{BR}}\).
  3. Why should \(\nu_{\text{BR}}\) be smaller than both \(\nu_{\text{old}}\) and \(\nu_{\text{obs}}\)?
  4. What is the main problem with reporting \(\nu_{\text{old}}\) alone when the complete-data analysis has few degrees of freedom?

Solutions. (1) \(\nu_{\text{com}}\) represents the degrees of freedom in the complete-data analysis and is used to calculate \(\nu_{\text{obs}}\). (2) \(\nu_{\text{BR}}=40(10)/(40+10)=400/50=8\) degrees of freedom. (3) Its reciprocal is the sum \(1/\nu_{\text{old}}+1/\nu_{\text{obs}}\), which is greater than either reciprocal separately, so the resulting degrees of freedom are below both inputs. (4) It ignores the finite complete-data contribution and can therefore overstate the pooled degrees of freedom.

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