Tutorials › Multiple Imputation by Chained Equations (MICE) › Pooling Log Odds Ratios And Reporting Odds Ratios

Rubin's Rules and Pooling · Tutorial 78 of 497

Pooling Log Odds Ratios And Reporting Odds Ratios

Pool the coefficient on the log-odds scale first, then transform the pooled estimate and interval for odds-ratio reporting.

Intermediate 11 min read

What You'll Learn

  • Identify the logistic-regression coefficient that represents a log odds ratio.
  • Pool that coefficient and its variance using Rubin’s rules.
  • Exponentiate the pooled log odds ratio to obtain the reported odds ratio.
  • Transform both limits of a pooled log-scale confidence interval.
  • Explain why averaging odds ratios across imputations is not Rubin pooling.
  • Check that the contrast, scale, and uncertainty are consistent across imputations.

Why Pool on the Log-Odds Scale?

A logistic-regression coefficient is naturally estimated on the log-odds scale, while readers often want an odds ratio. With missing data handled through multiple imputation, each completed data set gives a coefficient estimate and an estimated variance. The key is to pool those quantities on the coefficient’s scale first. Exponentiation is used afterward to express the pooled result as an odds ratio.

This order matters because exponentiation is nonlinear. The average of the odds ratios from separate imputations is generally not the odds ratio corresponding to the pooled coefficient, and it does not incorporate within- and between-imputation uncertainty according to Rubin’s rules. The target contrast must also be the same in every imputation: for example, the adjusted effect of a 10 mg increase in exposure, with the same covariate adjustment and reference category.

Definition: In imputation \(i\), let \(q_i\) be the estimated logistic-regression coefficient for a specified contrast, and let \(u_i\) be its estimated variance. For a binary predictor, \(q_i\) is commonly a log odds ratio. With \(m\) imputations, pool the coefficient as \(\bar{q}=m^{-1}\sum_{i=1}^{m}q_i\), calculate the average within-imputation variance \(\bar{U}=m^{-1}\sum_{i=1}^{m}u_i\), and calculate the between-imputation variance \(B=(m-1)^{-1}\sum_{i=1}^{m}(q_i-\bar{q})^2\). The total variance is \(T=\bar{U}+(1+1/m)B\). The pooled odds ratio is \(\exp(\bar{q})\).

The estimated variances \(u_i\) are variances of the log-scale coefficient, not variances of odds ratios. If a fitted model reports a standard error \(s_i\) for the coefficient, use \(u_i=s_i^2\). The pooled standard error on the log scale is \(\sqrt{T}\). For an interval, first form the appropriate pooled interval for \(q\), using the relevant degrees of freedom and confidence level; then exponentiate its lower and upper limits. The earlier material on confidence intervals from pooled estimates supplies the general interval construction.

Conditions for a Meaningful Pooled Odds Ratio

The arithmetic is only interpretable when the estimates refer to a common quantity. The logistic model and contrast must be defined consistently across imputations: the same outcome coding, predictor scale, reference group, covariates, and coefficient. If a coefficient is for a one-unit increase in a continuous predictor in one analysis but a five-unit increase in another, those are not estimates of the same contrast until they are placed on a common scale.

The usual pooling calculation also relies on the coefficient estimates being reasonably approximated by normal sampling distributions within completed data sets, with valid variance estimates. The imputation procedure must produce plausible completed data sets under its stated missing-data assumptions, and the analysis and imputation models should be sufficiently compatible for the target coefficient. Rubin’s rules account for sampling uncertainty within completed data sets and variation between them; they do not repair a misspecified outcome model, a poorly chosen imputation model, or an incorrectly defined estimand.

Logistic coefficients are conditional on the other variables in the model. Thus, the exponentiated coefficient is an adjusted odds ratio conditional on the included covariates, not a risk ratio and not automatically a marginal population effect. An odds ratio above 1 indicates higher odds for the numerator group or higher predictor value relative to the stated comparison; it does not by itself establish a causal effect.

Worked Example: Pooling an Adjusted Log Odds Ratio

Suppose five hypothetical imputations produce estimates for the adjusted log odds ratio comparing each 10 mg higher exposure with the same covariate adjustment. The estimates are \(0.40, 0.50, 0.30, 0.60, 0.20\). Their estimated variances are \(0.040, 0.044, 0.036, 0.049, 0.041\). Assume the coefficient and contrast are defined identically in all five analyses and the reported variances are valid.

First calculate the pooled coefficient:

$$ \bar{q}=\frac{0.40+0.50+0.30+0.60+0.20}{5} =\frac{2.00}{5}=0.400. $$

The average within-imputation variance is

$$ \bar{U}=\frac{0.040+0.044+0.036+0.049+0.041}{5} =\frac{0.210}{5}=0.0420. $$

The deviations from \(0.400\) are \(0, 0.10, -0.10, 0.20, -0.20\). Therefore,

$$ B=\frac{0^2+0.10^2+(-0.10)^2+0.20^2+(-0.20)^2}{5-1} =\frac{0.100}{4}=0.0250. $$

There are five imputations, so the total variance and pooled standard error are

$$ T=0.0420+\left(1+\frac{1}{5}\right)(0.0250) =0.0420+0.0300=0.0720, \qquad SE(\bar{q})=\sqrt{0.0720}=0.2683. $$

For illustration, use the ordinary Rubin degrees-of-freedom approximation with effectively infinite complete-data degrees of freedom. The relative increase in variance is \(r=(1+1/m)B/\bar{U}=0.0300/0.0420=0.7143\), giving \(df=(m-1)(1+1/r)^2=23.04\). The 95% pooled interval on the log scale is approximately \(0.400\pm 2.069(0.2683)=(-0.155,0.955)\). The pooled odds ratio is \(\exp(0.400)=1.492\), and the interval limits transform to \(\exp(-0.155)=0.856\) and \(\exp(0.955)=2.599\).

Thus, the estimated adjusted odds are about 1.49 times as high for each 10 mg higher exposure, conditional on the model covariates. The interval runs from about 0.86 to 2.60 for that odds ratio and includes 1, so it is compatible with no difference in odds at the 95% confidence level. It does not establish that the odds are higher.

Why the Interval Transforms Endpoint by Endpoint

Exponentiation is strictly increasing: if \(a<b\), then \(\exp(a)<\exp(b)\). Consequently, if a log-scale interval has limits \(L\) and \(U\), the corresponding odds-ratio interval has limits \(\exp(L)\) and \(\exp(U)\), in the same order. This transformation preserves the interval’s coverage under the same model and interval procedure; it does not change the uncertainty calculation that produced \(L\) and \(U\).

$$ P(L\leq q\leq U) = P\{\exp(L)\leq \exp(q)\leq \exp(U)\}. $$

The odds-ratio interval is usually asymmetric around its point estimate because exponentiation is nonlinear. That asymmetry is expected. It is not a reason to recenter the interval around the odds ratio or to calculate a new standard error by treating the transformed coefficient as though it were a normally distributed estimate.

Worked Example: Why Averaging Odds Ratios Is Not Pooling

Consider three hypothetical imputations with log odds ratio estimates \(-0.80, 0.00, 0.80\), all for the same contrast. Their pooled log-scale point estimate is

$$ \bar{q}=\frac{-0.80+0.00+0.80}{3}=0. $$

The odds ratio corresponding to that pooled estimate is \(\exp(0)=1.000\). The three imputation-specific odds ratios, however, are \(\exp(-0.80)=0.4493\), \(\exp(0)=1.000\), and \(\exp(0.80)=2.226\). Their arithmetic mean is

$$ \frac{0.4493+1.000+2.226}{3}=1.225. $$

The value 1.225 is not the exponentiated Rubin-pooled coefficient; it answers a different averaging question. In general, because the exponential function is convex, the average of exponentiated estimates can exceed the exponential of their average. Rubin pooling also requires the imputation-specific variances, so averaging odds ratios alone cannot provide the pooled uncertainty. Pool the log coefficients and their variances, then transform the resulting estimate and interval.

Keeping the Predictor Scale Consistent

A reported odds ratio is incomplete unless its contrast is clear. For a continuous predictor measured in years, \(\exp(\bar{q})\) describes the odds ratio for a one-year increase. If the desired report is per 10 years, multiply the log coefficient by 10 before exponentiating. Its variance must be multiplied by \(10^2\) if the contrast itself is redefined on the 10-year scale. Equivalently, the one-year pooled result can be exponentiated to the tenth power. The estimate and uncertainty must use the same scale.

Worked Example: Reporting an Odds Ratio per 10 Units

Four hypothetical imputations give log-odds coefficients per one-unit increase of \(0.03, 0.04, 0.05, 0.04\), with variances \(0.0004, 0.0005, 0.0004, 0.0005\). Suppose the scientific report calls for the odds ratio per 10-unit increase. The pooled one-unit coefficient is \(\bar{q}=(0.03+0.04+0.05+0.04)/4=0.0400\), so the pooled coefficient for 10 units is \(10(0.0400)=0.400\).

The pooled odds ratio per 10 units is

$$ \exp(10\bar{q})=\exp(0.400)=1.492. $$

For completeness, the variance for the 10-unit contrast is 100 times the variance for the one-unit contrast. On the one-unit scale, \(\bar{U}=(0.0004+0.0005+0.0004+0.0005)/4=0.00045\), and \(B=\{(-0.01)^2+0^2+(0.01)^2+0^2\}/3=0.00006667\). Hence \(T=0.00045+(1+1/4)(0.00006667)=0.0005333\). On the 10-unit scale, \(T_{10}=100(0.0005333)=0.05333\), and the standard error is \(\sqrt{0.05333}=0.2309\). The transformed coefficient and variance therefore refer to the same 10-unit contrast.

Diagnostics, Alternatives, and Common Pitfalls

Before reporting a pooled odds ratio, check that each imputation estimates the same coefficient and that each variance is on the log-odds scale. Review the coefficient and variance estimates across imputations for unusual values, large shifts, or extreme imputation-specific standard errors. Such patterns may indicate unstable model fitting, sparse outcome-by-predictor cells, separation, or inconsistent coding. Pooling can combine valid estimates; it cannot make an unstable or non-comparable set of estimates reliable.

A common mistake is to exponentiate each coefficient, average those odds ratios, and then report that average as the pooled result. This skips the Rubin pooling calculation and ignores the distinction between within- and between-imputation uncertainty. Another mistake is to exponentiate the pooled coefficient but leave confidence limits on the log scale, or to exponentiate only one limit. Transform the point estimate and both limits, and label the contrast and units in the report.

The alternative of reporting the pooled log odds ratio directly is statistically valid and can be useful when a model-scale result is desired, but it may be less intuitive for many readers. Reporting both the log coefficient and odds ratio can aid reproducibility. Neither format changes the pooling calculation. Do not interpret an odds ratio as a probability ratio: for instance, an odds ratio of 1.5 does not mean that the probability is 50% higher.

Key takeaway: For a logistic-regression odds ratio after multiple imputation, pool the coefficient and its variance on the log-odds scale using Rubin’s rules. Exponentiate the pooled coefficient for the point estimate and exponentiate each endpoint of the pooled log-scale interval for the reported interval.

Check Your Understanding

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

  1. A pooled log odds ratio is \(0.25\). What is the corresponding odds ratio, to three significant figures?
  2. A pooled log-scale confidence interval is \((-0.20, 0.60)\). What are the odds-ratio interval limits, and does the interval include the null value?
  3. Two imputations give log odds ratios \(-0.40\) and \(0.40\). Compare the exponentiated pooled estimate with the arithmetic mean of the two imputation-specific odds ratios. Why are they different?
  4. A coefficient is \(0.07\) per one-unit increase in a predictor. What is the log coefficient and odds ratio for a five-unit increase?

Solutions. (1) \(\exp(0.25)=1.284\), so the odds are estimated to be 1.28 times as high for the specified contrast. (2) Exponentiating both endpoints gives \((\exp(-0.20),\exp(0.60))=(0.819,1.822)\). The interval includes 1, the null odds ratio. (3) The pooled log estimate is \((-0.40+0.40)/2=0\), so its exponent is \(1.000\). The individual odds ratios are \(\exp(-0.40)=0.6703\) and \(\exp(0.40)=1.492\); their mean is \(1.081\). The difference reflects the nonlinear exponential transformation, and the arithmetic mean is not Rubin pooling. (4) The five-unit log coefficient is \(5(0.07)=0.35\), giving an odds ratio of \(\exp(0.35)=1.419\). The contrast is a five-unit increase, not a one-unit increase.

Clinical Biostats