Why Pool Ratios on the Log Scale?
Many effects of interest are ratios: a risk ratio, rate ratio, or ratio of two positive means, for example. A ratio is positive, and its sampling distribution can be skewed, especially when uncertainty is substantial. In many models, however, the log ratio is estimated on a scale where a normal approximation is more reasonable. With multiple imputation, pool those log-scale estimates and their variances first; then exponentiate the pooled result and the confidence limits.
The order matters. Exponentiation is nonlinear, so the exponential of an average log estimate is generally not the average of the ratios. Likewise, a confidence interval that is symmetric on the log scale becomes asymmetric after exponentiation. This tutorial develops the pooling and reporting steps for a positive ratio, including its equivalent expression as a percentage change.
Pooling and Back-Transformation
The imputations must target the same estimand and use compatible models and variance estimates. In particular, each \(Q_j\) must estimate the same log ratio, and each \(U_j\) must estimate its complete-data sampling variance. The imputation procedure and analysis also need to support valid multiple-imputation inference; pooling cannot repair a poorly specified analysis or an inappropriate imputation model.
For a scalar estimate, Rubin’s rules combine the estimates and variances as follows. The within-imputation variance is the average of the \(U_j\); the between-imputation variance measures how much the \(Q_j\) estimates vary; and the total variance includes both sources of uncertainty.
A confidence interval is formed on the log scale using the appropriate degrees of freedom, such as the Barnard–Rubin small-sample degrees of freedom when applicable. If the interval is \([L_Q,U_Q]\), monotonicity of the exponential function gives the ratio interval \([\exp(L_Q),\exp(U_Q)]\). A percentage change is obtained by subtracting 1 from the ratio and multiplying by 100, including at each endpoint:
The interval is generally asymmetric around the point estimate on the ratio or percentage-change scale. Its null value is 1 for a ratio and 0% for a percentage change. These are two ways of expressing the same comparison, not different analyses.
Why Pooling Happens Before Exponentiation
Rubin’s rules combine estimates and their uncertainty on the scale on which each \(Q_j\) and \(U_j\) are defined. A short variance argument clarifies why the within- and between-imputation components both enter. Variation in an estimate across repeated samples includes its average sampling variance within completed datasets and the variation in its underlying estimates across imputations. The latter is estimated by \(B\). Because only \(m\) imputations are used, the usual finite-imputation adjustment adds \(B/m\) to \(B\), yielding \(T=\bar{U}+B+B/m\).
This pooled variance describes uncertainty in \(\bar{Q}\), not in an average of exponentiated estimates. The natural-scale interval follows by transforming the endpoints of the log-scale interval. Since \(\exp(x)\) is strictly increasing, every value between \(L_Q\) and \(U_Q\) maps to a value between \(\exp(L_Q)\) and \(\exp(U_Q)\), with the order preserved. The transformation also explains the asymmetry: equal distances above and below \(\bar{Q}\) do not generally map to equal distances above and below \(\exp(\bar{Q})\).
Worked Example: Pooling Five Log Ratios
Suppose five hypothetical imputations give log ratio estimates \(Q_j=-0.10,-0.04,0.02,0.08,0.14\), each with estimated variance \(U_j=0.0100\). Assume the estimates target the same positive ratio, their variance estimates are appropriate, and the Barnard–Rubin degrees of freedom for this analysis are 20.
The pooled log estimate is \(\bar{Q}=[-0.10-0.04+0.02+0.08+0.14]/5=0.0200\), and \(\bar{U}=0.0100\). The squared deviations sum to \(0.0144+0.0036+0+0.0036+0.0144=0.0360\), so \(B=0.0360/4=0.00900\).
Thus \(T=0.0100+(1+1/5)(0.00900)=0.02080\), and \(SE(\bar{Q})=\sqrt{0.02080}=0.1442\). Using \(t_{0.975,20}=2.086\), the log-scale interval is \(0.0200\pm2.086(0.1442)=[-0.2808,0.3208]\).
The pooled ratio is \(\exp(0.0200)=1.020\). Exponentiating the interval endpoints gives \([\exp(-0.2808),\exp(0.3208)]\approx[0.7552,1.3782]\). The equivalent percentage change is \(100(1.0202-1)=2.020\%\), with interval \([100(0.7552-1),100(1.3782-1)]\approx[-24.48\%,37.82\%]\). The interval includes 1 for the ratio, or 0% for the percentage change; it therefore does not exclude no change at the 5% two-sided level.
Worked Example: A Ratio Above One With an Interval Crossing One
Consider four hypothetical imputations with log ratio estimates \(0.10,0.14,0.18,0.22\), each having variance \(0.00250\). Assume all four analyses estimate the same ratio, and suppose the applicable degrees of freedom are 12.
The mean is \(\bar{Q}=0.160\). The squared deviations are \(0.0036,0.0004,0.0004,0.0036\), so \(B=0.0080/3=0.002667\). With \(\bar{U}=0.00250\), the total variance is \(T=0.00250+(1+1/4)(0.002667)=0.005833\), giving \(SE(\bar{Q})=\sqrt{0.005833}=0.07638\).
Using \(t_{0.975,12}=2.179\), the margin is \(2.179(0.07638)=0.1664\), and the log interval is \([-0.00642,0.3264]\). Back-transforming gives a pooled ratio of \(\exp(0.160)=1.174\) and interval approximately \([0.9936,1.386]\). In percentage terms, the point estimate is a 17.35% increase, with interval approximately \([-0.64\%,38.60\%]\). Although the point estimate is above 1, this interval includes 1, so the data do not rule out no change at the 5% two-sided level.
Worked Example: Reporting a Decrease
Suppose three hypothetical imputations produce log ratio estimates \(-0.35,-0.30,-0.25\), with \(U_j=0.0100\) in each. Assume the ratio is defined as the outcome under a specified condition divided by the outcome in its reference condition, and the Barnard–Rubin degrees of freedom are 8.
Here \(\bar{Q}=-0.300\), \(\bar{U}=0.0100\), and \(B=[0.0025+0+0.0025]/2=0.00250\). Therefore \(T=0.0100+(1+1/3)(0.00250)=0.013333\), so \(SE(\bar{Q})=0.11547\). With \(t_{0.975,8}=2.306\), the log-scale interval is \(-0.300\pm2.306(0.11547)=[-0.56627,-0.03373]\).
The ratio estimate is \(\exp(-0.300)=0.7408\), with interval \([\exp(-0.56627),\exp(-0.03373)]\approx[0.5676,0.9668]\). Subtracting 1 and multiplying by 100 gives a 25.92% decrease, with interval approximately \([-43.24\%,-3.32\%]\). In context, the estimated outcome under the specified condition is 25.92% lower than the reference outcome; the interval indicates a decrease throughout its range. The percentage change is not 30%: the log ratio \(-0.300\) must be exponentiated before converting to a percentage.
Assumptions, Diagnostics, and Alternatives
The usual interval relies on an approximately normal sampling distribution for the pooled log estimate, with its variance adequately estimated. The log scale is appropriate only when the estimand is a positive ratio and the analysis supplies a meaningful log ratio and variance in every imputation. The imputation-specific estimates must refer to the same contrast, population, and model specification. If these conditions fail, a neatly computed interval can still be misleading.
Useful checks include confirming that all log estimates and variances are finite, that every variance is nonnegative, and that the ratio is defined in the same direction in every imputation. Inspect the \(Q_j\) values for extreme estimates and substantial between-imputation dispersion. Large \(B\) relative to \(\bar{U}\) signals that missing-data uncertainty is a substantial part of the total variance; it is not, by itself, proof of an error. Also check that reported ratio and percentage-change endpoints correspond exactly: ratio 1 must map to 0%, and endpoints must remain in the same order after transformation.
| Approach | What it does | Key caution |
|---|---|---|
| Pool log ratios, then exponentiate | Applies Rubin’s rules to the estimates and variances on the log scale, then transforms the pooled estimate and interval. | Requires a suitable log-ratio estimate and variance in each imputation. |
| Average the imputation-specific ratios | Calculates a simple arithmetic mean of \(\exp(Q_j)\). | Does not combine within- and between-imputation uncertainty using Rubin’s rules. |
| Average percentage changes | Averages \(100\{\exp(Q_j)-1\}\) across imputations. | Is not Rubin pooling and does not provide the required pooled variance or interval. |
If an estimate is supplied only on a natural scale, do not treat a standard error on that scale as though it were a log-scale variance. Use an analysis and variance estimate appropriate to the intended estimand. If ratios can be zero or negative, a log ratio is not defined; a different scale or inferential method is needed. When the log-scale normal approximation is poor, or an imputation produces undefined estimates, the simple procedure here may be unsuitable and the source of the instability should be investigated rather than concealed by back-transformation.
Common Mistakes
- Exponentiating before pooling. Averaging the \(m\) ratios does not generally equal \(\exp(\bar{Q})\), and it does not apply Rubin’s variance formula. Pool \(Q_j\) and \(U_j\) first.
- Exponentiating only the point estimate. A confidence interval must be transformed at both endpoints. Reporting a symmetric interval around the exponentiated estimate ignores the nonlinear transformation.
- Using the wrong percentage formula. A ratio of 1.20 means a 20% increase, not a 120% increase: calculate \(100(1.20-1)=20\%\).
- Reversing the comparison. A ratio defined as condition A divided by condition B is not interchangeable with B divided by A. State the numerator and reference clearly.
- Overstating an interval that crosses the null. If a ratio interval includes 1, its corresponding percentage-change interval includes 0%; the point estimate alone does not establish a change.
Check Your Understanding
Use the pooling and back-transformation rules to answer each question. Solutions follow each item.
- If the pooled log ratio is \(0.10\), what is the pooled ratio and percentage change? Solution: The ratio is \(\exp(0.10)=1.105\). The percentage change is \(100(1.1052-1)=10.52\%\), an increase relative to the reference.
- If a log-scale confidence interval is \([-0.20,0.30]\), give its ratio interval and percentage-change interval. Solution: The ratio interval is \([\exp(-0.20),\exp(0.30)]\approx[0.8187,1.350]\). The percentage-change interval is approximately \([-18.13\%,34.99\%]\). It includes the no-change values 1 and 0%, respectively.
- Why is averaging the imputation-specific ratios not the same procedure as pooling log ratios? Solution: The exponential is nonlinear, so \(\frac{1}{m}\sum_j\exp(Q_j)\) is generally not \(\exp(\bar{Q})\). In addition, a simple average of ratios does not combine within-imputation and between-imputation uncertainty using Rubin’s rules.
- A reported ratio interval is \([0.90,1.25]\). What is the corresponding percentage-change interval, and what does it imply about no change? Solution: Transform both endpoints: \(100(0.90-1)=-10\%\) and \(100(1.25-1)=25\%\). The interval is \([-10\%,25\%]\), which includes 0%; it does not exclude no change at the corresponding confidence level.