Why the Scale of the Interval Matters
In logistic regression, a coefficient for a predictor is expressed on the log-odds scale. Exponentiating that coefficient gives an odds ratio, which is often easier to interpret. But a confidence interval should respect the scale on which the coefficient’s uncertainty was estimated. In multiple imputation, the usual approach is to pool the coefficient and its variance on the log-odds scale, form a t interval there, and exponentiate the two endpoints.
The order matters. Exponentiation is nonlinear, so exponentiating a point estimate and attaching a symmetric margin on the odds-ratio scale does not, in general, reproduce the interval obtained from the pooled coefficient. The transformed interval is usually asymmetric around the reported odds ratio.
When the Transformation Is Appropriate
For a logistic regression coefficient \(q\), \(\exp(q)\) is an odds ratio. For a one-unit increase in a quantitative predictor, it compares the odds of the outcome at predictor values one unit apart, conditional on the other modeled covariates. For a binary predictor coded 0 and 1, it compares the odds in the group coded 1 with the odds in the group coded 0, conditional on those covariates. The direction and units of the comparison therefore depend on the model’s coding and predictor scale.
The interval procedure is appropriate when each completed-data analysis supplies an estimate and a valid estimated variance for the same log-odds coefficient, and the imputation and analysis procedures support the usual multiple-imputation inference. In particular, the imputation model must be suitable for the intended analysis and its uncertainty; the completed-data estimates should not be dominated by separation or other serious model-fitting failures. The pooled t interval is an approximate inferential procedure, with its reference degrees of freedom accounting for uncertainty from missing data as well as estimation uncertainty.
Let \(c=t_{1-\alpha/2,df}\). The confidence limits on the coefficient scale are \(L_q=\bar{q}-cSE\) and \(U_q=\bar{q}+cSE\). Since the exponential function is strictly increasing, applying it to every value in the interval preserves the order of the endpoints. Thus the odds-ratio limits are \(L_{OR}=\exp(L_q)\) and \(U_{OR}=\exp(U_q)\). No standard error needs to be exponentiated.
Worked Intervals
Worked Example: A Pooled Interval for an Adjusted Odds Ratio
Suppose a hypothetical analysis estimates a pooled log-odds coefficient \(\bar{q}=0.60\), with \(SE=0.25\) and pooled \(df=18\). The coefficient is for a one-unit increase in a predictor, and the model adjusts for age and treatment group. Use \(t_{0.975,18}=2.101\) for a 95% interval.
First form the interval on the log-odds scale:
The limits are \(L_q=0.07475\) and \(U_q=1.12525\). Exponentiating each limit gives:
The point estimate is \(\exp(0.60)\approx1.822\). Under the fitted model, the estimated odds are about 1.82 times as large for a one-unit increase in the predictor, conditional on age and treatment group. The 95% interval ranges from about 1.08 to 3.08 for that conditional odds ratio. In repeated applications under the procedure’s assumptions, 95% of intervals constructed this way would cover the true odds ratio; this is not a statement that the fixed parameter has a 95% probability of lying in this particular interval.
Worked Example: A Negative Coefficient
For a second hypothetical analysis, suppose \(\bar{q}=-0.35\), \(SE=0.16\), and \(df=30\). The coefficient represents a one-unit increase in a predictor, with other modeled covariates held fixed. Using \(t_{0.975,30}=2.042\), calculate a 95% odds-ratio interval.
The log-scale limits are:
So \(L_q=-0.67672\) and \(U_q=-0.02328\). Exponentiating gives:
The estimated odds ratio is \(\exp(-0.35)\approx0.705\), and the 95% interval is approximately 0.508 to 0.977. The interval lies below 1, the value corresponding to no change in odds. This describes an estimated association under the model; it does not establish that the predictor causes a change in the outcome.
Why the Transformed Interval Is Asymmetric
An equal-width interval around \(\bar{q}\) becomes unequal distances from \(\exp(\bar{q})\) after exponentiation. In fact, if the log-scale limits are \(\bar{q}-cSE\) and \(\bar{q}+cSE\), then the lower and upper odds-ratio limits are \(\exp(\bar{q})\exp(-cSE)\) and \(\exp(\bar{q})\exp(cSE)\). The multiplicative factors are reciprocal, but their additive distances from 1 are not generally equal. Because odds ratios are positive, this approach also ensures that the lower limit cannot be negative.
This calculation provides a short justification for the interval transformation. If a confidence interval for the log coefficient covers its true value \(q\), then, because exponentiation is increasing, the exponentiated interval covers \(\exp(q)\). The coverage probability is preserved by this one-to-one transformation; what changes is the interval’s shape and scale.
Worked Example: Comparing With a Symmetric Approximation
Use the estimates from the negative-coefficient example: \(\bar{q}=-0.35\), \(SE=0.16\), and \(df=30\). The transformed interval was approximately (0.508, 0.977), around a point estimate of \(\exp(-0.35)\approx0.7047\).
A first-order delta-method approximation on the odds-ratio scale uses the derivative of \(\exp(q)\), which is \(\exp(q)\). It therefore approximates the odds-ratio standard error by \(\exp(\bar{q})SE\):
Using the same \(t\) critical value gives the symmetric approximate interval:
This differs from (0.508, 0.977), the interval obtained by exponentiating the pooled log-scale limits. The delta-method interval is a local, first-order approximation; it may be reasonable when uncertainty is small, but it is not the exact transformed t interval. It can also produce an impossible negative lower limit when uncertainty is large. For example, an estimate near 0.5 with an approximate standard error of 0.4 and critical value 2 would give a symmetric lower limit of \(0.5-2(0.4)=-0.3\), despite an odds ratio being necessarily positive.
Pooling and Checking the Result
Worked Example: From Imputation-Specific Estimates to an Odds-Ratio Interval
Consider four hypothetical imputations with log-odds estimates \(0.42, 0.58, 0.51, 0.49\), and estimated within-imputation variances \(0.040, 0.050, 0.045, 0.045\). Suppose the applicable pooled degrees of freedom are \(12\). Calculate the pooled interval and exponentiate it.
The mean estimate is \(\bar{q}=(0.42+0.58+0.51+0.49)/4=0.50\), and the mean within-imputation variance is \(\bar{U}=(0.040+0.050+0.045+0.045)/4=0.045\). The between-imputation variance is:
Thus the total variance and standard error are:
With \(t_{0.975,12}=2.179\), the log-scale interval is \(0.50\pm2.179(0.2245)\), or approximately \((0.0107,0.9893)\). The odds-ratio interval is:
The point estimate is \(\exp(0.50)\approx1.649\). The imputation-specific estimates vary, and their variation contributes to \(T\) through \(B\); using only \(\bar{U}\) would omit that component of uncertainty. The example takes \(df=12\) as the result of the applicable pooled degrees-of-freedom calculation.
Before reporting an interval, check that both endpoints came from the same pooled coefficient, standard error, confidence level, and degrees of freedom. Also check that exponentiation preserves the endpoint order and yields positive limits. An extremely wide interval can be a sign of substantial uncertainty, a very small pooled degrees of freedom, unstable coefficient estimates, or limited information in some imputations. Exponentiation does not fix those problems; it only changes the scale on which the result is reported.
A frequent mistake is to exponentiate the coefficient and then attach the original log-scale standard error as a symmetric margin: \(\exp(\bar{q})\pm cSE\). This mixes scales and is not the transformed pooled interval. The correction is to form \(\bar{q}\pm cSE\) first and exponentiate both limits. A delta-method standard error may support a separate approximation, but it should be identified as such rather than substituted silently for the transformed interval.
The exponentiated point estimate is not itself a confidence interval, and a confidence interval that includes 1 does not prove that the null is true. It indicates that the data and model do not exclude an odds ratio of 1 at the stated confidence level using this interval procedure. Report the odds ratio, its interval, the predictor contrast and units, and the covariate adjustment so the estimate has a clear interpretation.
Check Your Understanding
Work through each question, then compare your reasoning with the solutions below.
- A pooled log-odds estimate is \(0.20\), with \(SE=0.10\) and \(df=20\). Using \(t_{0.975,20}=2.086\), what is the 95% odds-ratio interval?
- Why should both confidence limits be exponentiated rather than exponentiating the standard error?
- Suppose a transformed odds-ratio interval includes 1. What does this imply, and what does it not imply?
- Name one reason a symmetric delta-method interval might be less suitable than the transformed log-scale interval.
Solutions. (1) The log limits are \(0.20\pm2.086(0.10)=(-0.0086,0.4086)\). Exponentiating gives approximately \((0.9914,1.505)\). (2) The t interval estimates uncertainty on the log-coefficient scale; the exponential function must be applied to its endpoints to transform that interval. A standard error is not a confidence limit, and exponentiating it does not produce the required transformation. (3) It means the interval procedure does not exclude an odds ratio of 1 at the stated confidence level. It does not show that the null is true or that there is no association. (4) The delta method is a first-order approximation and can be inaccurate when uncertainty is large; a symmetric interval may even have a negative lower limit, which is impossible for an odds ratio.