The Pitfall Of Transforming Confidence Limits In The Wrong Order

Why the Order of Transformation Matters

Many regression effects are estimated on a scale that is convenient for modeling but reported on a different scale. A logistic regression coefficient, for example, is a log odds ratio; exponentiating it gives the odds ratio. After multiple imputation, the coefficient and its uncertainty are pooled on the log scale. The confidence limits for the reported odds ratio must then reflect that pooled uncertainty.

The central issue is that exponentiation is nonlinear. Exponentiating the two endpoints of a confidence interval is not generally the same as taking the exponentiated point estimate and adding and subtracting a symmetric margin on the reporting scale. The latter can be a useful approximation when uncertainty is small, but it is not the same interval and can be misleading when uncertainty is substantial.

Definition: Suppose \(\theta\) is an effect estimated on a log scale, so the reported effect is \(\exp(\theta)\). A pooled-scale confidence interval is an interval for \(\theta\) formed from the pooled estimate and its pooled variance. Its transformed interval is obtained by applying \(\exp\) separately to the lower and upper limits. Because \(\exp\) is strictly increasing, the transformed limits remain in the same order.

Forming and Transforming the Pooled Interval

Let \(m\) be the number of imputations. In imputation \(i\), let \(\theta_i\) be the estimate of the same log-scale estimand and \(U_i\) its estimated variance. The standard Rubin pooling quantities are

\bar{\theta}=\frac{1}{m}\sum_{i=1}^{m}\theta_i,\qquad W=\frac{1}{m}\sum_{i=1}^{m}U_i,\qquad B=\frac{1}{m-1}\sum_{i=1}^{m}(\theta_i-\bar{\theta})^2,\qquad T=W+\left(1+\frac{1}{m}\right)B.

Here \(W\) is the average within-imputation variance, \(B\) is the between-imputation variance, and \(T\) is the pooled variance. Write \(SE=\sqrt{T}\). With a suitable Rubin degrees-of-freedom approximation, a \(100(1-\alpha)\%\) confidence interval for \(\theta\) is \(\bar{\theta}\pm t_{1-\alpha/2,df}SE\). This tutorial assumes that each imputation estimates the same quantity on the same scale, that the log-scale estimates are approximately normal for inference, and that the imputation and variance estimates support Rubin pooling. It also uses the conventional large-complete-sample degrees-of-freedom approximation; small-sample adjustments may be needed when complete-data degrees of freedom are limited.

For a ratio \(R=\exp(\theta)\), the corresponding interval is obtained by exponentiating both pooled-scale limits. The justification is an event equivalence: since \(\exp\) is strictly increasing, \(\theta\) lies between two limits if and only if \(\exp(\theta)\) lies between their exponentials. Thus transforming the endpoints preserves the coverage of the original interval, subject to the validity of the pooled-scale interval.

\left[\bar{\theta}-t_{1-\alpha/2,df}SE,\ \bar{\theta}+t_{1-\alpha/2,df}SE\right] \quad\longrightarrow\quad \left[\exp\!\left(\bar{\theta}-t_{1-\alpha/2,df}SE\right),\ \exp\!\left(\bar{\theta}+t_{1-\alpha/2,df}SE\right)\right].

This transformed interval is usually asymmetric around \(\exp(\bar{\theta})\). That asymmetry is not a defect: it reflects the nonlinear transformation while retaining the interval's endpoint coverage.

Worked Example: Pooling a Log Odds Ratio and Reporting an Odds Ratio

Suppose a hypothetical analysis has \(m=20\) imputations. The pooled log odds ratio is \(\bar{\theta}=0.5878\), equivalent to an odds ratio of approximately \(1.800\). Suppose the average within-imputation variance is \(W=0.0400\), and the between-imputation variance is \(B=0.0100\). These values give

T=0.0400+\left(1+\frac{1}{20}\right)(0.0100) =0.0505,\qquad SE=\sqrt{0.0505}=0.2247.

Using the conventional approximation \(df=(m-1)\left[1+W/\{(1+1/m)B\}\right]^2\), the degrees of freedom are about \(439\), giving \(t_{0.975,df}\approx1.965\). The pooled-scale interval is

0.5878\pm1.965(0.2247) =0.5878\pm0.4416 =[0.1462,\ 1.0294].

Exponentiating each endpoint gives

[\exp(0.1462),\ \exp(1.0294)] \approx[1.157,\ 2.799].

The estimated odds ratio is \(1.800\), and the interval is about \(1.157\) to \(2.799\). In context, under the model and inference assumptions, this procedure gives a 95% confidence interval for the odds ratio. It is not a statement that there is a 95% probability the fixed odds ratio lies in this particular interval.

What Goes Wrong with a Symmetric Reporting-Scale Interval?

A common alternative is to transform the pooled estimate first and then form a symmetric interval around it using a delta-method standard error. For \(g(\theta)=\exp(\theta)\), the first-order delta method gives an approximate standard error \(\exp(\bar{\theta})SE\). The resulting interval is

\exp(\bar{\theta})\ \pm\ t_{1-\alpha/2,df}\exp(\bar{\theta})SE.

This approximation linearizes the transformation near \(\bar{\theta}\). It may be close when the pooled uncertainty is small, but it does not preserve the original interval by endpoint transformation. Its symmetry, and any resulting limits outside the possible range of the effect, are important warnings. The approximation should not be described as the exact transformed pooled-scale interval.

Worked Example: A Symmetric Approximation Produces an Impossible Limit

Consider another hypothetical pooled log odds ratio, \(\bar{\theta}=\log(1.5)=0.4055\), with \(m=20\), \(W=0.250\), and \(B=0.200\). Then

T=0.250+1.05(0.200)=0.460,\qquad SE=\sqrt{0.460}=0.6782.

The approximate degrees of freedom are \(19[1+0.250/(1.05\times0.200)]^2\approx91.2\), so \(t_{0.975,df}\approx1.986\). The pooled-scale limits are

0.4055\pm1.986(0.6782) =0.4055\pm1.347 =[-0.9415,\ 1.7525].

Exponentiating gives the interval for the odds ratio:

[\exp(-0.9415),\ \exp(1.7525)] \approx[0.390,\ 5.77].

The symmetric delta-method interval instead uses a reporting-scale estimate of \(1.5\) and an approximate standard error of \(1.5(0.6782)=1.017\):

1.5\pm1.986(1.017) \approx1.5\pm2.020=[-0.520,\ 3.520].

An odds ratio cannot be negative. The lower limit therefore signals that the symmetric reporting-scale interval is unsuitable here. The endpoint-transformed interval is positive and reflects the strong right-skew on the odds-ratio scale. The delta-method result is not evidence for a negative odds ratio; it is a poor approximation in this setting.

When the Approximation Is Close, and How to Check

Worked Example: Reporting a Percentage Change

Suppose a hypothetical pooled log-scale estimate is \(\bar{\theta}=0.0800\), with \(m=10\), \(W=0.000800\), and \(B=0.000100\). The reported quantity is percentage change, \(100\{\exp(\theta)-1\}\). First,

T=0.000800+1.1(0.000100)=0.000910,\qquad SE=\sqrt{0.000910}=0.03017.

The degrees of freedom are approximately \(9[1+0.000800/(1.1\times0.000100)]^2\approx616\), so \(t_{0.975,df}\approx1.964\). The pooled-scale interval is

0.0800\pm1.964(0.03017) =0.0800\pm0.05925=[0.02075,\ 0.13925].

Transform both endpoints to the percentage-change scale:

100\{\exp(0.02075)-1\}\approx2.10\%,\qquad 100\{\exp(0.13925)-1\}\approx14.94\%.

The point estimate is \(100\{\exp(0.0800)-1\}\approx8.33\%\). For comparison, the delta-method standard error on the percentage scale is \(100\exp(0.0800)(0.03017)\approx3.27\) percentage points. A symmetric interval is therefore \(8.33\%\pm1.964(3.27\%)\), or about \(1.91\%\) to \(14.75\%\). It is fairly close here, but not identical to the endpoint-transformed interval of \(2.10\%\) to \(14.94\%\).

When intervals are narrow, the transformation is nearly linear over the interval and the two answers can be similar. That agreement is a useful approximation check, not a reason to treat the two procedures as identical.

Diagnostics, Alternatives, and Limitations

Before reporting a transformed interval, check that the pooled estimate and variances refer to the same log-scale quantity across imputations. Calculate the pooled-scale limits first, then transform each endpoint using the reporting function. For exponentiation, the lower limit must be positive and less than the upper limit. If a symmetric interval on the reporting scale crosses zero for a ratio, that is a clear warning to reconsider the approximation rather than truncate the limit without justification.

The main alternative discussed here is a first-order delta-method interval on the reporting scale. It can be useful for a quick approximation when the standard error is small relative to the estimate and the transformation is nearly linear over the uncertainty range. Its performance should be checked against the endpoint-transformed interval, especially when uncertainty is large or the reporting scale has a restricted range. Transforming limits does not repair a poorly specified model, invalid imputation, or an invalid pooled-scale confidence interval; the assumptions supporting the original inference still matter.

A frequent mistake is to exponentiate the pooled point estimate and then assume that the transformed interval must be symmetric around it. The correction is to distinguish the point estimate from the interval procedure: for an interval built on the log scale, exponentiate its lower and upper limits separately. The resulting interval can be asymmetric and is still centered, in the relevant inferential sense, on the pooled log-scale estimate.

Key takeaway: For an exponentiated pooled estimate, form the confidence interval on the pooled log scale and transform both limits. A symmetric interval built after transformation is a separate approximation, not an equivalent rearrangement.

Check Your Understanding

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

A pooled log-scale interval is \([-0.10,0.30]\). What is the corresponding interval for the exponentiated effect, to three significant figures?
Why does applying a strictly increasing transformation to both confidence limits preserve the interval's coverage event?
A delta-method interval for a ratio has a negative lower limit. What does that indicate, and what should be checked?
In the percentage-change example, why is the symmetric delta-method interval close to, but not exactly, the endpoint-transformed interval?

Solutions. (1) Exponentiate both limits: \([\exp(-0.10),\exp(0.30)]=[0.905,1.35]\). (2) For a strictly increasing function \(g\), \(L\leq\theta\leq U\) if and only if \(g(L)\leq g(\theta)\leq g(U)\); therefore the two interval events are identical. (3) A ratio cannot be negative, so the limit warns that the symmetric approximation may be poor. Check the pooled-scale interval and compare it with the interval formed by transforming its endpoints; do not interpret the negative value as a possible ratio. (4) The uncertainty is relatively small, so the exponential function is nearly linear across the interval. The delta method uses that local linear approximation, while transforming endpoints applies the nonlinear function directly.