Back-Transforming Confidence Limits

Why Back-Transforming Confidence Limits Matters

A confidence interval is often easiest to construct after transforming a parameter. A transformation can make an estimator’s sampling distribution more nearly normal or stabilize its variance, so a familiar interval can be built on the transformed scale. But the parameter of scientific interest is usually on its original scale. The final step is to map the interval limits back without changing their order or misrepresenting what the interval says.

The key point is that the inverse transformation acts on the limits, not just on the point estimate. For a nonlinear transformation, the resulting interval generally is not centered on the back-transformed estimate and need not have equal distances from it to its two endpoints. That asymmetry is a geometric consequence of the transformation; it does not by itself imply a change in coverage.

Definition: Let \(\theta\) be a parameter in a parameter space \(\Theta\), and let \(g:\Theta\to\Gamma\) be one-to-one. Write \(\eta=g(\theta)\), and let \(h=g^{-1}\) denote the inverse defined on the image \(g(\Theta)\). If \([L,U]\) is a confidence interval for \(\eta\), with finite endpoints in \(g(\Theta)\), its back-transformed confidence set is \[ \{\vartheta\in\Theta:L\leq g(\vartheta)\leq U\}. \] If \(\Theta\) is an interval and \(g\) is strictly monotone, this set is an interval with endpoints \(h(L)\) and \(h(U)\) in the appropriate order.

The image \(g(\Theta)\), rather than an arbitrary codomain \(\Gamma\), is the relevant range for taking the inverse. The requirement that both endpoints lie in \(g(\Theta)\) ensures that \(h(L)\) and \(h(U)\) exist. If \(\Theta\) is an interval, strict monotonicity also ensures that the back-transformed set is an interval. Without that condition on \(\Theta\), the preimage need not be an interval.

Ordering the Back-Transformed Limits

A strictly increasing transformation preserves order, so its inverse does too. A strictly decreasing transformation reverses order. Thus, even when the transformed limits satisfy \(L\leq U\), the back-transformed lower limit is not always \(h(L)\).

Endpoint rule: Suppose \(L\leq U\) and \(L,U\in g(\Theta)\). If \(g\) is strictly increasing, the back-transformed interval is \([h(L),h(U)]\). If \(g\) is strictly decreasing, it is \([h(U),h(L)]\). Equivalently, the limits are \(\min\{h(L),h(U)\}\) and \(\max\{h(L),h(U)\}\).

This rule comes directly from the inequalities defining the set. For an increasing \(g\), \(L\leq g(\vartheta)\leq U\) is equivalent to \(h(L)\leq\vartheta\leq h(U)\). For a decreasing \(g\), applying \(h\) reverses both inequalities, giving \(h(U)\leq\vartheta\leq h(L)\). No additional standard error is needed at this step: the interval’s limits are transformed directly.

Worked Example: Back-Transforming an Interval on a Log Scale

Suppose \(\theta>0\), the transformed parameter is \(\eta=\log\theta\), and a confidence interval for \(\eta\) is \([1.0923,1.6803]\). The transformation \(g(\theta)=\log\theta\) is strictly increasing, with inverse \(h(\eta)=e^\eta\). Therefore the limits remain in their original order:

[e^{1.0923},e^{1.6803}]\approx[2.9811,5.3672].

Rounded to three decimal places, the interval for \(\theta\) is \([2.981,5.367]\). To see the asymmetry, the midpoint of the transformed limits is \(1.3863\), whose exponential is approximately \(4.000\). The distances from this back-transformed midpoint to the interval endpoints are approximately \(4.000-2.9811=1.0189\) and \(5.3672-4.000=1.3672\). They are unequal, even though the original limits are equally spaced about \(1.3863\).

Worked Example: Reversing Limits Under a Reciprocal Transformation

Let \(\theta>0\) and use the decreasing transformation \(g(\theta)=1/\theta\). Suppose the interval for \(\eta=1/\theta\) is \([0.3432,0.6568]\). The inverse is \(h(\eta)=1/\eta\), which is also decreasing. Consequently, the lower limit for \(\theta\) comes from the larger transformed endpoint:

\left[\frac{1}{0.6568},\frac{1}{0.3432}\right] \approx[1.5225,2.9138].

The endpoint checks are \(0.6568(1.5225)\approx1.0000\) and \(0.3432(2.9138)\approx1.0000\), allowing for rounding. Writing \([1/0.3432,1/0.6568]\) would put the larger value first and would not be a conventionally ordered interval. Reversing the endpoints is essential for a decreasing transformation.

Coverage Is Preserved by the Transformation

A confidence interval is a random set: its endpoints vary from sample to sample. Its coverage probability is the probability that it contains the fixed parameter value under repeated sampling. Back-transformation preserves this probability when the interval is mapped through a one-to-one transformation and the endpoints are valid inverse inputs.

Theorem: Let \(\Theta\) be an interval, let \(g:\Theta\to g(\Theta)\) be strictly monotone, and fix \(\theta\in\Theta\). Put \(\eta=g(\theta)\). Suppose \([L,U]\) is a random interval with finite endpoints satisfying \(L\leq U\) and \(L,U\in g(\Theta)\) with probability one. Define the back-transformed interval by ordering \(g^{-1}(L)\) and \(g^{-1}(U)\). Then \[ P\{\theta\text{ belongs to the back-transformed interval}\} = P\{\eta\in[L,U]\}. \]

This is an exact equality of probabilities, not an asymptotic approximation. In particular, if the transformed interval has coverage \(1-\alpha\), then its back-transformed interval has the same coverage \(1-\alpha\) under the stated conditions.

Write membership on the transformed scale.
By definition, \(\theta\) belongs to the back-transformed set exactly when \(L\leq g(\theta)\leq U\). Since \(g(\theta)=\eta\), this is \(L\leq\eta\leq U\).

Identify the two events.
The event that the back-transformed interval contains \(\theta\) is therefore exactly the event that \([L,U]\) contains \(\eta\). This equivalence holds for each sample outcome on which the endpoints satisfy the assumptions.

Take probabilities.
Equal events have equal probabilities, so the coverage probabilities are equal. \(\square\)

The proof uses event equality, not a claim that an estimate and its transformed version have identical sampling distributions. The interval on the transformed scale is treated as given; the result says what happens to its coverage when its set of parameter values is expressed on the original scale.

Worked Example: A Symmetric Square-Root-Scale Interval

Suppose \(\theta>0\), and an interval for \(\eta=\sqrt{\theta}\) is \([1.4,2.0]\). The inverse transformation \(h(\eta)=\eta^2\) is increasing on the positive range, so the back-transformed interval is

$$ [1.4^2,2.0^2]=[1.96,4.00]. $$

The transformed midpoint is \(1.7\), which maps to \(1.7^2=2.89\). The endpoint distances on the \(\theta\) scale are \(2.89-1.96=0.93\) and \(4.00-2.89=1.11\). Thus, equal distances around the transformed midpoint have become unequal distances on the original scale. If the interval \([1.4,2.0]\) covers the fixed value \(\sqrt{\theta}\) with probability \(0.95\), the membership equivalence in the theorem shows that \([1.96,4.00]\) covers \(\theta\) with that same probability, assuming the interval’s construction satisfies the theorem’s range conditions.

Why Nonlinear Transformations Create Asymmetry

A nonlinear inverse transformation stretches some parts of the transformed scale more than others. This changes the widths of the two sides of an interval as it is mapped back. A useful precise statement applies when the inverse transformation \(h=g^{-1}\) is convex on the relevant range.

Result: Let \(h\) be convex on an interval containing \([m-a,m+a]\), where \(a>0\). For the symmetric transformed interval \([m-a,m+a]\), the upper-side distance after back-transformation is at least the lower-side distance: \[ h(m+a)-h(m)\geq h(m)-h(m-a). \] If \(h\) is strictly convex on this range, the inequality is strict.

To prove the claim, convexity at the midpoint \(m=\tfrac12(m-a)+\tfrac12(m+a)\) gives

h(m)\leq\frac{h(m-a)+h(m+a)}{2}.

Multiplying by \(2\) and rearranging yields \(h(m+a)-h(m)\geq h(m)-h(m-a)\). Strict convexity makes the midpoint inequality strict, giving the strict version. For a concave inverse, the inequality is reversed. This result describes a geometric effect; it does not say that every back-transformed interval must be asymmetric. A linear transformation has a linear inverse and preserves symmetry.

Asymmetry should therefore be interpreted in relation to the parameter scale. It does not, by itself, indicate a mistake, an incorrect confidence level, or a biased estimator. Coverage concerns whether the interval contains the fixed parameter over repeated samples; symmetry concerns the distances between the interval’s endpoints and a chosen center. They are different properties.

Conditions and Common Mistakes

Before reporting a back-transformed interval, check that the inverse is defined at both limits and that the endpoint order reflects whether the transformation is increasing or decreasing. The theorem above assumes both limits lie in the image \(g(\Theta)\) with probability one. If a computed limit is outside that image, it cannot simply be passed to \(g^{-1}\). Determining what to do in such cases requires a separate treatment of the parameter range and interval construction.

Wrong: “The transformed limits are ordered, so applying the inverse leaves them ordered.” This is true only when the inverse is increasing. For a decreasing inverse, the order reverses.
Correction: Check the direction of monotonicity and report the smaller inverse value first.
Wrong: “The inverse is defined on the stated codomain, so every endpoint there can be transformed.” A map need not attain every value in its codomain.
Correction: Require the endpoints to lie in the image \(g(\Theta)\), where the inverse is defined.
Wrong: “An asymmetric interval must have a different coverage probability.” Asymmetry alone says nothing about coverage.
Correction: Use the event-equivalence result: under its conditions, the transformed and back-transformed intervals have exactly the same coverage.

Key takeaway: Back-transform both confidence limits, order the resulting values using the monotonicity of the inverse, and check that the endpoints belong to the transformation’s image. Under these conditions, the interval’s coverage is unchanged, although its shape and symmetry on the parameter scale may differ.

Check Your Understanding

Use the definitions and results above to answer the following questions.

If a strictly increasing transformation maps a parameter interval to \([L,U]\), in what order should the inverse-transformed limits be reported?
For a strictly decreasing transformation, why must the inverse-transformed limits be reversed?
State the event equivalence that establishes coverage preservation.
For a convex inverse transformation and a symmetric transformed interval, which side has the greater or equal back-transformed distance from the image of the midpoint?
Why is it insufficient to know only that an endpoint lies in the codomain named for \(g\)?