Coverage Under Monotone Back-Transformation

Why Back-Transformation Needs Care

A confidence interval is often easier to construct on a transformed scale than on the original parameter scale. The transformation may make a parameter unconstrained, or may give an approximation that behaves better for inference. Once an interval has been obtained on that scale, it is tempting simply to transform its two endpoints back. That is correct for a strictly monotone transformation, with the endpoint order handled appropriately. But the shape of the parameter space matters: an endpoint formula describes an interval only when the parameter space itself does not have gaps.

The underlying coverage argument is even simpler than the endpoint calculation. A parameter belongs to the back-transformed confidence set exactly when its transformed value belongs to the original set. This equivalence preserves coverage, whether or not the transformation is monotone. Monotonicity determines the shape and endpoint description of the resulting set, not the basic coverage identity.

Definition: Let $X$ denote the observed data, let $\Theta$ be the parameter space, and let $g:\Theta\to\mathcal{H}$ be a function. A random set $C_\eta(X)\subseteq\mathcal{H}$ for the transformed parameter $\eta=g(\theta)$ induces the back-transformed confidence set $$ C_\theta(X)=\{\vartheta\in\Theta:g(\vartheta)\in C_\eta(X)\}. $$ For a fixed true value $\theta$, the coverage probability of this set is $P_\theta(\theta\in C_\theta(X))$. A confidence set has coverage at least $1-\alpha$ if this probability is at least $1-\alpha$ for every parameter value under consideration.

The set is defined by a preimage: it contains every parameter value whose transformed value lies in $C_\eta(X)$. The symbol $\vartheta$ is used for candidate parameter values in the set, while $\theta$ denotes the fixed true value when coverage is evaluated. We assume the relevant membership events are measurable, so their probabilities are defined.

Endpoint Formulas for Monotone Transformations

Suppose the transformed confidence set is the closed interval $C_\eta(X)=[L(X),U(X)]$. If $g$ is strictly increasing, smaller parameter values map to smaller transformed values, so the lower and upper endpoints retain their order. If $g$ is strictly decreasing, the order reverses. In either case, inverse endpoints require that the endpoints lie in the range of $g$.

Theorem: Suppose $\Theta$ is an interval, $g:\Theta\to g(\Theta)$ is continuous and strictly monotone, and $C_\eta(X)=[L(X),U(X)]$, where $L(X)$ and $U(X)$ belong to $g(\Theta)$. Then the induced back-transformed set is an interval. If $g$ is strictly increasing, $$ C_\theta(X)=[g^{-1}(L(X)),g^{-1}(U(X))]. $$ If $g$ is strictly decreasing, $$ C_\theta(X)=[g^{-1}(U(X)),g^{-1}(L(X))]. $$

Proof. Fix an observed data value and abbreviate the endpoints by $L$ and $U$. If $g$ is strictly increasing, then $g(\vartheta)\in[L,U]$ holds exactly when $L\leq g(\vartheta)\leq U$. Applying the increasing inverse to both inequalities gives $g^{-1}(L)\leq\vartheta\leq g^{-1}(U)$. Intersecting with $\Theta$ does not remove any points between these endpoints because $\Theta$ is an interval and both endpoints belong to $\Theta$. This proves the increasing case. If $g$ is strictly decreasing, applying its inverse reverses both inequalities, giving $g^{-1}(U)\leq\vartheta\leq g^{-1}(L)$, again an interval in $\Theta$. $\square$

The range condition in the theorem keeps the inverse endpoint expressions well-defined. If an endpoint of $[L,U]$ lies outside $g(\Theta)$, the defining preimage still applies, but the endpoint must be clipped at the relevant boundary of the parameter space, or the set may be empty. For example, if $g(\Theta)=(0,\infty)$, an interval $[-1,2]$ on the transformed scale has preimage $\{\vartheta\in\Theta:0<g(\vartheta)\leq2\}$, not a preimage of the value $-1$.

Worked Example: Back-Transforming a Log-Scale Interval

Suppose $\theta>0$, the transformed parameter is $\eta=\log\theta$, and an observed-data interval for $\eta$ is $[0.2,0.8]$. Find the induced interval for $\theta$.

Here $g(\theta)=\log\theta$ is strictly increasing on $\Theta=(0,\infty)$, and its inverse is $g^{-1}(\eta)=e^\eta$. Both endpoints are in the range $\mathbb{R}$, so the theorem gives $$ C_\theta=[e^{0.2},e^{0.8}]\approx[1.221,2.226]. $$

The endpoint order is unchanged because the logarithm is increasing. The coverage interpretation is exact: the true value $\theta$ is in $[e^{0.2},e^{0.8}]$ if and only if $\log\theta$ is in $[0.2,0.8]$. Thus, if the log-scale set covers $\log\theta$ with probability at least $1-\alpha$, the back-transformed set covers $\theta$ with at least that probability.

Worked Example: Reversing Endpoints Under a Decreasing Transformation

Let $\Theta=\mathbb{R}$, take $g(\theta)=-\theta$, and suppose the confidence interval for $\eta=g(\theta)$ is $[-2,1]$. Find the confidence set for $\theta$.

The function is strictly decreasing and $g^{-1}(\eta)=-\eta$. The endpoint order therefore reverses: $$ C_\theta=[g^{-1}(1),g^{-1}(-2)]=[-1,2]. $$

Directly checking the defining inequality confirms the result. The condition $-\theta\in[-2,1]$ is equivalent to $-2\leq-\theta\leq1$; multiplying through by $-1$ reverses both inequalities, so $-1\leq\theta\leq2$. Keeping the original endpoint order would instead report $[2,-1]$, which is not the correct interval.

Coverage Is Preserved by the Preimage

Coverage preservation follows from set membership, not from an approximation or a special property of the estimator. In particular, monotonicity is not needed for this result. It is enough to define the parameter set as the preimage under the same transformation used to define the transformed target.

Theorem: Let $g:\Theta\to\mathcal{H}$ be any function, and define $C_\theta(X)=\{\vartheta\in\Theta:g(\vartheta)\in C_\eta(X)\}$. For every fixed $\theta\in\Theta$, $$ P_\theta(\theta\in C_\theta(X))=P_\theta(g(\theta)\in C_\eta(X)). $$ Consequently, if $C_\eta(X)$ covers $g(\theta)$ with probability at least $1-\alpha$ for every $\theta\in\Theta$, then $C_\theta(X)$ covers $\theta$ with probability at least $1-\alpha$ for every $\theta\in\Theta$.

Proof. By the definition of $C_\theta(X)$, for every data outcome, $$ \theta\in C_\theta(X) \quad\Longleftrightarrow\quad g(\theta)\in C_\eta(X). $$ The two events are therefore identical. Taking their probabilities under $P_\theta$ proves the equality. The coverage conclusion follows by applying the assumed lower bound to the probability on the right-hand side. $\square$

This result concerns the set obtained by taking the full preimage. A reported interval that is merely formed by transforming two endpoints need not be that full preimage if the function is non-monotone, or if the parameter space has gaps. Replacing the preimage by an interval approximation can therefore change coverage. Coverage is preserved by the exact set definition; the endpoint theorem identifies conditions under which that set can be written using two endpoints.

Worked Example: A Restricted Parameter Space With a Gap

Let $\Theta=\{1,3\}$, let $g(\theta)=\theta$, and suppose the transformed confidence set is $[1,3]$. What is the induced set for $\theta$?

By definition, the induced set is $$ \{\vartheta\in\{1,3\}: \vartheta\in[1,3]\}=\{1,3\}. $$

Although the transformed set is an interval and $g$ is increasing, the induced set is not the interval $[1,3]$ as a subset of the real line: it contains only the parameter values 1 and 3. The parameter space has a gap at 2, so the endpoint formula cannot assert that all values between the two endpoints are allowed. The correct general expression is the preimage within $\Theta$, namely $\Theta\cap[1,3]$. This qualification is essential whenever the parameter space is not itself an interval.

Why Non-Monotone Transformations Can Produce Disconnected Sets

A non-monotone function can map separated parameter values to the same transformed value. Consequently, the preimage of one interval can consist of multiple separated pieces. The coverage identity still holds for that full preimage, but there may be no single pair of endpoints that describes it as an interval.

Worked Example: A Squared Parameter Produces Two Pieces

Let $\Theta=\mathbb{R}$, set $g(\theta)=\theta^2$, and suppose a confidence interval for $\eta=g(\theta)$ is $[0.25,1]$. Find its preimage in the original parameter space.

The condition for membership is $$ 0.25\leq\theta^2\leq1. $$ The first inequality gives $|\theta|\geq0.5$, while the second gives $|\theta|\leq1$. Combining them yields $$ C_\theta=[-1,-0.5]\cup[0.5,1]. $$

The set has two connected pieces because both positive and negative parameter values in these ranges have squares in $[0.25,1]$. Reporting just $[-1,1]$ would add values such as $\theta=0$, whose square is $0$, outside the transformed interval. Reporting only one of the two pieces would omit valid parameter values and would no longer be the full preimage. The exact coverage identity applies to the union of both pieces.

Not every non-monotone transformation produces a disconnected set for every transformed interval. For instance, the preimage of $[0,1]$ under $g(\theta)=\theta^2$ is $[-1,1]$, which is connected. The correct conclusion is that non-monotonicity can require disconnected confidence sets; the shape must be determined from the particular transformation, transformed set, and parameter space.

Common Pitfalls and Practical Use

Leaving endpoints in the wrong order. For a decreasing transformation, the inverse reverses endpoint order. Write the preimage inequality and solve it, or apply the decreasing-case formula.
Assuming every preimage is an interval. This is not true for non-monotone transformations, and it is not true for arbitrary parameter spaces. State the parameter space and use the full preimage definition.
Transforming endpoints outside the range. If an endpoint is not in $g(\Theta)$, its inverse may not exist. Use the preimage inequality and account for the boundary or possible emptiness instead.
Confusing unchanged coverage with unchanged shape. Coverage follows from equivalent membership events. The resulting set can be an interval, a union of intervals, or a restricted set of parameter values.
Replacing a disconnected set by its hull. Filling gaps can include values whose transformed values are not in the original confidence set, so the new set is not the exact back-transformation and need not have the same coverage.

In practice, first write the transformed set and the parameter space. Then define the back-transformed set as a preimage. If the transformation is strictly monotone on an interval parameter space, use inverse endpoints and reverse their order when the function is decreasing. If either condition fails, solve the membership inequality directly and preserve all components of the resulting set.

Key takeaway: The exact back-transformed confidence set is the preimage of the transformed confidence set within the parameter space. This definition preserves coverage. Monotonicity and the structure of the parameter space determine whether the preimage can be reported as a single interval with inverse-transformed endpoints.

Check Your Understanding

Use the preimage definition and check the parameter space before applying an endpoint formula.

If $g$ is strictly decreasing and the transformed interval is $[a,b]$, which inverse endpoint is the lower endpoint for the original parameter?
Why does the coverage identity hold even if $g$ is not one-to-one?
For $g(\theta)=\theta^2$ on $\mathbb{R}$, find the preimage of $[4,9]$.
Why does an increasing transformation on $\Theta=\{1,3\}$ not guarantee that the preimage of an interval is an interval in $\mathbb{R}$?
What should be done if a transformed confidence interval has an endpoint outside $g(\Theta)$?