Why a Boundary Changes Endpoint Inference
A partially identified effect is often summarized by lower and upper endpoints. These endpoints may be restricted by the parameter space: for example, a lower endpoint may be no smaller than zero, or an endpoint may be confined to the known range of an outcome. Such restrictions are useful for defining the parameter, but they also affect how an endpoint estimate behaves in repeated samples.
A familiar symmetric normal interval is most straightforward when an estimator is approximately normal around the target and the target is away from restrictions. Near a boundary, an estimator may instead pile up at the constraint. Its sampling distribution can be skewed, have a point mass, or change with the target’s distance from the boundary. A symmetric interval may then have coverage different from its nominal level, even when the unconstrained estimator is well approximated by a normal distribution.
The distinction matters for inference. Seeing an estimate equal to a constraint does not establish that the true endpoint is exactly there. It may reflect sampling variation and the rule used to keep the estimate feasible. Conversely, if the true endpoint is at the boundary, normal approximations derived for an interior value may not describe its sampling distribution.
A Simple Boundary Model
Consider an endpoint \(\theta\) known to satisfy \(\theta\geq 0\). Suppose an unconstrained estimator \(\tilde{\theta}\) has the exact sampling model \(\tilde{\theta}\sim N(\theta,s^2)\), where \(s>0\) is known. A natural constrained estimate is \(\hat{\theta}=\max(0,\tilde{\theta})\). This simplified model isolates the effect of the restriction; it is not a claim that every estimated causal bound has this exact distribution.
At the boundary, \(\theta=0\), half of the normal draws for \(\tilde{\theta}\) are negative and become zero after imposing the constraint. The other half remain positive. Thus the constrained estimator has a point mass of one half at zero and a positive half-normal component. It is not normally distributed.
More generally, if \(\tilde{\theta}\sim N(\theta,s^2)\), then \(P(\hat{\theta}=0)=P(\tilde{\theta}\leq0)=\Phi(-\theta/s)\), where \(\Phi\) is the standard normal cumulative distribution function. This probability is near one half at the boundary and declines as the true endpoint moves away from it. Consequently, the estimator’s distribution changes across the parameter space, including for values close to the boundary.
Worked Example: Coverage of a Symmetric Interval at the Boundary
Suppose \(\theta\geq0\), the true value is \(\theta=0\), and \(\tilde{\theta}\sim N(\theta,s^2)\) with known \(s=0.10\). Use the constrained estimate \(\hat{\theta}=\max(0,\tilde{\theta})\). Consider the usual symmetric interval \(\hat{\theta}\pm1.96s\), with its lower endpoint truncated to zero if necessary. What is its coverage at the boundary?
The interval contains zero exactly when its lower endpoint is no greater than zero. Before truncation, this is the condition \(\hat{\theta}-1.96s\leq0\), or \(\hat{\theta}\leq1.96s\). Since \(\hat{\theta}=\max(0,\tilde{\theta})\), that event is equivalent to \(\tilde{\theta}\leq1.96s\). Therefore,
The interval has about 97.5% boundary coverage, not 95%, in this exact model. It is conservative at the boundary. This calculation does not show that every Wald interval undercovers; it shows that ordinary symmetric calibration need not give the advertised nominal coverage. Other estimators, estimated standard errors, and more complicated endpoint functions can behave differently, so coverage must not be assumed from the interval’s familiar form alone.
One-Sided Limits: Check Which Estimate Defines the Limit
For a one-sided upper limit, the sign and construction of the limit are especially important. In the exact normal model above, an upper limit based on the unconstrained estimator is \(U=\tilde{\theta}+z_{0.95}s\), where \(z_{0.95}\approx1.645\). For any \(\theta\), it covers \(\theta\) when \(\tilde{\theta}+1.645s\geq\theta\), which has probability 0.95 under the stated normal model.
If instead the limit is formed from the constrained estimate, \(U_c=\hat{\theta}+1.645s\), it is always positive because \(\hat{\theta}\geq0\) and \(s>0\). It therefore covers the boundary value zero with probability one. The unconstrained limit and the constrained limit are not interchangeable; in particular, the constrained limit is not a nominal 95% upper procedure at the boundary in this example.
Worked Example: Comparing Two Upper Limits
Suppose \(\theta\geq0\), \(s=0.10\), and a hypothetical sample gives \(\tilde{\theta}=-0.20\). The constrained estimate is \(\hat{\theta}=\max(0,-0.20)=0\). Calculate both one-sided 95% upper limits.
Using the constrained estimate gives \(U_c=\hat{\theta}+1.645s=0+1.645(0.10)=0.1645\). It is positive, so it includes zero. In fact, for every possible sample, this construction includes zero whenever zero is the true boundary value.
Using the unconstrained estimate gives \(U=\tilde{\theta}+1.645s=-0.20+1.645(0.10)=-0.0355\). At the boundary, this limit covers zero only if \(U\geq0\), equivalently if \(\tilde{\theta}\geq-1.645s\). The probability of that event is \[ P\left(\frac{\tilde{\theta}-0}{s}\geq-1.645\right) =\Phi(1.645)\approx0.95. \]
The negative observed upper limit may look awkward when the parameter is known to be nonnegative. It is nevertheless the result of the stated unconstrained procedure; replacing it by zero changes the procedure and its coverage. If the limit is projected to \(\max(0,U)\), it includes zero with probability one at the boundary, so its boundary coverage becomes conservative rather than 95%. The inferential construction and any projection should be reported clearly.
Boundary-Aware Inference and Endpoint Pairs
One way to account for a boundary is to use a procedure whose reference distribution reflects the constraint. For example, in the simple normal model, testing \(H_0:\theta=0\) against \(\theta\geq0\) leads to a one-sided likelihood-ratio statistic proportional to \(\max(0,Z)^2\), where \(Z=\tilde{\theta}/s\) under the null. Under \(H_0\), \(Z\sim N(0,1)\). With probability one half, \(Z\leq0\) and the statistic is zero; otherwise its positive part follows the corresponding half of a \(\chi^2\) distribution with one degree of freedom. Its null distribution is therefore a mixture, not an ordinary \(\chi^2_1\) distribution.
For this single regular inequality constraint, the asymptotic reference distribution is \(0.5\) point mass at zero plus \(0.5\chi^2_1\). The upper-tail probability for a positive observed statistic \(q\) is \(0.5P(\chi^2_1\geq q)\). Inverting tests calibrated for the constraint can produce confidence limits that account for the boundary. This mixture result has specific regularity conditions; it should not be applied automatically to multiple active constraints, irregular endpoint functions, or settings with additional nuisance-parameter complications.
Worked Example: A Boundary-Aware Likelihood-Ratio Test
Suppose a hypothetical endpoint satisfies \(\theta\geq0\), and the unconstrained estimate and known standard error are \(\tilde{\theta}=0.04\) and \(s=0.05\). Test \(H_0:\theta=0\) against \(\theta>0\) using the one-sided boundary likelihood-ratio statistic.
First calculate \(Z=\tilde{\theta}/s=0.04/0.05=0.8\). The constrained likelihood-ratio statistic on the normal scale is \(q=\max(0,Z)^2=0.8^2=0.64\). Its boundary-mixture p-value is \[ 0.5P(\chi^2_1\geq0.64) =0.5P(|Z_0|\geq0.8) =0.5\{2[1-\Phi(0.8)]\} \approx0.2119, \] where \(Z_0\sim N(0,1)\). This is also the one-sided normal tail probability \(1-\Phi(0.8)\). Since \(0.2119>0.05\), the test does not reject \(H_0\) at the 5% level. This conclusion is not evidence that the endpoint equals zero; the data do not provide sufficient evidence against the boundary value under this test.
For an identified interval \([L,U]\), inference must address both endpoints if the goal is a confidence set containing the entire identified interval. A lower confidence limit for \(L\) and an upper confidence limit for \(U\) can be combined. Their joint coverage is not generally the product of their individual coverages; without a justified joint calibration, Bonferroni control is one simple option. Construct each one-sided limit at level 97.5%, so each has error probability at most 0.025. Then the probability that either endpoint limit fails is at most \(0.025+0.025=0.05\), and the joint coverage is at least 95%.
Worked Example: Combining Limits for a Partially Identified Interval
Suppose a hypothetical identified interval has endpoints known to lie in \([0,10]\). Unconstrained endpoint estimators are approximately normal, with estimates \(\tilde L=0.12\) and \(\tilde U=9.70\), and standard errors \(0.10\) and \(0.15\). Construct a conservative 95% confidence set for the entire identified interval using Bonferroni one-sided limits.
For two one-sided limits with 97.5% coverage each, use \(z_{0.975}=1.96\). The unprojected lower limit for \(L\) is \(0.12-1.96(0.10)=0.12-0.196=-0.076\). Since \(L\geq0\), project this lower limit to zero. The unprojected upper limit for \(U\) is \(9.70+1.96(0.15)=9.70+0.294=9.994\), which is below the known maximum of 10. The resulting confidence set is \([0,9.994]\).
Under the stated one-sided normal approximations, each unprojected limit has approximately nominal 97.5% coverage for its endpoint. Projection to the known parameter range does not remove a true endpoint from a correctly covering one-sided limit: for example, if \(L\geq0\), replacing a lower limit by its maximum with zero cannot make it exceed \(L\). If each one-sided limit actually has coverage of at least 97.5%, Bonferroni guarantees that the probability either endpoint limit fails is no greater than 0.05, so the set contains \([L,U]\) with coverage at least 95%. Under the stated normal approximations alone, however, the individual and resulting joint coverage are approximate, not guaranteed. The procedure is conservative when those one-sided coverage guarantees hold; the two-sided interval should not be described as a 95% interval for a point effect.
Diagnostics, Alternatives, and Limitations
Boundary concerns are more serious when estimates frequently land exactly on a constraint, when the unconstrained estimate is only a few standard errors from it, or when the estimated standard error is itself unstable near the constraint. These patterns suggest that a normal approximation centered on the constrained estimate may be a poor description of repeated-sample behavior. For an endpoint in a known range, check both edges and keep track of whether a limit was truncated or projected.
The main approaches answer somewhat different needs:
| Approach | What it does | Main caution |
|---|---|---|
| Symmetric Wald interval | Centers a normal-based interval on an estimate. | Can have non-nominal coverage near a constraint and may extend outside the parameter space. |
| Unconstrained one-sided normal limit | Uses the regular approximation for an unconstrained estimator. | May yield a limit outside the feasible range; projecting it changes coverage. |
| Boundary-aware test inversion | Uses a reference distribution that accounts for active constraints. | Requires the right boundary law and appropriate regularity conditions. |
| Conservative simultaneous limits | Combines endpoint limits with a guaranteed error allocation, such as Bonferroni. | May be wider than a correctly calibrated joint procedure. |
The exact-normal calculations above clarify the mechanism, but estimated causal bounds need not be simple truncations of normal estimators. They may depend on estimated nuisance quantities, optimization, or several constraints becoming active together. In those settings, neither the half-normal example nor the single-constraint mixture automatically supplies a valid calibration. The appropriate method depends on how the endpoint is estimated and on the sampling assumptions that justify its approximation.
Boundary inference is not a cosmetic adjustment to an interval’s endpoints. It is a question about the estimator’s sampling distribution and the procedure’s repeated-sample coverage. State the parameter-space restriction, distinguish constrained from unconstrained estimates, and identify the calibration used. For a confidence set for a partially identified effect, also say whether the limits are for individual endpoints or are jointly calibrated to contain the identified interval.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In the model \(\hat{\theta}=\max(0,\tilde{\theta})\) with \(\tilde{\theta}\sim N(\theta,s^2)\), what is the probability that the estimate is exactly zero when \(\theta=0\)?
- At the boundary, does the symmetric interval \(\hat{\theta}\pm1.96s\), with its lower limit truncated at zero, have 95% coverage in the exact-normal example?
- For an unconstrained one-sided 95% upper limit \(U=\tilde{\theta}+1.645s\), state the condition for covering \(\theta=0\).
- Why is it not enough to give each endpoint of an identified interval a separate 95% confidence limit and call the resulting set a 95% confidence set for the whole interval?
Solutions: 1. It is \(P(\tilde{\theta}\leq0)=0.5\), because the unconstrained estimate is centered at zero and normally distributed. 2. No. It has boundary coverage \(\Phi(1.96)\approx0.975\), so it is conservative in this model rather than exactly 95%. 3. It covers zero when \(U\geq0\), equivalently \(\tilde{\theta}\geq-1.645s\); this event has probability approximately 0.95 under the boundary model. 4. Separate 95% coverage does not ensure 95% joint coverage: either endpoint limit may fail. A joint calibration or an error allocation such as Bonferroni is needed to control the probability that at least one fails.