Why Parameter Boundaries Matter
A parameter may be known in advance to lie within a restricted space. A probability is between zero and one, for example, and a scale parameter is positive. A confidence interval obtained without using that restriction can extend into values that are impossible under the model. Intersecting the interval with the allowed space removes those values, but it is important to distinguish this operation from changing the confidence procedure itself.
The key result is precise: if a confidence set already has the stated coverage for every allowed parameter value, intersecting it with the parameter space does not change its coverage. However, other operations sometimes called “truncation”—especially replacing an empty intersection with a boundary point—can change coverage. The distinction is aligned with the SOA VEE Mathematical Statistics syllabus.
Coverage is a repeated-sampling property: the parameter stays fixed while the data, and therefore the confidence set, vary. A particular observed interval either contains the fixed parameter or it does not. The confidence level describes the procedure across repeated samples, not a probability assigned to the parameter after observing the data.
Intersecting a Confidence Set With the Parameter Space
Suppose a procedure produces the interval \(C(X)=[L(X),U(X)]\), and the allowed parameter space is \(\Theta=[a,b]\). The constrained set is the ordinary set intersection:
provided the new lower endpoint is no greater than the new upper endpoint. If it is greater, the intersection is the empty set. When \(\Theta=[a,\infty)\), the same rule gives \([\max\{L(X),a\},U(X)]\) when that interval is nonempty.
This result applies to any confidence set, not just an interval. It also applies whether the original procedure has exact or conservative coverage. It does not assert that the interval looks the same after intersection: its endpoints and width may change, and some data sets may give an empty intersection. It says that, for an allowed true value, the event that the set covers that value is unchanged.
Why Intersection Preserves Coverage
The justification follows directly from set membership. Fix any possible true value \(\theta\in\Theta\). For each possible data outcome \(X\), membership in the intersected set requires membership in both sets:
\(\{\theta\in C(X)\cap\Theta\}=\{\theta\in C(X)\}\cap\{\theta\in\Theta\}\).
\(\theta\) is fixed and belongs to \(\Theta\), so \(\{\theta\in\Theta\}\) is certain under \(P_\theta\).
Thus \(P_\theta(\theta\in C(X)\cap\Theta)=P_\theta(\theta\in C(X))\), which is the original coverage.
For an interval, the same fact can be checked by inequalities. If \(\theta\in[a,b]\), then \(\max\{L,a\}\leq\theta\) is equivalent to \(L\leq\theta\), and \(\theta\leq\min\{U,b\}\) is equivalent to \(\theta\leq U\). Together, these are exactly the original condition \(L\leq\theta\leq U\). If the intersection is empty, it cannot contain \(\theta\); the original interval also cannot have contained \(\theta\), since \(\theta\) is in the parameter space.
Normal-Mean Examples With a Lower Bound
Worked Example: Intersecting a 95% Interval With a Nonnegative Mean Space
Assume \(X_1,\ldots,X_4\) are independent observations from \(N(\mu,4)\), where the second parameter is the variance, and the known parameter space is \(\mu\geq0\). The observed sample mean is \(\bar X=0.60\). Find the constrained 95% confidence interval, using the known population standard deviation.
Method. Under the stated normal model, \(\bar X\sim N(\mu,4/4)=N(\mu,1)\). Therefore \(Z=\bar X-\mu\) has a standard normal distribution. The equal-tailed 95% interval uses the standard normal critical value \(1.96\):
The standard error is \(2/\sqrt{4}=1\). Substituting \(\bar X=0.60\), the unbounded interval is \([0.60-1.96,0.60+1.96]=[-1.36,2.56]\). Intersecting with \([0,\infty)\) gives \([0,2.56]\).
Coverage check. For any \(\mu\geq0\), the intersected interval contains \(\mu\) exactly when the original interval does. The standardized error is standard normal, so the coverage is \(P(-1.96\leq Z\leq1.96)=0.9500\). The lower endpoint changed, but coverage did not. This avoids the mistake of treating a negative endpoint as evidence that the model’s coverage calculation must be recomputed.
Worked Example: A Bounded Parameter and an Estimate Below Zero
Let \(W\sim N(\theta,0.25^2)\), with known standard deviation \(0.25\) and parameter space \(0\leq\theta\leq1\). The observed value is \(w=-0.10\). Construct the equal-tailed 90% confidence set and check its coverage.
Method. The normal pivot \((W-\theta)/0.25\) is standard normal. The two-sided 90% critical value is \(1.645\), so the standard error for this single observation is \(0.25\), and the margin is \(1.645(0.25)=0.41125\). The unbounded interval is
The parameter space is \([0,1]\). The intersection is therefore \([0,0.31125]\). The data value being negative does not make the interval invalid; the constrained set reports only the part compatible with the known parameter space.
Coverage check. For any \(\theta\in[0,1]\), the constrained interval covers \(\theta\) precisely when the unbounded interval does. Under the stated model, that event is \(-1.645\leq(W-\theta)/0.25\leq1.645\), which has probability \(0.9000\). In particular, this remains true at either boundary, \(\theta=0\) or \(\theta=1\). The calculation uses the known standard deviation; it would not be justified to use this same normal interval if the sampling model or standard deviation assumption differed.
Empty Intersections and Coverage-Changing Shortcuts
A valid intersection can be empty. This is not a reason to replace it with a singleton boundary value. The empty set means that the observed, unconstrained confidence set contains no allowed parameter values. Such an outcome is possible even when the model restricts the true parameter to the allowed space; a confidence procedure does not have to include an allowed value for every data set.
Worked Example: An Empty Intersection Is Not a Boundary Point
Assume \(Y\sim N(\theta,0.10^2)\), with known standard deviation \(0.10\), parameter space \([0,1]\), and an observed value \(y=1.30\). Find the 90% interval based on the normal pivot and intersect it with the parameter space.
Method. Use the equal-tailed 90% critical value \(1.645\). The margin is \(1.645(0.10)=0.1645\), so the unbounded interval is \([1.30-0.1645,1.30+0.1645]=[1.1355,1.4645]\). Its intersection with \([0,1]\) is empty, because even its lower endpoint, \(1.1355\), exceeds the upper parameter bound, \(1\).
The correct constrained set for this observation is \(\varnothing\), not \(\{1\}\). Replacing it by \(\{1\}\) is a different procedure: it includes a value that was not in the original confidence set. That change can affect coverage, so the intersection-preservation theorem no longer applies.
The following calculation shows the coverage effect of a tempting but incorrect endpoint rule. Suppose \(X\sim N(\mu,1)\), \(\mu\geq0\), and the ordinary 95% interval is \([X-1.96,X+1.96]\). At the boundary \(\mu=0\), its coverage is \(P(-1.96\leq X\leq1.96)=0.9500\). The correct intersection with \([0,\infty)\) has the same coverage.
Instead, suppose each endpoint is separately replaced by its maximum with zero, even when the resulting lower endpoint exceeds the upper endpoint. If \(X<-1.96\), the original interval lies entirely below zero, but this rule turns it into \([0,0]\), which includes the true boundary value. The modified procedure covers zero whenever \(X\leq1.96\), so its boundary coverage is \(\Phi(1.96)=0.9750\), not \(0.9500\). The shortcut has increased coverage by counting outcomes that should have produced an empty intersection.
How to Use Parameter Constraints Reliably
When reporting a constrained confidence interval, keep the model, the original confidence procedure, and the intersection operation separate. First derive or specify the interval using the assumed sampling distribution. Then intersect that set with the allowed parameter space. Finally, check whether the intersection is nonempty. This sequence prevents a boundary rule from quietly replacing the original procedure.
| Operation | Effect on the set | Effect on coverage for allowed values |
|---|---|---|
| Intersect \(C(X)\) with \(\Theta\) | Removes values outside \(\Theta\); may produce an empty set | Preserves the original coverage |
| Replace an empty intersection with a boundary singleton | Adds a value that was not in \(C(X)\) | May change coverage |
| Use a new interval centered at a constrained estimate | Changes the interval construction | Must be evaluated as a new procedure |
A frequent exam trap is to argue that a shortened interval must have lower coverage because it is narrower. That conclusion is not valid here: intersection removes only parameter values that cannot be true, and for each allowed \(\theta\), it leaves the event \(\theta\in C(X)\) unchanged. A different construction—such as centering an interval at a boundary-constrained estimate, reflecting a negative endpoint, or filling an empty set—does not follow from the theorem and requires its own coverage analysis.
Check Your Understanding
For each problem, identify the allowed parameter space before intersecting the interval. Use the stated sampling assumptions and keep an empty intersection empty.
- Independent observations have distribution \(N(\mu,9)\), with known variance \(9\). For \(n=9\), \(\bar X=-0.20\), and \(\mu\geq0\), form the equal-tailed 90% interval using \(z=1.645\), then intersect with the parameter space. (Answer: \([0,1.445]\).)
- Suppose \(Y\sim N(\theta,0.20^2)\), with known standard deviation \(0.20\) and \(0\leq\theta\leq2\). For \(y=2.10\), find the equal-tailed 95% interval using \(z=1.96\), then intersect it with \([0,2]\). (Answer: \([1.708,2]\), to three decimals.)
- An exact confidence interval has coverage \(0.92\) for every \(\theta\in[3,8]\). What is the coverage after intersecting every reported interval with \([3,8]\)? (Answer: \(0.92\) at every allowed \(\theta\).)
- For \(X\sim N(\mu,1)\), use \([X-1.96,X+1.96]\) at \(\mu=0\). If an empty intersection with \([0,\infty)\) is incorrectly replaced by \(\{0\}\), calculate the resulting coverage. (Answer: \(\Phi(1.96)=0.9750\).)
- An unbounded interval is \([L,U]\), and the parameter space is \([a,b]\). State the condition for its intersection with the parameter space to be empty. (Answer: \(L>b\) or \(U<a\).)