Boundary Failures Of Wald Proportion Intervals

Why Wald Intervals Can Fail at the Boundaries

A proportion is constrained: its value must be between zero and one. The Wald interval can ignore that constraint in its reported endpoints, and its estimated standard error can behave badly when the observed proportion is close to either boundary. These are not merely cosmetic issues. With small samples, the interval can also have actual coverage far below its stated confidence level.

The Wald interval is familiar because it resembles a normal-approximation interval for a mean. Its simplicity, however, does not guarantee reliable performance for every sample. The key warning signs are a small sample, few observed successes or failures, and an observed proportion near zero or one.

Definition: Suppose $n$ independent individuals each have the same probability $p$ of a binary outcome, such as a positive screening result. Let $X$ be the number of outcomes and let $\hat p=X/n$. Under the binomial model, $X\sim\operatorname{Binomial}(n,p)$. For a two-sided confidence level $1-\alpha$, the Wald interval is $$ \hat p \mathbin{\pm} z_{1-\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}, $$ where $z_{1-\alpha/2}$ is the standard normal quantile with cumulative probability $1-\alpha/2$. Its endpoints are the estimate plus and minus the estimated standard error multiplied by the normal critical value.

This formula uses the observed proportion in place of the unknown $p$ in the binomial standard error. It is based on a normal approximation and is not an exact interval with guaranteed coverage. In particular, it does not automatically respect the parameter space $0\le p\le1$.

How Boundary Estimates Create Unusual Endpoints

The estimated standard error is $ \sqrt{\hat p(1-\hat p)/n} $. If $\hat p$ is near zero or one, the product $\hat p(1-\hat p)$ is small, so the interval can be very narrow. At either endpoint it is exactly zero: if $X=0$, then $\hat p=0$; if $X=n$, then $\hat p=1$. The resulting Wald interval is a single point, even though the sample may be too small to establish that the true proportion is exactly zero or one.

Away from the endpoints, subtracting the margin of error can produce a negative lower endpoint, or adding it can produce an upper endpoint greater than one. Such an interval is not a valid range of possible proportions as reported. The endpoint problem is visible, but it is important not to confuse it with the separate problem of poor coverage: an interval can stay within the parameter space and still miss the true proportion too often.

Worked Example: A Zero Count in a Small Screening Sample

A hypothetical clinic tests 18 randomly selected adults for a particular marker. Assume independent outcomes with the same unknown positive-marker probability $p$. No one tests positive. Calculate the 95% Wald interval, then quantify one way it can fail when the true probability is $0.08$.

Method. Use the binomial sample proportion and the 95% Wald formula, with $z_{0.975}=1.96$. The observed count is $X=0$ and the sample size is $n=18$, so $$ \hat p=\frac{0}{18}=0. $$

The estimated standard error is $$ \sqrt{\frac{0(1-0)}{18}}=0. $$ Therefore the margin of error is $1.96(0)=0$, and the interval is $$ [0-0,\;0+0]=[0,\;0]. $$

This interval gives the appearance of no uncertainty. But if the true probability were $p=0.08$, the probability of observing no positive outcomes would be $$ P(X=0)=(1-0.08)^{18}=0.92^{18}\approx 0.2229. $$ Whenever $X=0$, this Wald interval is $[0,0]$ and does not contain $0.08$. Thus, at least $0.2229$ of the sampling probability is already assigned to outcomes that produce an interval missing this true value. This is a lower bound on the noncoverage probability, not a calculation of the complete coverage probability.

Check. The interval is a singleton because the estimated standard error is zero, not because the data prove $p=0$. The probability calculation is consistent with the binomial model: an individual has probability $0.92$ of a negative result, and all 18 results are negative with probability $0.92^{18}$.

Common mistake avoided: Do not interpret a zero observed count as a zero population probability. A sample can contain no successes even when $p$ is positive.

Out-of-Range Endpoints and the Normal Approximation

For a fixed observed proportion, the Wald margin is $z_{1-\alpha/2}\sqrt{\hat p(1-\hat p)/n}$. The lower endpoint is negative precisely when this margin exceeds $\hat p$; the upper endpoint exceeds one precisely when the margin exceeds $1-\hat p$. Near zero, the first comparison is especially likely to fail; near one, the second is. This explains why out-of-range endpoints often appear when only a few outcomes fall in the less common category.

The normal approximation is most credible when the binomial distribution is not concentrated near a boundary. A common diagnostic is to check whether both expected counts, $np$ and $n(1-p)$, are sufficiently large. Since $p$ is unknown, practitioners often inspect the observed counts $X$ and $n-X$ as a warning diagnostic. This check is not a guarantee of nominal coverage, and a small observed count should prompt caution rather than a claim that a specific alternative method has been established here.

Worked Example: A Negative Lower Endpoint

A hypothetical survey records whether 1 of 20 sampled policyholders used a new online service. Assume independent binary outcomes with a common population-use probability $p$. Calculate the 95% Wald interval and check both endpoints.

Method. Here $X=1$, $n=20$, and $z_{0.975}=1.96$. The sample proportion is $$ \hat p=\frac{1}{20}=0.05. $$

The estimated standard error is $$ \sqrt{\frac{0.05(1-0.05)}{20}} =\sqrt{\frac{0.0475}{20}} =\sqrt{0.002375} \approx 0.048734. $$ The margin of error is $$ 1.96(0.048734)\approx 0.095519. $$

Thus the lower endpoint is $0.05-0.095519=-0.045519$, and the upper endpoint is $0.05+0.095519=0.145519$. Rounded to four decimal places, the interval is $$ [-0.0455,\;0.1455]. $$ The negative lower endpoint cannot be a population proportion. The sample also has only one observed service user and 19 nonusers, so the estimate is near the boundary and the normal approximation deserves particular caution.

Check. The endpoints are centered at $0.05$: their midpoint is $(-0.045519+0.145519)/2=0.05$. Their half-width is approximately $0.095519$, matching the calculated margin.

Common mistake avoided: Do not assume that a formally computed normal interval must lie in the parameter space. The formula can produce an impossible endpoint.

Coverage: A Finite-Sample Check

A confidence procedure has coverage $1-\alpha$ at a particular $p$ if, over repeated samples from the model with that fixed $p$, its random interval contains $p$ with probability $1-\alpha$. To examine the Wald procedure, consider each possible count $k=0,1,\ldots,n$. The interval produced by count $k$ contains $p$ exactly when $$ \left|\frac{k}{n}-p\right| \le z_{1-\alpha/2}\sqrt{\frac{(k/n)(1-k/n)}{n}}. $$ The count itself has probability $\binom{n}{k}p^k(1-p)^{n-k}$. Adding those probabilities over the counts whose intervals contain $p$ gives the actual coverage.

Justification: For each possible sample count $k$, the Wald interval is centered at $k/n$ with the margin shown above. The true $p$ is in that interval if and only if its distance from the center is no greater than the margin, which is the displayed inequality. Since the events $X=k$ are mutually exclusive and exhaustive, the coverage is the sum of their binomial probabilities over precisely the counts satisfying that inequality. Because the possible counts are discrete and the margin is estimated from each count, this sum need not equal the nominal confidence level.

Worked Example: Exact Coverage Near Zero

A hypothetical quality-control check records whether a randomly selected component has a specified defect. Suppose the true defect probability is $p=0.05$, and $n=10$ independent components are inspected. Find the actual coverage probability of the two-sided 95% Wald interval.

Method. Determine which observed counts $k$ yield intervals containing $0.05$, then sum their binomial probabilities. Here $z_{0.975}=1.96$. For a count $k$, the interval center is $k/10$ and its margin is $1.96\sqrt{(k/10)(1-k/10)/10}$.

For $k=0$, the interval is $[0,0]$, so it misses $0.05$. For $k=1$, the center is $0.1$ and the margin is $1.96\sqrt{0.09/10}\approx0.1859$; the interval contains $0.05$. For $k=2$, the center is $0.2$ and the margin is $1.96\sqrt{0.16/10}\approx0.2479$, which also contains $0.05$. For $k=3$, the lower endpoint is $0.3-1.96\sqrt{0.21/10}\approx0.0160$, so the interval still contains $0.05$. For $k=4$, the lower endpoint is $0.4-1.96\sqrt{0.24/10}\approx0.0964$, which is greater than $0.05$. For $k\ge4$, the lower endpoint increases with $k/10$, so none of those counts yields an interval containing $0.05$.

The covering counts are therefore $k=1,2,3$. Under the binomial model, $$ \begin{aligned} P(X=1)&=\binom{10}{1}(0.05)(0.95)^9\approx0.3151247,\\ P(X=2)&=\binom{10}{2}(0.05)^2(0.95)^8\approx0.0746348,\\ P(X=3)&=\binom{10}{3}(0.05)^3(0.95)^7\approx0.0104751. \end{aligned} $$ Thus the actual coverage is $0.3151247+0.0746348+0.0104751=0.4002346$, or $0.4002$ to four decimal places. This is far below the nominal $0.95$.

Check. The probabilities of the three covering outcomes sum to less than one, as they must. In particular, $X=0$, an outcome with probability $0.95^{10}\approx0.5987$, does not cover the positive true probability.

Common mistake avoided: Do not equate a 95% label with 95% actual coverage for every $n$ and $p$. The label comes from the normal critical value; the actual coverage depends on the binomial outcomes and their intervals.

What Clipping Can and Cannot Do

A common response to an out-of-range Wald interval $[L,U]$ is to report $[\max(0,L),\min(1,U)]$. This operation is called clipping the endpoints to the parameter space. It changes the interval endpoints as random quantities: across repeated samples, the clipped endpoints have a different distribution from the original endpoints. However, for any true proportion $p$ in $[0,1]$, clipping does not change whether that particular $p$ is covered.

Key result: For $p\in[0,1]$, membership of $p$ in the original Wald interval $[L,U]$ is equivalent to membership in the clipped interval $[\max(0,L),\min(1,U)]$. If $L<0$, then the original lower-end condition $p\ge L$ is automatically true for such a $p$, and the clipped lower-end condition $p\ge0$ is also true. If $U>1$, then $p\le U$ is automatically true, as is the clipped condition $p\le1$. If an endpoint already lies inside the parameter space, clipping leaves that endpoint unchanged. Hence coverage is unchanged, even though the endpoint distribution changes.

Clipping therefore makes the reported set stay within the permissible parameter space, but it does not repair poor coverage or guarantee nominal coverage. In particular, it cannot fix the zero-width interval from a zero or all-success count: clipping leaves $[0,0]$ and $[1,1]$ unchanged.

Worked Example: Checking Coverage Before and After Clipping

Use the survey result from the earlier calculation: $n=20$, $X=1$, and the 95% Wald interval $[-0.045519,0.145519]$, with endpoints displayed to six decimals here to show the comparison. Suppose the true service-use proportion is $p=0.03$. Compare coverage before and after clipping.

Method. First test whether $p$ lies between the original endpoints: $-0.045519\le0.03\le0.145519$, which is true. Clipping gives $$ [\max(0,-0.045519),\min(1,0.145519)]=[0,0.145519]. $$ The same true value satisfies $0\le0.03\le0.145519$, so the clipped interval also covers it.

For this sample, clipping changes the lower endpoint from $-0.045519$ to $0$ but not the coverage decision for $p=0.03$. Across repeated samples, endpoint values can be changed whenever they fall outside $[0,1]$, so the endpoint distribution does change. But for every possible sample, the coverage decision for any fixed $p\in[0,1]$ is unchanged. Thus the procedure's actual coverage at $p=0.03$ is unchanged by clipping.

Check. Both intervals contain $0.03$. The correction to the reported endpoint has not added or removed this permissible true value.

Common mistake avoided: Clipping is not a coverage correction. It changes the interval's endpoints, but it neither changes coverage for an admissible true proportion nor guarantees that coverage reaches 95%.

Practical Diagnosis

When a Wald proportion interval is requested or presented, inspect the data and the endpoints, not only the confidence-level label. A boundary estimate can make the standard error vanish; a near-boundary estimate can create an impossible endpoint; and both small counts and the discrete sampling distribution can undermine the normal approximation. These problems are especially important when an interval will be interpreted as evidence about a small event probability.

Calculate $\hat p=X/n$ and the Wald standard error explicitly.
Check whether either endpoint falls outside $[0,1]$.
Inspect both observed counts, $X$ and $n-X$, as warning signs for a weak normal approximation.
Do not interpret a zero-width boundary interval as proof that the population proportion is exactly zero or one.
Do not claim that clipping restores nominal coverage.

These checks diagnose the Wald interval's limitations; they do not transform it into a different inference method. When a problem specifies the Wald procedure, calculate and interpret that procedure as specified, while clearly recognizing its limitations. The central lesson is that a simple formula can have undesirable finite-sample behavior precisely where a proportion is most constrained.

Check Your Understanding

For each problem, assume independent binary outcomes with a common population proportion and use the stated Wald critical value. Round interval endpoints to four decimal places when needed.

A hypothetical pilot study observes zero positive responses among 25 participants. Find the 90% Wald interval using $z_{0.95}=1.645$, and state why its width is a warning sign. (Answer: $[0,0]$; zero width because the estimated standard error is zero.)
A hypothetical service survey records 2 users among 40 sampled customers. Calculate the 95% Wald interval using $z_{0.975}=1.96$. (Answer: $[-0.0175,0.1175]$.)
A hypothetical equipment audit finds 49 functioning units among 50 tested. Calculate the 95% Wald interval using $z_{0.975}=1.96$, and identify the boundary problem. (Answer: $[0.9412,1.0188]$; the upper endpoint exceeds one.)
In a hypothetical binomial study, $n=8$ and the true event probability is $p=0.10$. What is the probability that the Wald interval is $[0,0]$ and therefore misses $p$? (Answer: $P(X=0)=0.9^8=0.4305$, to four decimals.)
A hypothetical sample gives an original Wald interval $[-0.04,0.18]$. If the true proportion is $p=0.02$, does clipping change whether this interval covers $p$, and does clipping guarantee nominal coverage? (Answer: No; both intervals cover $0.02$, and clipping does not guarantee nominal coverage.)

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