Why Wald Intervals Have First-Order Coverage

Why a Wald Interval Can Have Nominal Limiting Coverage

A Wald interval is built from an estimate and an estimated standard error. Its familiar form is simple, but the coverage claim rests on a specific chain of reasoning: the centered estimator must have a suitable limiting distribution, the estimated standard error must consistently estimate the right scale, and the interval’s critical value must be a continuity point of that limiting distribution.

The earlier material on asymptotic normality and plug-in standard errors develops the ingredients separately. Here we connect them: we show precisely why, under stated regularity conditions, a scalar Wald interval has coverage approaching its nominal level as the sample size grows. This is a first-order statement, not a guarantee of exact coverage for a particular finite sample.

Setup and Regularity Conditions

Definition: Let $\theta$ be a fixed scalar parameter, and let $\hat{\theta}$ estimate it from a sample of size $n$. Suppose $a_n$ is a positive sequence tending to infinity, and $\tau(\theta)$ is a finite, strictly positive scale. A studentized estimator is $$ T_n=\frac{a_n(\hat{\theta}-\theta)}{\hat{\tau}_n}, $$ where $\hat{\tau}_n$ estimates $\tau(\theta)$. The corresponding two-sided Wald interval with fixed critical value $c>0$ is $$ \left[\hat{\theta}-c\frac{\hat{\tau}_n}{a_n},\ \hat{\theta}+c\frac{\hat{\tau}_n}{a_n}\right]. $$

The usual case is $a_n=\sqrt n$. The estimated standard error is then $\hat{\tau}_n/\sqrt n$, so $\hat{\tau}_n$ estimates the asymptotic standard deviation of $\sqrt n(\hat{\theta}-\theta)$. It is important to keep these two scales distinct: $\hat{\tau}_n$ estimates the limiting scale, while $\hat{\tau}_n/\sqrt n$ estimates the standard error of $\hat{\theta}$.

The central conditions used below are:

Asymptotic normality: $a_n(\hat{\theta}-\theta)/\tau(\theta)\xrightarrow{d}N(0,1)$.
Scale consistency: $\hat{\tau}_n\xrightarrow{p}\tau(\theta)$, where $0<\tau(\theta)<\infty$.
Well-defined studentization: $\hat{\tau}_n>0$ with probability tending to one. The statistic can be defined arbitrarily on the event where it is zero without changing its limiting distribution if that event’s probability tends to zero.

These are pointwise conditions at the fixed parameter value $\theta$. They do not by themselves assert uniform performance over a parameter space, and they do not require the estimator to be exactly unbiased in finite samples.

From Studentization to Coverage

Theorem: Suppose the three conditions above hold. Then $T_n\xrightarrow{d}N(0,1)$. For a fixed $c>0$, if the standard normal CDF is continuous at $-c$ and $c$, then $$ P\left(\theta\in\left[\hat{\theta}-c\frac{\hat{\tau}_n}{a_n},\hat{\theta}+c\frac{\hat{\tau}_n}{a_n}\right]\right) \longrightarrow \Phi(c)-\Phi(-c). $$ In particular, if $c=z_{1-\alpha/2}$, where $z_{1-\alpha/2}$ satisfies $\Phi(z_{1-\alpha/2})=1-\alpha/2$, the limiting coverage is $1-\alpha$.

Proof. Rewrite the studentized statistic as

T_n= \frac{a_n(\hat{\theta}-\theta)/\tau(\theta)} {\hat{\tau}_n/\tau(\theta)}.

The numerator converges in distribution to $N(0,1)$ by asymptotic normality. Since $\hat{\tau}_n\xrightarrow{p}\tau(\theta)>0$, the denominator converges in probability to $1$. Slutsky’s theorem therefore gives $T_n\xrightarrow{d}N(0,1)$.

Next, because $a_n>0$ and $\hat{\tau}_n>0$ with probability tending to one, the event that the interval contains $\theta$ is, up to an event whose probability tends to zero, equivalent to

\hat{\theta}-c\frac{\hat{\tau}_n}{a_n}\leq\theta \leq\hat{\theta}+c\frac{\hat{\tau}_n}{a_n} \quad\Longleftrightarrow\quad -c\leq T_n\leq c.

Convergence in distribution implies convergence of the probabilities of intervals whose boundary points are continuity points of the limiting CDF. The standard normal CDF is continuous everywhere, so

P(-c\leq T_n\leq c)\longrightarrow \Phi(c)-\Phi(-c).

For $c=z_{1-\alpha/2}$, symmetry of the standard normal gives $\Phi(-c)=\alpha/2$, while $\Phi(c)=1-\alpha/2$. Their difference is $1-\alpha$, as claimed. $\square$

The continuity condition is what allows a limiting distribution to determine the probability at the interval’s boundary. For a more general limiting CDF $G$, the same argument requires continuity at both $-c$ and $c$, and the limiting coverage is $G(c)-G(-c)$. When critical values are described through quantiles, a unique quantile is ensured if the limiting CDF is continuous and strictly increasing in a neighborhood of the relevant probability level. For the standard normal distribution, these properties hold at every finite quantile.

Worked Applications

Worked Example: A Wald Interval for a Bernoulli Probability

Suppose $X_1,\ldots,X_n$ are independent Bernoulli random variables with success probability $p$, where $0<p<1$. The estimator is $\hat p=\bar X$. Its variance is $p(1-p)/n$; the central limit theorem gives $$ \frac{\sqrt n(\hat p-p)}{\sqrt{p(1-p)}}\xrightarrow{d}N(0,1). $$ The plug-in scale $\hat{\tau}_n=\sqrt{\hat p(1-\hat p)}$ converges in probability to $\sqrt{p(1-p)}$: the sample proportion converges to $p$, and the square-root expression is continuous at $p$. The theorem therefore gives limiting $1-\alpha$ coverage for the interval using $\hat{\tau}_n/\sqrt n$.

For a numerical illustration, take $n=400$ and observe 128 successes. Then $\hat p=128/400=0.32$, and the estimated standard error is $$ \sqrt{\frac{0.32(1-0.32)}{400}} =\sqrt{\frac{0.2176}{400}} =\sqrt{0.000544} \approx 0.02332. $$ Using $c=1.96$, the margin is $1.96(0.02332)\approx0.04571$, giving the Wald interval $[0.2743,0.3657]$, rounded to four decimals. The calculation illustrates the interval’s form; the theorem concerns its repeated-sampling coverage as $n$ grows, not whether this particular interval contains the unknown $p$.

Worked Example: A Wald Interval for a Poisson Rate

Let $X_1,\ldots,X_n$ be independent Poisson observations with mean $\lambda>0$. The sample mean $\hat{\lambda}=\bar X$ has variance $\lambda/n$. The central limit theorem gives $$ \frac{\sqrt n(\hat{\lambda}-\lambda)}{\sqrt{\lambda}} \xrightarrow{d}N(0,1). $$ Also, $\bar X\xrightarrow{p}\lambda$, so $\hat{\tau}_n=\sqrt{\bar X}\xrightarrow{p}\sqrt{\lambda}$. Thus studentization meets the theorem’s conditions.

Suppose $n=100$ observations have total count 230. Then $\hat{\lambda}=230/100=2.3$, and the plug-in standard error is $$ \sqrt{\frac{2.3}{100}}=\sqrt{0.023}\approx0.15166. $$ With $c=1.96$, the margin is $1.96(0.15166)\approx0.29725$, so the Wald interval is approximately $[2.0028,2.5972]$. Its limiting nominal coverage follows from the theorem; the interval’s finite-sample coverage need not equal the nominal level exactly.

Worked Example: A Wald Interval for a Normal Mean

Let $X_1,\ldots,X_n$ be independent $N(\mu,\sigma^2)$ observations with $0<\sigma^2<\infty$, and estimate $\mu$ by $\bar X$. Then $$ \frac{\sqrt n(\bar X-\mu)}{\sigma}\sim N(0,1) $$ for every $n$. The sample standard deviation $S$, defined by $S^2=\sum_{i=1}^n(X_i-\bar X)^2/(n-1)$, converges in probability to $\sigma$. Consequently, $\sqrt n(\bar X-\mu)/S\xrightarrow{d}N(0,1)$, and the Wald interval using $S/\sqrt n$ has the stated first-order limiting coverage.

For example, if $n=25$, $\bar X=10.4$, and $S=2.5$, the estimated standard error is $2.5/\sqrt{25}=0.5$. With $c=1.96$, the Wald interval is $10.4\pm1.96(0.5)$, or $[9.42,11.38]$. In this normal model, an interval using the appropriate finite-sample $t$ critical value has exact coverage; this does not change the fact that the fixed normal-critical-value Wald interval is justified by its limiting coverage.

What First-Order Coverage Does and Does Not Say

The result matters because it explains the logic behind a widely used interval without requiring an exact finite-sample distribution. In many models, asymptotic normality and a consistent plug-in scale are easier to establish than a complete finite-sample coverage calculation.

The phrase first-order coverage means that the coverage probability approaches the nominal level as $n$ increases. It does not specify how quickly the coverage error shrinks. It also does not imply good performance at a particular sample size, guarantee that the endpoints respect the parameter space, or establish uniform coverage across parameter values. Those questions require further analysis.

A common mistake is to argue, “$\hat{\theta}$ is consistent, so the Wald interval has the right coverage.” Consistency alone only says $\hat{\theta}\xrightarrow{p}\theta$; it gives neither the distribution of the estimation error at the standard-error scale nor evidence that the estimated standard error has the right scale. The correction is to verify both asymptotic normality and scale consistency, then use the continuity of the limiting CDF at the interval boundaries.

Key takeaway: A scalar Wald interval has nominal limiting coverage when its standardized estimation error converges to the stated limiting distribution, its estimated standard error consistently estimates the corresponding scale, and its critical value is at continuity points of that distribution.

Check Your Understanding

Use the theorem’s conditions, not just the interval’s formula, to assess each claim.

In the usual $a_n=\sqrt n$ setup, what quantity does $\hat{\tau}_n$ estimate, and what quantity estimates the standard error of $\hat{\theta}$?
Why does $\hat{\tau}_n\xrightarrow{p}\tau(\theta)>0$ allow Slutsky’s theorem to be applied to the studentized statistic?
Which continuity points of the limiting CDF are needed to derive coverage for a symmetric interval with critical value $c$?
Does consistency of $\hat{\theta}$ alone imply nominal limiting coverage? Explain which additional conditions are needed.
What is the difference between first-order limiting coverage and exact finite-sample coverage?