Conservative variance estimation does not guarantee Wald interval coverage

A conservative variance estimator bounds an expectation, whereas Wald confidence interval coverage is a probability of a joint event involving the centre and its random width. These are different requirements. In the cited sampling-design result, conservativeness means $E_{\theta }\hat{V}\ge {\mathrm{Var}}_{\theta }(\hat{\theta })$; the proposition is not a normal-distribution theorem.^[ref] The rare-large-variance counterexample in conservative variance estimator already shows how little that expectation alone controls coverage.

A separate original example isolates the distributional issue even when the variance is known exactly. Let $Y=\theta +X$, where $X$ is zero with probability $0.90$ and is each of $\sqrt{10}$ and $-\sqrt{10}$ with probability $0.05$. Symmetry gives $E_{\theta }Y=\theta$, while ${\mathrm{Var}}_{\theta }(Y)=0.1(10)=1$. Use the deterministic estimate $\hat{V}=1$: it is even a samplewise variance upper bound. Nevertheless,because $\sqrt{10}>1.96$. Accurate variance alone does not supply the normal distribution of a standardized error. NIST likewise distinguishes multiplying a standard uncertainty by a coverage factor from establishing a confidence level without distributional information.^[ref]

The tail assumptions needed for conservative Wald coverage may instead come from an asymptotic normal approximation with a consistent standard-error estimate, or from a justified finite-sample model. If only unbiasedness and a samplewise variance upper bound are available, Chebyshev's inequality gives a different, usually wider, confidence interval. One must establish the centre's bias, the variance condition, and the critical-value argument together.

The tail assumptions needed for conservative Wald coverage concern the distribution of the standardized estimation error and the realized scale of its interval. An expected-value bound on a conservative variance estimator does not establish either property. The discrete and rare-large-variance counterexamples in conservative variance estimation does not guarantee Wald interval coverage isolate those failures. NIST similarly makes a normal-model qualification before interpreting an expanded uncertainty as approximately 95 percent coverage.^[ref]

A precise sufficient asymptotic argument is the following. Let $V_n={\mathrm{Var}}_{\theta }({\hat{\theta }}_n)>0$, take ${\hat{V}}_n\ge 0$, and supposeThe first condition includes centring at the true parameter; asymptotically material bias would change it. The second condition is convergence in probability of a variance ratio, substantially more than an expectation inequality. By Slutsky's theorem, the Wald confidence interval with normal critical value $z={\Phi }^{-1}(1-\alpha /2)$ then has limiting coverage probabilityThe limiting normal distribution has no mass on the interval boundary, which justifies passing to that probability. This is asymptotic conservative Wald coverage under variance-ratio convergence, not a finite-sample guarantee.

For an original exactly normal comparison, let a sample mean have true variance $V_n$ and use the deterministic scale ${\hat{V}}_n=1.44V_n$. A nominal 95 percent Wald confidence interval uses a radius $1.2$ times the usual one and has exact coverage probability $2\Phi (1.2\times 1.959964)-1\approx 0.9813$. The calculation uses an actual normal sampling law and a fixed ratio; it cannot be inferred from $E{\hat{V}}_n\ge V_n$ alone. The normal cdf and inverse-cdf implementations provide the numerical evaluation.^[ref]

A samplewise variance upper bound supplies a separate finite-sample route when the centre is exactly normal, because each reported interval contains the known-variance normal interval. Without that shape assumption, use a justified tail inequality as in confidence bounds from a samplewise variance bound.

Asymptotic conservative Wald coverage under variance-ratio convergence requires the realized interval scale to be large enough in probability, together with a correctly centred limiting error law. The constant-ratio result in tail assumptions needed for conservative Wald coverage follows from Slutsky's theorem. A useful stronger formulation of the coverage conclusion needs no independence between centre and scale.

Let $Z_n=({\hat{\theta }}_n-\theta )/\sqrt{V_n}$ converge in distribution to a standard normal random variable, take $R_n={\hat{V}}_n/V_n\ge 0$, and let $z={\Phi }^{-1}(1-\alpha /2)>0$. If $R_n$ converges in probability to a constant $c\ge 1$, then for each $0<ϵ<1$, $\mathrm{Pr}(R_n<1-ϵ)\to 0$. ThereforeTaking the lower limit and then $ϵ\downarrow 0$ proves limiting coverage probability at least $1-\alpha$. In fact this lower-tail condition on $R_n$ suffices without convergence to one fixed constant. An expected-value inequality does not supply it; coverage failure under random variance-ratio limits can persist when the scale retains a substantial chance of being too small.

For an original comparison, choose $c=1.21$. Under an exactly centred normal limit, a nominal 95 percent Wald confidence interval has limiting coverage $2\Phi (1.1z)-1\approx 0.96891$. But if the standardized estimator instead converges to $N(0.8,1)$, the same scale givesA ten-percent increase in standard-error radius does not automatically cover an asymptotically material bias. These values use the normal cdf/ppf definitions.^[ref]

The result is pointwise in the assumed parameter/model sequence; it does not itself prove uniform coverage over a family of models or finite-sample accuracy. NIST's uncertainty guidance likewise conditions a confidence interpretation on the distributional justification, rather than on the multiplier alone.^[ref]

Coverage failure under random variance-ratio limits can persist even when the centre is exactly normal, the variance estimate is conservative in expectation, and its absolute error tends to zero. What fails is relative-scale concentration. Let $Z\sim N(0,1)$ and a positive random scale $R$ be independent, and defineThe variance ratio is $R$ at every sample size. Its nondegenerate distribution need not approach the constant required in asymptotic conservative Wald coverage under variance-ratio convergence.

Conditional on $R$, a normal critical value $z>0$ gives interval coverage $f(R)=2\Phi (z\sqrt{R})-1$. Therefore unconditional coverage is $Ef(R)$ under the independence assumption. A useful shape calculation isThe function is strictly concave. If $ER=1$, Jensen's inequality puts $Ef(R)\le f(1)$, with strict inequality for a nonconstant scale. An expectation equality therefore does not reproduce the normal interval's nominal coverage.

For an original strictly conservative example, let $R$ take $0.25$ and $2.75$ with equal probabilities. Then $E{\hat{V}}_n=1.5V_n$, but at nominal 95 percent the exact coverage isThis value is unchanged as $n$ grows. Meanwhile ${\hat{V}}_n-V_n=(R-1)/n$ tends to zero, showing why absolute consistency at a vanishing variance scale is insufficient. The normal cdf evaluation uses the pinned primary numerical definition.^[ref]

If the scale depends on the standardized error, the conditional normal formula above cannot simply be retained; the joint law is needed. NIST likewise distinguishes a justified coverage probability from selecting a conventional uncertainty multiplier.^[ref]