Tail assumptions needed for conservative Wald coverage

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]