Asymptotic conservative Wald coverage under variance-ratio convergence

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]