= 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 $E_\theta\widehat V\ge\operatorname{Var}_\theta(\widehat\theta)$; the proposition is not a normal-distribution theorem.<Conservative variance estimation for sampling designs with zero pairwise inclusion probabilities by Peter M. Aronow and Cyrus Samii>[§3.1, Proposition 2, expected-value inequality]{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 $\operatorname{Var}_\theta(Y)=0.1(10)=1$. Use the deterministic estimate $\widehat V=1$: it is even a <samplewise variance upper bound>. Nevertheless,
$$
\Pr_\theta\{\theta\in[Y-1.96,Y+1.96]\}
=\Pr\{|X|\le1.96\}=0.90,
$$
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.<Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results by Barry N. Taylor and Chris E. Kuyatt>[§§6.2–6.3, conditions for approximately 95 percent expanded uncertainty]{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.
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