= 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.<Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results by Barry N. Taylor and Chris E. Kuyatt>[§§6.2–6.3, Expanded Uncertainty]{ref}
A precise sufficient asymptotic argument is the following. Let $V_n=\operatorname{Var}_\theta(\widehat\theta_n)>0$, take $\widehat V_n\ge0$, and suppose
$$
\frac{\widehat\theta_n-\theta}{\sqrt{V_n}}\xrightarrow{d}N(0,1),
\qquad \frac{\widehat V_n}{V_n}\xrightarrow{p}c,
\qquad 1\le c<\infty.
$$
The 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 probability>
$$
\lim_{n\to\infty}\Pr_\theta\!\left\{|\widehat\theta_n-\theta|\le z\sqrt{\widehat V_n}\right\}
=2\Phi(z\sqrt c)-1\ge1-\alpha.
$$
The 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 $\widehat V_n=1.44V_n$. A nominal 95 percent <Wald confidence interval> uses a radius $1.2$ times the usual one and has exact <coverage probability> $2\Phi(1.2\times1.959964)-1\approx0.9813$. The calculation uses an actual normal sampling law and a fixed ratio; it cannot be inferred from $E\widehat V_n\ge V_n$ alone. The normal cdf and inverse-cdf implementations provide the numerical evaluation.<SciPy 1.16.2 documentation>[scipy.stats.norm, cdf and ppf]{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>.
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