Asymptotic conservative Wald coverage under variance-ratio convergence (source code)

= 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 $Z_n=(\widehat\theta_n-\theta)/\sqrt{V_n}$ converge in distribution to a standard <normal random variable>, take $R_n=\widehat V_n/V_n\ge0$, and let $z=\Phi^{-1}(1-\alpha/2)>0$. If $R_n$ converges in probability to a constant $c\ge1$, then for each $0<\epsilon<1$, $\Pr(R_n<1-\epsilon)\to0$. Therefore
$$
\Pr\{|Z_n|\le z\sqrt{R_n}\}
\ge\Pr\{|Z_n|\le z\sqrt{1-\epsilon}\}
-\Pr\{R_n<1-\epsilon\}.
$$
Taking the lower limit and then $\epsilon\downarrow0$ 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 $2\Phi(1.1z)-1\approx0.96891$. But if the standardized estimator instead converges to $N(0.8,1)$, the same scale gives
$$
\Phi(1.1z-0.8)-\Phi(-1.1z-0.8)\approx0.91089.
$$
A ten-percent increase in standard-error radius does not automatically cover an asymptotically material <bias>. These values use the normal cdf/ppf definitions.<SciPy 1.16.2 documentation>[scipy.stats.norm, cdf and ppf]{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.<Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results by Barry N. Taylor and Chris E. Kuyatt>[§§6.2–6.3]{ref}