= Conservative variance estimator
{title2=$\mathbb E_p\widehat V\ge V_p$}
A \b[<conservative variance estimator>], in the finite-sample expectation sense used here, is an <estimator> $\widehat V$ of a <sampling variance> $V_\theta$ satisfying
$$
\boxed{\mathbb E_\theta\widehat V\ge V_\theta}
$$
for every <statistical parameter> value $\theta$ in the stated <statistical model>. Assume $V_\theta$ and $\mathbb E_\theta|\widehat V|$ are finite. In <design-based inference>, the same definition applies for every permitted fixed population, with the <sampling design> supplying the randomness. The <bias of an estimator> is then nonnegative for this target. Aronow and Samii use this expectation criterion for their correction to the <Horvitz-Thompson variance estimator> when some pairs cannot be sampled together.<Conservative variance estimation for sampling designs with zero pairwise inclusion probabilities by Peter M. Aronow and Cyrus Samii>[Section3.1, Proposition2: expected corrected variance estimate is at least the true design variance]{ref}
A <samplewise variance upper bound> would require the stronger inequality $\widehat V\ge V_\theta$ <almost surely>. An expectation bound allows some estimates below $V_\theta$. It does not even require that every realized value be nonnegative; in practice a variance estimate is commonly chosen to be a <nonnegative random variable>.
One construction starts with an <integrable random variable> $W$ that is an <unbiased estimator> of $V_\theta$ and might take negative values. Its <positive part> $\widehat V=W_+=\max(W,0)$ satisfies $W_+\ge W$, hence is a <conservative variance estimator>. The amount of upward <bias of an estimator> is exactly
$$
\mathbb E_\theta W_+-V_\theta
=\mathbb E_\theta(-W)_+\ge0.
$$
The identity follows from the <positive part> and <negative part> decomposition $W=W_+-(-W)_+$. The <Horvitz-Thompson variance estimator> supplies a concrete setting in which the untruncated <unbiased estimator> can be negative.
A deliberately constructed example proves that <conservative variance estimation does not guarantee Wald interval coverage>. Observe $Y$ from a <normal distribution> $N(\theta,1)$ and an independent $B$ from a <Bernoulli distribution> with success <probability> $0<\varepsilon<1$. Use $\widehat\theta=Y$ and $\widehat V=B/\varepsilon$. This <nonnegative random variable> has
$$
\mathbb E_\theta\widehat V=1
=\operatorname{Var}_\theta(\widehat\theta).
$$
Thus it is even an <unbiased estimator> of the <sampling variance>. Yet the usual <Wald confidence interval>
$$
C(Y,B)=\left[Y-1.96\sqrt{\widehat V},\;Y+1.96\sqrt{\widehat V}\right]
$$
has <coverage probability>
$$
\mathbb P_\theta\{\theta\in C(Y,B)\}
=\varepsilon\left[2\Phi\!\left(\frac{1.96}{\sqrt\varepsilon}\right)-1\right]
\le\varepsilon,
$$
where $\Phi$ is the <standard normal distribution function>. When $B=0$, the interval is the <singleton> $\{Y\}$, which contains $\theta$ with <probability> zero because $Y$ is a <continuous random variable>. When $B=1$, its conditional <coverage probability> is the bracketed expression. At $\varepsilon=0.01$, overall <coverage probability> is at most 1 percent, despite the nominal 95 percent rule and the exactly normal centre.
\Image[/conservative-variance-low-coverage.png]
{title=An unbiased variance estimate can give a Wald interval with coverage at most one percent}
{description=Original randomized variance-estimation counterexample. The variance estimate is zero with probability 0.99 and 100 with probability 0.01, so its expected value is the true variance 1. The interval can cover its target only in the rare second case.}
This example concerns the logical strength of the expectation criterion. Controlling the distribution of an estimated <standard error>, for example through an appropriate <consistent estimator> in an asymptotic argument, is additional work needed to justify <confidence interval> coverage.
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