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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 $E_{\theta }\hat{V}\ge {\mathrm{Var}}_{\theta }(\hat{\theta })$; the proposition is not a normal-distribution theorem.^[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 ${\mathrm{Var}}_{\theta }(Y)=0.1(10)=1$. Use the deterministic estimate $\hat{V}=1$: it is even a samplewise variance upper bound. Nevertheless,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.^[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.

A conservative variance estimator, in the finite-sample expectation sense used here, is an estimator $\hat{V}$ of a sampling variance $V_{\theta }$ satisfyingfor every statistical parameter value $\theta$ in the stated statistical model. Assume $V_{\theta }$ and ${\mathbb{E}}_{\theta }|\hat{V}|$ 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.^[ref]

A samplewise variance upper bound would require the stronger inequality $\hat{V}\ge V_{\theta }$ 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 $\hat{V}=W_{+}=\max (W,0)$ satisfies $W_{+}\ge W$, hence is a conservative variance estimator. The amount of upward bias of an estimator is exactlyThe 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 $0<\epsilon <1$. Use $\hat{\theta }=Y$ and $\hat{V}=B/\epsilon$. This nonnegative random variable hasThus it is even an unbiased estimator of the sampling variance. Yet the usual Wald confidence intervalhas coverage probabilitywhere $\Phi$ 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 $\epsilon =0.01$, overall coverage probability is at most 1 percent, despite the nominal 95 percent rule and the exactly normal centre.

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.

A one-sided Cantelli confidence bound with a bias budget combines a validated bound on the estimator's systematic displacement with a one-sided variance inequality. Suppose $|E_{\theta }\hat{\theta }-\theta |\le B$, where $B\ge 0$ is deterministic, and a nonnegative statistic satisfies $\hat{V}\ge V_{\theta }={\mathrm{Var}}_{\theta }(\hat{\theta })$ almost surely. For $0<\alpha <1$, defineThen ${\mathrm{Pr}}_{\theta }(\theta \ge L)\ge 1-\alpha$. This is a lower confidence bound, with an unbounded upper side.

For the proof, put $Y=\hat{\theta }-E_{\theta }\hat{\theta }$. If $\theta <L$, the bias bound impliesFor positive $V_{\theta }$, Cantelli's inequality bounds this event by $V_{\theta }/(V_{\theta }+a^2)=\alpha$ at the last displayed threshold $a$. For zero variance, $Y=0$ almost surely and the conclusion follows directly. A random $\hat{V}$ may depend on the centre: the samplewise comparison is what supports the event inclusion.

For an original numerical bound, let $\hat{\theta }=2.0$, $B=0.2$, $\hat{V}=0.04$, and $\alpha =0.1$. The square-root term is $0.6$, so $L=1.2$ is a 90 percent lower bound under these assumptions. Reversing the inequality gives the separately valid upper bound $U=2.8$. Each has 90 percent one-sided coverage; reporting both endpoints together only guarantees 80 percent by the union bound. To guarantee a joint 90 percent interval this way, allocate error probabilities summing to $0.1$ across its two sides.

This construction complements the two-sided result in confidence bounds from a samplewise variance bound. It needs a bias bound and an almost-sure variance bound, not merely a conservative variance estimator in the expected-value sense. A coverage interpretation must follow the actual tail argument, as emphasized by NIST's discussion of expanded uncertainty.^[ref]

Peter M. Aronow develops statistical methods for causal inference, particularly design-based inference. Yale's institutional biography identifies research on complex experimental designs, network settings and incomplete survey data alongside appointments spanning political science and statistics.^[ref]

His joint work with Cyrus Samii studies how to estimate sampling variance when a sampling design never observes some pairs together. A missing product cannot simply be deleted from the population variance formula: doing so can produce either sign of bias. Their construction uses Young inequality to bound unobserved products by individually estimable powers and obtain a conservative variance estimator.^[ref]

For an original exact illustration, choose one of two fixed values $y_1,y_2$ with equal probabilities. As in the single-draw variance identifiability obstruction, the Horvitz-Thompson estimator $2y_J$ has true variance $V_p=(y_1-y_2{)}^2$. Retaining only diagonal contributions gives $D=2y_J^2$, withOpposite signs therefore produce downward bias. The elementary inequality $-2ab\le a^2+b^2$, obtained from $(a+b{)}^2\ge 0$, instead givesConsequently $C=4y_J^2$ is a conservative variance estimator for this stipulated sampling design; its excess in expected value is exactly $(y_1+y_2{)}^2$.

With $(y_1,y_2)=(3,-1)$, the true variance is16. The two possible values of $D$ are18 and2, averaging10; those of $C$ are36 and4, averaging20. Although ${\mathbb{E}}_{p}C\ge 16$, a realized value of $C$ can be below16. Conservativeness here is an expectation inequality, rather than a bound for every sample. By itself it also does not prove a specified confidence interval coverage probability.

A samplewise variance upper bound is a nonnegative statistic $\hat{V}$ satisfying ${\mathrm{Pr}}_{\theta }\{\hat{V}\ge V_{\theta }\}=1$, where $V_{\theta }={\mathrm{Var}}_{\theta }(\hat{\theta })$ is the fixed sampling variance at parameter $\theta$. It controls each realized value outside a probability-zero exceptional set. This implies the expected-value condition for a conservative variance estimator when the expectation exists; the converse fails, as the parent article's counterexample shows. The primary sampling-design definition only asserts the weaker expectation condition.^[ref]

Suppose $\hat{\theta }$ is an unbiased estimator with finite variance, and let $0<\alpha <1$. Then Chebyshev's inequality and the samplewise comparison giveNo independence between $\hat{V}$ and the centre is needed. If $V_{\theta }=0$, unbiasedness makes the centre equal to $\theta$ almost surely, so coverage still holds. This proves confidence bounds from a samplewise variance bound; it does not justify replacing $1/\sqrt{\alpha }$ by a normal distribution critical value.

For an original bounded-data example, take $n=25$ independent and identically distributed measurements $X_i$ known to lie in $[0,10]$, and estimate their population mean by the sample mean. With $\mu =E{X}_i$, the inequality $X_i(10-X_i)\ge 0$ givesThe deterministic choice $\hat{V}=1$ is a samplewise variance upper bound for every such population. Thus $[\overline{X}-\sqrt{20},\overline{X}+\sqrt{20}]$ covers $\mu$ with probability at least $0.95$. The radius $\sqrt{20}\approx 4.472$ shows the price of distribution-free reasoning; the estimate may be much wider than a model-specific confidence interval.

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.^[ref]

A precise sufficient asymptotic argument is the following. Let $V_n={\mathrm{Var}}_{\theta }({\hat{\theta }}_n)>0$, take ${\hat{V}}_n\ge 0$, and supposeThe 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 probabilityThe 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 ${\hat{V}}_n=1.44V_n$. A nominal 95 percent Wald confidence interval uses a radius $1.2$ times the usual one and has exact coverage probability $2\Phi (1.2\times 1.959964)-1\approx 0.9813$. The calculation uses an actual normal sampling law and a fixed ratio; it cannot be inferred from $E{\hat{V}}_n\ge V_n$ alone. The normal cdf and inverse-cdf implementations provide the numerical evaluation.^[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.