Samplewise variance upper bound ($\hat{V}\ge V_{\theta }\text{ almost surely}$)

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.

Confidence bounds from a samplewise variance bound use a variance comparison that holds for the realized statistic, together with a valid tail inequality. The unbiased two-sided construction is proved in samplewise variance upper bound. A useful extension includes a separately justified bias budget: assumewhere $B\ge 0$ is a validated deterministic bound and the estimator has finite variance. For $0<\alpha <1$, defineIf $\theta$ lies outside this interval, the triangle inequality implies $|\hat{\theta }-E_{\theta }\hat{\theta }|>\sqrt{\hat{V}/\alpha }$. The samplewise bound and Chebyshev's inequality then give ${\mathrm{Pr}}_{\theta }(\theta \notin I)\le \alpha$. No independence between the centre and $\hat{V}$ is required. Zero true variance is covered because the centre then equals its expectation almost surely.

For an original measurement example in a fixed unit, suppose the reported estimate is $2.00$, a validated residual-bias bound is $B=0.20$, and a deterministic samplewise variance upper bound is $0.04$. With $\alpha =0.05$, the radius isso the interval is approximately $[0.9056,3.0944]$. The conclusion applies over repeated measurements under those assumptions. The particular realized interval is not assigned a frequentist probability that the fixed parameter is inside it. Replacing the radius by $1.96\sqrt{0.04}=0.392$ would discard both the bias allowance and the tail justification.

The bound $B$ is an assumption about the centre's possible systematic displacement; a guessed correction uncertainty is not automatically such a bound. NIST distinguishes the coverage interpretation from merely applying a conventional multiplier to a standard uncertainty.^[ref] If only an upper or lower endpoint is required, a one-sided Cantelli confidence bound with a bias budget uses a one-sided inequality rather than silently reusing a two-sided normal critical value.

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]