Confidence Sets, Pivotal Quantities, and Test-Interval Duality

Intuition Beginner

A point estimate hands you a single number — the average bolt length is $10.2$ millimeters — and then quietly hopes you forget it is almost certainly a little bit wrong. A confidence interval is the honest version of the same report. Instead of one number it hands you a range, like "between $9.8$ and $10.6$ millimeters," together with a promise about how often a range built this way contains the real value.

The promise is the whole point, and it is easy to misread. A $95\%$ confidence interval does not mean "there is a $95\%$ chance the truth is in this particular range." The truth is a fixed number; it is either in your range or it is not. What is random is the range, because it was computed from random data. The promise is about the recipe: if you repeated the whole experiment many times, building a fresh interval each time by the same rule, about $95$ out of every $100$ of those intervals would capture the true value. The number $95\%$ describes the long-run hit rate of the procedure, not the status of one interval.

There is a beautiful trick for building these ranges, and it ties two ideas together. Suppose for every candidate value you could run a fair test of "is this the truth?" and either reject it or let it stand. Collect all the candidate values your data does not reject. That collection is your confidence interval. A value survives exactly when it is plausible, and the interval is the set of survivors. Testing and interval-building are two views of one machine.

Visual Beginner

Picture a vertical number line of candidate values for the true average. For each candidate you imagine a small hypothesis test: does the data argue against this value? Some candidates get rejected (the data says "no, not that one"); the rest survive. The survivors form a connected band, and that band is the confidence interval. Slide the candidate up or down and you trace out where rejection starts and stops — the two edges of the band are the interval's endpoints.

candidate value	does a 5% test reject it?	in the interval?
far below the estimate	yes	no
just below the estimate	no	yes
at the estimate	no	yes
just above the estimate	no	yes
far above the estimate	yes	no

The takeaway: the confidence interval is the set of candidate values that a level-$5\%$ test refuses to reject. Building an interval and running a family of tests are the same act, seen from two sides. The contiguous band of survivors is what you report as the range.

Worked example Beginner

A lab weighs the same reference mass $n=9$ times. The scale's noise is known to have a standard deviation of $\sigma =3$ milligrams, and the nine readings average to $\bar{x}=502$ milligrams. You want a $95\%$ confidence interval for the true mass.

Step 1. The spread of the average. Averaging nine independent readings shrinks the noise: the standard deviation of the average is $\sigma /\sqrt{n}=3/\sqrt{9}=3/3=1$ milligram.

Step 2. The magic quantity that does not depend on the unknown. Form $(\bar{x}-\mu )/1$, where $\mu$ is the true mass. Whatever the true mass is, this standardized gap behaves like a standard bell curve — its distribution does not depend on $\mu$ at all. That is what makes it useful.

Step 3. Bracket the bell curve. A standard bell curve puts $95\%$ of its area between $-1.96$ and $+1.96$. So with probability $0.95$, the standardized gap lands in that range: $-1.96\le (\bar{x}-\mu )/1\le 1.96$.

Step 4. Solve for the unknown. Rearranging, $\bar{x}-1.96\le \mu \le \bar{x}+1.96$. Plugging in $\bar{x}=502$: the interval is $502-1.96$ to $502+1.96$, that is $500.04$ to $503.96$ milligrams.

What this tells us: the data is consistent with any true mass between about $500$ and $504$ milligrams, at the $95\%$ confidence level. Report a single number and you hide the uncertainty; report this range and you show exactly how much wiggle the nine readings leave. Take more readings and $\sigma /\sqrt{n}$ shrinks, the bracket tightens, and the range narrows.

Check your understanding Beginner

Formal definition Intermediate+

Fix a sample space $(\mathcal{X},\mathcal{B})$ and a family $\{P_{\theta }:\theta \in \Theta \}$ as in 45.02.01. Interval estimation replaces a point estimator with a set-valued rule.

Definition (confidence set). A confidence set is a measurable set-valued map $C:\mathcal{X}\to 2^{\Theta }$, assigning to each outcome $x$ a subset $C(x)\subseteq \Theta$, with $\{x:\theta \in C(x)\}\in \mathcal{B}$ for every $\theta$. When $\Theta \subseteq \mathbb{R}$ and $C(x)$ is an interval, $C$ is a confidence interval, written $C(x)=[L(x),U(x)]$ for measurable endpoint statistics $L\le U$.

Definition (coverage probability, confidence coefficient). The coverage probability of $C$ at $\theta$ is

$${\mathrm{cov}}_C(\theta )=P_{\theta }(\theta \in C(X))={\int }_{\mathcal{X}}1\{\theta \in C(x)\}d{P}_{\theta }(x).$$

The confidence coefficient of $C$ is the infimum coverage, ${\inf }_{\theta \in \Theta }{\mathrm{cov}}_C(\theta )$. The set $C$ is a $1-\alpha$ confidence set if its confidence coefficient is at least $1-\alpha$. The frequentist content is exactly this: $\theta$ is fixed and non-random; the randomness is in $X$, hence in the set $C(X)$, and ${\mathrm{cov}}_C(\theta )$ is the long-run fraction of realizations whose set contains the fixed $\theta$.

Definition (acceptance region). For each ${\theta }_0\in \Theta$, let ${\phi }_{{\theta }_0}$ be a level-$\alpha$ test of $H_0:\theta ={\theta }_0$ (in the sense of 45.02.01). Its acceptance region is $A({\theta }_0)=\{x\in \mathcal{X}:{\phi }_{{\theta }_0}(x)=0\}$ for a non-randomized test, i.e. the outcomes that do not reject ${\theta }_0$. The family $\{A({\theta }_0){\}}_{{\theta }_0}$ has the level property $P_{{\theta }_0}(X\notin A({\theta }_0))\le \alpha$ for every ${\theta }_0$.

Definition (pivotal quantity). A pivotal quantity (or pivot) is a function $Q(X,\theta )$ of the data and the parameter whose distribution under $P_{\theta }$ is the same for every $\theta \in \Theta$. For a location family $f(x-\mu )$ the quantity $X-\mu$ is a pivot; for a scale family ${\sigma }^{-1}f(x/\sigma )$ the quantity $X/\sigma$ is a pivot; for the normal model with the sample mean and sample standard deviation, $\sqrt{n}(\bar{X}-\mu )/S$ is the $t$ pivot with a $t_{n-1}$ law, and $(n-1)S^2/{\sigma }^2$ is the ${\chi }^2$ pivot with a ${\chi }_{n-1}^2$ law.

Construction (the pivot method). Given a pivot $Q$, choose constants $a,b$ with $P_{\theta }(a\le Q(X,\theta )\le b)=1-\alpha$ (the same for all $\theta$, since the law of $Q$ is parameter-free). Then $C(x)=\{\theta :a\le Q(x,\theta )\le b\}$ is a $1-\alpha$ confidence set, because ${\mathrm{cov}}_C(\theta )=P_{\theta }(a\le Q(X,\theta )\le b)=1-\alpha$ for every $\theta$.

Counterexamples to common slips

• Coverage is the worst case over $\Theta$, not the value at one point. A set can have coverage $1-\alpha$ at the true parameter you happen to consider yet drop below $1-\alpha$ elsewhere; the confidence coefficient is the infimum, and a procedure advertised as $95\%$ must clear $0.95$ at every $\theta$.
• The confidence level is not the probability the parameter lies in the realized interval. Once $x$ is observed, $C(x)$ is a fixed set and $\theta$ is a fixed number; the statement "$P(\theta \in C(x))=0.95$" is a category error in the frequentist framework. The probability lives in the procedure, before the data are seen.
• A pivot is not the same as a statistic. A pivot $Q(X,\theta )$ depends on $\theta$; a statistic does not. The point of the pivot is precisely that it mixes data and parameter so that the dependence on $\theta$ cancels out of its distribution, which is what lets you invert the probability bracket into a parameter set.

Key theorem with proof Intermediate+

The organizing result is the exact correspondence between tests and confidence sets: inverting a family of level-$\alpha$ acceptance regions produces a $1-\alpha$ confidence set, and slicing a $1-\alpha$ confidence set produces a family of level-$\alpha$ tests. The two objects carry the same information.

Theorem (test-inversion duality). Let $\{A({\theta }_0){\}}_{{\theta }_0\in \Theta }$ be a family of acceptance regions of level-$\alpha$ tests of $H_0:\theta ={\theta }_0$, so $P_{{\theta }_0}(X\in A({\theta }_0))\ge 1-\alpha$ for every ${\theta }_0$. Define

$$C(x)=\{{\theta }_0\in \Theta :x\in A({\theta }_0)\}.$$

Then $C$ is a $1-\alpha$ confidence set: $P_{\theta }(\theta \in C(X))\ge 1-\alpha$ for every $\theta$. Conversely, if $C$ is a $1-\alpha$ confidence set, then the tests with acceptance regions $A({\theta }_0)=\{x:{\theta }_0\in C(x)\}$ have level $\alpha$, and the two constructions are mutually inverse.

Proof. Membership unfolds symmetrically: for any $x$ and any ${\theta }_0$,

$${\theta }_0\in C(x)⟺x\in A({\theta }_0),$$

directly from the definition of $C$. Fix an arbitrary $\theta \in \Theta$ and take ${\theta }_0=\theta$ in this equivalence; intersecting with the event over the random $X$,

$$\{\theta \in C(X)\}=\{X\in A(\theta )\}.$$

Apply $P_{\theta }$ to both sides:

$$P_{\theta }(\theta \in C(X))=P_{\theta }(X\in A(\theta ))\ge 1-\alpha ,$$

the inequality being the level property of the test at ${\theta }_0=\theta$. Since $\theta$ was arbitrary, ${\inf }_{\theta }{\mathrm{cov}}_C(\theta )\ge 1-\alpha$, so $C$ is a $1-\alpha$ confidence set.

For the converse, suppose $C$ has confidence coefficient $\ge 1-\alpha$ and set $A({\theta }_0)=\{x:{\theta }_0\in C(x)\}$. Then the type I error of the corresponding test at ${\theta }_0$ is

$$P_{{\theta }_0}(X\notin A({\theta }_0))=P_{{\theta }_0}({\theta }_0\notin C(X))=1-{\mathrm{cov}}_C({\theta }_0)\le \alpha ,$$

so the test has level $\alpha$. Applying the forward map to this acceptance family returns $\{{\theta }_0:x\in A({\theta }_0)\}=\{{\theta }_0:{\theta }_0\in C(x)\}=C(x)$, and applying the backward map to an acceptance family returns $\{x:{\theta }_0\in \{\theta :x\in A(\theta )\}\}=A({\theta }_0)$; the two maps are inverse. $\square$

Bridge. This duality builds toward the optimal-interval theory of the Master section, where the optimality of a test transfers to the dual set, and it appears again in the large-sample Wald, score, and likelihood-ratio intervals of 45.04.09, each the inversion of an asymptotic level-$\alpha$ test. This is exactly the same membership bookkeeping that the pivot method exploits: a pivot $Q(X,\theta )$ supplies, for every ${\theta }_0$, the acceptance region $\{x:a\le Q(x,{\theta }_0)\le b\}$, and inverting that family is the pivot construction read through the duality, so the pivot method is dual to a family of pivot-based tests. The central insight is that a confidence set and a family of tests are one object indexed two ways — by outcomes giving sets of parameters, or by parameters giving sets of outcomes — and the foundational reason the coverage equals one minus the size is the pointwise identity $\{\theta \in C(X)\}=\{X\in A(\theta )\}$. Putting these together, every method of finding tests (Neyman-Pearson thresholds, likelihood ratios, score statistics) becomes a method of finding intervals, and every optimality criterion for tests generalises to an accuracy criterion for sets; the bridge is the inversion map $C(x)=\{{\theta }_0:x\in A({\theta }_0)\}$.

Exercises Intermediate+

Advanced results Master

Uniformly most accurate sets from uniformly most powerful tests

False coverage is the interval analogue of type II error. For a confidence set $C$ and a wrong value ${\theta }^{\prime }\ne \theta$, the false-coverage probability is $P_{\theta }({\theta }^{\prime }\in C(X))$, the chance the set captures a value that is not the truth. A $1-\alpha$ set $C$ is uniformly most accurate (UMA) if it minimizes $P_{\theta }({\theta }^{\prime }\in C(X))$ over all $1-\alpha$ sets, simultaneously for every pair $\theta ,{\theta }^{\prime }$ in the relevant range. The duality theorem turns UMA-ness into a statement about tests: by the identity $\{{\theta }^{\prime }\in C(X)\}=\{X\in A({\theta }^{\prime })\}$, minimizing false coverage at $(\theta ,{\theta }^{\prime })$ is minimizing the acceptance probability $P_{\theta }(X\in A({\theta }^{\prime }))$, that is, maximizing the power of the dual test of $H_0:\theta ={\theta }^{\prime }$ at the alternative $\theta$. Inverting a family of uniformly most powerful tests therefore yields a UMA confidence set ^{[Lehmann & Romano §3.5]}. In a one-parameter model with monotone likelihood ratio, the one-sided UMP tests of 45.02.02 invert to the UMA one-sided confidence bounds; for two-sided problems no UMP test exists, and the construction is repaired by restricting to unbiased tests.

Unbiased sets and the shortest-length problem

A $1-\alpha$ confidence set is unbiased if its false-coverage probability never exceeds its coverage: $P_{\theta }({\theta }^{\prime }\in C(X))\le 1-\alpha$ for all ${\theta }^{\prime }\ne \theta$, so the set is at least as likely to cover the truth as any falsehood. Unbiased sets are exactly the inversions of unbiased tests, and inverting the UMP-unbiased tests produces UMA-unbiased sets, the canonical optimum for two-sided normal problems where no unrestricted UMP test exists ^{[Lehmann & Romano Ch. 5]}. Shortest length is a different criterion and need not coincide with UMA-unbiasedness. By the Pratt identity (Exercise 8), expected length equals integrated false coverage, so a UMA-unbiased set minimizes expected length within the unbiased class whenever the dominated family makes the two criteria agree; the equal-tailed ${\chi }^2$ interval for the normal variance is not the shortest, and the minimum-length variance interval solves the stationarity $\varphi (a)=\varphi (b)$ at the pivot density rather than splitting the tail probability evenly.

The relation of the Cramér-Rao bound to interval width

The information floor on variance is an information floor on length. For a regular model the score-based and likelihood-based intervals of 45.04.09 have asymptotic half-width governed by the inverse Fisher information: the Wald interval is $\hat{\theta }\pm z_{\alpha /2}{\hat{I}}_n^{-1/2}$, whose squared half-width scales as $z_{\alpha /2}^2/I_n(\theta )=z_{\alpha /2}^2/(n{I}_1(\theta ))$. Because the Cramér-Rao lower bound of 45.01.05 caps the precision of any unbiased estimator at ${\mathrm{Var}}_{\theta }(\hat{\theta })\ge 1/(n{I}_1(\theta ))$, no asymptotically-efficient interval centered at an unbiased estimator can be narrower in expectation than the scale $z_{\alpha /2}/\sqrt{n{I}_1(\theta )}$ permits; the inverse information is simultaneously the variance floor and, through the Pratt identity, the expected-length floor. Exact finite-sample attainment of both floors is the exponential-family rigidity of 45.01.05: only there does an affine-in-the-score statistic give both a minimum-variance estimator and the corresponding minimal-length pivot interval.

Bayesian credible sets contrasted with frequentist confidence sets

A Bayesian credible set $C_{\pi }(x)$ at level $1-\alpha$ satisfies $\Pi (\theta \in C_{\pi }(x)\mid X=x)\ge 1-\alpha$, where $\Pi (\cdot \mid x)$ is the posterior; the probability statement is over $\theta$, conditional on the observed $x$. A frequentist confidence set's statement is over $X$, conditional on a fixed $\theta$. These are distinct probabilities on distinct sample spaces: the credible set answers "given this data, where is the parameter likely to be?" while the confidence set answers "does this procedure, run repeatedly, capture the fixed parameter at the advertised rate?". The highest-posterior-density credible set is the Bayesian shortest-set analogue, minimizing Lebesgue measure at fixed posterior mass exactly as the $\varphi (a)=\varphi (b)$ condition minimizes length at fixed coverage. The two coincide numerically in special conjugate cases — a flat prior on a normal mean returns the credible set $\bar{x}\pm z_{\alpha /2}\sigma /\sqrt{n}$, identical to the frequentist interval — but the agreement is a coincidence of that model, and the interpretations remain incompatible: the frequentist quantity is calibrated over hypothetical repetitions, the Bayesian quantity over the prior-to-posterior update, and a credible set need not have $1-\alpha$ frequentist coverage nor conversely.

Synthesis. The central insight is that a confidence set is a family of tests indexed the other way, so the entire optimality theory of testing maps onto interval estimation through one membership identity, $\{\theta \in C(X)\}=\{X\in A(\theta )\}$. This is exactly why coverage is one minus size and why false coverage is one minus power: the dual test's two error rates are the set's coverage and accuracy, and the foundational reason uniformly most accurate sets exist precisely where uniformly most powerful tests exist is that inversion is a bijection preserving these probabilities pointwise. Putting these together, the one-sided MLR theory of 45.02.02 generalises to UMA one-sided bounds, the two-sided unbiased theory to UMA-unbiased intervals, and the asymptotic tests of 45.04.09 to the Wald, score, and likelihood-ratio intervals, every optimality crossing the duality intact. The Pratt identity is dual to the power calculation — expected length is integrated false coverage — so short intervals and powerful tests are one demand, and the Cramér-Rao bound of 45.01.05 caps both the estimator variance and, through that identity, the achievable expected length, with exact attainment confined to the exponential family. The Bayesian credible set sits beside this structure as a different probability over a different space, numerically tangent in conjugate-flat-prior models yet interpretively distinct; the bridge across the whole chapter is that inversion turns optimal tests into optimal sets, so testing and estimation are not two subjects but one viewed through two indices.

Full proof set Master

Proposition 1 (the inversion bijection). The map $\{A({\theta }_0){\}}_{{\theta }_0}\mapsto C$, $C(x)=\{{\theta }_0:x\in A({\theta }_0)\}$, is a bijection between families of acceptance regions and set-valued maps $C:\mathcal{X}\to 2^{\Theta }$, with inverse $C\mapsto \{A({\theta }_0)\}$, $A({\theta }_0)=\{x:{\theta }_0\in C(x)\}$. It carries level-$\alpha$ test families to $1-\alpha$ confidence sets and back.

Proof. Both maps are defined by the same biconditional ${\theta }_0\in C(x)⟺x\in A({\theta }_0)$, read in the two directions, so each is determined by the relation $\mathcal{R}=\{(x,{\theta }_0):x\in A({\theta }_0)\}\subseteq \mathcal{X}\times \Theta$. A family $\{A({\theta }_0)\}$ is the ${\theta }_0$-slices of $\mathcal{R}$, and a map $C$ is the $x$-slices of the same $\mathcal{R}$; the correspondence $\{A({\theta }_0)\}\leftrightarrow \mathcal{R}\leftrightarrow C$ is a bijection because slicing a fixed subset of a product in either coordinate is invertible. The level/coverage transfer is the Key theorem: $P_{\theta }(\theta \in C(X))=P_{\theta }(X\in A(\theta ))\ge 1-\alpha$ holds iff $P_{\theta }(X\notin A(\theta ))\le \alpha$ for every $\theta$. $\square$

Proposition 2 (the pivot construction has exact coverage). If $Q(X,\theta )$ is a pivot with law $\nu$ (independent of $\theta$) and $B$ is a Borel set with $\nu (B)=1-\alpha$, then $C(x)=\{\theta :Q(x,\theta )\in B\}$ has coverage ${\mathrm{cov}}_C(\theta )=1-\alpha$ for every $\theta$.

Proof. By the definition of $C$, $\{\theta \in C(X)\}=\{Q(X,\theta )\in B\}$ for the fixed $\theta$ under consideration. Hence ${\mathrm{cov}}_C(\theta )=P_{\theta }(Q(X,\theta )\in B)=\nu (B)=1-\alpha$, the middle equality because the law of $Q(X,\theta )$ under $P_{\theta }$ is $\nu$ for every $\theta$ (the pivot property). Coverage is exact and constant, not merely bounded below. $\square$

Proposition 3 (UMA from UMP). Suppose for each ${\theta }_0$ the test ${\phi }_{{\theta }_0}$ is uniformly most powerful at level $\alpha$ for $H_0:\theta ={\theta }_0$ against the relevant alternatives, with acceptance region $A({\theta }_0)$. Then $C(x)=\{{\theta }_0:x\in A({\theta }_0)\}$ is uniformly most accurate among $1-\alpha$ sets: for any other $1-\alpha$ set $C^{\ast }$ and any $\theta ,{\theta }^{\prime }$ in range, $P_{\theta }({\theta }^{\prime }\in C(X))\le P_{\theta }({\theta }^{\prime }\in C^{\ast }(X))$.

Proof. Let $C^{\ast }$ be any $1-\alpha$ set, with dual acceptance regions $A^{\ast }({\theta }_0)=\{x:{\theta }_0\in C^{\ast }(x)\}$, which by Proposition 1 are the regions of level-$\alpha$ tests ${\phi }_{{\theta }_0}^{\ast }$. Fix $\theta ,{\theta }^{\prime }$. By the membership identity,

$$P_{\theta }({\theta }^{\prime }\in C(X))=P_{\theta }(X\in A({\theta }^{\prime }))=1-{\beta }_{{\phi }_{{\theta }^{\prime }}}(\theta ),$$

where ${\beta }_{{\phi }_{{\theta }^{\prime }}}(\theta )$ is the power of the test of $H_0:\theta ={\theta }^{\prime }$ evaluated at the true $\theta$; likewise $P_{\theta }({\theta }^{\prime }\in C^{\ast }(X))=1-{\beta }_{{\phi }_{{\theta }^{\prime }}^{\ast }}(\theta )$. Because ${\phi }_{{\theta }^{\prime }}$ is UMP at level $\alpha$ and ${\phi }_{{\theta }^{\prime }}^{\ast }$ is some level-$\alpha$ test, ${\beta }_{{\phi }_{{\theta }^{\prime }}}(\theta )\ge {\beta }_{{\phi }_{{\theta }^{\prime }}^{\ast }}(\theta )$, so $1-{\beta }_{{\phi }_{{\theta }^{\prime }}}(\theta )\le 1-{\beta }_{{\phi }_{{\theta }^{\prime }}^{\ast }}(\theta )$, giving $P_{\theta }({\theta }^{\prime }\in C(X))\le P_{\theta }({\theta }^{\prime }\in C^{\ast }(X))$. Uniformity over $(\theta ,{\theta }^{\prime })$ is the uniformity of the UMP family. $\square$

Proposition 4 (Pratt expected-length identity). For an interval-valued $1-\alpha$ set $C(X)=[L(X),U(X)]$ with ${\mathbb{E}}_{\theta }|U|,{\mathbb{E}}_{\theta }|L|<\infty$,

$${\mathbb{E}}_{\theta }[U(X)-L(X)]={\int }_{\mathbb{R}}{P}_{\theta }({\theta }^{\prime }\in C(X))d{\theta }^{\prime }.$$

Proof. For each $x$, $U(x)-L(x)={\int }_{\mathbb{R}}1\{L(x)\le {\theta }^{\prime }\le U(x)\}d{\theta }^{\prime }={\int }_{\mathbb{R}}1\{{\theta }^{\prime }\in C(x)\}d{\theta }^{\prime }$. The integrand $(x,{\theta }^{\prime })\mapsto 1\{{\theta }^{\prime }\in C(x)\}$ is nonnegative and jointly measurable, so Tonelli's theorem permits exchanging ${\mathbb{E}}_{\theta }$ (integration in $x$ against $P_{\theta }$) with the Lebesgue integral in ${\theta }^{\prime }$:

$${\mathbb{E}}_{\theta }[U-L]={\int }_{\mathcal{X}}{\int }_{\mathbb{R}}1\{{\theta }^{\prime }\in C(x)\}d{\theta }^{\prime }d{P}_{\theta }(x)={\int }_{\mathbb{R}}{P}_{\theta }({\theta }^{\prime }\in C(X))d{\theta }^{\prime }.\square$$

Proposition 5 (shortest interval from a unimodal symmetric pivot). Let $Q(X,\theta )$ be a pivot with a continuous density $q$ that is symmetric about $0$ and strictly decreasing in $|q$-argument$|$, and let the confidence set be $\{\theta :a\le Q\le b\}$ with ${\int }_a^{b}q=1-\alpha$. Among all such $(a,b)$ the length-minimizing choice in the parameter (when $Q$ is affine and decreasing in $\theta$ with constant slope) is the symmetric $a=-b$.

Proof. When $Q(x,\theta )=(\hat{\theta }(x)-\theta )/{\sigma }_0$ is affine in $\theta$ with constant scale ${\sigma }_0$, the inverted set is $[\hat{\theta }-b{\sigma }_0,\hat{\theta }-a{\sigma }_0]$ of length $(b-a){\sigma }_0$, so minimizing parameter length is minimizing $b-a$ subject to ${\int }_a^{b}q(t)dt=1-\alpha$. Introduce a multiplier: stationarity of $b-a-\lambda ({\int }_a^{b}q-(1-\alpha ))$ gives $1-\lambda q(b)=0$ and $-1+\lambda q(a)=0$, so $q(a)=q(b)$. Strict unimodality and symmetry of $q$ force $a=-b$, and the constraint then determines $b$ as the upper $\alpha /2$ quantile. The second-order condition holds because $q$ decreasing away from $0$ makes the constrained length strictly convex at the symmetric point. For a skewed pivot density $q(a)=q(b)$ no longer forces symmetry, and the equal-tail set (${\int }_{-\infty }^{a}q={\int }_b^{\infty }q=\alpha /2$) differs from the shortest set. $\square$

Connections Master

• The most-powerful and uniformly-most-powerful test theory 45.02.01 is the source of every optimal interval here: a level-$\alpha$ test of $H_0:\theta ={\theta }_0$ inverts to a $1-\alpha$ confidence set, and the most-accurate sets are exactly the inversions of the most-powerful tests, so the Neyman-Pearson and Karlin-Rubin optimality proved upstream becomes minimal false-coverage probability downstream by the membership identity $\{\theta \in C(X)\}=\{X\in A(\theta )\}$.

• The Cramér-Rao lower bound 45.01.05 caps interval width: the inverse Fisher information that floors the variance of any unbiased estimator also floors, through the Pratt expected-length identity, the achievable expected length of any asymptotically-efficient interval centered at that estimator, since the Wald interval's squared half-width scales as $z_{\alpha /2}^2/(n{I}_1(\theta ))$ and exact attainment of both floors is the same exponential-family rigidity.

• The large-sample Wald, score, and likelihood-ratio intervals 45.04.09 are the asymptotic instances of the test-inversion construction: each inverts the corresponding asymptotic level-$\alpha$ test of $H_0:\theta ={\theta }_0$, the Wald interval from the asymptotically-normal estimator, the score interval from the mean-zero score, and the likelihood-ratio interval from the chi-squared limit of $2\log$ the ratio, and the duality theorem is what licenses reading each test off as an interval.

• The one-sided composite testing via monotone likelihood ratio 45.02.02 supplies the one-sided uniformly-most-powerful tests whose inversions are the uniformly-most-accurate one-sided confidence bounds; the rejection region $\{T>c({\theta }_0)\}$ with $c$ monotone is exactly the structure that makes the inverted set a single bound rather than a disconnected region.

Historical & philosophical context Master

The confidence-set framework is due to Jerzy Neyman, who introduced the confidence interval and the frequentist confidence coefficient in a 1934 discussion and developed the theory fully in his 1937 paper Outline of a theory of statistical estimation based on the classical theory of probability in the Philosophical Transactions of the Royal Society ^{[Neyman 1937]} (Phil. Trans. R. Soc. Lond. A 236, 333-380). Neyman framed the interval as a procedure with a guaranteed long-run coverage rate, deliberately separating his frequentist account from R. A. Fisher's contemporaneous fiducial argument, which sought to assign a probability distribution to the parameter directly from the data without a prior. The fiducial program was never made consistent and is now largely abandoned, but the Fisher-Neyman dispute over whether an interval statement is about the parameter or about the procedure shaped the foundations of estimation for decades.

The duality between tests and confidence sets was made explicit in the development running from Neyman and Pearson through Erich Lehmann, whose Testing Statistical Hypotheses organized uniformly most accurate and unbiased confidence sets as the inversions of uniformly most powerful and unbiased tests ^{[Lehmann & Romano Ch. 5]}. John W. Pratt's 1961 identity relating the expected length of an interval to the integrated probability of false coverage ^{[Pratt 1961]} (J. Amer. Statist. Assoc. 56, 549-567) tied the geometric notion of a short interval to the operational notion of a powerful dual test. The contrast with Bayesian credible sets traces to the Bayesian revival of the mid-twentieth century, where the posterior-probability interval was advanced as the answer to the question Neyman's coverage statement does not address, and the conditions under which the two numerically agree — flat priors in location models, matching priors more generally — remain an active subject in the theory of probability-matching priors.

Bibliography Master

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}

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