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The coverage probability is the repeated-sampling probability that a random confidence interval contains the fixed true statistical parameter. A finite-sample level-$1-\alpha$ procedure has coverage probability at least $1-\alpha$ at every admissible statistical parameter; exact nominal coverage means equality. An asymptotically calibrated procedure approaches the nominal value as sample size grows, under its stated assumptions. Higher-order Edgeworth expansions can quantify the discrepancy from nominal coverage probability when their regularity conditions hold.

Observed coverage in finitely many repetitions is itself random. For $m$ independent repetitions of the exact normal mean confidence interval with known variance, the count $K$ of covering confidence intervals has binomial distribution $\mathrm{Binomial}(m,0.95)$. At $m=40$, its expectation is $38$, butConsequently, the 39 successes in the figure are compatible with exact 95% coverage probability. The guarantee follows from the pivotal quantity's probability distribution; the finite observed fraction is an illustration. Also distinguish a confidence interval for a fixed mean from a prediction interval for a future observation: the latter must accommodate the new observation's variation as well as estimation uncertainty.

In a parametric statistical model, the likelihood function compares parameter values, while sampling distributions calibrate tests and confidence intervals. A sufficient statistic retains the parameter information. Conditioning on an observed ancillary statistic restricts that calibration to experiments with the same ancillary configuration, instead of averaging over configurations whose distribution is unrelated to the parameter. Since the ancillary's marginal law is parameter-free, conditional and unconditional likelihoods have the same parameter-dependent factor. With a nuisance parameter, a conditional law or pivotal quantity must remove it before inference for the parameter of interest can use a nuisance-free reference law.

Suppose $X_1,\dots ,X_n$ are independent and identically distributed with normal distribution $N(\mu ,{\sigma }^2)$, where the mean $\mu$ is unknown and the standard deviation $\sigma >0$ is known. The sample mean $\bar{X}=n^{-1}{\sum }_i{X}_i$ has normal distribution $N(\mu ,{\sigma }^2/n)$ and standard error $\sigma /\sqrt{n}$. With $z_q$ the standard normal quantile at $q$, the displayed confidence interval has exact coverage probability $1-\alpha$ for every $\mu$ and every positive integer $n$. No central limit theorem approximation is needed.

Take $\sigma =2$, $n=16$ and nominal confidence 95%. The standard error is $2/\sqrt{16}=0.5$, and $z_{0.975}=1.9599639845\dots$. ThereforeRounded for reporting, this is $\bar{x}\pm 0.98$. If $\bar{x}=1.2$, the realized confidence interval is approximately $[0.22,2.18]$. Exact 95% calibration refers to the unrounded standard normal quantile; using $1.96$ literally gives coverage probability $0.9500042097\dots$, almost but not exactly 0.95.

The confidence interval's width scales as $n^{-1/2}$: increasing sample size from 16 to 64 halves its half-width to $0.4899909961\dots$. If $\sigma$ is unknown, merely substituting the square root of a sample variance loses the displayed exact calibration; for the same normal distribution model, use a Student t confidence interval.

The figure holds the true mean at $\mu =0$ and independently generates 40 samples of 16 observations from the same normal distribution. NumPy's PCG64 generator with seed 42 gives 39 covering confidence intervals and one miss, at sample 32: observed coverage is $39/40=97.5\%$. Every bar uses the unrounded standard normal quantile. This finite collection illustrates the moving endpoints; it does not establish the analytic coverage probability or require exactly two misses.

The beauty of the construction is that a familiar standard normal distribution calibrates every possible unknown mean at once. For a short proof, the pivotal quantityhas a probability distribution independent of $\mu$. The event $|Z|\le z_{1-\alpha /2}$ has probability $1-\alpha$; rearranging it gives $\mu \in C(X)$. The right panel shows the equivalent event $|\bar{X}|\le 0.9799819923\dots$ when $\mu =0$.

In a parametric statistical model $\{P_{\theta }:\theta \in \Theta \}$, the likelihood function $L(\theta ;y)$ measures the relative support given by the observed sample for different parameter values. Maximum likelihood estimation supplies a point estimate; the sampling distributions of suitable statistics calibrate significance tests and confidence intervals. A sufficient statistic retains all parameter-dependent information in the likelihood function, so one first seeks a reduction to such a statistic rather than discarding information arbitrarily.

The characteristic conditioning principle in Fisherian conditional inference is to calibrate inference conditional on the observed value of an ancillary statistic. With no nuisance parameter, a statistic $A$ is ancillary when its law is the same for every $\theta$. Its observed value can nevertheless describe the experimental configuration or precision. Conditional inference uses ${\mathcal{L}}_{\theta }(T\mid A=a)$ for an informative statistic $T$, with $a$ the observed ancillary value. Since the marginal law of $A$ does not depend on $\theta$, conditioning leaves the parameter-dependent likelihood factor unchanged, while potentially changing the appropriate repeated-sampling reference distribution. Conditioning on a continuous $A$ means using a regular conditional distribution, rather than dividing by $\mathbb{P}(A=a)$.

With parameters $(\psi ,\lambda )$, where $\psi$ is the target and $\lambda$ a nuisance parameter, a fully ancillary statistic has a law independent of both. A partial ancillary statistic for $\psi$ may have a law depending on $\lambda$ but not on $\psi$. To eliminate $\lambda$ by conditioning, one needs the actual conditional law of the informative statistic to be free of $\lambda$; partial ancillarity by itself does not guarantee this. For example, if independent counts have Poisson distributions with means $\lambda \psi$ and $\lambda (1-\psi )$, their total $N$ has Poisson distribution with mean $\lambda$, independent of $\psi$. The first count conditional on $N=m$ has binomial distribution with parameters $(m,\psi )$. Conditioning removes the nuisance intensity $\lambda$ exactly. More generally, nuisance-free pivotal quantities or conditional likelihoods can be used where available; there is no universal conditioning device that removes every nuisance parameter.

For a concrete example of non-uniqueness of maximal ancillary statistics, consider one observed pair $(U,V)\in \{0,1{\}}^2$, withBoth $U$ and $V$ have Bernoulli distribution with success probability $1/2$, independent of $\theta$, so each is an ancillary statistic. Their generated partitions are different and incomparable. Moreover, each is maximal: refining either two-point cell singles out a point whose probability is $\theta /2$ or $(1-\theta )/2$, which is parameter-dependent. Their joint statistic is not ancillary, since $P_{\theta }(U=V)=\theta$. Thus there is no common finer ancillary partition that reconciles the two choices. The conditional laws differ as labelled experiments: $P_{\theta }(V=U\mid U)=\theta$ and $P_{\theta }(U=V\mid V)=\theta$, but they condition on different observed information. This illustrates why “condition on a maximal ancillary” need not specify a unique procedure.

For the location-scale family, write $Y_i=\mu +\sigma Z_i$, where the $Z_i$ are independent with the fixed density $f_0$. Let $s_Z^2=n^{-1}{\sum }_i(Z_i-\bar{Z}{)}^2$. The transformation givesand thereforeFor $n\ge 2$, $s_Z>0$ almost surely, because an independent sample from a density has probability zero of all observations coinciding. The right side depends only on a sample from the fixed base law, so the entire vector $A$ has a law independent of $(\mu ,\sigma )$: $\boxed{A\text{ is ancillary}.}$ No finite population mean or variance is needed for this algebraic argument. With one observation the displayed standardization is undefined, so at least two observations are implicitly required.

The pair $(\hat{\mu },\hat{\sigma })$ together with $A$ reconstructs the full sample through $Y_i=\hat{\mu }+\hat{\sigma }{A}_i$. Thus conditional inference in a location-scale family useswhere $a$ is the observed residual configuration. Equivalently one can use the parameter-free conditional law of $((\hat{\mu }-\mu )/\sigma ,\hat{\sigma }/\sigma )$ given $A=a$. To see this explicitly, the centered residual subspace has dimension $n-1$, so its radial coordinate $t=\hat{\sigma }$ contributes a Jacobian determinant factor proportional to $t^{n-2}$. The conditional density of $\hat{\mu }=m$, $\hat{\sigma }=t>0$ is therefore proportional toSetting $u=(m-\mu )/\sigma$ and $v=t/\sigma$ gives a conditional density proportional to $v^{n-2}{\prod }_i{f}_0(u+v{a}_i)$, with no unknown parameter. The normalizing factor can depend on $a$, but not on $\mu$ or $\sigma$. These statements concern almost every attainable ancillary value.

For a new independent tumour grown for ten days in B, use $x_0=(1,10,10,0{)}^T$. Its fitted mean isWrite $h_0=x_0^T(X^{T}X{)}^{-1}{x}_0$. The prediction interval in a normal linear model iswhere $s^2=\mathrm{RSS}/26$ and $t_{26,1-\alpha /2}$ is the indicated Student's t-distribution quantile. The design information needed to evaluate $h_0$ is not supplied, so a fully numerical interval cannot be recovered from the coefficient standard errors alone: the coefficient covariances are also needed.

To prove coverage, write $Y_0=x_0^T\beta +{\epsilon }_0$, with ${\epsilon }_0\sim N(0,{\sigma }^2)$ independent of the original sample. The error $Y_0-x_0^T\hat{\beta }$ has a normal distribution with mean zero and variance ${\sigma }^2(1+h_0)$. By the joint distribution of least-squares and variance estimators, $26s^2/{\sigma }^2\sim {\chi }_{26}^2$ and is independent of $\hat{\beta }$; it is also independent of ${\epsilon }_0$. Hence the pivotal quantityConsequently $\mathrm{Pr}(-t_{26,1-\alpha /2}\le T\le t_{26,1-\alpha /2})=1-\alpha$, which is exactly the claimed prediction interval coverage. This is repeated-sampling coverage conditional on the design, assuming the model; it need not hold conditional on the particular observed training responses. The extra $1$ represents new-observation noise and must not be omitted as in a mean confidence interval.

The sample mean $\overline{X}$ has normal distribution $N(\mu ,1/n)$ because the observations are independent with known variance 1. The pivotal quantity $Z=\sqrt{n}(\overline{X}-\mu )$ has the standard normal distribution. Consequentlyand inversion gives the confidence intervalReplacing 1.96 by the exact $0.975$ normal quantile gives exact 95% coverage; the printed numerical value gives the usual rounded interval. No variance estimation or Student distribution is needed.

For $0\le t\le 1$, independence and the uniform distribution giveTherefore the cumulative distribution function isIt contains no $\theta$, so $M/\theta$ is a pivotal quantity.

The pivotal quantity $U=M/\theta$ lies in $[0,1]$ and satisfies $\mathbb{P}(U\le t)=t^n$. ThusRearranging these inequalities yields the exact confidence intervalFor every $\theta >0$, its coverage probability is exactly $1-\alpha$; the result uses the finite-sample probability distribution, with no normal approximation.

The independent uniform choices from the student list describe sampling with replacement. Under the stated independent normal distribution model for the heights, the population variance is unknown and the given $8.5$ cm is the sample standard deviation $s$. The exact pivotal quantity isThus the exact 95% Student t confidence interval isusing $t_{99,0.975}\approx 1.9842$. The supplied approximation $\Phi (2)=0.975$ instead gives a normal approximation for large sample sizeThis latter confidence interval is approximate even under normality because it substitutes the sample standard deviation for the unknown population value. There is no finite population correction under the specified independent selections. Interpreting the finite student list literally as a discrete population makes normality an idealized model; it does not justify changing the stated sampling design.

Independence of the slope estimate and residual variance, together with their normal and chi-square laws, gives the Student pivotal quantityThus an exact Gaussian-model two-sided 95% confidence interval isUsing a normal quantile without estimating the variance uncertainty would not give this exact finite-sample coverage.

Each before-and-after observation belongs to the same individual, so pairing removes individual baseline variation and allows dependence within a person's two readings. Under the stated independent normal model for the differences,The independence of the normal sample mean and sample variance theorem gives independence of $\overline{D}$ and $S_D^2$, andAlso $(\overline{D}-\mu )/(\sigma /\sqrt{n})$ is standard normal. Dividing it by the square root of the independent ${\chi }_{n-1}^2/(n-1)$ variable givesIts law contains neither unknown parameter, providing the pivotal quantity for a paired t-test.