A directrix of a conic is the fixed line in its focus-directrix description: distance to the focus of a conic is a constant multiple of distance to this line. For a parabola the multiple is one. The directrix of a conic belongs to the geometry of the conic section, rather than being a tangent or an asymptote.

For an odd integer $N>1$, a nontrivial square root of unity modulo $N$ is a residue $z$ with $z^2\equiv 1(\mathrm{mod}N)$ and $z\not\equiv \pm 1(\mathrm{mod}N)$. Then both $\gcd (z-1,N)$ and $\gcd (z+1,N)$ are proper divisors. For $N=77$, $z=34$ works: $34^2\equiv 1(\mathrm{mod}77)$, $\gcd (33,77)=11$, and $\gcd (35,77)=7$.

A congruence of squares with $\gcd (b,N)=1$ converts to this form by $z\equiv a{b}^{-1}(\mathrm{mod}N)$, using a modular multiplicative inverse. In the $a=20,b=13$ example, $13^{-1}\equiv 6(\mathrm{mod}77)$, so $z\equiv 20\cdot 6\equiv 43(\mathrm{mod}77)$. It yields $\gcd (42,77)=7$ and $\gcd (44,77)=11$. Nontriviality requires mixed sign choices across prime factors, which an odd prime number modulus cannot provide.

A comparison can have the same direction within every background stratum but reverse after pooling strata with different group compositions. For example, let unexposed event risks be $0.001$ in a low-risk stratum and $0.05$ in a high-risk stratum, and multiply the odds by $2.7$ for exposure within each stratum. The exposed risks are then approximately $0.002695$ and $0.124424$. If 80% of unexposed subjects but only 1% of exposed subjects belong to the high-risk stratum, the pooled unexposed risk is $0.0402$ while the pooled exposed risk is about $0.003913$. Every conditional odds ratio is above one, yet the pooled exposed risk is smaller. The differing background distributions create the reversal; this example does not identify a causal effect without assumptions on how exposure was assigned.

Case mix describes the distribution of patient characteristics affecting prognosis, such as age, severity, comorbidity and urgency. Comparing provider outcomes without accounting for case mix may confuse differences in patients with differences in care. Adjustment requires comparable, reliably measured baseline variables and does not remove all possible confounding.

For a positive conformal factor in dimension $n\ge 3$, set $S_{ij}=-u_{ij}+u_i{u}_j-\frac{1}{2}|\nabla u{|}^2{\delta }_{ij}$. These are the coordinate components of the Schouten tensor. In the convention $R_{ijkl}=g(e_i,R(e_k,e_l)e_j)$, the Riemann curvature tensor is $R_{ijkl}=e^{2u}({\delta }_{ik}{S}_{jl}+{\delta }_{jl}{S}_{ik}-{\delta }_{il}{S}_{jk}-{\delta }_{jk}{S}_{il})$. The orthonormal coframe $e^{u}d{x}^i$ has connection 1-forms ${\omega }^i{}_j=u_{j}d{x}^i-u_{i}d{x}^j$; substituting into Cartan's second structure equation proves this expression directly. Its contractions give $R_{ij}=(n-2)S_{ij}+{\delta }_{ij}{\sum }_k{S}_{kk}$ and $R=e^{-2u}[-2(n-1)\Delta u-(n-1)(n-2)|\nabla u{|}^2]$.

For $\lambda >0$, this four-dimensional Euclidean metric has two asymptotically flat ends. Inversion $\rho =\lambda /r$ puts the metric near the $r=0$ end into the same form with $\rho \to \infty$. On $r>0$, its Ricci scalar vanishes because $\Delta (r^{-2})=0$, but its Ricci tensor is $R_{ij}=4\lambda ({\delta }_{ij}-4x_i{x}_j/r^2)/(r^2+\lambda {)}^2$. It therefore does not satisfy the Vacuum Einstein equations unless $\lambda =0$. For negative $\lambda$ the conformal factor vanishes at a finite radius, so that metric is not a smooth wormhole across that sphere.

In a flat universe containing pressureless matter and a positive cosmological constant, normalize the scale factor to one today. The Friedmann equation and $d\tau =dt/a$ giveThe change of variable is $a=({\Omega }_m/{\Omega }_{\Lambda }{)}^{1/3}y$. Including radiation in cosmology makes this an approximation, with its omitted early contribution of order $H_0^{-1}\sqrt{{\Omega }_r}/{\Omega }_m$. When the upper limit is large, the integral approaches $I=\Gamma (1/6)\Gamma (1/3)/(3\sqrt{\pi })$; this follows by setting $z=y^3$ and using the beta function integral.

For a flat Conformal Killing vector field, let $\rho ={\partial }_{\mu }{\omega }^{\mu }$. Differentiate the Conformal Killing equation in three index orders, add two equations and subtract the third to obtain the displayed identity. Tracing gives $\Delta {\omega }_{\alpha }=(2-d){\partial }_{\alpha }\rho /d$, whose divergence implies $\Delta \rho =0$ for $d>1$. Taking the divergence of the original identity gives $(d-2){\partial }_{\beta }{\partial }_{\gamma }\rho =-{\delta }_{\beta \gamma }\Delta \rho$. Thus $\rho$ is affine and $\omega$ is at most quadratic when $d>2$.

A conformal compactification realizes a noncompact metric space as a dense open subset of a compact space whose metric on that subset is conformally related to the original one. Its smoothness and boundary interpretation depend on the example. Conformal compactification of Euclidean space adds one point to four-space; Projective Minkowski compactification adds a null cone. A regular null-boundary conformal completion and a projective compactification need not have the same global topology or causal covering convention.

An equivalence class of Riemannian metrics under multiplication by positive smooth functions. Such rescalings preserve angles. On an oriented smooth surface, a conformal structure is equivalent to a complex structure, with local holomorphic coordinates in which its metrics are positive multiples of $d{x}^2+d{y}^2$.

A classical path joining prescribed endpoints is a stationary point of the action, satisfying the Euler-Lagrange equation and the endpoint boundary conditions. Expanding the path integral about each such path separates its classical phase from its quadratic fluctuation determinant. There may be multiple classical paths or none within a chosen real saddle class; the semiclassical propagator must respect that saddle structure rather than assuming a unique trajectory for every potential.

A confidence band provides simultaneous bounds on an unknown function over its entire index set. Its coverage event is the intersection over all indices, unlike separate pointwise confidence intervals. For a continuous cumulative distribution function, the Kolmogorov-Smirnov theorem gives an asymptotic band $F_n\pm c_{\alpha }/\sqrt{n}$, clipped to $[0,1]$, where $c_{\alpha }$ is a $(1-\alpha )$-quantile of the absolute Brownian bridge maximum. The simultaneous event equals $\sqrt{n}\Vert F_n-F{\Vert }_{\infty }\le c_{\alpha }$, so convergence in distribution at this continuity point proves coverage tending to $1-\alpha$.

Invert a likelihood-ratio test for a scalar parameter, maximizing over nuisance parameters for each fixed candidate value. Under regularity, a level $1-\alpha$ interval includes values whose profile log-likelihood is at most ${\chi }_{1,1-\alpha }^2/2$ below its maximum. This explains the horizontal cutoff on a Box–Cox transformation likelihood plot.

A lower confidence bound $L$ gives a one-sided confidence interval with the displayed coverage requirement for every admissible parameter. An upper bound $U$ instead satisfies ${\mathrm{Pr}}_{\theta }\{\theta \le U\}\ge 1-\alpha$. Coverage is a repeated-sampling property; it does not assign a posterior probability to the fixed parameter.

For jointly approximately normal estimates $(\hat{C},\hat{E})$ with estimated variances $v_C,v_E$ and covariance $v_{CE}$, a confidence interval for the ratio $C/E$, where $E\ne 0$, can be obtained by inverting tests of $C-rE=0$. For each proposed ratio $r$, the contrast has estimate $\hat{C}-r\hat{E}$ and variance $v_C-2r{v}_{CE}+r^2{v}_E$. Retain $r$ when its absolute standardized contrast is at most the chosen critical value $q$. This gives the displayed quadratic inequality, with coefficients $A={\hat{E}}^2-q^2{v}_E$, $B=-2(\hat{C}\hat{E}-q^2{v}_{CE})$, $D={\hat{C}}^2-q^2{v}_C$. When $A>0$ and the discriminant is positive, retain the interval between the roots. When $A<0$, a confidence set can instead be two unbounded rays or the entire real line. Degenerate linear cases are interpreted directly from the inequality. A bounded interval must not be forced when the denominator is poorly separated from zero. Known covariance matrix and a joint normal distribution give exact normal critical values; estimated covariance usually gives approximate coverage, unless an appropriate exact common-scale pivot exists.

For an exponential distribution with mean $\theta$, the maximum-likelihood estimator is $\bar{Y}$ and $\sqrt{n}(\bar{Y}-\theta )/\theta$ converges to a standard normal distribution. Inverting the event that this pivot lies between $-z$ and $z$ gives the displayed confidence interval, provided $n>z^2$. Its two-sided coverage probability has error $O(n^{-1})$, because the first Edgeworth expansion term is even and cancels between the two tails.

Let $X_1,\dots ,X_n$ be independent and identically distributed observations from a continuous cumulative distribution function $F$, with $n\ge 2$. The interval from the sample minimum to maximum contains random population probability $C=F(X_{(n)})-F(X_{(1)})$. The probability integral transform identifies $C$ with the uniformized sample range, soFor small $\epsilon$, setting $n\epsilon \approx \vartheta$ makes this approximately $(1+\vartheta )e^{-\vartheta }$. Solve $(1+\vartheta )e^{-\vartheta }=\delta$ for the unique positive $\vartheta$ when $0<\delta <1$, then use the exact displayed inequality to choose an integer sample size.

This is a distribution-free tolerance interval for population content under the continuous iid model. It differs from a confidence interval for a fixed parameter. Atoms or dependent observations require their own treatment; uniformizing them does not automatically preserve the same ordinary sample-range calculation.

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.

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 $\mu =0$.

Electrostatic shielding decouples interior and exterior electric fields using a conductor in electrostatic equilibrium. Exterior sources cannot produce an electric field in an empty closed cavity in a conductor, whether grounded or isolated. Outward independence requires grounding: internal sources can affect an isolated enclosure's exterior, as in Faraday's ice pail experiment. Grounding makes the exterior solution independent of internal sources, while allowing exterior sources' own response. Openings and changing electromagnetic fields need additional analysis.^[ref]