Varying the canonical scalar action gives $\square \varphi +V_{,\varphi }=0$. In a homogeneous flat Friedmann-Lemaître-Robertson-Walker metric written in conformal time, $\sqrt{-g}=a^4$ and $g^{00}=a^{-2}$, soThe quadratic potential therefore givesEquivalently $\ddot{\varphi }+3H\dot{\varphi }+m^2\varphi =0$ in cosmic time. The intrinsic roll/oscillation time is $m^{-1}$ and the damping time is $H^{-1}$. If $m\gg H_0$, the scalar has long since begun rapid oscillations: averaging over a quadratic oscillation gives equal kinetic and potential energies, $⟨p⟩\simeq 0$, so it behaves as matter rather than accelerating dark energy. To remain nearly frozen or only begin rolling today requires the mass scale of quadratic quintessenceIf its energy is predominantly potential, equating $m^2{\varphi }_0^2/2$ to ${\rho }_{X0}=3M^2{H}_0^2{\Omega }_X$, with $M$ the reduced Planck mass, givesTaking the natural onset estimate $m\sim H_0$ and ${\Omega }_X\sim 0.7$ gives $|{\varphi }_0|\sim 2M\sim 5\times 10^{18}\mathrm{GeV}$, of order the Planck mass. More strongly frozen, negative-pressure evolution requires a smaller mass and correspondingly larger field. The density fixes the product $m|{\varphi }_0|$ at this accuracy, rather than a unique mass and field separately.

Without substantial annihilation or decay, monopole number in a comoving volume is conserved. Hence $n_M\propto a^{-3}$ and, for constant nonrelativistic mass, ${\rho }_M\propto a^{-3}$. Radiation redshifts as $a^{-4}$, so the monopole-to-radiation ratio grows with $a$.

Use comoving entropy conservation rather than assuming the same relativistic content at every epoch:With $g_{\ast s,i}\simeq 100$, $g_{\ast s,0}\simeq 3.91$ and $T_0\simeq 2.35\times 10^{-13}\mathrm{GeV}$, the surviving energy density isTaking $m_{\mathrm{pl}}\simeq 1.22\times 10^{19}\mathrm{GeV}$ and $H_0\simeq 70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ gives ${\rho }_{\mathrm{crit},0}\simeq 4.0\times 10^{-47}{\mathrm{GeV}}^4$. Requiring ${\rho }_{M0}<0.01{\rho }_{\mathrm{crit},0}$ yieldsUnknown geometric and mass prefactors change this estimate only through a fourth root. Conventional grand-unification scales near $10^{15}$–$10^{16}\mathrm{GeV}$ are far above the bound: this is the cosmological monopole problem, not a universal prohibition on unification. Inflation after monopole formation can dilute the relics; reheating must then stay below their regeneration temperature. Later entropy production, efficient annihilation, or a symmetry-breaking history avoiding stable monopoles can also weaken the bound. These mechanisms change the assumed conserved initial relic abundance.

Take the transition temperature $T_i\sim \eta$ and radiation-dominated universe. With $g_{\ast }\simeq 100$ relativistic degrees of freedom, ${\rho }_{r,i}=({\pi }^2/30)g_{\ast }{T}_i^4$ and the Friedmann equation givesA saturated causal correlation length is of order the radiation-era horizon $H_i^{-1}=2t_i$. Write $C_n\sim 1$ for the geometric/formation efficiency and $C_M=M_M/\eta$. One monopole per horizon volume gives the horizon-density monopole abundance bound estimateThe first quantity is number density and the second is defect energy density. Relative to the initial radiation critical density,This can initially be small even when the surviving monopole abundance is unacceptable today. Causality supplies an order-of-magnitude lower abundance estimate under the stipulated saturation; it is not a precise statistical count of defects.

The SU(5) grand unified theory has simply connected space $G=SU(5)$ and a connected Standard Model subgroup. More precisely that subgroup is $(SU(3)\times SU(2)\times U(1))/{\mathbb{Z}}_6$; the product notation suppresses a finite central quotient. The long exact sequence of homotopy groups of a fibration $H\to G\to G/H$ givesThus stable magnetic monopoles form, whereas this transition does not require walls or strings. The monopole core and mass scales in an SU(5) transition are set by the massive gauge and scalar fields:For order-one couplings, the core size is of order ${\eta }^{-1}$ and the mass is of order $\eta$ up to the often substantial factor $4\pi /g$. These heavy objects are nonrelativistic soon after formation and can survive as stable relics.

Finite-energy configurations approach the vacuum manifold outside defect cores. The surrounding boundary in the transverse directions is a sphere, and noncontractible boundary data prevent a smooth deformation to the uniform vacuum.

Nontrivial ${\pi }_0(M)$ means disconnected vacuum components. Choosing different components on the two sides of an interface produces domain walls. Here ${\pi }_0$ labels components and need not itself carry a group structure. Nontrivial ${\pi }_1(M)$ means noncontractible loops: a circle surrounding a line can wind around the vacuum manifold, producing cosmic strings. Nontrivial ${\pi }_2(M)$ means noncontractible maps from an enclosing two-sphere, producing pointlike magnetic monopoles in three spatial dimensions.During the transition, initially uncorrelated vacuum choices in causally separated regions can generate these boundary classes through the Kibble mechanism. Nontrivial topology allows stable defects; it does not specify their exact formation efficiency or abundance.

Choose a vacuum order parameter ${\Phi }_0$ and let $H$ be its stabilizer in $G$. Acting with $G$ produces all symmetry-related minima; two elements produce the same minimum precisely when they differ by an element of $H$. The vacuum manifold is thereforeThis assumes a single transitive orbit of minima; additional independent vacuum sectors would require more than one orbit. For a local gauge symmetry, $G/H$ describes the vacuum directions used in finite-energy boundary conditions and defect classification, rather than distinct observable vacua after quotienting by every gauge redundancy.

The acceptance mass for an unnormalized rejection envelope follows by integrating the proposal-specific acceptance probability:The inequality $f\le Mg$ also gives $C\le M$, so this is a valid probability. It is the same at every independent trial.

If $J_i$ indicates acceptance of the $i$th proposal, then $J_1,\dots ,J_n$ are independent Bernoulli distribution variables with parameter $C/M$. Hence the accepted count $K={\sum }_i{J}_i$ satisfiesIts variance is $n{p}_{\mathrm{acc}}(1-p_{\mathrm{acc}})$. Conditional on any acceptance pattern, the accepted values have independent target probability density function $h$; mixing over patterns retains this product law conditional on their count.

Write $C={\int }_{\mathbb{R}}f(x)dx$, so $0<C<\infty$. The finite envelope condition means that $f\le Mg$ almost everywhere; in particular, the proposal must cover the support of the target.

For each independent proposal $Y\sim g$, generate an independent $U\sim \mathrm{Unif}(0,1)$ and accept $Y$ precisely whenDiscard a rejected proposal and continue. The ratio is in $[0,1]$ and is needed only where $g(Y)>0$. Neither evaluation of $C$ nor prior knowledge of the normalized target is required.

To prove the output law, for any measurable set $D$ the joint probability of acceptance and a proposal in $D$ isDividing by the total acceptance probability $C/M$ gives ${\int }_{D}f(y)/Cdy$, the target probability. Independent proposal-uniform pairs give independent accepted observations. With a finite proposal list there can be no accepted value; with continued independent trials, acceptance eventually occurs almost surely because $C/M>0$.

Two reasons are selection bias and unequal ascertainment.

The one-day slaughter-inspection yield isor 500 detections per million inspected adult cattle. The routine field detections correspond, using a stable herd size of five million, toper adult-cattle day, or 8 detections per million adult-cattle days. Therefore the numerical one-day comparison isThe annual field rate is $14600/(5\times 10^6)=0.00292$ per cattle-year; comparing the inspection probability directly with this annual proportion would mix observation periods. Applying the random inspection-sample proportion to all 150000 slaughtered adults would predict about 75 detected cases in that slaughter cohort. These arithmetic comparisons describe detection yields, not a 62.5-fold biological disease-risk difference: a cross-sectional inspection estimates a prevalence yield, whereas daily field reporting is an ascertainment rate over animal-time.

Use a concurrent randomized comparison of the new leaflet with the existing release information. Random allocation makes the groups comparable in expectation and avoids attributing a secular change in drug supply, release populations or post-release services to the leaflet. Both arms should use the same outcome ascertainment and two-week follow-up. This improves the causal interpretation over a historical before-and-after comparison; adequate sample size remains necessary given the low statistical power calculated above. If contamination prevents individual allocation, a carefully designed cluster randomized controlled trial is an alternative, with its sample-size calculation allowing for within-cluster dependence.

Treat each release as an independent Bernoulli observation with a two-week event probability, giving two independent binomial distributions. Under the specified effect, the probabilities are $p_0=0.0016$ and $p_1=0.0008$ with $n=15000$ in each period. The expected event counts are therefore 24 and 12. For a conventional two-sided 5% comparison of proportions, let $D={\hat{p}}_0-{\hat{p}}_1$, $\delta =p_0-p_1=0.0008$ and $\bar{p}=(p_0+p_1)/2=0.0012$. UseThe approximate null rejection region is $|D|>1.96s_0$, and under the alternative $D$ is approximately $N(\delta ,s_1^2)$. Thus the power of a two-sample rare-event comparison isHere $\Phi$ is the standard normal distribution function. Hence the normal approximation gives $\boxed{\text{two-sided power approximately }52\%}$: detecting the proposed halving is far from assured.

For the small counts, a discrete calculation is also appropriate. Approximate the two counts by independent Poisson distributions with means 24 and 12. Conditional on their total $K$, the pre-intervention count is $\mathrm{Binomial}(K,1/2)$ under equal rates and $\mathrm{Binomial}(K,2/3)$ under the halving alternative; $K\sim \mathrm{Poisson}(36)$ under that alternative. If ${\mathcal{R}}_k$ is the rejection set of the symmetric exact two-sided 5% binomial test, its power isThis conservative discrete test has about 45% power, lower than the normal approximation because its achieved level can be below 5%. If a decrease-only one-sided 5% test had been specified in advance, replace 1.96 by 1.645; the normal-approximation power is about 64%. The tail convention and the chosen test must accompany any quoted statistical power.

Insert $f_i=h_i{F}_i$ and $F_i=e^{-H_i}$ into the survival likelihood. Each individual's log contribution simplifies toThereforeFailures contribute a log hazard function term as well as exposure, whereas both failures and censored subjects contribute the negative cumulative hazard function over their observed follow-up.

Assume independent individuals and independent censoring conditional on their explanatory variables, with the censoring mechanism carrying no parameter $\theta$. A failure contributes its failure-time probability density function $f_i(x_i;\theta )$, while a censored observation contributes the survival function $F_i(x_i;\theta )$ because its failure time exceeds $x_i$. Terms from the censoring distribution then factor out of the survival-parameter likelihood. Thus the survival likelihood and its logarithm areThe individual subscripts allow the same parameter vector to act through different covariate values.

The continuous failure-time probability density function is the negative derivative of the survival function:The minus sign is essential because survivor probability decreases with time.

Writing $F$ for the survival function here, the relation $F^{\prime }(t)=-h(t)F(t)$ with $F(0)=1$ givesThis notation uses $F$ for survival rather than for the cumulative distribution function.

For a continuous nonnegative event time with hazard function $h$, the cumulative hazard function isIts derivative is the instantaneous hazard wherever the derivative exists.

Let $S_A,S_B$ be the distinct group survival functions, and let $H_{\ast }$ denote the common pooled cumulative hazard function used for the residuals. At a point where its inverse is defined, residual survival in group $g$ isThus the group residual curves generally differ. On intervals where $S_A(t)>S_B(t)$, the same common increasing transformation gives $S_{u,A}(u)>S_{u,B}(u)$: the better-surviving group has more large residual failure times. If the original survival curves cross, the residual curves can also cross, so no fixed ordering or proportional hazards assumption is implied.

Distinct group distributions generally produce distinct residual survival curves, even when the pooled residual curve looks exponential. Separate correctly fitted group hazards would restore the unit-exponential target within each group, but that is a different transformation from the common fit under discussion.

If both groups share the same survival function and the common fit is adequate, their complete Cox–Snell residuals have the same unit exponential distribution. Under independent censoring in each group, both groupwise Kaplan–Meier estimators should therefore be close to $e^{-u}$ and to each other, subject to ordinary sampling variation and the number of subjects still at risk. Different censoring patterns can change precision without changing the residual survival target.

The two projective lines in the $i$th pair form the fibre of a morphism $f^{-1}([s_i:t_i])$ of the extended conic bundle from a line on a smooth cubic surface constructed above. Distinct points of the projective line have disjoint inverse images, soThe extension of $f$ across $l$ matters: arguing only that distinct planes through $l$ meet along $l$ would not by itself exclude intersections there. The formula $f{|}_l=[-B:A]$ excludes them as well. Lines within a single pair do intersect; the assertion concerns different pairs.