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.

Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic aswith homogeneous polynomials $Q_2,Q_3$ of degree two. Their restrictions $A=Q_2{|}_l$, $B=Q_3{|}_l$ have no common zero: at such a zero all first partial derivatives of $F$ would vanish. The rational projection $[x_2:x_3]$ therefore extends to a morphism of algebraic varietiesIndeed, wherever $Q_2\ne 0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne 0$.

For the plane with $x_2=sw$, $x_3=tw$, its plane section is $l$ together with the residual plane conicThese residual plane conics are exactly the fibres of a morphism of $f$, giving the conic bundle from a line on a smooth cubic surface. The family itself is smooth: off $l$ it is the graph of projection, and over $l$ the condition $sA+tB=0$ chooses the unique value $[-B:A]$.

Write $q_{s,t}=z^{\mathsf{T}}M(s,t)z$ with $z=(u,v,w{)}^{\mathsf{T}}$ and $M$ symmetric. The degrees of its entries in $(s,t)$ areThus the conic discriminant $\Delta (s,t)=\det M(s,t)$ is a homogeneous polynomial of degree five.

Smoothness also shows that every zero of $\Delta$ is simple and has matrix rank two. To see the first point, at a rank-two singular fibre let $z_0$ span the kernel and use a local base coordinate $\tau$. All fibre-coordinate derivatives vanish at $z_0$, so smoothness of the total family requires $z_0^{\mathsf{T}}{M}^{\prime }(\tau )z_0\ne 0$. Since the adjugate matrix of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0{z}_0^{\mathsf{T}}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf{T}}{M}^{\prime }z$ has a zero over the algebraically closed field; that point would be singular in the total family. This proves the reduced singular fibres of a smooth conic bundle assertion and also rules out $\Delta$ vanishing identically.

The discriminant quintic of a cubic surface conic bundle consequently has five distinct zeros in ${\mathbb{P}}^1$. At each, a rank-two plane conic splits into two distinct projective lines $l_i,l_i^{\prime }$. Neither equals $l$: that would require $sA+tB$ to vanish identically, making $A,B$ proportional, contrary to their having no common zero. The original plane section is therefore $l\cup l_i\cup l_i^{\prime }$, proving the required coplanarity and giving five distinct pairs.

Every unitary matrix is unitarily diagonalizable. Part (ii) therefore proves the stronger matrix statement $\varphi (A(t))=I$, not merely a statement about its eigenvalues. The kernel of a group homomorphism is a normal subgroup, and it contains every upper unipotent. By part (iii) their normal closure is the entire group. ThusThis proves the triviality of finite-dimensional unitary representations of SL2R. In fact the argument used only the algebraic homomorphism property and finite-dimensional unitarity; it did not need continuity.

Let $N$ be the normal closure of the upper unipotents $U(t)=A(t)$. Conjugation by $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ givesThus $N$ also contains every lower unipotent $L(s)=\left(\begin{array}{cc}1& 0\\ s& 1\end{array}\right)$. We prove explicitly that these elementary unipotent generators of SL2R generate the full group.

For $a\ne 0$, multiplication givesSo all determinant-one diagonal matrices lie in $N$. If $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ has $a\ne 0$ and $ad-bc=1$, thenbecause its lower-right entry is $(1+bc)/a=d$. If $a=0$, then $c\ne 0$ and left multiplication by $U(1)$ makes the upper-left entry equal to $c$. The resulting matrix lies in $N$ by the preceding factorization, and $U(1{)}^{-1}\in N$ restores $g$. Therefore

Put $U=\varphi (A(t))$. Applying the group homomorphism $\varphi$ to the identity just proved shows that $U$ is conjugate to $U^{m^2}$ for every positive integer $m$. Thus the multiset of its eigenvalues is unchanged by taking their $m^2$th powers.

Take $m=2$. The fourth-power operation permutes this finite multiset. For each eigenvalue $z$, iterating its permutation cycle gives $z^{4^r}=z$ for some positive integer $r$. Since $U$ is unitary, $z\ne 0$, so $z^{4^r-1}=1$: every eigenvalue is a root of unity. Choose a positive integer $d$ divisible by all their orders. Every eigenvalue of $U^{d^2}$ is then one. Because its eigenvalue multiset is also the original one, we obtainThis is square-power rigidity of a finite nonzero spectrum. Finiteness of the dimension is essential to the permutation-cycle argument.

Let $E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, so $A(t)=I+t{E}_{12}$ and $E_{12}^2=0$. Multiplication givesThe nilpotence also gives $(I+t{E}_{12}{)}^k=I+kt{E}_{12}$ for every nonnegative integer $k$, since all higher binomial terms vanish. Hence, for each positive integer $m$,These are unipotent matrices in SL2R. The integer is positive, since the displayed diagonal matrix is undefined at $m=0$.

For short and long simple roots $\alpha ,\beta$ of $G_2$, the long roots form an $A_2$ root system with simple roots ${\delta }_1=\beta$ and ${\delta }_2=3\alpha +\beta$. If ${\omega }_1,{\omega }_2$ are the corresponding $A_2$ fundamental weights, then $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$.

For $E:y^2=x^3-x-1$ over ${\mathbb{F}}_3$, the Trace of Frobenius is three andIn particular, $\#E({\mathbb{F}}_{3^r})=3^r+1$ exactly when $r\equiv 3(\mathrm{mod}6)$.

The effective conductance between two terminals is the reciprocal of their effective resistance. Under a unit voltage difference it equals the total current flowing from one terminal to the other.