For mutually exclusive event types, let $T$ be the first event time and $J$ its type. The cause-specific hazard for cause $j$ isIt is a rate conditional on having had no event of any cause. The cumulative incidence function, also called the cumulative risk function, is the actual probability $F_j(t)=P(T\le t,J=j)$.

The total hazard function is $h(t)={\sum }_k{h}_k(t)$, giving $S(t)=\exp \{-{\int }_0^t{\sum }_k{h}_k(u)du\}$. Surviving every cause to time $u$ and then experiencing cause $j$ givesIn competing risks, $F_j$ is generally not $1-e^{-{\int }_0^t{h}_j}$, because competing events remove individuals before they can experience cause $j$. This formula does not require assuming independent latent failure times for the different causes.

For an event at $t_j$ from subject $i_j$, with covariate vector $z_i$, define the Schoenfeld functionThe second term is the hazard-weighted mean covariate in the risk set just before the event. The Schoenfeld residual is this function evaluated at the fitted coefficient, $r_j=s_j(\hat{\beta })$. Calculate one residual vector per event, using every at-risk subject, including those who will subsequently be censored. There is no ordinary event residual assigned at a right censoring time.

The Cox partial likelihood score function is ${\sum }_j{s}_j(\beta )$. At the true constant coefficient in a Cox proportional-hazards model, the conditional event subject is selected with weights proportional to $e^{{\beta }^T{z}_i}$, so each Schoenfeld function has conditional mean zero. If the coefficient varies with time, that centering changes. Plot residuals against event time or a transformation of it, smooth them, and investigate departures from zero. Scaled Schoenfeld residuals account for the risk-set covariate variance and can display departures in coefficient units; score function tests based on residual-time association provide a formal check. Risk-set composition affects unscaled residual variance, and the total residual score function can be zero by fitting even when a time trend is present.

Write the year-$j$ count as a sum of its $m_j$ individual counts, with $m_j>0$. The specified conditional independence within that year givesDividing by $m_j$ and scaling the conditional variance by $m_j^{-2}$ therefore givesThis is the Bühlmann–Straub model: larger exposures reduce the process noise in a year's average.

Define the population parameters and total observed exposure byHere $v$ is the expected process variance and $a$ the variance of hypothetical means. Assume finite second moments. The law of total variance gives $\mathrm{Var}(X_j)=a+v/m_j$, and conditional independence between years gives $\mathrm{Cov}(X_j,X_k)=a$ for $j\ne k$: their only shared variation is the latent conditional mean.

To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of $\mu (\Theta )$. For coefficients $b_j$ and $B={\sum }_j{b}_j$, the optimal intercept is $m_0(1-B)$. Centering at $m_0$ and conditioning on $\Theta$ shows that the resulting error isThe cross term vanishes because $X_j-\mu (\Theta )$ has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.

For fixed $B$, the Cauchy-Schwarz inequality giveswith equality when $b_j=B{m}_j/W$. Minimize the remaining quadratic $a(1-B{)}^2+(v/W)B^2$. Its minimizing value is the Bühlmann–Straub credibility factorThe experience term is exposure-weighted:Since the future conditional mean count is $m_{n+1}\mu (\Theta )$, multiplying the optimal estimate by the known future exposure givesIt is the best affine linear least-squares projection, not a claim that the exact conditional expectation given all data is always affine. If $a=0$, the conditional mean is a known constant almost surely and one takes $Z=0$; if $v=0<a$, the observations reveal it without process noise and $Z=1$. The ordinary case has $a,v>0$.

The complex tautological line bundle over Complex projective space isOn the standard chart $U_i=\{[z_0:\cdots :z_n]:z_i\ne 0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The mapis a local trivialization; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial complex line bundle.

Its unit sphere bundle is $S^{2n+1}$: a unit vector $v\in {\mathbb{C}}^{n+1}$ corresponds to $([v],v)$. The projection is the Hopf fibration. Regard $\lambda$ as an oriented real rank-two vector bundle using its complex orientation, and put $t=e(\lambda )\in H^2({\mathbb{CP}}^n;\mathbb{Z})$. The Gysin sequence of a sphere bundle gives, for $2\le q\le 2n$,as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1({\mathbb{CP}}^n)=0$. The CW complex structure has one cell in dimensions $0,2,\dots ,2n$ and none above $2n$, so these groups and products giveThe Euler class of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb{Z}$.

Now let $\alpha$ generate $H^2(S^2;\mathbb{Z})$ and let $u_i$ be its pullback from the $i$th factor of $(S^2{)}^n$. The Künneth theorem gives $H^2((S^2{)}^n;\mathbb{Z})={⨁}_i\mathbb{Z}{u}_i$. If a map $f$ were invariant under all factor permutations, write $f^{\ast }\alpha ={\sum }_i{a}_i{u}_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta :S^2\to (S^2{)}^n$ be the diagonal. ThenThe second condition is $f\circ \Delta =\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. For $n>1$ this is impossible in $\mathbb{Z}$, and $\boxed{\text{no such continuous map exists}.}$ This is the diagonal-degree obstruction to a symmetric sphere retraction.

A metric space is complete if every Cauchy sequence in it converges to a point of that space. The stated shrinking-set characterization is the Cantor intersection theorem together with its converse.

First assume completeness, and choose $x_n\in A_n$. If $m,n\ge N$, then both points belong to $A_N$, so $d(x_m,x_n)\le \mathrm{diam}(A_N)$. The vanishing diameters imply that $(x_n)$ is a Cauchy sequence, hence it converges to some $x\in X$. For every fixed $N$, its tail belongs to the closed set $A_N$, so $x\in A_N$. Thus $x\in {\bigcap }_N{A}_N$. In fact the intersection contains only one point, since any two of its points have distance at most every diameter.

Conversely assume the stated intersection property and let $(x_n)$ be a Cauchy sequence. PutThese closures are nonempty, closed and nested. Taking a closure does not increase the diameter: approximate two points of the closure by points of the set and use continuity of distance. The Cauchy sequence therefore makes $\mathrm{diam}(A_N)\to 0$. Choose $x$ in the intersection. For $n\ge N$, both $x_n$ and $x$ belong to $A_N$, and $d(x_n,x)\le \mathrm{diam}(A_N)\to 0$. Every Cauchy sequence converges in $X$, proving completeness.

Replace the autocovariance in the finite-lag linear prediction equations by the sample autocovariance function. The last regression coefficient at order $k$ estimates the lag-$k$ partial autocorrelation function.

For $\dot{A}=D^{2}f(t)B$ and $\dot{B}=iA$, with $f$ alternating between $+1$ and $-1$ for intervals of length $T$, the monodromy matrix of a periodic linear system has determinant one and trace $\cosh (\sqrt{2}DT)+\cos (\sqrt{2}DT)>2$. Its reciprocal Floquet multipliers give positive dominant Floquet growth rate for $D,T>0$, despite the zero mean of $f$. An exceptional initial state in the stable eigenspace decays.

A parallel frame along a curve consists of a basis of vector fields $E_a$ satisfying $D_t{E}_a=0$ for its covariant derivative along a curve. Transporting a basis at the initial point constructs the frame. If the connection preserves the metric, an initially orthonormal basis stays orthonormal. In this frame the Jacobi equation becomes a linear matrix ordinary differential equation.

Along a smooth curve $\gamma :I\to M$, a horizontal lift of a linear connection on a vector bundle has fibre equation $v^{\prime }=-\Gamma (\dot{\gamma })v$. On each compact time interval the coefficient matrix is bounded, and the Gronwall inequality prevents finite-time blowup. A finite cover by vector bundle trivializations and ODE uniqueness patch local solutions. Exhausting the open interval gives a unique lift on all of $I$. Its endpoint map is linear, and reversal of the path gives its inverse; no completeness of the base manifold is needed.

Jointly time-change the radial and area martingales by the inverse of $A=\int R^{2}ds$. The bracket matrix becomes $u{I}_2$, so the Lévy characterization of multidimensional Brownian motion gives two independent Brownian coordinates. The clock is strictly increasing; if it were finite at infinity, the radial martingale would converge and $R_t^2=2t+O(1)$ would force its integral to diverge. Thus $R_t^2=2W_{A(t)}+2t$ and $Z_t=B_{A(t)}$ with $W,B$ independent.

Let $M_1,\dots ,M_N$ be continuous local martingales starting at zero, with $⟨M_i,M_j⟩=0$ for $i\ne j$ and $⟨M_j{⟩}_{\infty }=\infty$ almost surely. The individual Dambis-Dubins-Schwarz theorem time changes produce independent standard Brownian motions $B_j$ with $M_j(t)=B_j(⟨M_j{⟩}_t)$. Independence is between the Brownian motions; each may still depend on its own clock.

Equating the optically thin cooling time to the uniform-sphere free-fall time gives a critical gas density proportional to $T^2/{\Lambda }^2$. With hydrogen mass fraction $X$, particle ratio $\chi$ and gravitating gas fraction $f_g$, it is $[32G{m}_p/(3\pi X{f}_g)][3\chi k_{B}T/(2\Lambda ){]}^2$. At fixed temperature, gas above this density cools faster than free fall. A smaller gravitating gas fraction moves the boundary upward because gravity is faster at a fixed gas density.

The one-triangle adjunction idempotent splits if and only if $G$ has a left adjoint. From a splitting $F\stackrel{r}{\to }H\stackrel{s}{\to }F$, define the new unit $Gr\eta$ and counit $\epsilon s_G$; the absorption identities imply both triangle identities for an adjunction. Conversely, for $H⊣G$ with unit $\psi :1\Rightarrow GH$ and counit $\phi :HG\Rightarrow 1$, the splitting maps are $r={\epsilon }_{H}F\psi$ and $s={\phi }_{F}H\eta$. Their composites are $rs=1_H$ and $sr=e$.

For a planar nematic director $\mathbf{n}=(\cos \chi ,\sin \chi )$ and constant amplitude ${\lambda }_0$, the nematic order parameter isDifferentiation gives $\nabla \cdot Q={\lambda }_0(-\sin 2\chi {\chi }_x+\cos 2\chi {\chi }_y,\cos 2\chi {\chi }_x+\sin 2\chi {\chi }_y)$. The coefficient matrix is orthogonal, proving the identity. Hence $K|\nabla \cdot Q{|}^2/2$ is the angle-gradient energy with effective modulus $K{\lambda }_0^2$. The full tensor-gradient norm instead obeys ${\sum }_{ijk}({\partial }_k{Q}_{ij}{)}^2=2{\lambda }_0^2|\nabla \chi {|}^2$ in this normalization; the two local densities must not be confused.