The affine null cone is the set of vectors whose orbit closure contains the origin. For a reductive algebraic group it is also the common zero locus of all positive-degree homogeneous invariants. A nonzero vector in this set represents an unstable projective point. The Hilbert-Mumford criterion for the affine null cone detects membership by a algebraic one-parameter subgroup with only positive active weights.

A point is stable when a positive-degree homogeneous invariant is nonzero there and its orbit is closed in that invariant's nonvanishing affine open set, with finite stabilizer. This is stronger than merely having a closed orbit or a nonzero invariant. For binary forms of degree four, the stable points have four distinct projective roots.

For a complex reductive algebraic group acting linearly on $V$, decompose a nonzero vector into weights for each algebraic one-parameter subgroup. With the displayed sign convention, semistability is equivalent to $\mu \ge 0$ for every such subgroup, and stability is equivalent to strict positivity for every nontrivial subgroup. One must test conjugates of the subgroups as well as those diagonal in a single fixed coordinate system.

Independently retain each event of a zero-based geometric distribution count with probability $\alpha$. Conditional on the original count, the retained count is binomial, so its probability generating function is $p/[1-(1-p)(1-\alpha +\alpha z)]$. This is again a zero-based geometric distribution, with success parameter $p_{\alpha }$ and mean $\alpha (1-p)/p$. In excess of loss reinsurance, $\alpha$ is the probability that a claim exceeds the retention.

Let $i$ be a state outside an absorbing set $A$ of a Markov chain. Write $T_A$ for its first hitting time of $A$ and $T_i^{+}=\inf \{n\ge 1:X_n=i\}$. Assume each excursion ends: ${\mathbb{P}}_i(T_A\wedge T_i^{+}<\infty )=1$, and put $p={\mathbb{P}}_i(T_A<T_i^{+})>0$. Then the number $K$ of visits to $i$ before absorption, including the initial visit, has the positive-support geometric distributionAfter a return, the Strong Markov property restarts the same experiment. Thus $\mathbb{P}(K\ge k)=(1-p{)}^{k-1}$ and $\mathbb{E}K=1/p$.

The excursion-ending assumption matters: on an infinite state space, an excursion may escape forever without either absorption or return. In that case the geometric parameter for the number of visits is ${\mathbb{P}}_i(T_i^{+}=\infty )$, which need not equal the absorption-before-return probability.

For a continuous-time Markov chain, use the embedded jump chain and assume independent holding times at $i$ with exponential distribution of rate ${\lambda }_i>0$, independent of the jump destinations. The total occupation time before absorption is a geometric sum of exponential variables, with exponential distribution of rate $p{\lambda }_i$ and expected value $1/(p{\lambda }_i)$.

For the positive-integer geometric distribution with success probability $0<p\le 1$, the displayed identity holds for nonnegative integers $m,r$ whenever the conditioning event has positive probability. The first $m$ attempts have failed; future independent attempts use the same success probability. Equivalently, divide the tail $(1-p{)}^{m+r}$ by $(1-p{)}^m$. For a zero-based failures count $F=T-1$, use $\mathbb{P}(F\ge m+r\mid F\ge m)=\mathbb{P}(F\ge r)$. Constant success probability and independence are essential: accumulated wear or learning would require a different model.

The full remaining-count law is unchanged, not only one tail probability. With $0<p<1$, after $m$ failures set $R=T-m$ on the event $T>m$. For $r=1,2,\dots$, its probability mass function under this conditional probability isHere independence makes the future trials unaffected by the observed failures, and the unchanged success probability makes their law identical to the original one. Consequently $\mathbb{E}[T-m\mid T>m]=1/p$, while $\mathbb{E}[T\mid T>m]=m+1/p$. For the constructed $p=0.3$ mechanism, five failures leave a mean of $10/3$ more attempts, making the conditional mean total $25/3$; the five already spent are still counted.

The same mechanism has constant discrete hazard:Thus each reached attempt has the same chance of success. This differs from the decreasing unconditional probability $\mathbb{P}(T=k)$: later success requires surviving earlier attempts. At $p=1$, only conditioning on an event of positive probability is allowed: $T>m$ permits only $m=0$, and $T\ge k$ only $k=1$.

Among proper laws on the positive integers, memorylessness of the geometric distribution also characterizes the geometric distribution. Write $S(n)=\mathbb{P}(T>n)$. The identity gives $S(m+r)=S(m)S(r)$; if $S(m)=0$, both sides vanish anyway. Since $S(0)=1$, induction yields $S(n)=S(1{)}^n$. Properness excludes $S(1)=1$, so setting $p=1-S(1)>0$ recovers the geometric distribution, including $p=1$.^[ref]

The retry example stipulates independent Bernoulli trials with a known constant $p$; it is not evidence that a real retry system behaves this way. Shared changing conditions or adapting attempts can alter the next success probability. Even constant individual success probabilities alone do not imply independence.

For an original contrasting model, uniformly shuffle one success card and three failure cards, then draw by sampling without replacement. Its success position $U$ has the discrete uniform distribution on $\{1,2,3,4\}$. Initially the success probability is $1/4$; after one failure it is $1/3$. In particular,This shows concretely why sampling without replacement need not be independent and why the remaining wait can change. Every draw still has unconditional success probability $1/4$ when all four positions are considered; the relevant discrete hazard changes after failures.

If the oriented coordinate tangent vectors meet at a constant angle with cosine $k$, the first fundamental form has $F=k\sqrt{EG}$. The geodesic coordinate grid identities reduce to $k(\sqrt{E}{)}_v=(\sqrt{G}{)}_u$ and $k(\sqrt{G}{)}_u=(\sqrt{E}{)}_v$. Since positive definiteness gives $|k|<1$, both derivatives vanish. Thus $E=E(u)$ and $G=G(v)$ locally. Changing coordinates to $U=\int \sqrt{E}du$ and $V=\int \sqrt{G}dv$ gives the constant metric $d{U}^2+2kdUdV+d{V}^2$. It is locally Euclidean after a linear change of coordinates, hence has zero Gaussian curvature.

// Target: geometry-and-topology.bigb

An element $k$ of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$ is a generator of a group precisely when $\gcd (n,k)=1$. The order of an element of a finite cyclic group proves this immediately: its cyclic subgroup has size $n/\gcd (n,k)$, and this equals $n$ exactly when the greatest common divisor is one. There are therefore $\phi (n)$ choices, where $\phi$ is the Euler totient function.

For $C_6$, the two choices are $1$ and $5$, so $\phi (6)=2$. For $C_8$, they are $1,3,5,7$. Starting from zero and repeatedly adding $3$ modulo eight gives $0,3,6,1,4,7,2,5,0$, visiting every element once before returning. By contrast, adding $2$ gives $0,2,4,6,0$ and misses the odd integers represented in the other coset.

If the cyclic group is written multiplicatively as $⟨g⟩$ with $g$ of order $n$, the same statement says that $g^k$ is a generator of a group exactly when $k$ and $n$ are coprime. This changes notation without changing the calculation.

Assume $1\le t\le r\le n$, $n>2r-t$, and $j\ge 0$. A maximum-cardinality left-compressed set family of $r$-sets that is t-intersecting has a generating family for a uniform set family supported on $[t+2j]$ under the displayed strict inequality. Reflection gives a right-end supporting interval for a right-compressed optimum. The strict threshold is important: at a tie between consecutive candidate families, use the next larger support bound. This auxiliary generating result, combined with the compatibility bound for complementary generating families, supplies the upper bound in the Ahlswede-Khachatrian theorem.

Let $n>2r-t$ and let $G,H$ generate respectively a t-intersecting family of $r$-sets and its complement family. If their union had size at most $n-r+t-1$, contain it in a set $T$ of that size. Since $|T|\ge r,n-r$, extend $G,H$ within $T$ to an $r$-set $A$ and an $(n-r)$-set $B$. Both $A$ and $B^c$ belong to the original family, but $|A\cap B^c|=|A\cup B|-|B|\le t-1$. This contradiction proves the bound.

Suppose a left-compressed set family of $k$-sets has an antichain of generators on $[m]$, closed under left shifts followed by taking inclusion-minimal members. For a generator $G$ containing $m$, the members generated only by $G$ have intersection with $[m]$ exactly $G$. If another prefix coordinate $i$ occurred, the left shift $G-\{m\}+\{i\}$ would supply another generator. Conversely, an exact prefix $G$ contains no other generator by the antichain condition. The free tail gives the displayed binomial coefficient.

For a t-intersecting family with left-compressed generators supported on $[m]$, a tight pair containing $m$ cannot both miss an earlier coordinate $i$. Replacing $m$ by $i$ in one generator gives a shifted set containing a generator, whose intersection with the other has at most $t-1$ points. Thus the union is $[m]$, and their sizes sum to $m+t$. This identifies precisely which size classes conflict when the largest support coordinate is removed.

For a finite abelian group $G$, put $t_j=|\{x\in G:jx=0\}|$. If $|f_i|\le 1$, then $|{\mathbb{E}}_{x,d}{\prod }_{i=0}^3{f}_i(x+id)|\le (t_2{t}_3{)}^{1/8}{\min }_i\Vert f_i{\Vert }_{U^3}$. Three applications of Cauchy-Schwarz, eliminating the factors at slopes $3,2,1$, bound the eighth power by ${\mathbb{E}}_{h,k}|{\mathbb{E}}_x{\Delta }_{-3h}{\Delta }_{-2k}{f}_0(x){|}^2$, where ${\Delta }_{h}f(x)=f(x)\overline{f(x+h)}$. The integrand is nonnegative; averaging over $3G$ and $2G$ costs at most their indices $t_3$ and $t_2$ compared with the unrestricted average defining $\Vert f_0{\Vert }_{U^3}^8$. Reversal treats slope three. For slope one, elimination at slopes $3,0,2$ produces increments $-2h,k,-r$, costing only $t_2$; reversal treats slope two. In a cyclic group, $t_2\le 2,t_3\le 3$, so the constant is at most $6^{1/8}$ independently of group size.

A subject has a random intercept $B_i\sim N(0,{\tau }^2)$, with responses conditionally independent given $B_i$ and their covariates. Integrating the product of Bernoulli likelihoods over $B_i$ defines the joint subject likelihood and induces within-subject dependence. Its coefficients describe conditional odds ratios for a fixed latent subject effect. Under missing at random with distinct parameters, the observed-response likelihood integrates over the same random effect and sums out unobserved responses. Dependence of dropout on $B_i$ is not generally ignorable merely because conditional independence holds given $B_i$.

Inversion in a circle of centre $a$ and radius $R$ is an orientation-reversing involution of the Riemann sphere fixing that circle pointwise and exchanging its interior and exterior. It is an anti-Möbius transformation. Reflection in a line is the corresponding inversion for a generalized circle through infinity. A composition of two such inversions is a Möbius transformation.

An additive regression model represents a conditional mean as a sum of univariate functions and an intercept. It permits nonlinear effects without estimating a general high-dimensional response surface. The backfitting algorithm smooths each component's partial residual. Centering separates constants from the intercept, while concurvity can still prevent component identification.

// Target: probability-and-statistics.bigb

For an isotropic initial ensemble in Gaussian white-noise magnetic stretching, $F(B)=4\pi B^{2}P(B)$ is the magnitude probability density function. The radial vector equation $P_t=D(B^2{P}_{BB}+4B{P}_B)$ gives the displayed conservative Fokker-Planck equation. Its integral is conserved under zero endpoint probability flux. For an anisotropic initial vector distribution, use the angularly averaged magnitude density instead; isotropic noise alone does not make arbitrary initial data isotropic.

Twice integrating the radial magnetic-field Fokker-Planck equation by parts gives the displayed law when moments and endpoint terms are controlled. In particular $⟨B^2⟩$ grows at rate $10D=2\gamma$, with $\gamma =5{\kappa }_2/4$. The corresponding magnetic energy density is $⟨B^2⟩/(8\pi )$ in Gaussian units.

Set $x=\ln (B/B_0)$ and $f=BF$ in the radial magnetic-field Fokker-Planck equation. Then $f_t=D{f}_{xx}-3D{f}_x$, so $f$ is a Gaussian distribution drifting at speed $3D$. For initial magnitude $B_0>0$, $F=[B\sqrt{4\pi Dt}{]}^{-1}\exp [-(\ln (B/B_0)-3Dt{)}^2/(4Dt)]$. This lognormal distribution has moments $B_0^q{e}^{Dq(q+3)t}$; the width of its logarithm grows as $\sqrt{2Dt}$.

The lattice theta series of a full-rank Euclidean lattice records the multiset of squared lengths of lattice vectors. The Gaussian form shown here converges for $t>0$. For a flat torus ${\mathbb{R}}^d/\Lambda$, its heat trace is ${\Theta }_{{\Lambda }^{\ast }}(4\pi t)$, by the spectrum of a flat torus. Equal theta series do not in general imply orthogonal equivalence of lattices.