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Write $H={\int }_{\mathbb{R}}h(x)dx$, and assume $0<H<\infty$; a uniform distribution on the region requires positive area. Set $x=V/U$, $s=U$, so the inverse transformation is $U=s$, $V=sx$. Its Jacobian determinant has absolute value $s$ for $s>0$. The region becomes $x\in \mathbb{R}$, $0<s\le \sqrt{h(x)}$, and its area isThe original joint probability density function is therefore $2/H$ on $C_h$. The change of variables formula givesThe boundary $U=0$ has probability zero, so division by $U$ causes no ambiguity for the resulting random variable.

Independently generate $U$ uniform on $(0,1)$ and $V$ uniform on $(b_1,b_2)$, using the bounds above. Accept the proposal ifotherwise generate a fresh pair. On acceptance return $X=V/U$. The bounding rectangle contains the entire ratio region: $U\le \sqrt{h(X)}\le a$, and $V=XU$ lies between the extrema of $x\sqrt{h(x)}$. Conditioning a uniform rectangle point on acceptance makes it uniform on $C_h$, so the preceding marginal distribution calculation proves that the returned value has the desired Laplace distribution.

For this target $H=4$, so the region has area two. The rejection sampling acceptance probability isEndpoints have zero probability and can be excluded to avoid divisions by zero in implementation.

Take the indicator function $\theta (x)=1_{\{x\ge k\}}$. Its expectation under the standard Cauchy distribution is the desired upper-tail probability. To use the proposed tail probability density function in the remaining parts, assume $k>0$: a probability density function proportional to $x^{-2}$ on $[k,\infty )$ is not normalizable when that interval contains zero.

For $X_i=k/U_i$, the indicator is always one, and the importance weight reduces toAveraging these independent contributions givesAs a direct normalization check, its expectation is ${\int }_0^{1}k/[\pi (u^2+k^2)]du={\pi }^{-1}\arctan (1/k)$. For $k>0$ this equals $1/2-{\pi }^{-1}\arctan k$, the exact standard Cauchy distribution tail probability. The substitution $X=k/U$, rather than $k/(1-U)$ with the same symbol left unchanged, explains why the displayed denominator contains $U^2$.

The order prior and probabilities of choosing upward or downward moves are not specified, so include them explicitly. Let ${\rho }_k$ be the model-order prior probability, $b_k$ the birth-selection probability at order $k$, and $d_{k+1}$ the reverse death-selection probability. Denote the correctly normalized coefficient prior by $p_k(a)$ and the probability density function of $z\sim N(0,{\sigma }_a^2)$ by $q(z)$. Common variance-prior factors cancel since $v={\sigma }^2$ is unchanged.

Put $m=n^{-1}{\sum }_i{x}_i^{k+1}$, $r=y-X_{k}a$, and $w_i=x_i^{k+1}-m$. The centered polynomial birth move in reversible-jump sampling gives $X_{k+1}{a}^{\prime }=X_{k}a+zw$, so the residual-sum-of-squares change isThe map $(a_0,a_1,\dots ,a_k,z)\mapsto (a_0-mz,a_1,\dots ,a_k,z)$ is triangular with determinant one. Thus the reversible-jump Markov chain Monte Carlo ratio isFor an entirely explicit Gaussian expression, define $Q_k=(a-{\mu }_k{)}^T{\Sigma }_k^{-1}(a-{\mu }_k)$ and $Q_{k+1}=(a^{\prime }-{\mu }_{k+1}{)}^T{\Sigma }_{k+1}^{-1}(a^{\prime }-{\mu }_{k+1})$. Dividing the two normalized Gaussian priors and the proposal probability density function givesEqual model-order priors and equal birth/death selection probabilities remove the first ratio, if that is the intended convention. Dropping the order-dependent normal-prior constants instead would give the wrong acceptance rule.

At temperature $T>0$, target a probability density function proportional to $\exp (f(\theta )/T)$ on a parameter region where it is normalizable, possibly with a specified proper base measure. A Metropolis–Hastings algorithm proposal $q(\theta ,{\theta }^{\prime })$ has acceptance probabilityFor a symmetric proposal, improving moves are accepted and a loss $\Delta$ is accepted with probability $e^{-\Delta /T}$. Run updates while gradually lowering $T$ toward zero. This simulated annealing scheme initially permits exploration and increasingly favors high objective values. Appropriate cooling and mixing conditions are needed for a general global convergence theorem; a finite arbitrary schedule does not guarantee finding a global maximum.

To maximize a positive likelihood function, take $f=\log L$, which has the same maximizers. The resulting likelihood-power annealing target is $L^{1/T}$, and available Gibbs sampler conditional distributions can replace Metropolis proposals.

The E-step uses the posterior class probabilitiesLet $n_j^{(m)}={\sum }_i{w}_{ij}^{(m)}$, so $n_1^{(m)}+n_2^{(m)}=n$. Up to constants, the expected complete-data log-likelihood isMaximizing the first term with ${\alpha }_1+{\alpha }_2=1$ gives ${\alpha }^{(m+1)}=n_1^{(m)}/n$. Differentiating in each ${\mu }_j$ gives ${\mu }_j^{(m+1)}={\sum }_i{w}_{ij}^{(m)}{x}_i/n_j^{(m)}$. After these updates, maximizing in the shared variance givesThe residuals use the new component means, while the weights remain from the old parameters. The divisor is $n$, not $n-2$, because this is a maximum-likelihood update. Starting with $0<\alpha <1$ and $v>0$ gives positive responsibilities and positive $n_j$; a zero effective component size at a boundary would leave its mean unidentified. These are the EM for Gaussian mixtures with a common variance transitions.

Put ${\alpha }_1=\alpha$, ${\alpha }_2=1-\alpha$, and introduce a categorical class label for each observation, encoded by $Z_{i1},Z_{i2}\in \{0,1\}$ with $Z_{i1}+Z_{i2}=1$. Its prior class probability is ${\alpha }_j$, and conditional on class $j$ the observation probability density function is $f_j(x_i)$. Thus its joint observation-label probability density function isIndependence across observations yields the finite Gaussian mixture with a common variance complete-data likelihood functionSumming over the two possible labels at each observation recovers ${\prod }_i[\alpha f_1(x_i)+(1-\alpha )f_2(x_i)]$, the observed-data likelihood function. The product requested in the PDF is therefore a joint complete-data likelihood function. Strictly conditioning on $Z$ would remove the factors ${\alpha }_j$; the PDF's conditional notation for this joint product is imprecise.

A system containing a red giant and another evolved component is either a giant–giant pair or an unordered giant–white dwarf pair. At common age $y$, these disjoint possibilities have conditional probabilities $g(y{)}^2$ and $2g(y)w(y)$. The conditional evolved-companion fraction in a coeval binary population therefore has numeratorDivide by the fraction $299/30000$ of systems containing at least one giant:Thus just over $3\%$ of giant-containing binaries have an evolved companion. The $0.03\%$ numerator is a fraction of all binaries; it is not itself the requested conditional fraction.

Condition on the common age $Y=y$ of the coeval binary population. Each component is a red giant with conditional probability $g(y)=y/100$, independently of the other component at that age. The conditional probability of at least one giant is $1-(1-g{)}^2=2g-g^2$. Integrating over the uniform age distribution givesThus the fraction is $299/30000=0.996\overline{6}\%$, just below $1\%$. The subtracted term removes the systems counted twice because both components are red giants. Using the square of the marginal giant fraction would incorrectly ignore their shared age.

A constant number star formation rate over $10\mathrm{Gyr}$ produces a uniform present-age distribution. The birth-age distribution under constant star formation therefore givesOutside this age interval the cumulative distribution function is zero or one. The census here retains stellar remnants as objects, as required by the population model.

Write $m=M/M_{\odot }$. The initial mass function is proportional to $m^{-3}$ on $m\ge 0.2$. Normalizing the tail integral gives the power-law initial-mass-function tail coordinate:For lower thresholds $X=1$. Because the birth mass and formation time have independence, the probability integral transform makes $(X,Y)$ uniform on the unit square. $X$ is an upper-tail coordinate, so larger masses have smaller $X$.

A red giant of current age $t$ lies in the stellar lifetime regions in a mass-age diagramThere are no red giants below $m=\sqrt{0.8}$ in this model. For $\sqrt{0.8}<m<1$ the upper boundary is the population's age limit, while for $m\ge 1$ both lifetime boundaries are present. A white dwarf lies above $t/\mathrm{Gyr}=10/m^2$.

In uniform probability coordinates the main-sequence boundary is $Y=20X$ and the end of the giant phase is $Y=25X$. The giant region is therefore $20X\le Y<25X$, clipped to the unit square: it is the triangle with vertices $(0,0),(1/25,1),(1/20,1)$. Its area, and hence the individual giant fraction, isThe white dwarf region $Y\ge 25X$ is the triangle with vertices $(0,0),(0,1),(1/25,1)$, givingThe remaining $97.5\%$ are on the main sequence. Boundaries have zero probability and do not affect these counts. Treating every post-giant object as a white dwarf is a stipulated toy-model assumption; real high-mass stars have other stellar remnants.

For the binary calculations, use a coeval binary population: components share their birth time even though their birth masses are independently drawn and their subsequent evolution is independent. Hence $(X_1,X_2,Y)$ is uniform in the unit cube. At a fixed age coordinate $Y=y$, a component is a giant when $y/25<X<y/20$, and a white dwarf when $0<X<y/25$. Define the conditional state fractionsComponents have conditional independence given $y$; averaging over the common age must be done after forming the conditional pair probabilities. The continuous-birth binary state fractions below are consequently volumes in this cube, not products of the marginal single-star fractions.

Let $a,b,e$ be fractions of patches occupied by the two competitors and empty, with $a+b+e=1$. Assume well-mixed propagule dispersal, equal independent fire extinction rate $m>0$, colonization constants $c_A>c_B>0$, and net replacement advantage $q=k_B-k_A>0$ for the poorer colonizer. The Levins metapopulation model with replacement isReplacement fluxes $k_{B}ab$ and $k_{A}ab$ cancel in the total occupancy equation. At every face the outward flux is zero or points into the patch probability simplex, so these equations preserve valid fractions. The constants include both propagule arrival and establishment probability; the model does not require unequal fire susceptibility.

Put $\Delta c=c_A-c_B$. Solving the two occupied-patch balances and their sum constraint givesThus strict coexistence is equivalent toToo little disturbance lets the better competitor exclude the colonizer; too much disturbance favors colonization or destroys occupancy. A positive equilibrium automatically has $m<c_A$. If the poorer colonizer cannot survive alone, coexistence can nevertheless be possible because it repeatedly replaces established patches of the better colonizer.

These bounds also have an invasion interpretation. The $A$-only equilibrium is $a=1-m/c_A$ when $c_A>m$; the growth rate of rare $B$ there is $q-m(q+\Delta c)/c_A$, positive at the upper-bound condition above. At a feasible $B$-only equilibrium the rare-$A$ growth is $-q+m(q+\Delta c)/c_B$, giving the lower bound. At coexistence the occupied-patch Jacobian matrix has trace $-c_A{a}_{\ast }-c_B{b}_{\ast }<0$ and determinant $a_{\ast }{b}_{\ast }q(q+\Delta c)>0$, proving local asymptotic stability. The Bendixson-Dulac criterion with multiplier $1/(ab)$ gives divergence $-c_A/b-c_B/a<0$ in the interior, excluding periodic competition cycles.

To specify selection, a monotone tradeoff alone is insufficient: one must say how two different trait values determine replacement. A natural model uses a trait $x$ with increasing competitive ability $B(x)$ and decreasing colonization $c(x)$, and net replacement $B(y)-B(x)$ by a mutant $y$ against a resident $x$. The resident metapopulation occupies $1-m/c(x)$, requiring $c(x)>m$. The rare-mutant per-capita growth, or invasion fitness, isIt is zero at $y=x$. Differentiating with respect to the mutant trait gives the selective pressureHigher colonization is valuable in empty habitat; higher competitive ability is valuable where the resident already occupies patches. This is colonization-competition tradeoff invasion fitness.

At an interior candidate for a local evolutionarily stable strategy, $g(x_{\ast })=0$. A strict, nondegenerate local evolutionarily stable strategy follows ifsince nearby distinct mutants then have negative invasion fitness. For a strict global evolutionarily stable strategy, every feasible distinct mutant must satisfyThis is the supporting-line condition on the tradeoff curve. At a trait boundary replace the vanishing-gradient condition by the appropriate one-sided inequality. Equality against another mutant, or zero local curvature, needs the second condition of evolutionary stability or higher-order terms; the strict tests above are sufficient and are the usual nondegenerate conditions, not a claim that every tie is unstable.

For evolutionary approach to a singular point, the selection gradient must also point toward it, locally $g^{\prime }(x_{\ast })<0$. This is logically distinct from testing mutant fitness. In this particular additive-ability model with $c^{\prime }<0<B^{\prime }$, negative mutant curvature implies negative $g^{\prime }$ becauseIf replacement instead has a general kernel $\kappa (y,x)$, replace $B(y)-B(x)$ in $f$ by that kernel and differentiate it; monotonicity does not justify importing the displayed additive formula without that modeling assumption. The tradeoff shape, disturbance rate and replacement law together determine selection and local or global uninvadability.

Take “cousins” to mean first cousins with unrelated, outbred founders and unrelated spouses in the intervening generation. Interpret a homozygous chromosome as two copies identical by descent, and first use an idealization of independent, nonrecombining chromosome blocks. Ordinary sequence homozygosity is a different quantity: unrelated copies can have the same allele, so its probability also requires allele frequencies. Similarly, identity of an entire recombining chromosome requires a genetic map, not just a pedigree.

At a single autosomal locus, the probability of autozygosity in their child is the parents' kinship coefficient. Each of the two common grandparents contributes a path of length two from each parent, with probability $(1/2{)}^{2+2+1}$. Thus the first-cousin inbreeding coefficient isIf there are $L$ independent autosomal blocks, the probability that none is autozygous is $(1-F{)}^L$. Hence the usual 23-block idealization givesThe exponent describes the stated independent-block approximation; it is not a claim that all 23 human chromosome pairs inherit like autosomes.

Humans have 22 autosomal pairs. A son has one X and one Y and cannot have two homologous X copies. A daughter can have an autozygous X, with probability $F_X$ determined by the sex-linked cousin kinship calculation in part (b). Thus a sex-aware version iswhere $p_D$ is the probability of a daughter, or is one or zero for a specified daughter's or son's sex. The independent assortment approximation applies between the autosomes and the X. Opposite-sex parentage alone does not determine $F_X$: the parents may be cousins through their mothers, their fathers or one of each. Without these pedigree and chromosome-block assumptions, the wording does not specify a unique probability of an entire homozygous human chromosome.

Use the conventional additive pedigree relatedness $r_{UV}=2{\varphi }_{UV}$, where ${\varphi }_{UV}$ is the kinship coefficient, the probability that randomly selected gene copies from $U$ and $V$ are identical by descent. For noninbred individuals this convention gives ordinary full siblings relatedness $1/2$, as required. It is important that this is not the probability of sharing at least one copy: that probability is $3/4$ for ordinary full siblings.

Let the parents be $M,F$. They are otherwise outbred first cousins, so ${\varphi }_{MF}=1/16$ and ${\varphi }_{MM}={\varphi }_{FF}=1/2$. For two children $C,D$, independent Mendelian segregation givesTherefore the autosomal additive relatedness isThe additional $1/16$ beyond $1/2$ comes from maternal-paternal cross-origin matches. Each child has inbreeding coefficient $1/16$. The additive coefficient is a sum of descent probabilities rather than a literal bounded probability in all inbred pedigrees; a relationship normalized to self-relatedness would be $r_{CD}/(1+F)=9/17$. Likewise, identity-by-descent sharing of inbred full siblings gives a different matched-copy statistic. The exam's calibration to $1/2$ supports the additive convention, but the quoted probability wording needs this qualification.

For sex chromosomes the route of cousinship matters, not merely the fact that the parents have opposite sex. Let $P_M$ be the mother-cousin's parent in the shared sibling pair, and $P_F$ the father-cousin's parent in that pair. Let $F_X$ be the probability that the mother's randomly transmitted X matches the father's single X by descent. With no founder inbreeding, the possibilities areIn the first two rows the father-cousin is related through his father, whose X he does not inherit. In the third row his mother is a sister of the mother-cousin's father: the relevant brother and sister match their maternal X copies with probability $1/4$, and the mother-cousin transmits her paternal X with probability $1/2$, giving $1/8$. In the last row the two sisters' random X gametes match with probability $3/8$: their common father's X contributes $1/4$, and their common mother's two X copies contribute $1/8$. Transmission by the mother-cousin halves this to $3/16$. This derives the sex-linked cousin kinship values.

For an unweighted comparison by chromosome pair, two daughters have sex-pair additive relatedness $3/4+F_X$: their paternal X always matches, maternal X matches with probability $1/2$, and the two cross-origin terms each contribute $F_X$. Two sons have mean sex-pair relatedness $3/4$, from an X match of probability $1/2$ and a common paternal Y with probability one. A daughter-son pair has sex-pair value $1/4+F_X/2$, with the Y contribution zero. These calculations treat chromosomes as unrecombined blocks and weight X and Y equally; chromosome-length-weighted or locus-specific coefficients are different.

For two children with independent equally likely sexes, these three cases have weights $1/4,1/4,1/2$. Their mean sex-pair relatedness is $1/2+F_X/2$, so averaging the 22 autosomal pairs and the sex pair givesIf the cousins' route has $F_X=0$, this is $103/184$, slightly below $9/16$. If, as an additional sampling assumption, the four parent-sex routes in the table are equally likely, $\mathbb{E}{F}_X=5/64$ and the answer is $1653/2944$. These are different legitimate numerical corrections under different specified pedigrees. There is no unique sex-chromosome correction from opposite-sex cousinhood alone. The table and formula give it once the missing pedigree and offspring-sex information is supplied.

Assume a closed population with even size $N$, unrelated founders, monogamous mating, nonoverlapping generations and exactly two offspring per couple. Sex labels are exchangeable for the autosomal calculation, with enough individuals of each sex to form the stipulated pairs. A randomly chosen distinct offspring pair are siblings with probability $1/(N-1)$: among the $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ pairs, exactly $N/2$ are within-family pairs. This fixed-family-size fact, rather than independent sampling of all parents, determines the recurrence.

First give the elementary chromosome-sharing idealization consistent with taking self-relatedness as one and ordinary sibling relatedness as $1/2$. In this model each individual carries one ancestral label, passed through one randomly chosen parent. Let $r_G$ be mean sharing between distinct individuals, and let $m_G$ be the mean sharing between mates. Siblings share their parental label with probability $(1+m_G)/2$. The average relatedness between parents from different couples isTo see the latter count, the ordered pairs from different couples are all ordered distinct pairs minus the $N$ ordered mating pairs. Each pair of parents from different couples contributes equally to the four possible cross-family offspring relationships. Combining sibling and nonsibling offspring givesWith random mating, $m_G=r_G$ in expectation, so $r_0=0$ givesThe mean sharing tends to one, on a generation scale of order $2(N-1)$. It is not linear indefinitely: repeated ancestry saturates the probability. This is an exact result for the stated one-label approximation, not an exact diploid formula after substantial inbreeding has developed.

Now prohibit full-sibling matings, with $N\ge 4$ and representative sampling of allowed nonsibling mating pairs. The following closed recurrences assume that opposite-sex mating pairs have the same mean autosomal sharing as all nonsibling pairs; a sex-specific or assortative scheme requires separate mating means. In the exchangeable one-label model write $u_G$ for mean relatedness of a nonsibling pair, so $m_G=u_G$. The two required recurrences areThe second relation includes sharing through grandparents and more distant ancestors; setting $u_G=0$ forever would be wrong. For $v_G=1-r_G$, $w_G=1-u_G$, the linear recurrence matrix isIts eigenvalues areboth of magnitude less than one for $N\ge 4$. Thus $r_G\to 1$ even when siblings never mate. Explicitly, $1-r_G=A{\lambda }_{+}^G+(1-A){\lambda }_{-}^G$, where $A=[1-1/(2(N-1))-{\lambda }_{-}]/({\lambda }_{+}-{\lambda }_{-})$. Sibling avoidance prevents immediate close-relative matings, but it does not make the remaining population unrelated. In a fixed-two-offspring model it need not reduce average pairwise sharing at every generation: a smaller mating mean actually increases the displayed increment in $r_G$.

For a fully diploid formulation, self-kinship also changes with inbreeding and cannot remain artificially fixed. Let ${\Phi }_G$ be mean kinship coefficient between distinct individuals, $F_G$ mean inbreeding coefficient, and ${\Phi }_{m,G}$ mean kinship of mates. A sibling pair has kinship $(1+F_G)/4+{\Phi }_{m,G}/2$, because parental self-kinship is $(1+F_G)/2$. Repeating the pair count yields finite-population chromosomal coancestry:Initially ${\Phi }_0=F_0=0$. For random mating take ${\Phi }_{m,G}={\Phi }_G$. An explicit solution is the affine matrix recurrenceIts characteristic polynomial is ${\lambda }^2-[1-1/(2(N-1))]\lambda -1/[4(N-1)]$; its two roots have magnitude less than one. Hence ${\Phi }_G,F_G\to 1$. The normalized relationship $2{\Phi }_G/(1+F_G)$ tends to one, while unnormalized additive relationship $2{\Phi }_G$ tends to two, demonstrating why that coefficient cannot always be read literally as a probability.

With sibling avoidance replace ${\Phi }_{m,G}$ by the nonsibling mean $U_G$ and addThese equations compute every generation's kinship and inbreeding, including delayed consanguinity through cousins. Under random allowed mating in a connected closed population, backward ancestral gene-copy lineages have a positive chance to meet in each sufficiently long block of generations; repeated blocks force eventual coalescence. Thus sibling avoidance alone does not prevent long-term identity by descent and genetic drift. Maintaining distinct isolated breeding groups, introducing immigrants, or changing offspring-number variance gives a different model. The elementary exponential formula answers the fixed-self-sharing approximation; the diploid recurrences give the consistent inbreeding-aware answer.

Write $K_s(t)$ for a balanced complete multipartite blow-up with $s$ classes of size $t$. In the nonvacuous range $0<\epsilon <1/r$, the answer isWe prove both bounds, including the quantitative lower bound.

First, suppose a bipartite graph with parts $A,B$, of sizes $m,N$, has at least $pmN$ edges. For any integer $s\le pm/2$, counting pairs consisting of an $s$-subset of $A$ and a common neighbour in $B$ givesThe inequality follows from discrete convexity: $\left(\genfrac{}{}{0ex}{}{x+1}{s}\right)-\left(\genfrac{}{}{0ex}{}{x}{s}\right)=\left(\genfrac{}{}{0ex}{}{x}{s-1}\right)$ is nondecreasing, so balancing integer degrees of vertices minimizes their sum. Consequently some $S$ hasThe product estimate uses $\lfloor pm\rfloor -j\ge pm-s$ for $0\le j<s$. This is a common neighbourhood from bipartite density bound.

We next prove by induction on $q\ge 1$ that, for every fixed $\alpha >0$, a graph $J$ of order $N$ with minimum degree at least $(1-1/q+\alpha )N$ contains $K_{q+1}(\lfloor c\log N\rfloor )$, for some $c=c(q,\alpha )>0$ and all sufficiently large $N$. Impossible minimum-vertex degree parameters need no argument. For $q=1$, take two copies of $V(J)$ with adjacency given by $J$. The preceding count with $p=\alpha$ and $s=\lfloor c\log N\rfloor$, where $c\log (2/\alpha )<1/2$, gives at least $\sqrt{N}$ common neighbours. The selected vertices and their common neighbours are disjoint in $J$, since a vertex is not its own neighbour. Selecting $s$ common neighbours gives $K_2(s)$.

For $q\ge 2$, the minimum degree hypothesis implies the induction hypothesis for $q-1$, for example with excess $1/[q(q-1)]$. Thus $J$ contains $K_q(m)$ with $m=\lfloor a\log N\rfloor$ for a positive constant $a$. Pair the vertices in its classes into $m$ disjoint transversal cliques $C_1,\dots ,C_m$, each of order $q$. By the union bound, every $C_i$ has at leastcommon neighbours. Form an auxiliary bipartite graph between these $m$ cliques and the $N$ vertices, using common-neighbour incidence, and put $p=q\alpha$. Choose $c>0$ small enough that $c<pa/3$ and $c\log (2/p)<1/2$. The count above, with $s=\lfloor c\log N\rfloor$, gives $s$ cliques with at least $\sqrt{N}$ simultaneous common neighbours. Their vertices form $q$ classes of size $s$, and $s$ of those common neighbours form the last class. None of the common neighbours belongs to a selected clique, because that would require a loop. This proves the logarithmic clique blow-up from minimum degree assertion.

To pass from an edge count to minimum degree, put $b=1-1/r+\epsilon$ and $a=b-\epsilon /2$. Repeatedly remove a vertex whose current degree of a vertex is below $a$ times the current order. If the process reaches order $m$, its deleted edges and remaining edges giveThus $(1-a)m^2\ge (\epsilon /2)n^2-O(n)$. In particular, for a fixed sufficiently small $\gamma >0$, the process cannot reach $m=\lfloor \gamma n\rfloor$: the displayed inequality would fail. It stops earlier with a subgraph of order $N\ge \gamma n$ and minimum degree at least $(1-1/r+\epsilon /2)N$. The induction just proved supplies $K_{r+1}(\lfloor c\log N\rfloor )$, hence a lower bound $c^{\prime }\log n$ on $t(r,\epsilon ,n)$. In particular, the guaranteed part size tends to infinity.

For the upper bound, choose a fixed $p$ with $b<p<1$ and take a binomial random graph $G(n,p)$. Its edge count has expectation $p\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and variance $p(1-p)\left(\genfrac{}{}{0ex}{}{n}{2}\right)$. The Chebyshev inequality shows that $e(G)\ge b\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ with probability tending to one. The expected number of labeled copies of $K_{r+1}(t)$ is at mostFor $t=\lceil C\log n\rceil$ and $C>2/[r\log (1/p)]$, this tends to zero. The Markov inequality therefore shows that with positive probability both the required edge density holds and there is no such copy. This proves $t(r,\epsilon ,n)=O(\log n)$ and completes the random obstruction to larger clique blow-ups argument.

There is a necessary qualification to the density parameter. If $\epsilon =1/r$, the only eligible graph is the complete graph, and $t(r,1/r,n)=\lfloor n/(r+1)\rfloor$. If $\epsilon >1/r$ and $n\ge 2$, there are no eligible simple graphs; the universal assertion is vacuous for every $t$, so the printed maximum has no finite value in that range.

Independence requires each joint probability to equal the product of the relevant marginals. In particular it implies $d=(c+d)(b+d)$, equivalent to $ad=bc$.

Conversely let $p=c+d$ and $q=b+d$, and assume $ad=bc$. The covariance calculation gives $d=pq$. ThenAll four joint probabilities therefore factor, proving independence. HenceThis is Bernoulli independence from zero covariance, including degenerate marginals. It is special to two-point variables and is not a general equivalence for arbitrary random variables.

The probabilities satisfy $a,b,c,d\ge 0$ and $a+b+c+d=1$. Since the variables are indicators,Their covariance isThus the necessary and sufficient condition for uncorrelated random variables isThis uses zero covariance as the definition. If one variable is constant, a normalized correlation coefficient is undefined, but zero covariance and this criterion remain meaningful.

An uneventful day leaves one aphid, a one-offspring day leaves the parent plus its offspring, and a two-offspring day leaves three aphids. Death leaves none. Thus the daily descendant counts are $0,1,2,3$ with probabilities $(1-q)t$, $q$, $(1-q)r$, $(1-q)s$, respectively. The mean and probability generating function areIndependence across individuals and days makes the daily population a Galton-Watson process, with each surviving parent counted as one of the next day's descendants.

The standard Galton-Watson extinction fixed point result states that the eventual extinction probability of a branching process $e$ from one ancestor is the smallest root in $[0,1]$ of $G(e)=e$. Moreover a nondegenerate offspring law of mean at most one has extinction probability of a branching process one. Here, since $q<1$, the fixed-point equation reduces towhich contains no $q$. This proves laziness invariance of branching extinction: uneventful days change the timescale, not eventual extinction. The excluded case $q=1$ would keep one aphid forever.

If $2r+3s\le 1$, the offspring mean is at most one. In this parameter range $t>0$: if $t=0$, then $r+s=1$ would imply $2r+3s\ge 2$. Thus the offspring law is not deterministically one, and the extinction criterion applies. Extinction is certain.

For the specified probabilities, the fixed-point equation factors asIts roots in $[0,1]$ are $1/2$ and $1$, and the smaller root is the extinction probability of a branching process:

By rotational invariance, fix the first landing point at the north pole of a sphere of radius $R$. A band of central angle $\theta$ and width $d\theta$ has area $2\pi R^2\sin \theta d\theta$. Dividing by total area gives the central angle between independent uniform sphere points densityand zero outside this range. The radius cancels.

Given the first pair's central angle $\gamma$, the third point can communicate with each of them exactly when it lies in that point's positive hemisphere. Choose their intersection line as a polar axis. Each hemisphere permits an azimuth interval of length $\pi$, and the intervals' centres are separated by $\gamma$. Their overlap is a spherical lune of opening $\pi -\gamma$, with area $2R^2(\pi -\gamma )$. ThereforeThe union probability follows from inclusion-exclusion for two hemispheres:Thus the requested conditional answers are“Either” here is inclusive: a third ship linked to both is also linked to at least one.

For network connectivity, if $\gamma <\pi /2$ the first pair is already linked, and the third must link to at least one of them. If $\gamma >\pi /2$, the first pair is not linked, and both must link through the third. Consequently the connectivity of three acute-angle sphere points isUsing ${\int }_0^{\pi /2}\gamma \sin \gamma d\gamma =1$ and ${\int }_{\pi /2}^{\pi }\gamma \sin \gamma d\gamma =\pi -1$ givesAngles exactly equal to the threshold have probability zero. Connectivity allows relaying and is weaker than requiring direct links between every pair.