The kinship coefficient ${\varphi }_{XY}$ is the probability that a randomly selected allele copy from individual $X$ and an independently selected copy from individual $Y$ at the same genetic locus are identical by descent. The sampling is of gene copies, not of whole genotypes. For a noninbred individual $X$, sampling its two copies with replacement gives ${\varphi }_{XX}=1/2$; in general ${\varphi }_{XX}=(1+F_X)/2$, where $F_X$ is its inbreeding coefficient.

For two full siblings in an outbred pedigree, the two sampled copies are both paternal with probability $1/4$, both maternal with probability $1/4$, and of different parental origin with probability $1/2$. Given that both are paternal, their independent Mendelian segregation transmissions select the same paternal ancestral copy with probability $1/2$; likewise for both maternal. The different-parent case cannot be identical by descent when the parents are unrelated. ThereforeEquivalently, conditional on an IBD-sharing count $J$, the chance that the two random copies match is $J/4$, so ${\varphi }_{\mathrm{sib}}=E[J]/4=1/4$. The expected shared proportion $E[J/2]=1/2$ and the usual additive relationship coefficient $2\varphi =1/2$ are twice this kinship coefficient; confusing these conventions would double the subsequent inbreeding coefficient.

Put $N=n_0+n_1+n_2$ and $s=2x=2\theta (1-\theta )$. The usual recombination fraction range $0\le \theta \le 1/2$ gives $0\le s\le 1/2$, and the preceding likelihood function becomesTo see its relation to nonparametric linkage analysis, use Bayes' theorem and the constant disease probability $1/16$ to obtain the affected-pair marker-sharing probabilitiesTheir multinomial likelihood has exactly the same $s$-dependent kernel, since its extra factor $2^{n_1}$ is constant. Set $A=2n_0+n_1$ and $B=2n_2+n_1$. Then $A+B=2N$, and $B$ is the total number of shared ancestral copies. At fixed $N$, the Fisher-Neyman factorization theorem makes $B$ a sufficient statistic. The observed mean shared proportion is $\hat{p}=B/(2N)$, and the binomial likelihood kernel givesThe unlinked null is $s=1/2$, whereas linkage corresponds to $s<1/2$, or expected shared proportion $1-s>1/2$. If $\hat{p}\le 1/2$, the constrained likelihood is maximized at the unlinked null. For $\hat{p}>1/2$, its log likelihood ratio iswhose derivative in $\hat{p}$ is $2N\log [\hat{p}/(1-\hat{p})]>0$. Hence it orders samples exactly by excess IBD sharing. Under this fully penetrant recessive model, parametric linkage testing and a one-sided test of the overall IBD-sharing proportion use the same information. This is allele-sharing sufficiency for recessive linkage, not a general equivalence for all disease models. Use limiting values for terms such as $0\log 0$, and assume $N>0$ for the estimators.

Take the affected sib pairs from independent families with the same disease model. Conditional on their marker sharing states, multiply the recessive affected-sib-pair likelihoods:Write $x=\sqrt{L_0}=\theta (1-\theta )$, so that $\sqrt{L_2}=1/2-x$ and $L_1=x(1/2-x)$. ConsequentlyFor an unordered count sample there can also be a multinomial likelihood coefficient, and the Mendelian probabilities of the observed marker sharing states supply further factors. All these factors are independent of $\theta$.

If genetic-study ascertainment explicitly conditions on both children being affected, the normalizing disease probability is constant:Thus the conditional sharing likelihood differs from the displayed likelihood function only by factors independent of the recombination fraction. Multiple pairs drawn from the same larger sibship would not, in general, justify treating these pair likelihoods as independent.

Let $a$ denote the disease allele. Complete penetrance with recessive inheritance makes each affected child $aa$. The unaffected parents must therefore both be $Aa$ genetic carriers. Condition on these parental genotypes and the fully informative marker; the distinct paternal and maternal marker copies identify which parent's copies are shared. Average over the two equally likely marker-disease haplotype phases in each parent. This is the phase-symmetric model implicit in the requested pedigree likelihood identities.

For one parent, suppose first that both children receive the same marker copy. If that marker copy is on the parental chromosome carrying $a$, both children receive $a$ with probability $(1-\theta {)}^2$; in the opposite phase the probability is ${\theta }^2$. The phase-averaged probability isIf the children receive different marker copies, one disease-allele transmission is recombinant and the other is nonrecombinant in either phase. The probability that both receive $a$ is thereforePaternal and maternal transmissions are independent. Sharing zero marker copies means the different-copy case for both parents; sharing two means the same-copy case for both; sharing one means one of each. Thus the recessive affected-sib-pair likelihood isHere $L_j$ is the probability that both children are affected conditional on the marker sharing state $J=j$, not the probability of $J=j$ conditional on their being affected. Since $u,v\ge 0$,At $\theta =1/2$, all three conditional disease probabilities equal $1/16$, as expected from two independent children of an $Aa\times Aa$ mating. At $\theta =0$, only the two-copy-sharing state can produce two affected children. Phase weights other than $1/2$ would require a different likelihood function and need not obey these identities.

For the Mendelian identity by descent count $J$, the expected shared proportion isThe corresponding expected number of shared copies is one. These are different scales for the same random variable.

With genetic linkage to a disease locus, selecting two affected full siblings preferentially selects transmissions of disease-associated ancestral copies. A nearby genetic marker may then show excess identity by descent: the expected shared proportion can exceed $1/2$, which is the basis of affected-sib-pair analysis. The precise distortion depends on penetrance, inheritance and the recombination fraction; disease status alone does not prescribe a universal alternative distribution. In the fully penetrant recessive model derived below, setting $s=2\theta (1-\theta )$ givesThus the expected shared proportion rises from $1/2$ at $\theta =1/2$ to one at complete genetic linkage, $\theta =0$.

For the nilpotent matrix $T$, define the polynomial $E(z)={\sum }_{k=0}^{m-1}{z}^k{T}^k/k!$. Because $T^m=0$, differentiating gives $E^{\prime }(z)=TE(z)=E(z)T$. The derivative of the preserved bilinear form matrix is thereforeIt is a constant matrix polynomial, equal to $A$ at $z=0$. Evaluation at one provesAlso $E(-1)$ is the inverse of $E(1)$ by the finite exponential product identity, so this really is an invertible isometry, even when $A$ is degenerate.

Use the standard global convention that ${\mathcal{M}}_c^2$ consists of L2-bounded continuous martingales, here starting at zero. This matters for arbitrary unbounded stopping times in part (iv); square-integrability at each separate deterministic time alone is weaker. Such martingales have terminal $L^2$ limits and satisfy the optional sampling theorem at arbitrary finite stopping times. Infinite stopping times can also be included by evaluating at the terminal limit. The weakly orthogonal continuous martingales condition compares cross-time expectations. The following four parts prove its equivalence with the equal-time, bracket-mean and forward-stopping tests.

The capacity and quality targets alone do not determine flow counts: the per-flow traffic statistics must also be supplied. The following explicit model gives a computable sufficient region, with genuine upper bounds on both loss fractions.

Measure queue workload and buffer size in bits. Denote capacity by $C$, shared buffer size by $B$, voice delay limit by $D$, target voice loss by ${\ell }_v$, and target data loss by ${\ell }_d$, with $0<{\ell }_v,{\ell }_d<1$. Let there be $n_v$ voice flows and $n_d$ data flows. Model each class-$j$ flow, $j\in \{v,d\}$, by a stationary Compound Poisson process with packet rate ${\nu }_j$ and positive independent packet sizes $S_j$, and assume independence across flows and marks. Use a work-conserving preemptive-resume priority queue with a shared buffer, with voice priority and first come first served within each class. This preemptive or fluid approximation makes the predicted voice waiting time exactly $V/C$, where $V$ is the remaining voice queue workload just before arrival. Assume packet sizes are bounded by a known $s_{\max }<B$ and interpret the buffer gate as a packet-fit test; rejecting packets which do not fit is the packet version of the full-buffer rule.

For one flow the Compound Poisson process formula givesFor fixed-size packets of size $s_j$, substitute $e^{\theta s_j}$. Estimate the packet rate and size law from measurements or a declared traffic contract. Also require the strict stability inequalityDefine $B_0=B-s_{\max }>0$ and $K=CD$. Couple the switch with an unthinned infinite-buffer aggregate queue and an unthinned infinite-buffer voice-only queue, both driven by the same offered packets and both served at rate $C$. Under the preemptive service discipline, dropping arrivals can only decrease both the total queue workload and the voice queue workload. Thus total switch queue workload $Q$ and voice queue workload $V$ satisfy $Q\le W_T$ and $V\le W_V$ in the stationary coupling. A packet-fit loss implies $Q>B-S_j\ge B_0$, and a voice-delay rejection implies $V\ge K$.

Apply the continuous-time exponential workload bound to the two comparison queues. For any positive ${\theta }_T,{\theta }_V$ satisfyingthe stationary loss-event probabilities obeyThe second inequality is a union bound: voice can be lost either at the shared buffer or at its delay gate. Data congestion therefore contributes to voice loss despite its lower service priority. Poisson arrivals see time averages applies to each offered packet stream, whose future arrivals are independent of the prearrival state; the stationary event bounds are consequently bounds on the offered-packet loss fractions. Voice packets that are admitted have queueing delay below $D$, because later voice packets join behind them and data work cannot delay them in this model.

To turn the bounds into flow counts, allocate loss budgetsFor example, ${\eta }_B=\min ({\ell }_d,{\ell }_v/2)$ and ${\eta }_V={\ell }_v-{\eta }_B$ also work; equality with ${\ell }_d$ is harmless. The required sufficient effective-bandwidth admission region with a voice delay gate isIt ensures $p_d\le {\ell }_d$ and $p_v\le {\ell }_v$. A simple enumeration isdiscarding negative bounds and pairs failing strict mean-load stability. The formulas assume positive-rate nonzero-size flows. Optimize a chosen revenue or flow-count objective over these integer pairs. Searching over the budget split ${\eta }_B$ can enlarge the sufficient region. For a bit-fluid buffer gate with no packet-fit issue, $B_0$ can be replaced by $B$.

Real nonpreemptive packet service needs an additional residual-service allowance or an explicit priority queue workload analysis: a voice packet may have to wait for the data packet already in service. Burst-correlated voice sources also require richer descriptors, such as measured finite-time effective bandwidths ${\alpha }_j(\theta ,t)$ or a Markov-modulated source model. For such inputs, additivity still holds for independent flows, but the independent-increment supermartingale bound is not automatic; use a valid sample-path large deviation principle or a model-specific overflow bound, with prefactors or a safety margin when using asymptotics. If packet arrivals do not form a Poisson process, loss fractions must be assessed at arrival epochs rather than inferred solely from time-average queue workload probabilities. These qualifications explain both how to calculate an admission region from specified traffic statistics and why the five switch/QoS parameters by themselves cannot supply a numerical answer.

Let $\kappa =2(1-H)$ and fix $a>0$. Starting with part (b), the contraction principle for large deviations under the map $q\mapsto q/a$ gives an LDP for $r(X)/(aN)$ with speed $N^{\kappa }$ and rate function $q\mapsto J(aq)$.

Alternatively, replace $N$ by $aN$ in parts (a) and (b), using the real-scale justification in part (a). The same family has large-deviation speed $(aN{)}^{\kappa }$ and rate function $J(q)$. When expressed at speed $N^{\kappa }$, that rate function is $a^{\kappa }J(q)$. The two descriptions of the same large deviation principle must agree:For clarity, uniqueness here is a general property of lower semicontinuous rate functions: at a point $x$, the shrinking-neighborhood logarithmic bounds recover $I(x)$, since ${\inf }_{B(x,\eta )}I\to I(x)$ as $\eta \downarrow 0$. Thus two rate functions for the same family and speed cannot differ.

Taking the base argument $q=1$ and replacing $a$ by any positive queue size givesFor $q<0$ the rate function is infinite. In the usual nondegenerate case $J(1)$ is finite, so the formula also holds at $q=0$. This is the self-similar Gaussian workload rate; the exponent comes from the Hurst parameter, while the constant comes from the path rate function and the drift and noise scales.

Let $\delta =C-\mu >0$, and first suppose $\sigma >0$. For the infinite-horizon contraction, use the usual sublinear path space for fluid queues,We interpret the supplied path-space large deviation principle in this norm, or in a topology with the same workload-continuity property. The paper does not specify its norm explicitly; mere convergence on compact time intervals would not justify an infinite-horizon contraction.

Define the queue workload functionalIt is finite and nonnegative on the sublinear path space for fluid queues. Here is the required continuity argument. Fix $f$ and choose $T\ge 1$ so that $|\sigma f(t)|\le \delta t/4$ for $t\ge T$. If $\Vert g-f{\Vert }_{\mathrm{sl}}<\delta /(4\sigma )$, then for $t\ge T$,The corresponding expression for $f$ is also negative there, whereas at zero both are zero. Each supremum can therefore be restricted to $[0,T]$, givingThus $R$ is a continuous map; this proof also shows why the negative drift is essential.

Substituting $t=Ns$ into the queue workload givesThe contraction principle for large deviations says that a continuous image of a family with good rate function $I$ has the same large-deviation speed and good rate function obtained by minimizing $I$ over each fibre. ConsequentlyIn particular $J(0)=0$: $Z/\sqrt{L}\to 0$ in the path norm almost surely, so the given large deviation principle and lower semicontinuity give $I(0)=0$, and $R(0)=0$. If $\sigma =0$, the queue workload is identically zero and the rate function is zero at zero and infinite elsewhere. No unspecified variance normalization of the fractional Brownian motion is needed for this variational answer.

Put $\kappa =2(1-H)>0$. For the Gaussian process $Z$, self-similarity of a stochastic process gives the identity in distributionApply the given small-noise large deviation principle with $L=N^{\kappa }$. The answer isNo Legendre-Fenchel transform needs to be recomputed: the scaled paths have exactly the small-noise law at this scale.

If the stated small-noise large deviation principle is indexed only by integer $L$, use $L_N=\lfloor N^{\kappa }\rfloor$. This gives speed $N^{\kappa }$ as well, since $L_N/N^{\kappa }\to 1$. The multiplying factor $\sqrt{L_N/N^{\kappa }}\to 1$ does not change the large deviation principle. To justify this last step, the error in the path norm is $o(1)\Vert Z/\sqrt{L_N}\Vert$. For each fixed $M$, good rate functions have bounded sublevel sets, so ${\inf }_{\Vert f\Vert \ge M}I(f)\to \infty$ as $M\to \infty$. The closed-set upper bound then makes the probability of an error greater than any fixed $\eta >0$ superexponentially small. Thus exponential equivalence applies. This also justifies positive real, rather than only integer, rescalings used in part (c).

Let $Z_t$ denote the generation-$t$ type-count vector, and let $p_i^{(t)}={\mathrm{Pr}}_i(|Z_t|>0)$ from a single type-$i$ ancestor. Every type-$j$ child independently has probability $p_j^{(t)}$ of having descendants in generation $t$. For a Poisson random variable $N$ with mean $\lambda$, $\mathbb{E}{s}^N=\exp [\lambda (s-1)]$. Independence across child types therefore givesThe events of nonempty successive generations decrease. Each individual has finitely many children almost surely, so the total tree is infinite precisely when no generation is empty. Consequently $p^{(t)}\downarrow p$, the actual survival probability of a branching process. Continuity of $T$ yieldsThe map $T$ is coordinatewise increasing on nonnegative vectors. Every nonnegative fixed point $p^{\prime }$ automatically lies in $[0,1{]}^l$, since its coordinates equal $1-e^{-z}$ with finite $z\ge 0$. Hence $p^{\prime }\le 1$ and, by induction, $p^{\prime }=T^t(p^{\prime })\le T^t(1)=p^{(t)}$. Taking limits proves $p^{\prime }\le p$ in every coordinate, so $p$ is the greatest nonnegative solution, not merely some fixed point. This argument applies without irreducibility assumptions to the multitype Poisson branching process.

Here the $L^1$ difference is ${\int }_n^m{x}^{-2}dx\le 1/n$, and the squared $L^2$ difference is ${\int }_n^m{x}^{-4}dx\le 1/(3n^3)$. Both vanish uniformly for $m>n$, so the sequence is Cauchy in both spaces, with limit $x^{-2}$ in each.

For $m>n$, the $L^1$ difference is ${\int }_n^m{x}^{-1}dx=\log (m/n)$. Choosing $m=2n$ makes it $\log 2$, so the sequence is not Cauchy in $L^1$. Its squared $L^2$ difference is ${\int }_n^m{x}^{-2}dx\le 1/n$, which tends uniformly to zero for $m>n$. Hence it is Cauchy in $L^2$, with norm limit $x^{-1}$ on $(1,\infty )$.

Completeness means that every norm-Cauchy sequence has a limit belonging to the same Lp space, with $\Vert f_n-f{\Vert }_p\to 0$. To prove it, choose a subsequence $f_{n_j}$ with $\Vert f_{n_{j+1}}-f_{n_j}{\Vert }_p\le 2^{-j}$. The functionshave $p$-norm bounded by $\Vert f_{n_1}{\Vert }_p+1$, by Minkowski inequality. They increase to $H$, and Fatou's lemma gives $\int H^p\le (\Vert f_{n_1}{\Vert }_p+1{)}^p$. Thus $H$ is finite almost everywhere, so the subsequence telescopes absolutely there to a measurable limit $f$ with $|f|\le H$ and $f\in L^p$.

The same argument on the tails gives $\Vert f-f_{n_j}{\Vert }_p\le {\sum }_{k\ge j}{2}^{-k}\to 0$. Given $ϵ$, choose such a subsequence term beyond the Cauchy threshold and use $\Vert f_n-f{\Vert }_p\le \Vert f_n-f_{n_j}{\Vert }_p+\Vert f_{n_j}-f{\Vert }_p$. This proves convergence of the full sequence and completeness for every $1\le p<\infty$, on an arbitrary measure space.

For $3\pi /2<2\alpha <2\pi$, $1/2<{\lambda }_0<2/3$ and $3/2<{\lambda }_1<2$, whereas ${\lambda }_2>5/2$. Hence both of the first two harmonic modes precede the $r^2$ particular term in the corner expansion; the first still dominates as $r\to 0$. The first-mode gradient is singular. These are reentrant wedges, and the global arc condition determines the important homogeneous coefficients.

For $2\alpha <\pi /2$, ${\lambda }_0>2$ and ${\lambda }_1>6$. Every harmonic correction decays faster than $r^2$ at the corner, so the $r^{2}f(\theta )$ particular solution is the leading corner velocity. It is positive inside this acute wedge and the velocity gradient tends to zero at the vertex.

A useful description of the low-column Lyman-alpha forest at $z\simeq 3$ is a power law $f_X(N,z)=A(z)N^{-\beta }$, with $\beta$ around $1.4$–$1.5$ over the ordinary unsaturated forest range. The same column dependence holds for $f_z$ at a fixed redshift. It is not a universal unbroken power law: saturation, increased neutral fraction and changing absorber populations produce breaks. Lyman limit systems become optically thick at $N\simeq 10^{17.2}{\mathrm{cm}}^{-2}$ because $N{\sigma }_{912}\simeq 1$ with ${\sigma }_{912}\simeq 6.3\times 10^{-18}{\mathrm{cm}}^2$. Damped Lyman-alpha systems have $N\gtrsim 2\times 10^{20}{\mathrm{cm}}^{-2}$ and strong radiative damping wings. They are rare in number but contain substantial neutral mass. The highest-column distribution steepens or cuts off rather than extending indefinitely.

This sketch shows representative shapes and transition scales, not an observational fit; its ordinate is distribution per unit column, not counts per logarithmic bin. In an Einstein-de Sitter universe, $H=H_0(1+z{)}^{3/2}$, so a fixed comoving population with fixed physical cross-section above a fixed column threshold has $d\mathcal{N}/dz\propto (1+z{)}^{1/2}$. ThusThe observed redshift evolution of the Lyman-alpha forest is much stronger for commonly selected forest lines. For example, Kim, Cristiani and D'Odorico's absorption survey finds an index about $2.19$ in $d\mathcal{N}/dz\propto (1+z{)}^{\gamma }$ for $10^{13.64}<N/{\mathrm{cm}}^{-2}<10^{16}$ over $1.5<z<4$, corresponding to an increase by approximately $2^{2.19}\simeq 4.56$. This number is selection-dependent. Expansion changes the gas density, and the evolving cosmic ionizing background changes its neutral fraction through photoionization equilibrium, moving systems across a fixed neutral-column threshold. Structure also evolves. A non-evolving population with fixed physical cross-section is therefore not an adequate physical model of the observed forest.

An isolating integral is a single-valued integral of motion $I(\mathbf{x},\mathbf{v})$ whose fixed value confines an orbit to a lower-dimensional level set of phase space. Along the flow, $dI/dt=0$; for an autonomous Hamiltonian, this means $\{I,H\}=0$. Several isolating integrals are independent when their gradients are linearly independent on the regular region considered. Counting them requires a specified domain: merely local orbit labels in a flow box do not establish global single-valued integrals. The following cases concern regular globally meaningful orbit constraints.

Each independent run uses one oracle call and gives a uniformly sampled equation $z\cdot s=0$. Collect $n-1$ linearly independent equations and use Gaussian elimination over ${\mathbb{F}}_2$. Their solution space has dimension one and consists of $\{0,s\}$, so the promised nonzero solution identifies $s$ uniquely.

To quantify the expected query count for Simon sampling, write $d=n-1$. If the existing equations have rank $k<d$, their span contains $2^k$ of the $2^d$ possible samples. A fresh equation raises the rank with probability $1-2^{k-d}$. The mean waiting time is the corresponding geometric distribution mean, givingThe last sum is less than $2$: its first term is $1$ and its remaining terms are bounded by ${\sum }_{j=2}^{\infty }{2}^{1-j}=1$. Hence the expected number of quantum queries is $O(n)$. For a fixed high-confidence budget, use $m=d+c$ samples. Failure to span means that a nonzero linear functional on the $d$-dimensional sampling space annihilates every sample. Each such functional has probability $2^{-m}$, so the union bound gives failure probability at most $(2^d-1)2^{-m}<2^{-c}$. Thus $O(n+\log (1/\delta ))$ queries suffice for failure probability at most $\delta$. For $n=1$, the nonzero string is known already and no samples are needed.

Classically, a collision between distinct queried arguments immediately yields $s=x+y$. Random queries produce this collision after order $2^{n/2}$ calls, by the birthday problem. Conversely, choose the nonzero period uniformly and label the fibres randomly. Before a collision, $q$ queries can exclude at most $\left(\genfrac{}{}{0ex}{}{q}{2}\right)$ periods, their pairwise argument differences; every remaining period has the same no-collision transcript distribution. When $q=o(2^{n/2})$, those exclusions occupy a vanishing fraction of the candidate periods, and neither collision discovery nor guessing a remaining period has a constant success probability. More quantitatively, conditioned on no previous collision, the next query tests at most $q$ of at least $2^n-1-\left(\genfrac{}{}{0ex}{}{q}{2}\right)$ remaining periods; summing these conditional probabilities proves the same square-root lower bound. Therefore the optimal bounded-error classical query count is $\Theta (2^{n/2})$, exponentially larger in $n$ than the quantum count.

This comparison presumes a hidden nonzero period. Under a literal interpretation asking merely for some $s$ satisfying the printed identity, returning $s=0$ requires no queries at all, quantum or classical. Nonzero $s$ is the necessary intended promise, not a consequence of two-to-one behavior alone.