Work with real-valued functions and put $\mathrm{sign}(0)=0$. Let $r=f-p_{\ast }$, and write an arbitrary competitor as $p=p_{\ast }+v$ with $v\in U$. Pointwise,For $r\ne 0$ this is $|r-v|\ge \mathrm{sign}(r)(r-v)$; for $r=0$ the right side is zero. Integrating and using the assumed orthogonality givesThus $p_{\ast }$ is a best $L^1$ approximation from $U$. This sign criterion for best L1 approximation is sufficient without any finite-dimensionality or uniqueness assumption; possible residual zeros do not invalidate the inequality.

Use the approximants just chosen. The weak type (1,1) bound givesThe measures of these exceptional sets are summable. Indeed, the measure of their union for $n\ge m$ is at most $C{\sum }_{n\ge m}{2}^{-n}\to 0$. Hence their limsup is null; this is the Borel-Cantelli first lemma. Outside it, $S(g-f_n)\to 0$.

Remove also the countable null set from part (b). At each remaining point and for every $n$, the estimate in part (a) yieldsLetting $n\to \infty$ makes the right side zero. ThereforeThis proves almost-everywhere convergence from weak-type domination. The argument uses countably many measurable exceptional sets and does not assume that a supremum over all real parameters is measurable.

The Metropolis–Hastings algorithm is useful when a target posterior can be evaluated up to a normalizing constant but direct independent sampling or simple conditional sampling is unavailable. It constructs a Markov chain with that posterior as invariant law. After adequate convergence, averages of functions of the chain estimate posterior expectations, and the retained draws approximate marginal distributions, quantiles and credible intervals.

For the random-walk Metropolis algorithm, at current $\theta$ propose ${\theta }^{\prime }=\theta +Z$, where $Z$ has a symmetric distribution, often $N(0,\Sigma )$. With a symmetric proposal the acceptance probability isAccept the proposal when an independent uniform is below this probability; otherwise retain the current value. For a nonsymmetric proposal $q({\theta }^{\prime }\mid \theta )$, include $q(\theta \mid {\theta }^{\prime })/q({\theta }^{\prime }\mid \theta )$ in the ratio. Detailed balance follows because the accepted flow is the minimum of the two proposed directional flows; rejection supplies the remaining self-transition probability. Irreducibility, appropriate recurrence and aperiodicity are also needed for convergence.

In the displayed trace plots, the first parameter moves in small increments with pronounced long-term wandering and a shift from its early region to a later region. This is consistent with an undersized proposal or a strongly correlated slow direction. Measure its acceptance rate: if it is very high, increase that coordinate's proposal variance. The second parameter moves rapidly within regions but changes its level around iteration 4500; that is evidence against treating the full run as a settled stationary sample, and it suggests investigating dependence with the first parameter or separated posterior regions. Its local scale may already be reasonable, so a blanket reduction of all proposal variances would be unwarranted. The third parameter has conspicuous long plateaus, indicating many repeated states: if the recorded acceptance is low, reduce its proposal variance; also check constraints and dependence before deciding that scale alone is the cause.

Rerun with coordinate-specific scale changes, inspect joint dependence, consider a covariance-based block proposal, and run longer chains from dispersed starting points. The MCMC trace diagnosis and proposal tuning should include actual acceptance rates and effective sample sizes; the three marginal trace plots do not by themselves identify correlations or prove convergence. Further practical details are in the three subparts.

The plug-in estimate isFor a concrete bootstrap construction, suppose the rows $(x_i,y_i)$ are independent and identically distributed, $\mathbb{E}(e_i\mid x_i)=0$, the design moment matrix is nonsingular, and the required moments exist. Use a paired bootstrap: sample $n$ rows with replacement, form $X^{\ast },Y^{\ast }$, refit by ordinary least squares, and compute ${\hat{\theta }}^{\ast }={\hat{\beta }}_1^{\ast }{\hat{\beta }}_2^{\ast }$. Repeat many times. Singular resampled designs require a suitable rank-preserving implementation or replacement replicate; with a regular continuous design their frequency vanishes asymptotically.

The bootstrap-t confidence interval method also needs a standard error in the original sample and every replicate. One asymptotically valid heteroscedasticity-robust covariance estimate isThe delta method for the coefficient product givesRecompute ${\hat{V}}^{\ast },d^{\ast },{\hat{s}}^{\ast }$ in each paired replicate. This is bootstrap inference for a product of regression coefficients.

With fixed rather than random design, resample centered residuals if errors are identically distributed with homoskedasticity, or use a properly specified wild bootstrap for heteroscedastic errors, keeping $X$ fixed and refitting after generating $Y^{\ast }$. The printed assumption $\mathbb{E}{e}_i=0$ alone does not establish independent observations or validity of a bootstrap confidence interval. The formulas below give asymptotic 95% coverage under the stated regularity conditions, not an exact finite-sample guarantee. In particular the ordinary first-order studentization needs $({\beta }_1,{\beta }_2)\ne (0,0)$ and a nonzero asymptotic variance; when both coefficients vanish the product has nonstandard behavior.

Assume the parents are unaffected, the disease has complete penetrance with a single recessive causal allele, and there are no phenocopies, new disease mutations or alternative causal loci. An affected first child has genotype $aa$, so each unaffected parent must be a heterozygote $Aa$. Conditional on these parental genotypes, Mendelian segregation in the second child's paternal and maternal meioses is independent of the transmissions to the first child. ThereforeThis is the same sibling recurrence risk as for unrelated, unaffected parents under the same recessive model. Parental kinship increases the prior probability that both parents carry the same rare disease allele; after both are identified as carriers, it does not alter their Mendelian transmission probabilities. The factor $63.4$ from the preceding part is therefore not an additional multiplier on the sibling recurrence risk.

If unaffected parents are not explicitly required, the rare-disease assumption is an approximation supporting the usual carrier-by-carrier calculation, rather than a logically exact consequence of an affected child alone. Under complete penetrance, the compatible parental matings are $Aa\times Aa$, $Aa\times aa$ and $aa\times aa$, with second-child risks $1/4,1/2,1$ respectively. Let their prior probabilities be $w_{11},w_{12},w_{22}$, including both orders in $w_{12}$. Conditioning on the affected first child givesWithout negligible or excluded affected-parent matings, this mixture need not equal $1/4$ and may depend on consanguinity. Likewise incomplete penetrance or disease heterogeneity invalidates the assertion that an affected child of unaffected parents necessarily identifies an $Aa\times Aa$ mating. The stated $1/4$ answer uses the unaffected-parent, fully penetrant single-locus assumptions.

Assume a single disease allele $a$, complete penetrance of $aa$, no disease in $Aa$ or $AA$, and Hardy-Weinberg equilibrium in the outbred population. If its population frequency is $q$, the disease prevalence is $q^2=10^{-6}$, so $q=10^{-3}$.

For a child with inbreeding coefficient $F$, the two inherited copies are identical by descent with probability $F$. In that component there is just one ancestral allele draw, and it is $a$ with probability $q$. In the remaining component there are two independent ancestral draws, with probability $q^2$ of $aa$. Thus the inbreeding genotype frequencies giveUsing the first-cousin inbreeding coefficient $F=1/16$, the disease-risk ratio isThe absolute risk in this model is $6.34375\times 10^{-5}$. This calculation averages over the parental genotypes in first-cousin marriages before observing an affected child. It is not the risk conditional on both parents being known genetic carriers, and assumes the specified cousin relationship is independent of disease status rather than conditioning on a selected disease pedigree.

Let the first cousins be $X,Y$, with parents $(P,U)$ and $(Q,V)$ respectively, where $P,Q$ are full siblings and $U,V$ have no ancestral relationship to one another or to that sibling family. To compute ${\varphi }_{XY}$, choose one allele copy from each cousin. They originate from $P$ and $Q$ with probability $1/4$. All other parental combinations have zero kinship coefficient under the otherwise outbred pedigree assumption. HenceThe two gene copies within a child of $X,Y$ are random transmissions from these parents. They are identical by descent precisely in the event defining parental kinship, so the child's inbreeding coefficient isFor a direct ancestral-path check, there are two shared noninbred grandparents. Through either grandparent, the two cousin-to-grandparent paths have two meioses each, and matching the ancestral grandparental copy supplies a further factor $1/2$. Each contributes $2^{-2-2-1}=1/32$, giving the same first-cousin inbreeding coefficient. Additional ancestral relationships or inbred grandparents would change the result.

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.