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Write $H(R)$ and $V(R)$ for the horizontal and vertical open-crossing increasing events of a rectangle in the rotated square lattice. At parameter $1/2$, planar duality for rectangle crossings says that failure of $H$ is exactly a closed dual vertical crossing. Quarter-turn symmetry and the boundary convention of the rotated rectangles identify the two alternatives on a square. Thus its crossing probability is $1/2$; a rectangle obtained by a harmless one-step boundary adjustment has crossing probability at least $1/2$. This is the finite-domain exact self-dual rectangle crossing probability, not an assumption about infinite percolation clusters.

We first prove a RSW reflection extension lemma. In coordinate units adapted to the rotated drawing, take $S=[0,n{]}^2$, $T=[-n,n{]}^2$, and $R=[-n,2n]\times [-n,n]$. Reveal the rightmost open vertical crossing $\Gamma$ of $S$, when one exists, using only its edges and the region to its right. Reflect $\Gamma$ across the horizontal axis. The part $D$ of $T$ to the left of these two paths remains unexposed and has the independent parameter-$1/2$ law. A horizontal crossing of $T$ must reach their union. Its probability is at least $1/2$. Reflection symmetry and the union bound give a probability at least $1/4$ of reaching the upper path $\Gamma$ through $D$. Conditional opening of the edges on $\Gamma$ can only help. Since $V(S)$ has probability at least $1/2$, the increasing event $B$ that a vertical crossing of $S$ is connected inside $T$ to its left side has probability at least $1/8$.

Reflect this construction in the vertical midline of $S$ to obtain an increasing event $B^{\prime }$ of the same probability, connecting a vertical crossing of $S$ to the right side of $[0,2n]\times [-n,n]$. On $H(S)\cap B\cap B^{\prime }$, the horizontal crossing of $S$ meets both vertical crossings, so all three graph paths join and give $H(R)$. The Harris-FKG inequality now yieldsThe exploration only conditions the exposed side of the extremal graph path; reflecting a drawn path does not assert that its reflected edges are open. This distinction is what makes the extension argument valid.

To pass from aspect ratio $3/2$ to $3$, put four rectangles of width $3h/2$ and height $h$ next to each other, with successive left boundaries separated by $h/2$. Adjacent rectangles overlap in a square of side $h$. Require horizontal crossings of the four rectangles and vertical crossings of the three overlap squares. Every consecutive pair of horizontal crossings meets the vertical crossing in its overlap; together they form a horizontal crossing of width $3h$. The Harris-FKG inequality gives a strictly positive bound, for these compatible scales, ofThe construction works on the rotated square lattice with the boundary sites used in the source drawing. Integer roundings do not create a scale restriction: for an odd height use the largest smaller compatible even height, and use eight horizontal rectangles and seven overlap squares, whose union has auxiliary aspect ratio $5$, so that its width exceeds the desired $3m$ width for all sufficiently large $m$. Truncate its horizontal crossing on first reaching the desired right side. The number of extra rectangles and overlap-square crossings is bounded independently of $m$, and every factor is bounded below by the same positive constants. The finitely many smallest rectangles each have positive crossing probability, since a specified finite open graph path suffices. Therefore a constant $c>0$ works at every integer scale. Taking $\tau =c/2$ ensures the requested strict inequality:This proves the required instance of the Russo-Seymour-Welsh theorem through reflection, exploration and gluing. The parameter in this calculation is $1/2$.

First obtain a polynomial critical one-arm upper bound from part (a). Fix a sufficiently large integer radius $r$. Use rotated-coordinate boxes $B_r=[-r,r{]}^2$ for this geometric construction, in the axes of part (a). Surround $B_r$ by a square ring inside $B_{3r}$. It can be assembled from the four stripsThe first two require horizontal dual crossings; the last two require vertical dual crossings. Their corner overlaps are squares, so the adjacent long crossings intersect there and their union contains a dual graph cycle surrounding the inner box. Translate the strips by the half-lattice displacement of the planar dual graph and adjust the endpoints by one lattice unit if needed. All aspect ratios remain bounded. Part (a) and the same overlap gluing give a uniform lower bound $c_1>0$ for each dual crossing at $1/2$. Applying the Harris-FKG inequality to these decreasing events in the primal configuration gives a closed dual circuit with probability at least $\delta =c_1^4>0$. We may replace $\delta$ by a smaller number in $(0,1)$.

Use the independent annular barriers for percolation at radii $r_j=r_{0}10^j$. Their supporting edge sets are disjoint, so the circuit events are independent. In the original unrotated coordinates, $B_{3r}$ lies inside ${\Lambda }_{5r}$ and the ring still separates the origin from infinity. These fixed geometric comparisons let us bound the original $g_N$. An open graph path from $0$ to distance $N$ must avoid every such closed dual circuit inside ${\Lambda }_N$. Thus, for some constants $C<\infty$ and $\alpha >0$,where the first bound is used only when at least one of the indicated rings fits; increasing $C$ handles smaller $N$.

Now let $\epsilon =p-1/2>0$ and use the monotone coupling of Bernoulli percolation on edges. The event $\{0\leftrightarrow \partial {\Lambda }_N\}$ only needs edges whose endpoints lie in ${\Lambda }_N$, because a first boundary hit gives an internal graph path. There are exactly $4N(2N+1)$ such edges. Each changes state between parameters $1/2$ and $p$ with probability $\epsilon$. The union bound therefore gives the finite-box comparison for percolation parametersFor sufficiently small $\epsilon$, choose $N=\lfloor {\epsilon }^{-1/(\alpha +2)}\rfloor$, so that $N$ is at least half the unrounded value. Both terms then have order ${\epsilon }^{\alpha /(\alpha +2)}$. In particular,for a finite $A$. Increase $A$ to cover the remaining compact range of parameters by $\theta (p)\le 1$. This near-critical percolation power upper bound requires only a positive crossing constant, not the exact value of a critical exponent. It also implies $\theta (1/2)=0$ through the critical one-arm probability bound.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

For each ball, the event of landing in $S_j$ is a Bernoulli distribution trial. Under the null hypothesis its probability is $\epsilon$. Under the selected alternative $Q_j$, its probability is $p_1=\pi +(1-\pi )\epsilon =\epsilon +\pi (1-\epsilon )$. The placements are independent random variables conditional on that selected index, so the count has the binomial distribution:The second assertion concerns $Q_j$, rather than the unconditional mixture model $P_1$.

Two independent cyclic intervals intersect with probability at most $\min \{1,(2k-1)/d\}\le 2k/d$. For $k\le d/2$, precisely the $2k-1$ starting-point offsets $-(k-1),\dots ,k-1$ can overlap; for larger $k$, the upper bound is automatic. The cyclic interval overlap bound and the preceding chi-squared divergence estimate therefore giveAs in 1(e), the square-root hypothesis is meaningful when $\alpha d\ge 1$. The chi-squared testing lower bound bounds the total variation distance by $\sqrt{2{\alpha }^{\epsilon }}/2$. The worst individual Type II error dominates the error under the uniform mixture model. ThusTaking $\alpha$ small makes this lower bound informative. This argument uses the actual cyclic-interval alternative, without substituting a different family of subsets.

Decompose $h={\Pi }_{1}h+D^{†}Dh$ using the incidence pseudoinverse decomposition. The Cauchy-Schwarz inequality controls the constant component, and the duality of the ${\ell }_{\infty }$ and ${\ell }_1$ norms controls the remaining component:On the stipulated event, the second term contributes at most $\lambda \Vert Dh{\Vert }_1$ after multiplication by $2/n$. The triangle inequality gives $\Vert Dh{\Vert }_1\le \Vert D\hat{\theta }{\Vert }_1+\Vert D{\theta }^{\ast }{\Vert }_1$, so the negative penalty in the basic inequality for a penalized least-squares estimator cancels. We obtainon an event of probability at least $1-\delta$.

With the usual variance-proxy convention for a sub-Gaussian random vector, every column inner product is a sub-Gaussian random variable with proxy at most ${\tau }^2$. The two-sided Chernoff bound and union bound implywith probability at least $1-\delta$. No independence between these inner products is required.

The printed coefficient-one bound is false under this convention. Take $k=m=1$, $A=[1]$, and $g=\pm \tau$ with equal probability. This Rademacher random variable has the stipulated variance proxy, but for $\delta =0.99$ the printed threshold is less than $\tau$, whereas $|g|=\tau$ surely. Even a general sub-Gaussian assumption does not imply Gaussian concentration inequality for arbitrary Lipschitz functions. Under the stronger convention $\mathbb{E}{e}^{⟨a,g⟩}\le e^{{\tau }^2\Vert a{\Vert }_2^2/4}$, the same union bound proves the printed constants. That convention, however, differs from the binomial variance proxy used in 1(b).

Let $B_{\delta }=\sqrt{\log (2m)}+\sqrt{\log (1/\delta )}$. The columns of $D^{†}/L$ have Euclidean norm at most one. Under the usual sub-Gaussian random vector convention, 3(d) justifies the tuning choice $\lambda =2\sqrt{2}L{B}_{\delta }/n$, which is independent of ${\theta }^{\ast }$. Also $\Vert {\Pi }_{1}z{\Vert }_2=|⟨z,1/\sqrt{n}⟩|$, so the Chernoff bound gives $\Vert {\Pi }_{1}z{\Vert }_2^2\le c_{\delta }=2\log (2/\delta )$ with probability at least $1-\delta$.

On the intersection of these events, of probability at least $1-2\delta$, set $a=\Vert h{\Vert }_2/\sqrt{n}$ and $b=\Vert {\Pi }_{1}z{\Vert }_2/\sqrt{n}$. The weighted Young inequality gives $2ab\le t^2{a}^2+b^2/t^2$ for $0<t<1$. Applying it to 3(c) and rearranging yields the rigorous boundIf the coefficient-one concentration statement in 3(d) is taken as an extra hypothesis, instead choose $\lambda =2L{B}_{\delta }/n$; the identical argument gives precisely the printed coefficient $4$ in the first term. Thus the deterministic argument and its risk rate are sound; its concentration constants depend on correcting or strengthening 3(d).

In natural logarithms, Fano's inequality for a uniform index $J$ among $N$ alternatives states $\mathbb{P}(\hat{J}\ne J)\ge 1-(I(J;G)+\log 2)/\log N$. The mutual information is at most the average Kullback-Leibler divergence to any fixed reference law. With the supplied packing, nearest-neighbour decoding turns partition error below half the separation into correct decoding. Thus the packing alone proves a bound at radius $n(1/4-\epsilon /2)$, with the unavoidable $\log 2$ term. It does not automatically prove the much larger printed radius.

A list-decoding Fano inequality supplies the intended near-half error threshold. Assume $0<\epsilon <1/4$, since otherwise that threshold is nonpositive and its event is automatic. Take a uniform prior on all $N=\left(\genfrac{}{}{0ex}{}{n}{n/2}\right)/2$ unlabelled balanced partitions, and let $r=n(1/2-2\epsilon )$. A Hamming ball of radius less than $r$ around any estimated partition contains at most $B={\sum }_{j<r}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ such partitions. This follows by selecting the representative closer to the estimated group; each unlabelled partition contributes at most one such representative because $r<n/2$. The entropy bound for a Hamming ball gives $\log B\le nh(r/n)$, where $h$ is the binary entropy function measured in natural logarithms. Its curvature gives $\log 2-h(1/2-2\epsilon )\ge 8{\epsilon }^2$. Also $\left(\genfrac{}{}{0ex}{}{n}{n/2}\right)\ge 2^n/(n+1)$, soUse the all-$1/2$ independent-edge law $P_{\ast }$ as a reference. Only the $n^2/4$ crossing pairs differ; the preceding Bernoulli distribution divergence estimate gives $D_{\mathrm{KL}}(P_S\Vert P_{\ast })\le n{t}^2$. Hence $I(J;G)\le n{t}^2$. Whenever the denominator is positive, the list-decoding Fano inequality proves the finite-sample answerFor $4{\epsilon }^{2}n\ge \log (2(n+1))$, this is at least $1-t^2/(4{\epsilon }^2)-\log 2/(4{\epsilon }^{2}n)$. If in addition $n{t}^2\ge \log 2$, it implies the intended form with $c^{\prime }=1/2$. This proves the claimed information-theoretic scale with its needed finite-sample qualifications.

As a uniform finite-sample assertion for every positive $t$, the printed form is false. At $t=0$ every partition has the same data law, and a uniformly random balanced guess has a positive chance of being correct, giving error strictly below one. Such randomness can also be extracted from the finite graph sample: assign distinct graph outcomes to the finitely many balanced partitions; under the common law every selected outcome has positive probability. The worst error remains strictly below one for sufficiently small positive $t$, by continuity. The printed lower bound tends to one as $t\downarrow 0$ for any fixed universal $c^{\prime }$, which contradicts this. The entropy correction cannot be dropped without an asymptotic or signal-size condition.

Let $M={\sup }_{\gamma >1}{F}_{\gamma }(H_{\gamma })<\infty$. Choose ${\gamma }_j\to \infty$. Boundedness of the minimizers and the Bolzano-Weierstrass theorem yield a subsequence, still indexed by $j$, with $H_{{\gamma }_j}\to H^{\ast }\in {\mathbb{R}}^n$. Each $H_{{\gamma }_j}$ is deterministic, hence so is $H^{\ast }$. The first exponential term givesand passing to the limit gives $H^{\ast }\cdot p\le x$.

To obtain the superhedging inequality, suppose instead that $\mathbb{P}(X>H^{\ast }\cdot P)>0$. There is then an $\epsilon >0$ such that the event $E=\{X-H^{\ast }\cdot P\ge \epsilon \}$ has positive probability. Let $\Vert P\Vert \le B$ almost surely. For all sufficiently large $j$,Consequently $X-H_{{\gamma }_j}\cdot P\ge \epsilon /2$ on $E$, givinga contradiction. The limiting portfolio meets both constraints:Bounded $P$ is used precisely to turn convergence of deterministic holdings into a uniform bound on the payoff error.

Let $h$ be the stock holding and let $c$ be the initial cost, so the cash holding is $b=c-10h$. Terminal portfolio wealth at stock price $s$ is $c+h(s-10)$. Dominating the European call option payoff at the three possible prices givesAdding the two endpoint inequalities yields $2c\ge 1$. Equality is attained by $h=1/2$ and $c=1/2$, which imply $b=-9/2$. The corresponding terminal wealth is $0,1/2,1$ at $s=9,10,11$, respectively, compared with the required payoff $0,0,1$. The cheapest super-replication strategy isThe middle-state excess shows why this is superhedging rather than exact claim replication. The endpoint bound proves global minimality, without relying on the physical state probabilities.

An open migration process is a continuous-time Markov chain recording the populations $n=(n_1,\dots ,n_J)$ of finitely many colonies. Independent external Poisson processes bring individuals to colony $j$ at rate ${\nu }_j$. When that colony contains $k>0$ individuals, a departure opportunity occurs at total rate ${\varphi }_j(k)>0$, with ${\varphi }_j(0)=0$. The departing individual moves to colony $l$ with probability $p_{jl}$, or leaves the system with probability $p_{j0}=1-{\sum }_l{p}_{jl}$. Routing choices and the clocks are independent. A return to the same colony does not change the state.

The word open means that the routing matrix $P=(p_{jl})$ is transient: every individual eventually leaves, or equivalently the spectral radius of $P$ is less than one. The traffic equations of an open migration process, with row-vector convention, give the total arrival rates:These rates include migrations as well as external arrivals. On the active colonies assume ${\alpha }_j>0$ and the usual irreducible Markov chain condition. PutThe product form and its normalization condition areTo establish the product-form stationary distribution of an open migration process, first use its unnormalized weight $b(n)={\prod }_j{b}_j(n_j)$. Its ratios satisfy $b_j(k){\varphi }_j(k)={\alpha }_j{b}_j(k-1)$. The incoming external arrivals into $n$, divided by $b(n)$, contribute ${\sum }_j{\nu }_j{\varphi }_j(n_j)/{\alpha }_j$. Incoming exits contribute ${\sum }_i{\alpha }_i{p}_{i0}$. Incoming migrations from $i$ to $j\ne i$ contribute ${\alpha }_i{p}_{ij}{\varphi }_j(n_j)/{\alpha }_j$. Terms involving an empty destination are zero. Thus the full incoming rate divided by $b(n)$ isHere the traffic equations of an open migration process give the second equality, including ${\sum }_i{\alpha }_i{p}_{i0}={\sum }_j{\nu }_j$. The right side is precisely the total outgoing rate. This proves global balance for a continuous-time Markov chain; pairwise detailed balance is not required for general routing.

For finite rates at each finite population, the continuous-time Markov chain is nonexplosive Markov chain: its total population is bounded over a finite time interval by the initial population plus the finitely many external arrivals, so it visits a finite set on which all jump rates are bounded. Consequently finite $Z_j$ normalize the invariant weight to a stationary distribution. Under the stated irreducible Markov chain assumptions, this is the unique stationary distribution, and it is a positive recurrent Markov chain. Conversely, the unique invariant measure of an irreducible recurrent chain is proportional to $b$, so a positive recurrent Markov chain requires its normalization to be finite. A sufficient condition is ${\mathrm{liminf}}_{k\to \infty }{\varphi }_j(k)>{\alpha }_j$; the series criterion itself is the exact condition. Colonies with ${\alpha }_j=0$ have zero stationary population and can be removed from the active class.

For the single-server M/M/1 queue specialization, ${\varphi }_j(k)={\varphi }_j$ for $k>0$, each stationary colony has the geometric distribution of an M/M/1 queue:The traffic equations of an open migration process do not involve the service rates, so the ${\alpha }_j$ remain fixed during this service-capacity allocation in an open migration process. Write $A={\sum }_j{\alpha }_j$, $B=F-A$ and $b_j={\varphi }_j-{\alpha }_j$. A stationary allocation requires $B>0$, with $b_j>0$ and ${\sum }_j{b}_j=B$. The Cauchy-Schwarz inequality givesEquality holds exactly when $b_j$ is proportional to $\sqrt{{\alpha }_j}$. Hence, for positive arrival rates, the unique optimal service allocation and the minimum expected value of the total population areIf $F\le A$ and some arrival rate is positive, no stable service allocation exists. Zero-arrival colonies need no service in their stationary empty class: allowing zero service there gives the same formula restricted to active colonies. If strictly positive service is demanded even at inactive colonies, the displayed minimum is an infimum approached as that unused allocation tends to zero. If all arrivals vanish, the stationary mean population is zero for every allocation that drains the initial population.

In a loss network with fixed routing, calls of type $r$ arrive as independent Poisson processes with rates ${\nu }_r$. Every accepted call occupies $A_{jr}$ units of resource $j$ throughout its holding time. The link-route incidence matrix $A$ has nonnegative integer entries, $C$ is the resource-capacity vector, $n_r$ counts calls currently present, and $e_r$ is the $r$th coordinate unit vector. The feasible occupancy space isThe inequalities are coordinatewise. A call that would leave this space is rejected immediately, without queuing or changing $n$. With independent unit-rate exponential distributions for holding times, the memoryless property makes $n$ a continuous-time Markov chain, whose nonzero off-diagonal transition rates areThere are $n_r$ independent clocks of rate one for the second transition. Rejected arrivals are not jumps and do not appear in the off-diagonal transition rates.

Let $a(n)={\prod }_r{\nu }_r^{n_r}/n_r!$. For every feasible adjacent pair,Thus detailed balance for a continuous-time Markov chain gives the product-form stationary distribution of a loss network:In this convention $G(C)$ is the reciprocal normalizing sum. If every route consumes some resource, finite capacities make $S(C)$ finite. Even zero-resource routes cause no normalization difficulty: the sum is at most $\exp ({\sum }_r{\nu }_r)$. With positive arrival rates the reachable feasible class is an irreducible Markov chain, since calls can depart to the empty state and any feasible occupancy can be built from it. The stationary distribution is therefore unique on that class.

The vector $m=C-An$ records the free units of each resource. Its probability distribution ${\pi }^{\prime }(m)$ sums $\pi (n)$ over all occupancies with the same free resources. The map from $n$ to $m$ generally loses information: different call mixtures may consume exactly the same resources.

For ${\pi }^{\prime }(m)>0$, the conditional expectation can be computed by removing one call. The factorial identity $n_r\pi (n)={\nu }_r\pi (n-e_r)$ givesIndeed, $n\mapsto n-e_r$ is a bijection from the occupancies in this sum with $n_r>0$ to occupancies using $C-m-A{e}_r$ units. Conversely, adding one call to any such occupancy is feasible because its final resource use is $C-m$. Consequentlyand the conditional expectation is zero otherwise. Conditioning the identity ${\sum }_r{A}_{jr}{n}_r=C_j-m_j$ and multiplying by ${\pi }^{\prime }(m)$ now yieldsAlternatively, multiply the preceding unnormalized factorial identity by $A_{jr}$ and sum over $r$ directly; this proves the same formula even when ${\pi }^{\prime }(m)=0$, without dividing by a zero probability. This gives both the conditional expectation argument and the direct summation argument.

Use an ideal carrier-sense multiple access scheme on the interference graph. An active station finishes its transmission after an independent unit-rate exponential distribution waiting time. An inactive station has an independent attempt clock with an exponential distribution of rate $e^{{\theta }_r}$; an attempt succeeds only when all its neighbors are inactive. Attempts that are blocked do not change the state. Thus, on the independent sets represented by $S$, the continuous-time Markov chain has transition ratesThe empty independent set can be reached by successive deactivations, and any other independent set can be reached from it by successive activations. Hence this finite chain is an irreducible Markov chain. The weight $a_{\theta }(n)=e^{\theta \cdot n}$ satisfiesBy detailed balance for a continuous-time Markov chain, its unique stationary law isThis is a finite exponential family. Its natural parameter of an exponential family is $\theta$ and its mean parameter of an exponential family is the vector of active fractions.

The throughput region of an interference graph is exactly $\Lambda =\mathrm{conv}S$. To see the possibly less immediate inclusion, suppose a random feasible schedule has mean $v\ge \lambda$. Retain each of its active vertices $r$ independently with probability ${\lambda }_r/v_r$, interpreting zero coordinates as always deleted. A subset of an independent set remains an independent set; the thinned schedule has mean exactly $\lambda$. Thus $\lambda$ lies in the convex hull of $S$. The opposite inclusion follows from the definition using equality. Moreover $0,e_1,\dots ,e_R\in S$, so this convex hull has full dimension.

Now choose $\epsilon >0$ such that $\mu =\lambda +\epsilon 1$ remains in the interior of $\mathrm{conv}S$. Use interior moment matching in a finite exponential family. Consider the convex functionThere is a ball of radius $\delta >0$ around $\mu$ contained in $\mathrm{conv}S$. Maximizing a linear functional over the convex hull is the same as maximizing over its generating set, so, for every $\theta$,The last inequality uses the point $\mu +\delta \theta /\Vert \theta \Vert$ when $\theta \ne 0$. Hence $F$ is a coercive function and has a finite minimizer ${\theta }^{\ast }$. Differentiating the finite normalizing sum givesThis covariance matrix is a positive-definite matrix: a linear functional constant on all states of positive weight must be constant on $0,e_1,\dots ,e_R$, so its coefficients vanish. Thus $F$ is a strictly convex function; its minimizer is unique. At that minimizer the desired strict service margins areFor a backlogged station that transmits at unit speed whenever active, the long-run throughput is ${\mathbb{E}}_{\theta }[n_r]$, by the ergodic theorem for a finite continuous-time Markov chain. Therefore the ideal carrier-sense multiple access scheme can supply a strict service margin for every arrival vector in the interior of the throughput region of an interference graph. The local attempt rates can realize the whole interior capacity region in this sense. Boundary points may require parameters tending to infinity, and the argument gives no uniform delay or mixing-time bound near the boundary.

Let $T=10\mathrm{Gyr}$ and let $t$ denote present stellar age, rather than time measured from the galaxy's birth. With a fixed initial mass function, constant star formation rate also gives a constant number of births per unit time, as in the birth-age distribution under constant star formation. Counting the supplied stellar evolution model's remnants as part of the population, ages have a uniform distribution on $[0,T]$. Integrating the constant birth rate therefore givesOutside this interval the cumulative fraction is zero or one as appropriate.

Write $m=M/M_{\odot }$ for the birth-mass parameter and $M_{\min }=0.2M_{\odot }$. The normalization cancels when integrating the initial mass function:Thus the mass-tail fraction isIt is one at and below the lower cutoff. The normalization constant has no effect on any number fraction.

For evolutionary-state selection by stellar lifetimes, a red giant has completed its main sequence lifetime but not its giant lifetime. Its stellar lifetime regions in a mass-age diagram are bounded byThe lower boundary meets $t=T$ at $m=\sqrt{0.8}$ and the upper boundary meets it at $m=1$. A white dwarf lies above the upper lifetime curve, within the same age interval.

The probability integral transform, applied to the decreasing mass-tail coordinate, makes $X$ uniform on $(0,1)$: $\mathrm{Pr}(X<x)=x$. The independent birth age gives $Y=t/T$ uniform on $(0,1)$, so number fractions are areas in a unit square. Since $m^{-2}=25X$, the red giant region becomes $20X<Y<25X$, a triangle with vertices $(0,0)$, $(1/25,1)$ and $(1/20,1)$. Its area is $\frac{1}{2}(1/20-1/25)=1/200$. The white dwarf region $Y>25X$ is a triangle of area $\frac{1}{2}(1/25)=1/50$. Hence the present individual-star fractions areThe rest are on the main sequence under the given toy lifetime model.

For the binary stars, draw two independent mass-tail coordinates $X_1,X_2$, but only one shared age coordinate $Y$: components born together are coeval. The coeval binary population therefore occupies a uniform unit cube $(X_1,X_2,Y)$. At a fixed age $Y=y$, the state probabilities for either component areTheir evolutionary states have conditional independence given $y$. They generally do not have unconditional independence, because sharing an age correlates their states. The binary fractions below are slice areas averaged over $0<y<1$.

For the three unordered evolved pairs, conditional independence gives the slice probabilities $g^2$, $2gw=8g^2$ and $w^2=16g^2$. Averaging over the shared age in the coeval binary population yieldsThus their system-number ratio is $1:8:16$, as required.

For the unheaded instantaneous-burst continuation, put $\tau =t/\mathrm{Gyr}$, with $0<\tau <10$. All binary stars now have the same age, so the individual red giant fraction is $g=\tau /1000$. Independent component masses give the giant-pair fraction in an instantaneous stellar burst, the two-giant system fractionThe final comparison has an ambiguity about which population is being counted. Taken literally as a comparison with the first galaxy's individual-star giant fraction $0.005$, it would require $\tau >\sqrt{5000}\simeq 70.71$. There is no allowed burst age: throughout $0<\tau <10$, the two-giant fraction is less than $10^{-4}$, well below $0.005$.

If the intended comparison instead concerns two-giant systems in both galaxies, the comparison fraction is $1/30000$. Then ${\tau }^2/10^6>1/30000$ givesFor completeness, a comparison of individual red giant fractions in both galaxies would give $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. These three comparisons have different denominators and should not be conflated.

For the evolved-companion fraction, a red giant has another evolved component precisely for unordered $GG$ or $GW$ pairs. By conditional independence at fixed age, their combined slice probability is $g^2+2gw=9g^2$. ThereforeDivide by the system fraction containing at least one red giant to obtain the requested conditional fraction:A double-giant system is counted once in this denominator, and the factor two for a giant–white dwarf pair represents its two possible component orderings.