Proof. Send mass one along the inward half-edge at \(a\). At each newly visited vertex split it between the two possible continuations, multiplying by \(\rho e^{i\lambda}\) and \(\rho e^{-i\lambda}\), respectively. The identity \(2\rho\cos\lambda=1\) conserves total mass. Stop a branch when it exits \(D\) or is about to enter a vertex already visited. This produces a finite tree because \(D\) has finitely many vertices.

For a stopped collision, the newly closed cycle is simple and has a disjoint incoming stem. The stem lies outside the cycle: it starts on \(\partial D\) and cannot cross the cycle before their first common vertex. Consequently the unused branch at that vertex is exterior to the cycle. The cycle turn there is \(\pi/3\) in the counterclockwise orientation, rather than \(-\pi/3\); the clockwise statement is reversed. The turn from the stem into the cycle replaces this closing turn by its negative. The two ways to traverse the cycle therefore have winding increments \(4\pi/3\) and \(-4\pi/3\) relative to the stem. They have the same unsigned weight and cancel, since \(e^{it(4\pi/3)}+e^{-it(4\pi/3)}=0\). Cycle reversal is an involution on all stopped collisions. The remaining terminal mass gives (5).

For a path ending at \(b\), close it using the counterclockwise boundary arc from \(b\) to \(a\). The turns at the two ports are \(\pi/2\), so the turning theorem for this simple closed curve gives \[W(\gamma)=\pi-\operatorname{turn}_{\partial D}(b\to a).\] For a convex boundary the latter turn lies in \([0,2\pi]\). Thus \(|W(\gamma)|\le\pi\) and \(\cos(tW(\gamma))\ge\cos(3\pi/8)=c\). Taking real parts proves the last assertion. ◻

Proof. First take a homogeneous triangular parallelogram of height \(H>0\) and width \(M\). The total weight of paths from a fixed left port to its right side tends to zero as \(M\to\infty\). To see this, append a column immediately to the right of the exit and continue, entirely inside this new column, to its bottom port. The continuation has at most a fixed number of vertices depending only on \(H\), so its weight is bounded below by a positive constant depending only on \(H\). For every \(M<L\) these extensions are paths in a parallelogram of width \(L\). Their bottom endpoints identify \(M\), and their first exit into the appended column recovers the original path. They are therefore distinct. Writing \(q_H(M)\) for the crossing mass and \(C_H\) for the maximum number of added vertices, Lemma 2 gives \[\rho^{C_H}\sum_{M<L}q_H(M)\le 1/c.\] Thus \(q_H(M)\to0\). The same lemma uniformly bounds paths joining ports on the same side.

For a mixed list of the four allowed parameters, collapse the flat rows. The row vectors belong to \(\{1,e^{i\pi/3},e^{2i\pi/3},-1\}\), so outside two end pieces of horizontal size \(O(N)\) the remaining triangles form a homogeneous parallelogram of height \(H\), the number of nonflat rows, and width \(M-O(N)\). This description does not depend on which neighboring triangles were grouped into rhombi. A collapsed deterministic route drifts by at most \(N\) columns. The end pieces have bounded size for fixed \(N\) and contribute bounded weights.

Every nonzero diagram contributing to either restricted transfer power has a component connecting the left and right cuts. In the vacuum sector, otherwise the nonempty input pairing would close into a loop; in the source sector, the source must reach the output. Cut all components at the two ends of the homogeneous core. There are only boundedly many endpoint patterns, depending on \(N\), and at least one segment spans the core. Ignoring avoidance between these segments bounds their total weight by a product of single-path partition sums. Each nonspanning factor is at most \(1/c\). For \(H>0\), choose the core width \(M-\kappa_X\), where the fixed offset satisfies \(0\le\kappa_X=O(N)\). If \(Q_N\) is the nonempty vacuum block and \(\|\cdot\|_{\max}\) is the maximum absolute matrix entry, then \[\|Q_N^M\|_{\max}+\|(T_N^{(1)})^M\|_{\max} \le C_N\sum_{a\in\mathcal E_H}q_{H,a}(M-\kappa_X)\longrightarrow0.\] Here \(\mathcal E_H\) is the finite set of left ports of the core, \(q_{H,a}\) is its crossing mass from \(a\), and \(C_N\) absorbs the bounded end pieces and endpoint patterns. All constants are for fixed \(N\); no uniform spectral gap is needed. If \(H=0\), no spanning component is possible once \(M\) is sufficiently large. Thus both restricted matrix powers tend to zero. Since these are finite matrices, their spectral radii are less than one.

At the specified parameters all tile weights are nonnegative. The matrices obtained by taking absolute tile weights vary continuously, so their restricted spectral radii remain less than one in a small complex neighborhood. This also proves the asserted absolute convergence. The reflected argument is identical.

Writing the vacuum matrix in the empty/nonempty decomposition gives \(T_N=\left(\begin{smallmatrix}1&0\\v&Q\end{smallmatrix}\right)\). Its normalized stationary vector is \((1,(I-Q)^{-1}v)\), which is regular in the stated neighborhoods and rational in the spectral variables. This proves both uniqueness and the continuation claim. Equivalently, \(P_N\) is the sum of finite path diagrams in the vacuum half-strip. ◻

Proof. After multiplication by all \(D\) factors, the Pfaffian is an alternating polynomial, hence divisible by the Vandermonde product in (18). Counting degrees before and after division gives degree at most \(N-1\) in each variable. Letting a last argument tend to infinity proves the leading-coefficient statement: its kernel entries tend to \(-1\). In odd size first add the augmenting column and row to that argument’s column and row. At the flat value, \(D(z,c)=(z-c)(z+C)\) and the kernel entry equals one. This proves (19).

At \(D(f_0,f_6)=0\) only Pfaffian terms pairing these two arguments survive after denominators are cleared. The identity \(D(f_0,z)=(z-f_{-6})(z-f_6)\) then gives (20).

Prove the merge identity by induction on \(r=|Y|\). The case \(r=0\) follows from \(p_2(a,b)=a+b+\rho\). If \(r\ge1\), distinguish a spectator \(z\). Both sides have degree at most \(r+1\) in \(z\), and their leading coefficients agree by induction. They vanish at \(z=-f_0\): delete this argument with \(f_{-2}\), and the remaining factor involving \(f_2\) vanishes. At \(z=c\) they agree by the flat formula and \[(C+f_{-2})(C+f_2)=(c+f_0)(C+f_0).\] They also agree at each deletion value of \(z\) with another spectator. To check the factors, write that pair as \(g_0,g_6\), where \(g_j=\cos(v+j\lambda)\), and use \[\prod_{\ell=\pm2}(f_\ell-g_{-6})(f_\ell-g_{12}) =(f_0+g_0)(f_0+g_6)(f_0-g_{-6})(f_0-g_{12}).\] This is the product-to-sum identity obtained from shifts by \(\pm2\lambda\). There are \(2(r-1)\) such deletion values, in addition to \(-f_0\) and \(c\). For generic arguments these are distinct and give at least \(r+1\) roots for a difference of degree at most \(r\), after its leading coefficient has canceled. Polynomial continuation concludes the proof. ◻

Proof. The assertion for \(N=1\) is immediate. We construct the asserted polynomial by induction and then prove its stationarity. Throughout the construction, \(W\) denotes the candidate at size \(N\), whereas every smaller \(W_j=p_jP_j\) is already known by induction. We first remove the candidate’s possible poles and bound its degree, then establish its special-row relations, and finally use those relations to prove stationarity.

Deleting a vacant last slot identifies that sector with \(\mathcal L_{N-1}^0\). Deleting an occupied last slot leaves its partner as the exposed defect of a state in \(\mathcal L_{N-1}^1\). We use these identifications for the subscripts \(v\) and \(o\) below.

Proof. We identify \(m\) from its row reductions and a separate degree bound of \(2N\) for \(p_N^2m\) in each \(a_i\). The source-sector limit supplies that bound; the scalar reductions then determine the coefficient in (29).

Proof. Apply Lemma 2 to parallelogram truncations of the strip. The left and right boundary terms tend to zero by Lemma 4: after the source column, propagate in the one-source sector to the distant cut, where the exit is the only occupied boundary port. The remaining vacuum states are bounded. The other sums increase to their infinite-strip limits. Bottom exits have winding \(\pm\pi\) and top exits winding zero, so the real part is \(c\mathcal A_N+\mathcal B_N=1\). Increasing strip height increases the arch sum by inclusion, hence decreases \(\mathcal B_N\).

Add one triangular row above the right half of \(\mathcal S_N\), in columns \(1,2,\ldots\), creating a stepped strip with a single side port \(p\) at the left end of the new row. Denote its bottom arch kernel by \(\overline K(i,j)\); see Figure 5.

From a bottom source \(i\), the imaginary part of the boundary identity is \[ \sin(3\lambda)\left( \sum_{j<i}\overline K(i,j)-\sum_{j>i}\overline K(i,j) \right)+\sin\lambda\,Z(i,p)=0. \tag{33}\] Indeed horizontal top exits have winding zero, and closing along the leftward boundary gives winding \(\pi/3\) at \(p\). Lateral truncation errors vanish as before, dominated by a strip of height \(N+1\).

Sum (33) over \(-L\le i\le L\). Reversal pairs and cancels all internal arches. The uncanceled terms cross the two ends of this interval. After translating each end to the origin, the stepped domains converge locally to strips of heights \(N\) and \(N+1\), respectively. All crossing sums are dominated by the first moment in the strip of height \(N+1\), which is finite by transfer decay. This also controls paths leaving an arbitrarily large lateral window for each fixed pair of ports. Dominated convergence therefore gives the exact identity \[ m_{N+1}-m_N=\frac{\sin\lambda}{\sin(3\lambda)} \sum_i Z(p,i). \tag{34}\]

Turning down immediately from \(p\) and following an arbitrary bridge in the base strip yields \(\sum_iZ(p,i)\ge\rho\mathcal B_N\). The top-source bridge mass in the base strip equals its bottom-source mass by translation and reversal.

For the upper bound, order the path’s visits to the interface between the base strip and the new row from left to right. In the new row, whose top is vacant, disjoint intervals pair \(p\) with the first visit and then pair subsequent visits consecutively. The base portions are a noncrossing pairing with one unpaired visit connected to the bottom; that defect cannot be enclosed. Since the union is a single path, the first base visit must pair with its next neighbor: pairing it farther away would leave a closed component inside. Removing these two visits and repeating proves that base visits also pair consecutively, with the last visit the defect. Thus the path encounters the interface in increasing horizontal order.

At \(u=\lambda\), the initial new-row segment has total weight at most \(A/(1-B)\). A one-sided base return arch has total mass at most \(1/(2c)\), using the boundary identity from the top and equality of the two directions by translation and reversal. A one-sided return through the new row has total mass \(AU/(1-B)=2cB\). The final base segment has total mass \(\mathcal B_N\). Ignoring avoidance between pieces only increases the sum, since the base weights are products over vertices. Therefore \[\sum_iZ(p,i) \le\frac{A}{1-B}\sum_{r\ge0} \left(\frac1{2c}\,2cB\right)^r\mathcal B_N =\frac{A}{(1-B)^2}\mathcal B_N.\] Here \(B=B(\lambda)=\rho^2<1\), so the constants are independent of \(N\). Combine this with (34). ◻

Proof. Fix \(i\) and write \(f_j=\cos(v+j\lambda)\) with \(f_0=a_i\). The product in (35) gives \(\Phi_X(v\pm2\lambda)=0\). The deletion quotient also gives \(\Phi_X(v+6\lambda)+\Phi_X(v-6\lambda)=0\). To verify its sign and factors, cancel the contributions of the other \(a_j\) in the ratio: on both sides these contributions involve exactly \(f_6,f_{-6},f_4,f_{-4},-f_0\). The remaining expressions are \[R_\pm=\sin(v\pm6\lambda)(f_{\pm6}-C) (f_0+f_{\pm6}+\rho)(f_{\pm6}-f_0)(f_{\pm4}-f_0).\] Product-to-sum and \(\rho=2cd\) give \[R_\pm=\cos(2v)(f_{\pm6}-C) \sin(v\pm3\lambda)(\cos(v\pm3\lambda)+d).\] The last two expressions after removing \(\cos(2v)\) are trigonometric polynomials of degree three. They have the same six zeros \(v=\pm3\lambda,\pm5\lambda,\pm7\lambda\) modulo \(2\pi\), and opposite values at zero; hence they are negatives of each other. This proves the claimed cancellation as a polynomial identity, including where canceled factors vanish.

Consider the sum of the four shifts of \(\Phi_X\) by \(\pm2\lambda,\pm6\lambda\). On frequency \(k\) its multiplier is \(4\cos(4k\lambda)\cos(2k\lambda)\), nonzero precisely when four divides \(k\). Its maximum frequency is \(4N+2\). Dividing this shifted sum by \(\sin4w\) therefore gives a polynomial in \(\cos4w\) of degree at most \(N-1\). The preceding identities give \(N\) distinct roots, for generic \(a_i\), so it is zero. This proves the harmonic assertion.

Every polynomial in \(b\) has a unique expression \(P_0(h)+bP_1(h)+b^2P_2(h)+b^3P_3(h)\), by repeatedly using \(b^4=b^2-h\). The first three terms, after multiplication by \(\sin w\), have no frequency divisible by four. For the last term use \[\sin w\,b^3=\frac14\sin2w+\frac18\sin4w.\] Its projection onto multiples of four is \(\frac18\sin4w\,P_3((1-\cos4w)/8)\), so it vanishes only if \(P_3=0\). The degree bounds follow from \(\deg Q_X=4N+1\). ◻

Proof. We first turn the zero-multiplicity conditions into Cauchy-transform moment equations. A positive determinant will make those equations uniquely solvable and control the coalescing limit. We then identify the resulting coefficient with the expectation in (40). The proposed probability law is well defined: \(\omega(x)=O(x^{-1/4})\) at zero and each pair factor is at most \(y_j\le1\), so its normalizing integral is finite; it is positive because the density is positive inside the ordered region.

Cauchy moment equations. The two prescribed zeros satisfy \(h(C)=h(c)=1/8\). To place them at infinity, set \(z=(8h-1)^{-1}\), so that the inverse relation is \[b^2(1-b^2)=\frac{1+1/z}{8}.\] We use the two branches near infinity tending to \(C\) and \(c\), respectively. Their differences from these limits have order \(z^{-1}\), since \(h'(C)\) and \(h'(c)\) are nonzero. For each polynomial in (37), multiply by \(z^N\) after the substitution \(h=(1+1/z)/8\). Denote the resulting polynomials by \(\widetilde{\mathcal P}\), \(\widetilde{\mathcal J}\), and \(\widetilde{\mathcal E}\); the last satisfies \(\widetilde{\mathcal E}(0)=0\) because \(\deg\mathcal E\le N-1\). The branch tending to \(C\) and the sum of the two branches are, respectively, \[f(z)=\sqrt{\frac{1+\sqrt{(1-1/z)/2}}2},\qquad l(z)=\sqrt{1+\sqrt{(1+1/z)/2}}.\] These formulas continue \(f\) analytically off \([0,1]\) and \(l\) off \([-1,0]\). On the \(C\) branch, the zero of order \(2N+1\) in (37) becomes \(O(z^{-2N-1})\); on the \(c\) branch, the zero of order \(N\) becomes \(O(z^{-N})\). After multiplication by \(z^N\) this gives \[\begin{align*} \mathcal F(z)&=\widetilde{\mathcal P}(z) +f(z)\widetilde{\mathcal J}(z) +f(z)^2\widetilde{\mathcal E}(z)=O(z^{-N-1}), \tag{41}\\ \mathcal D(z)&=\widetilde{\mathcal J}(z) +l(z)\widetilde{\mathcal E}(z)=O(1). \tag{42}\end{align*}\] For the second statement, evaluate the same polynomial expression on the branch tending to \(c\), where multiplication by \(z^N\) makes it \(O(1)\), and subtract the expression on the \(C\) branch; the quotient by their nonzero difference is \(\mathcal D\).

With subscripts \(+\) and \(-\) denoting boundary values from the upper and lower half-planes, the jumps across the cuts are \[f_+(x)-f_-(x)=2i\omega(x),\qquad l_+(-x)-l_-(-x)=-2\sqrt2\,i\omega(x),\qquad f_+(x)+f_-(x)=l(x).\] At the outer endpoints these functions are bounded. At zero, \(\mathcal F=O(|z|^{-1/4})\) and \(\mathcal D(z)\to D_0:=\widetilde{\mathcal J}(0)\); the latter uses \(\widetilde{\mathcal E}(0)=0\). Cauchy’s formula, with small endpoint circles shrinking to zero, yields \[\begin{align*} \mathcal F(z)&=\frac1\pi\int_0^1 \frac{\mathcal D(x)\omega(x)}{x-z}\,dx, \tag{43}\\ \int_0^1x^j\mathcal D(x)\omega(x)\,dx&=0, \qquad 0\le j<N, \tag{44}\\ \mathcal D(z)&=D_0-\frac{\sqrt2}{\pi}z \int_0^1\frac{\widetilde{\mathcal E}(-y)/y}{z+y} \omega(y)\,dy. \tag{45}\end{align*}\] The moment equations are the first \(N\) coefficients of the expansion of (43) at infinity. To obtain (45), subtract the bounded-at-infinity representation at zero. Its integrand is integrable there because \(\widetilde{\mathcal E}(-y)/y\) is a polynomial.

To control the endpoint at zero, use \(\omega(x)=O(x^{-1/4})\) and \(\widetilde{\mathcal E}(z)=O(z)\) near zero. The increment in (45), split at \(y=x\), is bounded by \[C\left(\int_0^x y^{-1/4}\,dy +x\int_x^1y^{-5/4}\,dy\right)=O(x^{3/4}).\] This estimate will also justify the singular moment integral below.

Nonsingularity and coalescence. The polynomial \(\widetilde{\mathcal E}(-y)/y\) has degree at most \(N-1\), and the moment equations determine it uniquely from \(D_0\). Indeed the corresponding matrix, apart from nonzero scalar factors, is \[ M_{jk}=\int_0^1\int_0^1 \frac{x^{j+1}y^k}{x+y}\omega(x)\omega(y)\,dx\,dy, \qquad 0\le j,k<N. \tag{46}\] Its determinant is strictly positive. This is the Cauchy-bimoment positivity argument used in [3]; we give the finite determinant calculation for the present weights. Expand the determinant and antisymmetrize successively in the \(x\) and \(y\) variables. On the ordered integration regions the integrand becomes the product of two positive Vandermonde determinants, \(\prod_i x_i\), positive weights, and the Cauchy determinant \[\det\left[\frac1{x_i+y_j}\right] =\frac{\prod_{i<j}(x_j-x_i)(y_j-y_i)}{\prod_{i,j}(x_i+y_j)}>0.\] For completeness this identity follows by multiplying through by the denominator: the numerator is alternating in each family and has exactly the degree of the two Vandermondes. The remaining constant is one, recursively evaluating at \(x_N=-y_N\).

Consider now a solution of the homogeneous linear problem for \(g\), so \(\deg g\le N-1\). The coefficient of \(b^{4N+1}\) in \(Q\) is zero. Since \[[b^{4N+1}]Q=(-8)^N D_0,\] we obtain \(D_0=0\). Positivity of (46) forces \(\widetilde{\mathcal E}=0\); then (45) gives \(\widetilde{\mathcal J}=0\), and (41) gives \(\widetilde{\mathcal P}=0\). Thus \(g=0\). For the monic problem, write \(g(b)=b^N+\sum_{r=0}^{N-1}\gamma_r b^r\). The excluded \(b^3P_3(h)\) part in the proof of Lemma 9 has \(N\) coefficients. Requiring them to vanish therefore gives an \(N\times N\) linear system for \((\gamma_0,\ldots,\gamma_{N-1})\), and the established injectivity proves that this system is invertible at the homogeneous parameters. This proves existence and uniqueness of \(g\). Apply the same coefficient conditions to (35), with \(g_X\) replaced by a monic unknown polynomial. Their matrix and right side depend polynomially on the \(a_i\), through the displayed zero factors, and converge to those of the homogeneous system. This coefficient matrix is consequently invertible for all \(a_i\) sufficiently close to \(C\). The generic \(g_X\) satisfies that system by Lemma 9, so it converges to \(g\). The moment itself is regular at the homogeneous point by Lemma 4.

The moment coefficient and its expectation. For this monic solution \(D_0=(-8)^{-N}\). Put \(G(z)=\mathcal D(z)/D_0\). The leading coefficient computation also gives \([b^{4N}]Q=(-8)^N\widetilde{\mathcal P}(0)\). Comparing this with the product in (37) gives the coefficient needed by the arch formula: \[[b^{N-1}]g(b)=\frac{\widetilde{\mathcal P}(0)}{D_0} +(2N+1)C+Nc.\] Substitution into (32), using \(C=\eta(2C+c)\), yields the first equality below: \[ m_N=-\eta\left(C+\frac{\widetilde{\mathcal P}(0)}{D_0}\right) =\frac\eta\pi\int_0^1(1-G(x))\frac{\omega(x)}x\,dx. \tag{47}\] For the second equality, use \[f(z)=C+\frac1\pi\int_0^1\frac{\omega(x)}{x-z}\,dx\] and subtract \(D_0f\) from (43), then let \(z\to0-\). Equation (45) gives \(\mathcal D(x)-D_0=O(x^{3/4})\), so the integrand after subtraction is \(O(x^{-1/2})\) at zero. This ensures integrability and justifies the limit. On the left, terms involving \(f(z)(\widetilde{\mathcal J}(z)-D_0)\) and \(f(z)^2\widetilde{\mathcal E}(z)\) tend to zero.

The moment equations now have a unique solution. It remains to exhibit their probabilistic solution, thereby identifying \(G\) with the expectation in (40). Let \(\mathcal W_n\) denote the product in (39) with \(n\) coordinates, viewed on \((0,1)^n\) without the ordering constraint. We use its normalized symmetric law for \(n=N\). Partial fractions give \[\prod_{i=1}^N\frac{y_i-x}{y_i+x} =1-x\sum_{i=1}^N\frac{\alpha_i}{x+y_i}, \qquad \alpha_i=2\prod_{j\ne i}\frac{y_j+y_i}{y_j-y_i}.\] After multiplication by the density, the factors in \(\alpha_i\) cancel one Vandermonde and all corresponding denominator factors. Each canceled term is integrable: its remaining polynomial factors and pair kernels are bounded on \((0,1)^N\), as is \(x/(x+y_i)\), and the one-variable weights are integrable. Integrating over the other variables therefore gives a representation \(1-x\int_0^1P(y)\omega(y)/(x+y)\,dy\) with \(\deg P\le N-1\). Its moments against \(x^k\omega(x)\) vanish for \(k<N\). To see this, adjoin \(x\) as an \((N+1)\)st variable and use the density identity \[x^k\omega(x)\mathcal W_N(y_1,\ldots,y_N) \prod_{i=1}^N\frac{y_i-x}{y_i+x} =\frac{x^k\mathcal W_{N+1}(x,y_1,\ldots,y_N)} {\prod_{i=1}^N(y_i-x)}.\] The left side is absolutely integrable, since the product has modulus at most one. Symmetrizing the right side in all \(N+1\) variables gives zero by the elementary interpolation identity \[\sum_{r=0}^N\frac{z_r^k}{\prod_{j\ne r}(z_j-z_r)}=0, \qquad 0\le k<N.\] Thus the expectation has the same polynomial Cauchy representation and vanishing moments as \(G\). The uniqueness argument using (46) now identifies this expectation with \(G\). Substitute into (47). ◻

Proof. The mixed derivative of each pair contribution is \[\frac{\partial^2}{\partial x\partial y}\log K(x,y) =\frac2{(y-x)^2}+\frac1{(x+y)^2}>0.\] We prove association by induction on the number of coordinates, initially for bounded increasing functions. Conditional on the largest coordinate \(t\), the remaining coordinates have a law of the same form on the shorter interval, with modified one-variable factors. They are associated by induction. If \(s<t\), the ratio of the conditional density at \(s\) to that at \(t\), on the former support, is coordinatewise decreasing by the mixed-derivative inequality. Extended by zero where a coordinate exceeds \(s\), it remains decreasing on the latter support. Association for the conditional law at \(t\) shows that a decreasing likelihood tilt decreases every increasing expectation. If the ratio is unbounded, first truncate it and pass to the limit by monotone convergence. Thus the conditional expectation of any increasing function of all the coordinates increases with \(t\).

The law of total covariance now has two nonnegative terms: conditional covariance by induction, and covariance of the two conditional expectations by the one-dimensional inequality \[\operatorname{Cov}(F(T),G(T)) =\frac12\mathbb E[(F(T)-F(T'))(G(T)-G(T'))]\ge0.\] Here \(T'\) is an independent copy of \(T\). This proves association; truncation and integrable approximation extend the assertion to the stated class. Finally for an increasing likelihood ratio \(L\), \(\mathbb E[FL]/\mathbb EL\ge\mathbb EF\) for increasing \(F\), and the reversed inequality holds for decreasing \(L\). This is the claimed stochastic comparison. ◻

Proof. Normalization. Write the pair product as a Vandermonde determinant times the Schur Pfaffian with entries \((s-r)/(s+r)\). The classical identity, in the form recorded in [19], is \[\operatorname{Pf}\left[\frac{x_j-x_i}{x_j+x_i}\right] =\prod_{i<j}\frac{x_j-x_i}{x_j+x_i}.\] For even size, clearing denominators leaves an alternating polynomial of Vandermonde degree, so only its constant factor needs checking. Pairing two opposite arguments gives that constant recursively as one. For odd size, add an argument tending to infinity; the resulting augmenting column consists of ones.

Expand the determinant and the Pfaffian and integrate over the unordered region, divided by \(n!\). This is de Bruijn’s determinant–Pfaffian integration formula [5]; the general antisymmetric-kernel form, including its odd-size border, is stated in [1]. The result is the Pfaffian of double moments, with the single moments as augmenting entries in odd size. Absolute integrability follows from \(|(s-r)/(s+r)|\le1\) and the finite gamma moments for \(a>0\). The double moment with indices \(i,j\) is \[\int_0^\infty\!\int_0^\infty r^{a+i-1}s^{a+j-1}e^{-r-s}\frac{s-r}{s+r}\,dr\,ds =\Gamma(a+i)\Gamma(a+j)\frac{j-i}{2a+i+j}.\] Use \(r+s\) and \(r/(r+s)\) to evaluate the integral; the latter beta integral has mean \((a+i)/(2a+i+j)\). Factoring out the gamma terms and using the Schur identity again, now at the arguments \(a+i\), proves (50).

Coordinate tails and inverse moments. For the maximum, condition on its value \(s\). Every factor involving a smaller coordinate is at most \(s\), since \(K(r,s)\le s\) for \(0<r<s\). Consequently \[\mathbb P_{n,a}(r_n>Kn) \le\frac{Z_{n-1}(a)}{Z_n(a)} \int_{Kn}^\infty s^{a+n-2}e^{-s}\,ds.\] The explicit normalizer gives \[\frac{Z_{n-1}(a)}{Z_n(a)} =\frac1{\Gamma(a+n-1)} \prod_{j=1}^{n-1}\frac{2a+2n-2-j}{j} \le\frac{e^a(2e)^{n-1}}{\Gamma(a+n-1)}.\] For \(n\ge2\), the inequality follows by replacing every numerator by \(2(a+n-1)\), using \((n-1)!\ge((n-1)/e)^{n-1}\), and then \((1+a/(n-1))^{n-1}\le e^a\); \(n=1\) is immediate. The exponential-moment bound for a gamma random variable gives \[\frac1{\Gamma(a+n-1)}\int_{Kn}^\infty s^{a+n-2}e^{-s}\,ds \le e^{-Kn/2}2^{a+n-1}.\] Choosing \(K\) large makes the desired tail uniformly small, and in fact exponentially small in \(n\).

For the small-coordinate estimate, integrate the first \(k\) coordinates below \(h\). Their mutual pair factors are at most \(h\); a factor between a small coordinate and a remaining one is at most the latter. Discarding order constraints between these two groups therefore gives \[ \mathbb P_{n,a}(r_k<h) \le\frac{h^{D_k}}{a^k k!}\frac{Z_{n-k}(a+k)}{Z_n(a)}, \tag{55}\] with \(Z_0=1\). From (50), \[\frac{Z_{n-k}(a+k)}{Z_n(a)} =\prod_{i=0}^{k-1}\Gamma(a+i)^{-1} \prod_{\substack{0\le i<j<n\\i<k}} \left(1+\frac{2a+2i}{j-i}\right).\] For gaps \(j-i\le k\), use \(2a+2i\le C_a k\). For each \(i<k\), \[\sum_{j=1}^k\log(1+C_a k/j) \le k\log(1+C_a)+k\log k-\log(k!)=O_a(k).\] Thus the short gaps contribute \(O_a(k^2)=O_a(D_k)\). For larger gaps, \(\log(1+x)\le x\) gives a contribution at most \[\sum_{i<k}(2a+2i)\sum_{j=k+1}^{n-1-i}\frac1j \le 2D_k\log(n/k).\] This includes \(k=n\), when all the latter sums are empty. Furthermore \[ \prod_{i=0}^{k-1}\Gamma(a+i)\ge k^{D_k}e^{-O_a(D_k)}. \tag{56}\] Here is a direct estimate. For \(q=a+i\), integration on \([q,q+1]\) gives \(\log\Gamma(q)\ge(q-1)\log q-q-O_a(1)\). Summing, separating the factor \(\log k\), and using \(\sum_{i<k}(a+i)\log((a+i)/k)\ge-k^2/e\) gives \(D_k\log k-O_a(D_k)\); the missing \(k\log k\) is absorbed in \(O_a(D_k)\). Adjust the constant for bounded \(k\). Combining these bounds with (55), and absorbing \(a^{-k}/k!\) into \(C_a^{D_k}\), proves (51).

For any \(0<p<D_k\), integration of that tail estimate gives \[ \mathbb E_{n,a}r_k^{-p} \le\frac{D_k}{D_k-p} \left(\frac{C_a n^2}{k^3}\right)^p. \tag{57}\] At \(a=1/2\), apply this with \((k,p)=(1,1/4)\) and then with \(p=1\), \(k\ge2\). It gives \(\mathbb Er_1^{-1/4}\lesssim n^{1/2}\) and \(\mathbb E\sum_{k\ge2}r_k^{-1}\lesssim n^2\). Subadditivity of the fourth root and Jensen’s inequality yield (52). At \(a=2\), use \(p=1\) for every \(k\) to obtain (53).

The minimum at \(a=1\). The density of the smallest coordinate at \(s>0\) is \[e^{-s}\frac{Z_{n-1}(2)}{Z_n(1)} \mathbb E_{n-1,2}\left[ \mathbf1_{\{r_i>s\ \forall i\}} \prod_i\frac{(1-s/r_i)^2}{1+s/r_i}\right].\] The normalizer ratio is exactly \(n(n+1)/2\). Set \(q_s(r)=\mathbf1_{\{r>s\}}(1-s/r)^2/(1+s/r)\). For all \(r>0\), including \(r\le s\), one has \(0\le q_s(r)\le1\) and \(1-q_s(r)\le3s/r\). Consequently \[\prod_iq_s(r_i)\ge1-3s\sum_i r_i^{-1}.\] By (53), the expectation is at least \(1/2\) whenever \(0<s<\varepsilon n^{-2}\) for a sufficiently small fixed \(\varepsilon>0\). The displayed density is then bounded below by a positive multiple of \(n^2\). An interval with endpoints fixed positive multiples of \(n^{-2}\) has probability bounded below; on it \(r_1^{-1/4}\gtrsim n^{1/2}\). This proves (54), with \(n=1\) also covered directly. ◻

Completion of Theorem 1. We compare (39) with the reference laws to estimate the two sides of (48). For the upper estimate, first condition the target law on \(y_N<1/2\). This decreasing event shifts all coordinates downwards, by Lemma 11, and therefore increases the inverse statistic. On this shorter interval compare it with \(y_i=r_i/(bN)\) under (49) with \(a=1/2\), conditioned on \(r_N<bN/2\). Up to a constant, the target-to-reference likelihood ratio is the product of \[\omega(y)y^{1/2}e^{bNy}.\] Direct differentiation gives \[\frac{d}{dy}\log(\omega(y)y^{1/2}) =\frac{1-2y-\sqrt{2y/(1+y)}}{4y(1-y)}\ge-4, \qquad 0<y\le\frac12.\] Indeed the numerator is positive for \(y\le1/8\); on the remaining interval it is at least \(-1\) and the denominator is at least \(1/4\). Thus the likelihood factors are increasing for every \(N\ge1\) once \(b\ge4\). Association shows that the conditioned reference law has the larger inverse statistic. Choosing also \(b/2\ge K_{1/2}\) makes its conditioning probability bounded below uniformly in \(N\), by Lemma 12. Hence \[\mathbb E\left(\sum_i y_i^{-1}\right)^{1/4} \lesssim (bN)^{1/4} \mathbb E_{N,1/2}\left(\sum_i r_i^{-1}\right)^{1/4} \lesssim N^{3/4}.\]

For the lower estimate, set \(y_i=2z_i/(1+z_i)\) in the target law, so \(0<z_1<\cdots<z_N<1\). The transformed pair kernel is \[K\left(\frac{2z_i}{1+z_i},\frac{2z_j}{1+z_j}\right) =\frac{2K(z_i,z_j)} {(1+z_i)(1+z_j)(1+2z_iz_j/(z_i+z_j))}.\] Together with the Jacobian this gives one-variable factors \(\omega(2z/(1+z))(1+z)^{-(N+1)}\) and additional pair factors \((1+2z_iz_j/(z_i+z_j))^{-1}\). Compare with \(z_i=r_i/(bN)\) under the reference law with \(a=1\) and fixed \(0<b<1/2\), conditioned on \(r_N<bN\). The ratio has individual factors \[\omega\left(\frac{2z}{1+z}\right)(1+z)^{-(N+1)}e^{bNz}\] and the additional pair factors just displayed. Each is decreasing in every coordinate: \(\omega\) is decreasing, and \(-(N+1)/(1+z)+bN<0\) for \(0<z<1\). Association therefore puts the target \(z\) coordinates below those of this conditioned reference, which are themselves below the unconditioned reference coordinates. Since \(y_1\le2z_1\), it follows that \[\mathbb Ey_1^{-1/4} \ge 2^{-1/4}(bN)^{1/4}\mathbb E_{N,1}r_1^{-1/4} \gtrsim N^{3/4}.\] Equation (48) now proves \(m_N\asymp N^{3/4}\).

It remains to extract the bridge estimate from Lemma 8. By monotonicity, summing the lower increment bound over \(1\le j<N\) gives \((N-1)\mathcal B_N\lesssim m_N\), so \(\mathcal B_N\lesssim N^{-1/4}\), with bounded \(N\) absorbed in the constant. Conversely choose a fixed integer \(L\) so large that the lower bound for \(m_{LN}\) exceeds twice the upper bound for \(m_N\). Then \[N^{3/4}\lesssim m_{LN}-m_N \lesssim\sum_{j=N}^{LN-1}\mathcal B_j \le(L-1)N\mathcal B_N.\] This proves \(\mathcal B_N\gtrsim N^{-1/4}\) and completes all assertions of the theorem. ◻

Proof. Equation (8) follows by direct substitution. For completeness we record a coefficient check of the splitting identity. Subscripts \(1,2\) mean arguments \(x+s\lambda,x\). Let \(q\) be the cup coefficient and \(k\) the strand coefficient of the splitter: \(q=k=\rho\) for \(s=2\), and \(q=1,k=0\) for \(s=3\). Unsubscripted weights refer to \(\widehat R_s\); in the deletion case use \(B=1\) and set its other nonempty two-slot coefficients to zero. The inputs are the passing auxiliary strand and the slot to be split; the latter is absent in the deletion case. Outputs \(1,2\) are the split slots and output \(3\) is the auxiliary slot. Input bits record occupancy. For one input strand, a bare output index specifies its destination and a parenthesized pair specifies an additional cup. The complete coefficient check is \[\begin{array}{c|c|c|c} \text{input}&\text{connections at output}&\text{left side}&(S_s\otimes I)\widehat R\\\hline 00&\emptyset&1&1\\ &12\ (\text{cup})& U_1 A_2+q B_1 B_2 &q\\ &13&U_1 B_2+q B_1 A_2 & k U\\ &23&U_2+q A_1 E_2&k U\\\hline 10&1&A_1+q F_1 U_2 & k A\\ &2&B_1 A_2+q U_1 B_2 &k A\\ &3&B_1 B_2+q U_1 A_2 & B\\ &3,(12)&q F_1 E_2 &q B\\ &1,(23)&A_1 U_2+q(E_1 E_2+F_1 F_2)&0 \end{array}\]

Here are the substitutions verifying these equalities. Put \(s_j=\sin(x+j\lambda)\). The weights at \(x+j\lambda\) have common denominator \(s_{j+2}s_{j+3}\) and numerators \[\begin{array}{c|ccccc} \text{weight}&A&B&U&E&F\\\hline \text{numerator}&d s_{j+5}&s_j s_{j+5}&d s_j& s_{j+5}s_{j+6}&-s_{j+7}s_j \end{array}\] Substitution uses the following product-to-sum identities: \[\begin{gathered} s_{j+8}=-s_j,\quad s_j s_{j+1}+s_{j+4}s_{j+5}=C,\quad s_2s_6=d^2-s_0^2,\quad s_1s_5=d^2-s_7^2,\\ \rho(s_1-s_7)=s_0,\quad \rho(s_6-s_0)=s_7,\quad \rho(s_1s_2-s_6s_7)=s_0s_4,\quad \rho(s_5s_6-s_0s_1)=s_7s_3,\\ \rho(s_7s_0+s_4s_5)=s_6s_2,\quad \rho(s_2s_3+s_7s_0)=s_1s_5,\quad s_1s_6-s_7s_0=\rho d^2,\\ s_1s_6s_5=s_7s_2s_3+d^2s_0,\qquad s_1s_6s_2=s_0s_4s_5+d^2s_7 \end{gathered}\] The last two trigonometric identities also follow from the preceding formulas for \(s_1s_5\) and \(s_6s_2\). Thus substitution verifies every entry in the tables.

For (10), the sector with one strand reduces to multiplication of two matrices with diagonal entries \(A\) and off-diagonal entries \(B\). In the even sector the required identities are \[E(x)E(-x)=1,\qquad U(x)+E(x)U(-x)=0,\] \[E(x)F(-x)+F(x)E(-x)+U(x)U(-x)=0.\] A cap composed with a cup creates a loop and therefore vanishes. These identities again follow from (6).

To prove (7), first set \(y=s\lambda\), \(s=2,3\). Use (8) and (9), together with their transposes and the reversal of slot order. The effective box is symmetric, and both sides become \((I\otimes S_s)\widehat R_s(x)(S_s^{\mathsf t}\otimes I)\). Unitarity transports these identities to \(x+y=s\lambda\), by applying the identity with arguments \((-x,x+y)\). The case \(y=0\) follows from \(R(0)=I\).

Multiply the difference of the two sides by their three scalar denominators, and call the resulting diagram sum \(\mathcal F(x,y)\). Each coefficient is a Laurent polynomial in \(e^{iy}\) with exponents between \(-4\) and \(4\). To see its parity, let \(D_j\) multiply a diagram by \(-1\) when slot \(j\) is occupied. Inspection of the nine boxes gives \[R_i(v+\pi)=D_{i+1}R_i(v)D_i.\] On either side of the braid relation the two internal \(D_2\) signs cancel. The remaining signs are at input slot \(1\) and output slot \(3\); the scalar denominators are \(\pi\)-periodic. Hence \[\mathcal F(x,y+\pi)=D_3\mathcal F(x,y)D_1.\] Every boundary coefficient has fixed parity. After multiplication by a Laurent monomial it is a polynomial of degree at most four in \(e^{2iy}\). We have proved its vanishing at \(y=0,2\lambda,3\lambda,2\lambda-x,3\lambda-x\); these five values are distinct modulo \(\pi\) for generic \(x\). Interpolation proves the identity, and rational continuation removes the genericity restriction. ◻