Universality of the Limit Shape of RandomConvex Lattice Polygons

Leonid V. Bogachev1 and Sakhavat M. Zarbaliev2

1 Department of Statistics, University of Leeds, Leeds LS2 9JT, UK.E-mail: bogachev@maths.leeds.ac.uk

2 International Institute of Earthquake Prediction Theory & MathematicalGeophysics, Moscow 117997, Russia. E-mail: szarbaliev@mail.ru

Abstract

Let Πn be the set of planar convex lattice polygons Γ (i.e., with vertices on Z2+

and non-negative inclination of all edges) with fixed endpoints 0 = (0, 0) andn = (n1, n2). We are concerned with the limit shape of a typical polygon Γ ∈ Πn

as n →∞with respect to a certain parametric family of probability measures {P rn}

(0 < r < ∞) on the space Πn, including the uniform distribution (r = 1). We showthat if 0 < C1 ≤ n2/n1 ≤ C2 < ∞ then, under the scaling (1/n1, 1/n2), the limitshape is universal in the class {P r

n} and thus coincides with that for the uniformdistribution P 1

n (found independently by Vershik, Barany, and Sinai). Our resultgives a partial affirmative answer to Vershik–Prokhorov’s universality conjecture.The measure P r

n is constructed, using Sinai’s approach, as a conditional distribu-tion induced by a suitable product measure Qr defined on the space Π = ∪nΠn

of polygons with a free right end. The proof involves subtle analytical tools in-cluding the Mobius inversion formula and properties of zeroes of the Riemann zetafunction.

Key words and phrases: Convex lattice polygons; Limit shape; Local limit theorem

2000 MSC: Primary 52A22; Secondary 05A17, 05D40, 60F05, 60G50

1. Introduction

1.1. Background: the limit shape

Convex lattice polygon Γ is a planar piecewise linear path starting at the origin, with verticeson the integer lattice Z2

+ = {(i, j) ∈ Z2 : i, j ≥ 0}, and such that the slope of its consecutiveedges is increasing. Let Π be the set of all such polygons with finitely many edges and Πn thesubset of polygons Γ ∈ Π with their right endpoint ξ = ξΓ fixed at n = (n1, n2) ∈ Z2

+.This paper is concerned with the problem of limit shape of a typical polygon Γ ∈ Πn, as

n →∞, with respect to some probability measure Pn on Πn. The limit shape is a curve γ0 onthe plane such that, under a suitable scaling transformation Sn : R2

+ → R2+ and for all large

enough n, scaled polygons Γn = Sn(Γ) with high Pn-probability lie within an arbitrarily smallneighborhood of γ0; that is to say, for any ε > 0,

limn→∞

Pn{d(Γn, γ0) ≤ ε} = 1, (1.1)

Research partially supported by DFG Grant 436 RUS 113/722.

1

where d(·, ·) is a suitable metric on the path space — for example, defined by the Hausdorffdistance between compact sets on the plane:

dH(A, B) :=1

2maxx∈A

miny∈B

|x− y|+ 1

2maxy∈B

minx∈A

|x− y|, A,B ⊂ R2. (1.2)

Of course, the limit shape γ0 and its very existence may depend on the chosen probability lawPn on the polygon space Πn. From the probabilistic point of view, the result (1.1) gives a lawof large numbers for random polygons Γ with respect to the distribution Pn on Πn.

The limit shape problem for convex lattice polygons under the uniform distribution on Πn

(i.e., such that Pn(Γ) = 1/#(Πn) for every Γ ∈ Πn) was solved independently by Vershik [24],Barany [3], and Sinai [21], who showed that under the scaling Sn : (y1, y2) 7→ (y1/n1, y2/n2),the limit shape exists and is given by the parabola arc γ∗ defined by the equation

√1− u1 +

√u2 = 1, 0 ≤ u1, u2 ≤ 1. (1.3)

More precisely, if n = (n1, n2) →∞ so that n2/n1 → c ∈ (0,∞) then, for any ε > 0,

limn1→∞

#{Γ ∈ Πn : dH(Γn, γ∗) ≤ ε}

#{Γ ∈ Πn}= 1. (1.4)

The proofs in papers [24, 3] involved a blend of combinatorial, functional and geometricarguments and were based on a direct analysis of the corresponding generating function usinga multivariate method of steepest descent for a Cauchy integral [24] or a suitable Tauberiantheorem [3]. Sinai [21] proposed an alternative, probabilistic method based on randomizationof the right endpoint of polygon Γ; we shall comment more on this approach below. Let uspoint out that the short paper [21] contained the basic ideas but only sketches of proofs. Someof these techniques were subsequently elaborated by Bogachev and Zarbaliev [5, 6] and byZarbaliev in his PhD thesis [32], however a complete proof has not been published yet. Notethat large deviations for random convex polygons were studied by Vershik and Zeitouni [29].

1.2. Main result

Vershik [24, p. 20] pointed out that it would be interesting to study asymptotic properties ofrandom convex polygons under other probability distributions on Πn, and conjectured thatthe limit shape might be universal for some classes of distributions. Independently, a simi-lar hypothesis was put forward by Prokhorov [20]. In the present paper, we prove Vershik–Prokhorov’s universality hypothesis for a parametric family of probability measures P r

n (0 <r < ∞) on Πn, defined by

P rn(Γ) :=

br(Γ)

Brn

, Γ ∈ Πn, (1.5)

with the coefficientsbr(Γ) :=

∏ei∈Γ

brki

, Brn :=

∑Γ∈Πn

br(Γ), (1.6)

where the product is over all edges ei of the polygon Γ ∈ Πn, ki is the number of lattice pointson the edge ei (except its left endpoint included in the previous edge), and

brk :=

(r + k − 1

k

)=

r(r + 1) · · · (r + k − 1)

k!, k = 0, 1, 2, . . . (1.7)

2

Note that for r = 1 the measure (1.5) is reduced to the uniform distribution on Πn. Quali-tatively, formulas (1.6), (1.7) introduce certain probability weights for random edges on Γ byencouraging (r > 1) or discouraging (r < 1) lattice points on each edge as compared to thereference case r = 1.

We assume throughout that 0 < C1 ≤ n2/n1 ≤ C2 < ∞. Consider scaled polygonsΓn := Sn(Γ) (Γ ∈ Πn), where Sn(y) := (y1/n1, y2/n2), y = (y1, y2) ∈ R2. It is convenientto work with a sup-distance between Γn and the limit shape curve γ∗, based on the tangentialparameterization of convex paths (see more details in the Appendix, Section 9.1). For any t ∈[0,∞], denote by ξn(t) the right endpoint of the part of Γn where the tangent slope (whereverit exists) does not exceed t. Similarly, the tangential parameterization of γ∗ is given by thevector function

g∗(t) =

(t2 + 2t

(1 + t)2,

t2

(1 + t)2

), 0 ≤ t ≤ ∞.

The tangential distance dT between Γn and γ∗ is defined as

dT (Γn, γ∗) := sup

0≤t≤∞|ξn(t)− g∗(t)|, (1.8)

where | · | denotes the Euclidean norm in R2. We can now state our main result about univer-sality of the limit shape γ∗ under the measures P r

n .

Theorem 1.1. For each r ∈ (0,∞) and any ε > 0,

limn→∞

P rn{dT (Γn, γ

∗) ≤ ε} = 1.

In fact, it can be shown that the Hausdorff distance dH (see (1.2)) is dominated by the tan-gential distance dT (however, these metrics are not equivalent; see the Appendix, Section 9.1).In particular, Theorem 1.1 with r = 1 recovers the limit shape result (1.4) for the uniformdistribution on Πn. As mentioned earlier, in the original paper by Sinai [21] the proof of thelimit shape result was only sketched, therefore even in the uniform case our proof seems tobe the first published complete realization of Sinai’s probabilistic method (which, as we try toexplain below, is far from straightforward). Let us also point out that Theorem 1.1 is a non-trivial extension of (1.4) since the measures P r

n (r 6= 1) are not close to the uniform distributionP 1

n in total variation, and moreover, ‖P rn − P 1

n‖TV → 1 as n → ∞ (see Theorem 9.4 in theAppendix, Section 9.2).

The result of Theorem 1.1 for “pure” measures P rn readily extends to mixed measures.

Theorem 1.2. Let ρ be a probability measure on (0,∞), and set

P ρn(Γ) :=

∫ ∞

0

P rn(Γ) ρ(dr), Γ ∈ Πn. (1.9)

Then, for any ε > 0,lim

n→∞P ρ

n{dT (Γn, γ∗) ≤ ε} = 1.

Proof. Follows from equation (1.9) and Theorem 1.1 by dominated convergence.

Remark 1.1. Theorem 1.2 shows that the limit shape result holds true (with the same limitγ∗) when the parameter r specifying the distribution P r

n is chosen at random according to aprobability law ρ(dr). Using the terminology of random media theory, Theorems 1.1 and 1.2can be interpreted as “quenched” and “annealed” statements, respectively.

3

1.3. Methods

Our proof of Theorem 1.1 relies on an elegant probabilistic approach first applied to convexpolygons by Sinai [21]. This method is based on the representation of a given (uniform)probability measure Pn as a conditional distribution induced on the space Πn by a suitableprobability measure Qz (depending on a “free”parameter z = (z1, z2)) defined on the “global”space Π = ∪nΠn of all polygons. To make the measure Qz closer to Pn on the subspaceΠn specified by the condition ξΓ = n, the parameter z can be chosen from the asymptoticrelation Ez(ξΓ) = n (1 + o(1)). Hence, asymptotic properties of polygons (e.g., a law of largenumbers) can be established first on (Π, Qz) and then transferred to (Πn, Pn) via conditioningon ξΓ = n and using an appropriate local limit theorem for the probability Qz{ξΓ = n}. Abig advantage of working with the measure Qz is that it may be chosen as a “multiplicativestatistic” (i.e., a direct product of one-dimensional probability measures), thus correspondingto the distribution of a sequence of independent random variables, which immediately bringsin insight and powerful techniques of the classic probability theory.

Let us point out that a similar idea is well known in statistical mechanics under the nameof “equivalence of ensembles” (see Khinchin [17]). In this context, the sets Πn and Π withthe corresponding measures (statistics) Pn and Qz represent the so-called canonical and grandcanonical ensembles, respectively, describing an ideal quantum gas of non-interacting parti-cles distributed over certain “cells” in the phase space. Deep connections between statisticalmechanics and asymptotic combinatorial problems are discussed in a series of papers by Ver-shik [26, 27]. Note also that a general idea of randomization has proved very efficient in alarge variety of combinatorial problems including the asymptotic theory of random partitions(see, e.g., [1, 2, 9, 11, 12, 25] and further references therein).

The probabilistic method is very insightful and heuristically efficient, as it makes the argu-ments transparent and natural. However, the practical implementation of this approach requiressubstantial work, especially in the two-dimensional context of random polygons as comparedto the one-dimensional case exemplified by integer partitions and the corresponding Youngdiagrams [25, 26]. To begin with, one is led to deal with various sums over the subset X ofpoints x = (x1, x2) ∈ Z2

+ with co-prime coordinates (see Section 2.1). Sinai [21] was able toobtain the limit of some basic sums of this kind by appealing to the known asymptotic densityof the set X in Z2

+ (i.e., given by 1/ζ(2) = 6/π2); however, this argumentation is insufficientfor more refined asymptotics. In the present paper, we handle this technical problem by usingthe Mobius inversion formula (see Section 3), which enables one to reduce sums over X tomore regular sums.

As already mentioned, another crucial ingredient required for the probabilistic method isa suitable local limit theorem that provides a “bridge” between the global distribution Qz andthe conditional one, Pn. Analytical difficulties encountered in the proof of such a result arealready significant in the one-dimensional case (for more details and concrete examples, see[2, 9, 10, 11, 12] and further references therein). The two-dimensional case (i.e., for planarpolygons) is notoriously tedious, even though the standard method of characteristic functionsis still applicable. To the best of our knowledge, after the original paper by Sinai [21] wherethe result was just stated (with a minor error in the determinant of the covariance matrix [21,p. 111]) full details have not been worked out in the literature (however, see [32]). We provethe following theorem in this direction (see Section 7).

4

Theorem 1.3. Let az := Erz(ξΓ), Kz := Ez(ξΓ− az)

>(ξΓ− az), and suppose that the param-eter z is chosen so that az = n (1 + o(1)). Then, as n →∞,

Qrz{ξΓ = n} ∼ 1√

det Kz

exp

(−1

2

∣∣(n− az)K−1/2z

∣∣2) . (1.10)

One can show that the covariance matrix Kz is of order of |n|4/3, and in particular det Kz ∼const (n1n2)

4/3 and ‖K−1/2z ‖ = O(|n|−2/3). From equation (1.10) it is then clear that one

needs to refine the error term in the asymptotic relation az = n (1 + o(1)) and estimate thedeviation az − n to at least the order of |n|2/3. We have been able to obtain the followingestimate (see Theorem 5.1).

Theorem 1.4. Suppose that az = Erz(ξΓ) = n (1 + o(1)). Then, as n →∞,

az = n + o(|n|2/3). (1.11)

The proof of this result is quite involved. The main idea is to use the inverse formula forMellin transform to furnish a suitable integral representation for the difference az − n. Forinstance, in the first coordinate we get

Erz(ξ1)− n1 =

r

2πi

∫ c+i∞

c−i∞

M(s)ζ(s + 1)

αs+11 ζ(s)

ds (1 < c < 2), (1.12)

where α1 ∼ const · n−1/31 , M(s) is an explicit function analytic in the strip 1 < Re s < 2,

and ζ(s) is the Riemann zeta function. As usual, in order to obtain a better estimate one hasto shift the integration contour in (1.12) as far to the left as possible. It turns out that to get anestimate of order of o(|n|2/3) one needs to enter the critical strip 0 < Re s < 1, which requiresinformation about zeroes of the zeta function in view of the denominator ζ(s) in (1.12).

Layout. The rest of the paper is organized as follows. In Section 2, we explain the basics ofthe probability method in the polygon context and define the parametric families of measuresQr

z and P rn (0 < r < ∞). In Section 3, we choose suitable values of the parameters z1, z2,

which implies convergence of expected paths to the limit curve γ∗ (Section 4). The refinederror estimate (1.11) is proved in Section 5, and various asymptotics for higher-order sumsare obtained in Section 6. Section 7 is devoted to the proof of the local central limit theorem.Finally, the limit shape results, with respect to both Qr

z and P rn , are proved in Section 8. Ap-

pendix includes necessary details of tangential parametrization and the tangential metric dT onthe space of convex paths (Section 9.1), as well as a discussion of the total variation distancebetween the measures P r

n (r 6= 1) and the uniform distribution P 1n (Section 9.2).

2. Probability measures on spaces of convex polygons

2.1. Encoding of convex lattice polygons

As observed by Sinai [21], one can encode convex lattice polygons via a certain integer-valuedfield. More specifically, let Z+ := {0, 1, 2, . . . }, Z2

+ := Z+ × Z+, and consider the set X ofall pairs of co-prime non-negative integers,

X := {x = (x1, x2) ∈ Z2+ : GCD(x1, x2) = 1}, (2.1)

5

where GCD is an acronym for “greatest common divisor”. In particular, pairs (0, 1) and (1, 0)are included in this set, while (0, 0) is not.

Let Φ := (Z+)X be the space of functions on X with non-negative integer values, andconsider the subspace of functions with finite support,

Φ0 := {ν ∈ Φ : #(supp ν) < ∞}, where supp ν := {x ∈ X : ν(x) > 0}.

Observe that Φ0 is in one-to-one correspondence with the space Π of finite polygons: Φ0 3ν ↔ Γ ∈ Π. Indeed, let us interpret points x ∈ X as radius-vectors (from 0 to x); then, givena configuration ν(x) ∈ Φ0, the (finite) collection of vectors {xν(x), x ∈ supp ν} arrangedin the order of increase of their slope τ(x) := x2/x1 ∈ [0,∞], will represent the consecutiveedges of a convex polygon Γ ∈ Π, and vice versa. (The special case where ν(x) = 0 forall x ∈ X , corresponds to the “trivial” polygon, Γ0, with coinciding endpoints.) That is tosay, each x ∈ X determines the direction of a potential edge, only used in the construction ifx ∈ supp ν, in which case the value ν(x) > 0 specifies the scaling factor.

Note that the right endpoint ξ = ξΓ of polygon Γ ∈ Π associated with configurationν(x) ∈ Φ0 is given by

ξ =∑x∈X

xν(x). (2.2)

In particular,Γ ∈ Πn ⇔ ξ = n. (2.3)

In what follows, we shall identify the spaces Π and Φ0. Hence, any probability measureon Π can be treated as the distribution of a random field ν(·) on X with a.s.-finite support andvalues in Z+.

2.2. Global measure Qz and conditional measure Pn

Let b0, b1, b2, . . . be a sequence of non-negative numbers, such that b0 > 0 (without loss ofgenerality, we put b0 = 1) and not all bk vanish for k ≥ 1. We assume that the generatingfunction

β(s) :=∞∑

k=0

bksk (2.4)

is finite for s ∈ [0, 1). Let z = (z1, z2) be a two-dimensional parameter, such that z1, z2 ∈(0, 1). Throughout, we will use the multi-index notation zx := zx1

1 zx22 , x = (x1, x2) ∈ X . Let

us define the “global” probability measure Qz on the space Φ as the distribution of a randominteger-valued field ν = {ν(x), x ∈ X} with independent components and the marginaldistributions

Qz{ν(x) = k} =bkz

kx

β(zx), k = 0, 1, 2, . . . (2.5)

Note that the probability generating function of ν(x) under the law Qz is given by

Gz(s; x) := Ez(sν(x)) =

1

β(zx)

∑x∈X

bkzkxsk =

β(szx)

β(zx), 0 ≤ s < 1. (2.6)

6

Remark 2.1. The coefficients bk introduce certain probability weights for possible numbers ofinteger points on random edges.

Consider the following condition:

β(z) :=∏x∈X

β(zx) < ∞. (2.7)

Remark 2.2. By mutual independence of ν(x) and from (2.5), we have

Qz{ν(x) ≡ 0, x ∈ X} =∏x∈X

Qz{ν(x) = 0} =∏x∈X

1

β(zx)=

1

β(z)> 0

whenever (2.7) holds. In other words, condition (2.7) is equivalent to the requirement that thetrivial polygon Γ0 ↔ ν ≡ 0 has a positive Qz-probability.

Proposition 2.1. Condition (2.7) is necessary and sufficient in order that the measure Qz beconcentrated on the set Φ0, that is,

Qz(Φ0) = 1, z ∈ (0, 1)2.

Proof. According to (2.5), Qz{ν(x) > 0} = 1− 1/β(zx). Since ν(x) are independent, Borel–Cantelli’s lemma implies that Qz{ν ∈ Φ0} = 1 if and only if

∑x∈X(1 − 1/β(zx)

)< ∞. In

turn, the latter condition is equivalent to (2.7).

That is to say, under condition (2.7) a realization of the random field ν(·) belongs to thespace Φ0 (Qz-a.s.) and therefore determines a (random) finite polygon Γ ∈ Π. The probabilityof such Γ under the law Qz is given by

Qz(Γ) =∏x∈X

bν(x)zxν(x)

β(zx)=

b(Γ)zξ

β(z), (2.8)

where ξ is defined by (2.2) and b(Γ) :=∏

x∈X bν(x). (Note that b(Γ) is well defined, since forx /∈ supp ν we have bν(x) = b0 = 1.)

On the subspace Πn of polygons Γ ∈ Π with the right endpoint fixed at n = (n1, n2), the“global” measure Qz induces the conditional distribution

Pn(Γ) := Qz(Γ |Γ ∈ Πn) =Qz(Γ)

Qz(Πn), Γ ∈ Πn. (2.9)

Proposition 2.2. The measure Pn in (2.9) does not depend on z = (z1, z2).

Proof. If Πn 3 Γ ↔ ν ∈ Φ0, then by (2.3) and (2.8)

Qz(Γ) =b(Γ)zn

β(z), Γ ∈ Πn.

Accordingly, using (2.7) and (2.9) we get

Pn(Γ) =b(Γ)∑

Γ′∈Πnb(Γ′)

, Γ ∈ Πn, (2.10)

which does not depend on z.

7

2.3. Parametric family {Qrz}

Let us consider a special parametric family of measures {Qrz, 0 < r < ∞}, determined by

formula (2.5) with the coefficients bk of the form

brk :=

(r + k − 1

k

)=

r(r + 1) · · · (r + k − 1)

k!, k = 0, 1, 2, . . . (2.11)

(note that br0 =

(r−10

)= 1, in accordance with our convention in Section 2.2). By the binomial

expansion formula, the generating function (2.4) of the sequence (2.11) is given by

βr(s) = (1− s)−r, 0 ≤ s < 1, (2.12)

and by (2.6) the probability generating function of ν(x) under the law Qrz is given by

Grz(s; x) =

βr(szx)

βr(zx)=

(1− zx

1− szx

)r

, 0 ≤ s < 1. (2.13)

Formula (2.5) then takes the form

Qrz{ν(x) = k} =

(k + r − 1

k

)zkx(1− zx)r, k = 0, 1, 2, . . . (2.14)

That is to say, under the measure Qrz the random variable ν(x) has a negative binomial distri-

bution with parameters r and p = 1− zx (see [8, § 6.8, pp. 155–156]). Note that if r = 1 thenb1k ≡ 1 and ν(x) has a geometric distribution:

Q1z{ν(x) = k} = zkx(1− zx), k = 0, 1, 2, . . . ,

while the corresponding conditional measure (see (2.10)) is reduced to the uniform distributionon Πn (cf. Sinai [21]):

P 1n(Γ) =

1

#(Πn), Γ ∈ Πn.

Remark 2.3. Since brk+1/b

rk = (r + k)/(k + 1), the sequence {br

k} is strictly increasing ordecreasing in k according as r > 1 or r < 1, respectively. That is to say, the measures Qr

z andP r

n encourage (if r > 1) or discourage (if r < 1) lattice points on edges, as compared to thereference case r = 1.

It is easy to see that condition (2.7) is satisfied and, by Proposition 2.1,

Qrz(Φ0) = 1, 0 < r < ∞. (2.15)

Indeed, using (2.12) we have

βr(z) =∏x∈X

1

(1− zx)r= exp

(−r∑x∈X

ln(1− zx)

)< ∞

whenever∑

x∈X ln(1− zx) > −∞, and the latter condition is fulfilled since

∑x∈X

zx ≤∑x∈Z2

+

zx =∞∑

x1=0

zx11

∞∑x2=0

zx22 =

1

(1− z1)(1− z2)< ∞.

8

Let Qz be the convex hull of the family {Qrz, 0 < r < ∞}; that is, a measure Q on Π

belongs to Qz if there is a probability measure ρ on (0,∞) such that

Q(·) =

∫ ∞

0

Qrz(·) ρ(dr). (2.16)

Clearly, each Q ∈ Qz is a probability measure on Φ, and moreover, it is concentrated on thespace Φ0 ↔ Π, since, according to (2.15),

Q(Φ0) =

∫ ∞

0

Qrz(Φ0) ρ(dr) =

∫ ∞

0

1 · ρ(dr) = 1.

The geometric structure of the set Qz can be characterized as follows.

Proposition 2.3. For each Q ∈ Qz, representation (2.16) is unique. In particular, the mea-sures Qr

z, 0 < r < ∞, constitute the set of extreme points of Qz.

Therefore, Qz is a Choquet simplex spanned by the set {Qrz, 0 < r < ∞} of its extreme

points (see [19, § 9]).

Proof of Proposition 2.3. Note that each Qrz admits a trivial representation of the form (2.16)

with Dirac measure δr. Suppose now that Q ∈ Qz admits the representation of the form (2.16)with two probability measures ρ1 and ρ2. Then (2.13) implies that for every x ∈ X ,∫ ∞

0

(1− zx

1− szx

)r

ρ1(dr) =

∫ ∞

0

(1− zx

1− szx

)r

ρ2(dr), 0 ≤ s < 1.

By setting λ := ln((1− szx)/(1− zx)

)this is reduced to∫ ∞

0

e−λrρ1(dr) =

∫ ∞

0

e−λrρ2(dr), 0 < λ ≤ λx, (2.17)

where λx := − ln(1 − zx). Since λx → ∞ as x → ∞, the identity (2.17) extends to allλ ∈ (0,∞). That is, the Laplace transforms of the measures ρ1 and ρ2 coincide, and by theuniqueness theorem it follows that ρ1 = ρ2.

3. Calibration of the parameter z

In what follows, asymptotic notation of the form an � bn means that

0 < lim infn1,n2→∞

an

bn

≤ lim supn1,n2→∞

an

bn

< ∞.

Throughout the paper, we shall work under the following convention about the limit n =(n1, n2) →∞.

Assumption 3.1. The notation n →∞ signifies that n1, n2 →∞ in such a way that n1 � n2;equivalently,

cn :=n2

n1

� 1. (3.1)

In particular, this implies that |n| =√

n21 + n2

2 →∞ as n →∞, and n1 � |n|, n2 � |n|.

9

The goal of this section is to use the freedom of the conditional distribution P rn(·) =

Qrz(· |Πn) from the parameter z = (z1, z2) (see Proposition 2.2) in order to better adapt the

measure Qrz to the subspace Πn determined by (2.3). A natural way to do this is to ensure

that condition (2.3) is satisfied “on average” (cf. [21, 5, 6]). More precisely, we will seek theparameter z = (z1, z2) as a solution to the following asymptotic equations:

limn→∞

n−11 Er

z(ξ1) = 1, limn→∞

n−12 Er

z(ξ2) = 1, (3.2)

where ξ = (ξ1, ξ2) is defined in (2.2) and Erz denotes expectation with respect to the distribution

Qrz. According to (2.5) we have

Erz [ν(x)] =

1

β(zx)

∞∑k=1

kbkzkx =

zxβ′(zx)

β(zx), (3.3)

and by (2.12) this gives

Erz [ν(x)] =

rzx

1− zx= r

∞∑k=1

zkx . (3.4)

Therefore, from (2.2) we have

Erz(ξi) =

∑x∈X

xiErz [ν(x)] = r

∞∑k=1

∑x∈X

xizkx (i = 1, 2). (3.5)

Let us seek the parameters z1, z2 in the form

zi = e−αi , αi = δin−1/3i (i = 1, 2), (3.6)

where the quantities δ1, δ2 > 0 (in general, depending on the ratio cn) are presumed to bebounded from above and separated from zero. Hence, (3.5) takes the form

Erz(ξi) = r

∞∑k=1

∑x∈X

xie−k〈α,x〉 (i = 1, 2), (3.7)

where 〈α, x〉 := α1x1 + α2x2. Set

κ :=

(ζ(3)

ζ(2)

)1/3

, (3.8)

where ζ(s) :=∑∞

m=1 m−s is the Riemann zeta function (recall that ζ(2) = π2/6).

Theorem 3.1. Conditions (3.2) are satisfied if δ1, δ2 in (3.6) are chosen as

δ1 = κ(rcn)1/3, δ2 = κ(r/cn)1/3. (3.9)

Proof. Settingf(x) := rx1e

−〈α,x〉, F ](h) :=∑x∈X

f(hx), (3.10)

10

we can rewrite (3.7) (for i = 1) as

Erz(ξ1) =

∞∑k=1

∑x∈X

f(kx)

k=

∞∑k=1

F ](k)

k. (3.11)

Let us also consider the function

F (h) :=∞∑

m=1

F ](hm) =∞∑

m=1

∑x∈X

f(hmx). (3.12)

Recalling the definition of the set X (see (2.1)), we note that⊔∞

m=0 mX = Z2+, that is, Z2

+ is acone generated by X . Hence, (3.12) is reduced to

F (h) =∑

x∈Z2+\0

f(hx) = rh∞∑

x1=1

x1e−hα1x1

∞∑x2=0

e−hα2x2

=rhe−hα1

(1− e−hα1)2(1− e−hα2).

(3.13)

By the Mobius inversion formula (see [14, § 16.5, Theorem 270]),

F (h) =∞∑

m=1

F ](hm) ⇔ F ](h) =∞∑

m=1

µ(m)F (hm), (3.14)

where µ(m) is a certain arithmetical function (known as the Mobius function) such that |µ(·)| ≤1. A sufficient condition for (3.14) is that the series

∑k,m |F ](hkm)| converges. Since

F ](·) ≥ 0, according to (3.12) and (3.13) we have

∞∑k,m=1

F ](kmh) =∞∑

k=1

F (kh) = rh∞∑

k=1

k e−hkα1

(1− e−hkα1)2(1− e−hkα2)< ∞,

so the above condition is satisfied.Using (3.13) and (3.14), we can rewrite (3.11) as

Erz(ξ1) =

∞∑k=1

k−1

∞∑m=1

µ(m)F (km) =∞∑

k,m=1

rmµ(m)e−kmα1

(1− e−kmα1)2(1− e−kmα2). (3.15)

Note that (3.6) and (3.9) imply

α21α2 =

rκ3

n1

, α1α22 =

rκ3

n2

, α2cn = α1 , (3.16)

where κ is defined in (3.8). Hence, we can rewrite (3.15) in the form

n−11 Er

z(ξ1) =1

κ3

∞∑k,m=1

mµ(m)α21α2 e−kmα1

(1− e−kmα1)2(1− e−kmα2). (3.17)

We now need an elementary estimate, which will also be instrumental later on.

11

Lemma 3.2. For any k > 0, θ > 0, there exists C = C(k, θ) > 0 such that

e−θt

(1− e−t)k≤ Ce−θt/2

tkfor all t > 0. (3.18)

Proof of Lemma 3.2. Set g(t) := tke−θt(1− e−t)−k and note that

limt→0+

g(t) = 1, limt→∞

g(t) = 0.

By continuity, the function g(t) is bounded on (0,∞), and (3.18) follows.

By Lemma 3.2, the general term of the series (3.17) is estimated, uniformly in k and m, byO(k−3m−2). Hence, by Lebesgue’s dominated convergence theorem one can pass to the limitin (3.17) termwise:

limn→∞

n−11 Er

z(ξ1) =1

κ3

∞∑k=1

1

k3

∞∑m=1

µ(m)

m2=

1

κ3· ζ(3)

ζ(2)= 1. (3.19)

Here the expression for the second sum (over m) is obtained using formula (3.14) with F ](h) =h−2, F (h) =

∑m(hm)−2 = h−2ζ(2) (cf. [14, § 17.5, Theorem 287]).

Similarly, for i = 2 we check that

n−12 Er

z(ξ2) =1

κ3

∞∑k,m=1

mµ(m)α1α22 e−kmα2

(1− e−kmα1)(1− e−kmα2)2→ 1, n →∞. (3.20)

The theorem is proved.

Remark 3.1. The term 1/ζ(2) appearing in formula (3.19) and the like, equals the asymptoticdensity of co-prime pairs x = (x1, x2) ∈ X among all integer points on Z2

+ (see [14, § 24.10,Theorem 459, p. 409]).

Assumption 3.2. Throughout the rest of the paper, we assume that the parameters z1, z2 arechosen according to formulas (3.6), (3.9). In particular, the measure Qr

z becomes dependenton n = (n1, n2), as well as all Qr

z-probabilities and mean values.

4. Asymptotics of the expectation

For t ∈ [0,∞], let Γn(t) denote the part of the polygon Γ ∈ Π where the slope of edges doesnot exceed tcn. Define the set

Xn(t) := {x ∈ X : τ(x) ≤ tcn}, (4.1)

where τ(x) = x2/x1. In particular, Xn(∞) = X . Recalling the association Γ ↔ ν describedin Section 2, the polygon Γn(t) is determined by the truncated configuration {1Xn(t)(x)ν(x)}.Denote by ξ(t) the right endpoint of Γn(t):

ξ(t) :=∑

x∈Xn(t)

xν(x). (4.2)

In particular, ξ(∞) ≡ ξ (see (2.2)). Similarly to (3.5) and (3.7) we have

Erz [ξ(t)] = r

∞∑k=1

∑x∈Xn(t)

xie−k〈α,x〉. (4.3)

12

4.1. Pointwise convergence of expected paths

Let us prove the following important extension of Theorem 3.1, in which we establish the“expected” limiting behavior of random paths ξ(·).

Theorem 4.1. Set

g∗1(t) :=t2 + 2t

(1 + t)2, g∗2(t) :=

t2

(1 + t)2(0 ≤ t ≤ ∞). (4.4)

Then, for each t ∈ [0,∞],

limn→∞

n−1i Er

z [ξi(t)] = g∗i (t) (i = 1, 2). (4.5)

Proof. Theorem 3.1 implies that (4.5) holds for t = ∞. Assume that t < ∞ and let i = 1 (thecase i = 2 can be considered similarly). Arguing as in the proof of Theorem 3.1 (see (3.7),(3.11), and (3.17)), we have

Erz [ξ1(t)] = r

∞∑k,m=1

mµ(m)∞∑

x1=1

x1e−kmα1x1

x2∑x2=0

e−kmα2x2

= r∞∑

k,m=1

mµ(m)

1− e−kmα2

∞∑x1=1

x1e−kmα1x1

(1− e−kmα2(x2+1)

), (4.6)

where x2 := [x1tcn] is the integer part of x1tcn, so that

0 ≤ tcnx1 − x2 < 1. (4.7)

Aiming to replace x2 + 1 by x1tcn in (4.6), we recall (3.16) and rewrite the sum over x1 as

∞∑x1=1

x1e−kmα1x1(1− e−kmα1tx1) + ∆k,m(t, α), (4.8)

where

∆k,m(t, α) :=∞∑

x1=1

x1e−kmα1x1(1+t)

(1− e−kmα2(x2+1−tcnx1)

).

Using that 0 < x2 + 1− tcnx1 ≤ 1 (see (4.7)) and applying Lemma 3.2, we obtain, uniformlyin k,m ≥ 1 and t ∈ [0,∞],

0 <∆k,m(t, α)

1− e−kmα2≤

∞∑x1=1

x1e−kmα1x1 =

e−kmα1

(1− e−kmα1)2= O(1)

e−mα1/2

(kmα1)2.

Substituting this estimate into (4.6), we see that the error resulting from the replacement ofx2 + 1 by x1tcn is dominated by

O(α−2

1

) ∞∑k=1

1

k2

∞∑m=1

e−mα1/2

m= O(α−2

1 ) ln(1− e−α1/2

)= O

(α−2

1 ln α1

).

13

Returning to representation (4.6) and computing the sum in (4.8), we find

Erz [ξ1(t)] = r

∞∑k,m=1

mµ(m)

1− e−kmα2· e−kmα1y

(1− e−kmα1y)2

∣∣∣∣ y=1

y=1+t

+ O(α−21 ln α1). (4.9)

Passing to the limit by Lebesgue’s dominated convergence theorem, similarly to the proof ofTheorem 3.1 (cf. (3.19)) we get

n−11 Er

z [ξ1(t)] →1

κ3

∞∑k=1

1

k3

∞∑m=1

µ(m)

m2

(1− 1

(1 + t)2

)=

t2 + 2t

(1 + t)2, (4.10)

which coincides with g∗1(t), as claimed.

4.2. Uniform convergence of expected paths

We can prove a stronger version of Theorem 4.1.

Theorem 4.2. Convergence in (4.5) is uniform in t ∈ [0,∞], that is,

limn→∞

sup0≤t≤∞

|n−1i Er

z [ξi(t)]− g∗i (t)| = 0 (i = 1, 2). (4.11)

For the proof, we need the following general lemma.

Lemma 4.3. Let {fn(t)} be a sequence of nondecreasing functions on a finite interval [a, b],such that for each t ∈ [a, b] one has limn→∞ fn(t) = f(t), where f(t) is a continuous(nondecreasing) function on [a, b]. Then fn(t) → f(t) uniformly on [a, b].

Proof of Lemma 4.3. Since f is continuous on a closed interval [a, b], it is uniformly contin-uous. Therefore, for any ε > 0 there exists δ > 0 such that |f(t′) − f(t)| < ε whenever|t′ − t| < δ. Let a = t0 < t1 < · · · < tN = b be a partition such that max1≤i≤N(ti −ti−1) < δ. Since limn→∞ fn(ti) = f(ti) for each i = 0, 1, . . . , N , there exists n∗ such thatmax0≤i≤N |fn(ti) − f(ti)| < ε for all n ≥ n∗. By monotonicity of fn and f , this implies thatfor any t ∈ [ti, ti+1] and all n ≥ n∗

fn(t)− f(t) ≤ fn(ti+1)− f(ti) ≤ fn(ti+1)− f(ti+1) + ε ≤ 2ε.

Similarly, fn(t) − f(t) ≥ −2ε. Therefore, supt∈[a,b] |fn(t) − f(t)| ≤ 2ε, and the uniformconvergence follows.

Proof of Theorem 4.2. Suppose that i = 1 (the case i = 2 is handled similarly). Note that foreach n the function

fn(t) :=1

n1

Erz [ξ1(t)] =

1

n1

∑x∈Xn(t)

x1Erz [ν(x)]

is nondecreasing in t. Therefore, by Lemma 4.3 convergence (4.5) is uniform on any interval[0, t∗]. Furthermore, since n−1

1 Erz [ξ1(∞)] → g∗1(∞) and the function g∗1(t) is continuous at

14

infinity (see (4.4)), for the proof of uniform convergence on a suitable interval [t∗,∞] it sufficesto show that for any ε > 0 one can choose t∗ such that for all large enough n1, n2 and all t ≥ t∗

n−11 Er

z |ξ1(∞)− ξ1(t)| ≤ ε. (4.12)

On account of (4.9) we have

Erz [ξ1(∞)− ξ1(t)] =

∞∑k,m=1

rmµ(m)

1− e−kmα2· e−kmα1(1+t)

(1− e−kmα1(1+t))2+ O

(ln α1

α21

). (4.13)

Note that by Lemma 3.2, uniformly in k, m ≥ 1,

e−kmα2

1− e−kmα2· e−kmα1(1+t)

(1− e−kmα1(1+t))2=

O(1)

α21α2(km)3(1 + t)2

.

Returning to (4.13), we obtain, uniformly in t ≥ t∗,

α21α2E

rz [ξ1(∞)− ξ1(t)] =

O(1)

(1 + t)2

∞∑k=1

1

k3

∞∑m=1

1

m2=

O(1)

(1 + t∗)2,

whence by (3.6) we get (4.12).

5. Further refinement

For future applications, we need to refine the asymptotical formulas (3.2) by estimating theerror term. The following theorem is one of the main technical ingredients of our method.

Theorem 5.1. As n →∞, we have Erz(ξ) = n + o(|n|2/3).

In order to prove this theorem, we have to make some preparations.

5.1. Approximation of sums by integrals

Let us be given a function f : R2+ → R+, continuous and absolutely integrable together with

its derivatives up to the second order. Set

F (h) :=∑x∈Z2

+

f(hx), h > 0 (5.1)

(as shown below, F (h) is well defined for all h > 0), and assume that for some β > 2

F (h) = O(h−β

), h →∞. (5.2)

Consider the Mellin transform of the function F (h) (see, e.g., [30, § 6.9])

MF (s) :=

∫ ∞

0

hs−1F (h) dh. (5.3)

15

Lemma 5.2. Under the above conditions, MF (s) is meromorphic in the strip 1 < Re s < β,with a single (simple) pole at s = 2. Moreover, MF (s) satisfies the identity

MF (s) =

∫ ∞

0

hs−1∆f (h) dh, 1 < Re s < 2, (5.4)

where∆f (h) := F (h)− 1

h2

∫R2

+

f(x) dx, h > 0. (5.5)

Remark 5.1. Identity (5.4) is a two-dimensional analogue of Muntz’s formula for functionsof one variable (see [23, § 2.11, pp. 28–29]).

Proof of Lemma 5.2. Let f be a function of real variable x ∈ [0,∞) with continuous derivativef ′, such that both f and f ′ are integrable on [0,∞). It follows that limx→∞ f(x) = 0. Indeed,note that ∫ ∞

0

f ′(x) dx = limx→∞

∫ x

0

f ′(y) dy = limx→∞

(f(x)− f(0)),

hence limx→∞ f(x) exists, and since f is integrable, the limit must equal zero. Then the Euler–Maclaurin summation formula states that

∞∑j=0

f(hj) =1

h

∫ ∞

0

f(x) dx +

∫ ∞

0

B1

(x

h

)f ′(x) dx, (5.6)

where B1(x) := x− [x]− 1 (cf. [4, §A.4, p. 254]).Applying formula (5.6) twice to the double series (5.1), we obtain

F (h) =1

h2

∫R2

+

f(x) dx +1

h

∫R2

+

(B1

(x1

h

)∂f(x)

∂x1

+ B1

(x2

h

)∂f(x)

∂x2

)dx

+

∫R2

+

B1

(x1

h

)B1

(x2

h

) ∂2f(x)

∂x1∂x2

dx.

(5.7)

Since |B1(·)| ≤ 1, the conditions on the function f imply that the integrals in (5.7) exist andhence F (h) is well defined for all h > 0. Moreover, from (5.7) it follows

F (h) = O(h−2), h → 0, (5.8)∆f (h) = O(h−1), h → 0. (5.9)

The estimates (5.2) and (5.8) imply that MF (s) as defined in (5.3) is a regular function for2 < Re s < β.

Let us now note that for such s we can rewrite (5.3) as

MF (s) =

∫ ∞

1

hs−1F (h) dh +

∫ 1

0

hs−1F (h) dh

=

∫ ∞

1

hs−1F (h) dh +

∫ 1

0

hs−3 dh

∫R2

+

f(x) dx +

∫ 1

0

hs−1∆f (h) dh

=

∫ ∞

1

hs−1F (h) dh +1

s− 2

∫R2

+

f(x) dx +

∫ 1

0

hs−1∆f (h) dh. (5.10)

16

According to the condition (5.2), the first term on the right-hand side of (5.10), as a function ofs, is regular for Re s < β, while the last term is regular for Re s > 1 by (5.9). Hence, formula(5.10) provides an analytic continuation of the function MF (s) into the strip 1 < Re s < β,where it is meromorphic and, moreover, has a single (simple) pole at point s = 2. Finally,observing that

1

s− 2= −

∫ ∞

1

hs−3 dh, Re s < 2,

and rearranging the terms in (5.10) using (5.5), we get (5.4).

Lemma 5.3. Under the conditions of Lemma 5.2,

∆f (h) =1

2πi

∫ c+i∞

c−i∞h−sMF (s) ds, 1 < c < 2. (5.11)

Proof. From (5.5), (5.7) we have ∆f (h) = O(h−2) as h → ∞. Combined with the estimate(5.9) established in the proof of Lemma 5.2, this implies that the integral (5.4) convergesabsolutely in the strip 1 < Re s < 2. The representation (5.11) then follows from (5.4) by theinversion formula for the Mellin transform (see [30, § 6.9, Theorem 9a, pp. 246–247]).

5.2. Proof of Theorem 5.1

Let i = 1 (for i = 2 the proof is similar). We shall split the proof into several steps.

Step 1. According to (3.15) we have

Erz(ξ1) =

∞∑k,m=1

µ(m)

kF (km), (5.12)

where (see (3.10), (3.13))

F (h) =∑x∈Z2

+

f(hx) =rhe−α1h

(1− e−α1h)2(1− e−α2h),

f(x) = rx1e−〈α,x〉.

Note that ∫R2

+

f(x) dx = r

∫ ∞

0

x1e−α1x1 dx1

∫ ∞

0

e−α2x2 dx2 =r

α21α2

. (5.13)

Moreover, using (3.16) we have

∞∑k,m=1

µ(m)

k· r

(km)2α21α2

=n1

κ3

∞∑k=1

1

k3

∞∑m=1

µ(m)

m2= n1 (5.14)

(cf. 3.19)). Subtracting (5.14) from (5.12) we obtain the representation

Erz(ξ1)− n1 =

∞∑k,m=1

mµ(m) ∆f(km), (5.15)

17

where ∆f (h) is defined in (5.5). Clearly, the functions f and F satisfy the hypotheses ofLemma 5.2 (with β = ∞). Using (3.16), the Mellin transform of F (h) defined by (5.3) can berepresented as

MF (s) = rα−(s+1)1 M(s), (5.16)

where

M(s) :=

∫ ∞

0

yse−y

(1− e−y)2 (1− e−y/cn)dy, Re s > 2. (5.17)

As a result, applying Lemma 5.3 we can rewrite (5.15) as

Erz(ξ1)− n1 =

r

2πi

∞∑k,m=1

mµ(m)

∫ c+i∞

c−i∞

M(s)

αs+11 (km)s

ds (1 < c < 2). (5.18)

Step 2. It is not difficult to find explicitly the analytic continuation of the function M(s) intothe domain 1 < Re s < 2. Indeed, let us represent the integral (5.17) as

M(s) = J(s) + cn

∫ ∞

0

ys−1e−y

(1− e−y)2dy +

1

2

∫ ∞

0

yse−y

(1− e−y)2dy, (5.19)

where

J(s) :=

∫ ∞

0

yse−y

(1− e−y)2

(1

1− e−y/cn− cn

y− 1

2

)dy. (5.20)

The last two integrals in (5.19) are easily evaluated:∫ ∞

0

ys−1e−y

(1− e−y)2dy =

∫ ∞

0

ys−1

∞∑k=1

ke−ky dy =∞∑

k=1

k

∫ ∞

0

ys−1e−ky dy

=∞∑

k=1

1

ks−1

∫ ∞

0

us−1e−u du = ζ(s− 1)Γ (s), (5.21)

where Γ (s) =∫∞

0us−1e−u du is the gamma function, and similarly,∫ ∞

0

yse−y

(1− e−y)2dy = ζ(s)Γ (s + 1). (5.22)

Substituting expressions (5.21) and (5.22) into (5.19), we obtain

M(s) = J(s) + cnζ(s− 1)Γ (s) +1

2ζ(s)Γ (s + 1). (5.23)

Since the expression in the square brackets in (5.20) is O(y) as y → 0 and O(1) as y →∞,the integral (5.20) is absolutely convergent (and therefore the function J(s) is regular) forRe s > 0. Furthermore, it is well known that Γ (s) is analytic for Re s > 0, while ζ(s) hasa single pole at point s = 1 (see [22, § 4.41, 4.43]). Hence, the right-hand side of (5.23) ismeromorphic in the semi-plane Re s > 0 with the poles at points s = 1 and s = 2.

18

Step 3. Let us estimate the function M(c + it) as t → ∞. Firstly, by integration by parts in(5.20) it is easy to show that uniformly in a strip 0 < c1 ≤ σ ≤ c2 < ∞

J(σ + it) = O(|t|−2), t →∞. (5.24)

The gamma function in such a strip is known to satisfy the uniform estimate

Γ (σ + it) = O(1) |t|σ−(1/2) e−π|t|/2, t →∞ (5.25)

(see [22, § 4.42, Eq. (4.12.2)]). Furthermore, the zeta function is obviously bounded in anysemi-plane σ ≥ c1 > 1:

|ζ(σ + it)| ≤∞∑

n=1

1

|nσ+it|=

∞∑n=1

1

nσ≤

∞∑n=1

1

nc1= ζ(c1) = O(1). (5.26)

We also have the following bounds, uniform in σ, on the growth of the zeta function as t →∞(see [15, § 1.5, Theorem 1.9, p. 25]):

ζ(σ + it) =

O(ln|t|) for 1 ≤ σ ≤ 2,

O(t(1−σ)/2 ln|t|) for 0 ≤ σ ≤ 1,

O(t1/2−σ ln|t|) for σ ≤ 0.

(5.27)

As a result, by (5.25), (5.26) and (5.27) the second and third summands on the right-hand sideof (5.23) give only exponentially small contributions as compared to (5.24), so that

M(c + it) = O(|t|−2), t →∞ (1 < c < 2). (5.28)

Step 4. In view of (5.28), for 1 < c < 2 there is an absolute convergence on the right-handside of (5.18),

∞∑k,m=1

m|µ(m)|∫ c+i∞

c−i∞

|M(s)|∣∣αs+11 (km)s+1

∣∣ |ds|

≤ 1

αc+11

∞∑k=1

1

kc+1

∞∑m=1

1

mc

∫ ∞

−∞|M(c + it)| dt < ∞.

Hence, the summation and integration in (5.18) can be interchanged to yield

Erz(ξ1)− n1 =

r

2πi

∫ c+i∞

c−i∞

M(s)

αs+11

∞∑k=1

1

ks+1

∞∑m=1

µ(m)

msds

=r

2πi

∫ c+i∞

c−i∞

M(s)ζ(s + 1)

αs+11 ζ(s)

ds. (5.29)

While evaluating the sum over m here, we used the Mobius inversion formula (3.14) withF ](h) = h−s, F (h) =

∑m(hm)−s = h−sζ(s) (cf. (3.19); see also [14, § 17.5, Theorem 287]).

Substituting (5.23) into (5.29), we finally obtain

Erz(ξ1)− n1 =

r

2πi

∫ c+i∞

c−i∞Ψ(s) ds (1 < c < 2), (5.30)

where

Ψ(s) :=ζ(s + 1)

αs+11

[J(s) + cnζ(s− 1)Γ (s)

ζ(s)+

1

2Γ (s + 1)

](5.31)

and the function J(s) is given by (5.20).

19

Step 5. By the La Vallee Poussin theorem (see [16, § 4.2, Theorem 5, p. 69]), there exists aconstant A > 0 such that ζ(σ + it) 6= 0 in the domain

σ ≥ 1− A

ln(|t|+ 2)=: η(t), t ∈ R. (5.32)

Moreover, it is known (see [23, § 3.11, Theorem 3.11 and in particular Eq. (3.11.8]) that in thedomain (5.32) the following uniform estimate holds:

1

ζ(σ + it)= O(ln|t|), t →∞. (5.33)

Without loss of generality, one can assume A < ln 2, so that (see (5.32))

η(t) ≥ η(0) = 1− A

ln 2> 0, t ∈ R. (5.34)

Therefore, Ψ(s) (see (5.31)) is regular for all s = σ + it such that 2 > σ ≥ η(t) (t ∈ R).Let us show that the integration contour Re s = c in (5.30) can be replaced by the curve

σ = η(t) (t ∈ R). By the Cauchy theorem, it suffices to check that

limT→±∞

∫ c+iT

η(T )+iT

Ψ(s) ds = 0.

We have ∣∣∣∣∫ c+iT

η(T )+iT

Ψ(s) ds

∣∣∣∣ ≤ ∫ c

η(T )

|Ψ(σ + iT )| dσ ≤∫ c

η(0)

|Ψ(σ + iT )| dσ. (5.35)

In view of the remark after formula (5.33), we have η(0) > 0, hence application of the estimate(5.26) gives, for s = σ + iT , η(T ) ≤ σ ≤ c,∣∣∣∣ζ(s + 1)

αs+11

∣∣∣∣ ≤ ζ(σ + 1)

ασ+11

≤ ζ(η(0) + 1)

αc+11

(5.36)

(since α1 → 0, we may assume that α1 < 1).To estimate the expression in the square brackets in (5.31), we use the estimates (5.24),

(5.25), (5.27) and (5.33). As a result we obtain

Ψ(σ + iT ) = O(|T |−2 ln|T |

), T → ±∞, (5.37)

whence it follows that the right-hand side of (5.35) tends to zero as T → ±∞, as required.Therefore, the integral in (5.30) can be rewritten in the form

Dn :=

∫ ∞

−∞Ψ(η(t) + it) d(η(t) + it). (5.38)

20

Step 6. It remains to estimate the quantity (5.38) as n →∞. Let us set

Ψ0(s) := αs+11 Ψ(s) = ζ(s + 1)

[J(s) + cnζ(s− 1)Γ (s)

ζ(s)+

1

2Γ (s + 1)

](5.39)

(see (5.31)), then equation (5.38) is rewritten as

Dn = α−21

∫ ∞

−∞α

1−η(t)−it1 Ψ0(η(t) + it) (η′(t) + i) dt. (5.40)

Using that α1 = δ1/n1/3 (see (3.6)), we get

|Dn| ≤ O(1) n2/31

∫ ∞

−∞α

1−η(t)1 |Ψ0(η(t) + it)| (|η′(t)|+ 1) dt

≤ O(1) n2/31

∫ ∞

−∞α

1−η(t)1 |Ψ0(η(t) + it)| dt, (5.41)

since by (5.32)

|η′(t)| = A

(|t|+ 2) ln2(|t|+ 2)≤ A

2 ln22= O(1).

Let us now note that, as n → ∞, the integrand function in (5.41) tends to zero for eacht, because α1 → 0 and 1 − η(t) > 0 (see (5.32)). Finally, eligibility of passing to the limitunder the integral sign follows from Lebesgue’s dominated convergence theorem. Indeed, theintegrand function in (5.41) is bounded by |Ψ0(η(t)+it)|, and integrability of the latter functionis easily checked by applying the estimates (5.24), (5.25), (5.27), (5.33) to the expression(5.39), which yields (cf. (5.37))

|Ψ0(η(t) + it)| = O(|t|−2 ln|t|

), t → ±∞.

Thus, we have shown that the integral in (5.41) is o(1) as n → ∞, hence Dn = o(|n|2/3).Substituting this estimate into (5.30), we obtain the statement of Theorem 5.1.

6. Asymptotics of higher order moments

6.1. The variance

For the random variable ν(x) with distribution (2.5), its variance is given by

Varrz[ν(x)] =

z2xβ′′(zx) + zxβ′(zx)

β(zx)−(

zxβ′(zx)

β(zx)

)2

. (6.1)

Substituting the expression (2.12) for β(s) into (6.1), after some calculations we find

Varrz[ν(x)] =

rzx

(1− zx)2= r

∞∑k=1

kzkx. (6.2)

Let Kz be the covariance matrix of ξ, that is, a (2 × 2)-matrix with elements Kz(i, j) :=Covr

z(ξi, ξj) (i, j ∈ {1, 2}). Using that the random variables ν(x) are independent for differentx ∈ X , from (2.2) we get

Kz(i, j) =∑x∈X

xixjVarrz[ν(x)] = r

∞∑k=1

∑x∈X

kxixjzkx. (6.3)

21

Theorem 6.1. As n →∞,

Kz ∼(n1n2)

2/3

r1/3κ

(2c−1

n 1

1 2cn

), (6.4)

where the constant κ is defined in (3.8) and cn := n2/n1.

Proof. Let us consider Kz(1, 1) (the other elements of Kz are analyzed similarly). Substituting(3.6) into (6.3), we obtain

Kz(1, 1) = r

∞∑k=1

∑x∈X

kx21e

−k〈α,x〉. (6.5)

Using the Mobius inversion formula (3.14), similarly to (3.17) expression (6.5) can be rewrittenin the form

Kz(1, 1) = r

∞∑k,m=1

km2µ(m)∑x∈Z2

+

x21e

−km〈α,x〉

= r∞∑

k,m=1

km2µ(m)∞∑

x1=1

x21e

−kmα1x1

∞∑x2=0

e−kmα2x2

= r∞∑

k,m=1

km2µ(m)

1− e−kmα2

∞∑x1=1

x21e

−kmα1x1 . (6.6)

Note also that∞∑

x1=1

x21e−kmα1x1 =

e−kmα1(1 + e−kmα1)(1− e−kmα1

)3 =O(1)

k3m3α31

. (6.7)

Returning to the representation (6.6) and using (6.7), we obtain

α31 α2Kz(1, 1) = r

∞∑k,m=1

km2µ(m)α3

1α2 e−kmα1(1 + e−kmα1)(1− e−kmα1

)3(1− e−kmα2)

. (6.8)

By Lemma 3.2, the general term in the series (6.8) admits a uniform estimate O(k−3m−2).Hence, by Lebesgue’s dominated convergence theorem one can pass to the limit in (6.8) toobtain

α31 α2Kz(1, 1) → 2rζ(3)

ζ(2)= 2rκ3 (α1, α2 → 0).

Using (3.6) and (3.9), this yields Kz(1, 1) ∼ 2(n1n2)2/3(r1/3κcn)−1 (cf. (6.4)).

6.2. Statistical moments of ν(x)

Denoteν0(x) := ν(x)− Er

z [ν(x)], x ∈ X , (6.9)

and for k ∈ N setmk(x) := Er

z [ν(x)k], µk(x) := Erz |ν0(x)k|. (6.10)

22

Lemma 6.2. For each k ∈ N and all x ∈ Xµk(x) ≤ 2kmk(x). (6.11)

Proof. Omitting x for shorthand, by Newton’s binomial formula and Lyapunov’s inequalitywe obtain

µk ≤ Erz

(ν + m1

)k=

k∑i=0

(k

i

)mi m

k−i1 ≤

k∑i=0

(k

i

)m

i/kk m

(k−i)/kk = 2kmk,

and (6.11) follows.

Lemma 6.3. For any k ∈ N there exist positive constants ck = ck(r) and Ck = Ck(r) suchthat for all x ∈ X

ckzkx

(1− zx)k≤ mk(x) ≤ Ckz

x

(1− zx)k. (6.12)

Proof. Fix x ∈ X and let ϕ(s) ≡ ϕν(x)(s) := Erz [e

isν(x)] be the characteristic function of therandom variable ν(x). From (2.5) and (2.12) it follows that

ϕ(s) =β(zxeis)

β(zx)=

(1− zx)r

(1− zxeis)r. (6.13)

Let us first prove that for any k ∈ N

(1− zx)−r dkϕ(s)

dsk= ik

k∑j=1

cj,k(zxeis)j

(1− zxeis)r+j, (6.14)

where cj,k ≡ cj,k(r) > 0. Indeed, if k = 1 then differentiation of (6.13) yields

(1− zx)−r dϕ(s)

ds=

irzxeis

(1− zxeis)r+1,

which is in accordance with (6.14) if we put c1,1 := r. Assume now that (6.14) is valid forsome k. Differentiating (6.14) once more, we obtain

(1− zx)−r dk+1ϕ(s)

dsk+1= ik+1

k∑j=1

cj,kj(zxeis)j

(1− zxeis)r+j+ ik+1

k∑j=1

cj,k(r + j)(zxeis)j+1

(1− zxeis)r+j+1

= ik+1

k+1∑j=1

cj,k+1(zxeis)j

(1− zxeis)r+j,

where we set

cj,k+1 :=

c1,k, j = 1,jcj,k + (r + j − 1)cj−1,k, 2 ≤ j ≤ k,(r + k)ck,k, j = k + 1.

Hence, by induction formula (6.14) is valid for all k.Now, by (6.14) we have

mk(x) = i−k dkϕ(s)

dsk

∣∣∣∣s=0

=k∑

j=1

cj,kzjx

(1− zx)j≤ zx

(1− zx)k

k∑j=1

cj,k,

since 0 < zx < 1. Hence, inequalities (6.12) hold with ck = ck,k , Ck =∑k

j=1 cj,k .

23

6.3. Asymptotics of moment sums

Lemma 6.4. For any k ∈ N and θ > 0,

∑x∈X

|x|k zθx

(1− zx)k� |n|(k+2)/3, n →∞. (6.15)

Proof. Using (3.6), by Lemma 3.2 we have

zθx

(1− zx)k=

e−θ〈α,x〉

(1− e−θ〈α,x〉)k≤ Ce−(θ/2)〈α,x〉

〈α, x〉k≤ Ce−(θ/2)〈α,x〉

αk0 |x|k

, (6.16)

where α0 := min{α1, α2}. On the other hand,

zθx

(1− zx)k=

e−θ〈α,x〉

(1− e−θ〈α,x〉)k≥ e−θ〈α,x〉

〈α, x〉k≥ e−θ〈α,x〉

|α|k |x|k. (6.17)

Since α0 � |n|−1/3 and |α| � |n|−1/3, from (6.16) and (6.17) we see that for the proof of (6.15)it remains to show ∑

x∈X

e−〈α,x〉 � |n|2/3, n →∞. (6.18)

Using the Mobius inversion formula (3.14), similarly as in Sections 3 and 4 we obtain

∑x∈X

e−〈α,x〉 =∞∑

m=1

µ(m)∑

x∈Z2+\{0}

e−m〈α,x〉

=∞∑

m=1

µ(m)

(1

(1− e−mα1)(1− e−mα2)− 1

)

=∞∑

m=1

µ(m)e−mα1 + e−mα2 − e−m(α1+α2)

(1− e−mα1)(1− e−mα2). (6.19)

By Lemma 3.2, the general term of the series (6.19) is bounded by O(α−11 α−1

2 m−2) (uniformlyin m). Hence, by dominated convergence the right-hand side of (6.19) is asymptotically equiv-alent to

1

α1α2

∞∑m=1

µ(m)

m2=

1

α1α2ζ(2)� |n|2/3, (6.20)

and (6.18) follows.

Lemma 6.5. For any k ∈ N∑x∈X

|x|kmk(x) � n(k+2)/3, n →∞.

Proof. Readily follows from the estimates (6.12) and Lemma 6.4.

24

Lemma 6.6. For any integer k ≥ 2,∑x∈X

|x|kµk(x) � |n|5/3, n →∞.

Proof. An upper bound follows (for all k ≥ 1) from the inequality (6.11) and Lemma 6.5. Onthe other hand, by Lyapunov’s inequality and formula (6.2), for any k ≥ 2

µk(x) ≥ µ2(x)k/2 =(Varr

z[ν(x)])k/2

=rk/2zkx/2

(1− zx)k,

and a lower bound follows by Lemma 6.4.

Lemma 6.7. For each k ∈ N and i = 1, 2

Erz

[ξi − Er

z(ξi)]2k

= O(|n|4k/3), n →∞. (6.21)

Proof. Let i = 1 (the case i = 2 is considered similarly). Using the notation (6.9), we have

Erz

[ξ1 − Er

z(ξ1)]2k

= Erz

(∑x∈X

x1ν0(x)

)2k

=2k∑

`=1

∑k1, . . . , k` ≥ 1,

k1 + · · ·+ k` = 2k

Ck1,...,k`

∑xi ∈ Xxi 6= xj

∏i=1

(xi1)

kiErz

[ν0(x

i)ki],

where Ck1,...,k`are combinatorial coefficients accounting for the number of identical terms in

the expansion. Using that Erz [ν0(x)] = 0, we can assume that ki ≥ 2 for all i = 1, . . . , `. Since

k1 + · · ·+ k` = 2k, this implies that ` ≤ k. Hence, using Lemmas 6.2 and 6.5 we obtain

∑xi ∈ X ,xi 6= xj

∏i=1

(xi1)

kiErz

[ν0(x

i)ki]≤∏i=1

∑x∈X

xki1 µki

(x) ≤ 22k∏i=1

∑x∈X

xki1 mki

(x)

= 22k∏i=1

O(|n|(ki+2)/3) = O(|n|2(k+`)/3) = O(|n|4k/3),

and the lemma is proved.

7. Local limit theorem

7.1. Notations and statement of the theorem

Let us introduce some general notations. For a row-vector y = (y1, y2) ∈ R2, its Euclideannorm is denoted by |y| =

√y2

1 + y22 , and 〈λ, y〉 = λ1y1 + λ2y2 is the inner product of vectors

λ, y ∈ R2. The transpose of a vector y is written as y>. If A is a square matrix then det Adenotes its determinant, and ‖A‖ = max|y|=1 |yA| its norm.

25

Set az := Erz(ξ) and recall that Kz := Covr

z(ξ, ξ) = Erz(ξ − az)

>(ξ − az) denotes thecovariance matrix of the random vector ξ. Let Vz be the (unique) square root of K−1

z (see [18,§ 5.7, p. 185]), that is, a symmetric positive definite matrix such that

V 2z = K−1

z , Kz = V −2z (7.1)

(the matrix Kz is invertible, as it is nonsingular by Theorem 6.1).Let f(y) ≡ f0,I(y) be the probability density function of standard two-dimensional normal

distribution (i.e., with zero mean and identity covariance matrix),

f(y) =1

2πe−|y|

2/2, y ∈ R2.

Normal distribution with mean az and covariance matrix Kz has the probability density func-tion given by

faz ,Kz(y) = (det Kz)−1/2f((y − az)Vz), y ∈ R2. (7.2)

Theorem 7.1. Uniformly in m ∈ Z2+,

Qrz{ξ = m} = faz ,Kz(m) + O(|n|−5/3), n →∞. (7.3)

Remark 7.1. Theorem 7.1 is a two-dimensional local central limit theorem for the sum ξ =∑x∈X xν(x) with independent terms whose distribution depends on large parameter n =

(n1, n2); however, the summation scheme is quite different from the classic one, since thenumber of non-vanishing terms is not fixed in advance, and moreover, the summands actuallyinvolved in the sum are determined by sampling.

One implication of Theorem 7.1 will be particularly useful.

Corollary 7.2. As n →∞,

Qrz{ξ = n} =

r1/3κ

2√

3 π(n1n2)

−4/3 (1 + o(1)), (7.4)

where κ := (ζ(3)/ζ(2))1/3 (see (3.8)).

7.2. Lemmas about the matrix norm

For the proof of Theorem 7.1, we need a few (some of them well known) facts about the matrixnorm.

Lemma 7.3 (cf. [13, § 22, Theorem 4]). If A is a symmetric matrix then ‖A2‖ = ‖A‖2.

Proof. From the definition of the norm, it readily follows that ‖A2‖ ≤ ‖A‖2. On the otherhand, for each test vector y the Cauchy–Schwarz inequality implies

|yA|2 = 〈yA, yA〉 = 〈yA2, y〉 ≤ ‖A2‖ · |y|2,

whence we get the reverse inequality, ‖A‖2 ≤ ‖A2‖.

26

Lemma 7.4. Let A be a symmetric nonsingular 2× 2 matrix, then

‖A−1‖ =‖A‖|det A|

.

Proof. If A =(

a bb c

)then A−1 = A / det A, where A =

(c −b

−b a

). Hence, ‖A−1‖ =

‖A‖/| det A|, and in order to prove the lemma, it suffices to show that ‖A‖ = ‖A‖. Supposethat the norm of A is attained on a vector y = (y1, y2) with |y| = 1, so that ‖A‖2 = |yA|2 =

(ay1 + by2)2 + (by1 + cy2)

2. Taking y = (y2,−y1), we have |y| = |y| = 1, yA = (cy2 +

by1,−by2− ay1) and so |yA|2 = (cy2 + by1)2 +(by2 + ay1)

2 = ‖A‖2. Therefore, ‖A‖ ≥ ‖A‖.Interchanging A and A yields the reverse inequality, ‖A‖ ≥ ‖A‖.

Lemma 7.5. Let A = (aij) be a d × d matrix, then ‖A‖2 ≤∑d

i,j=1 a2ij . If all aij ≥ 0 then

‖A‖2 ≥ d−1∑d

i,j=1 a2ij .

Proof. For y ∈ Rd, by the Cauchy–Schwarz inequality we have

‖yA‖2 =d∑

j=1

(d∑

i=1

yiaij

)2

≤d∑

j=1

(d∑

i=1

y2i

d∑i=1

a2ij

)= |y|2

d∑j=1

d∑i=1

a2ij ,

and the upper bound for ‖A‖ follows. For the second part, let 1 = (1, . . . , 1) ∈ Rd, then

‖A‖2 ≥ |1A|2

|1|2=

1

d

d∑j=1

(d∑

i=1

aij

)2

≥ 1

d

d∑j=1

d∑i=1

a2ij ,

and the lower estimate is proved.

7.3. Estimates for the covariance matrix

Lemma 7.6. As n →∞,

det Kz ∼3(n1n2)

4/3

r2/3κ2.

Proof. Follows from Theorem 6.1 applied to det Kz = Kz(1, 1) Kz(2, 2)−Kz(1, 2)2.

We can now estimate the norms of the matrices Kz and Vz = K−1/2z .

Lemma 7.7. As n →∞, one has ‖Kz‖ � |n|4/3.

Proof. Lemma 7.5 and Theorem 6.1 imply

‖Kz‖2 �2∑

i,j=1

[Kz(i, j)]2 ∼ 4

r2/3κ2(n1n2)

4/3 � |n|8/3 (n →∞),

and the required estimate follows.

Lemma 7.8. For the matrix Vz defined in (7.1), one has ‖Vz‖ � |n|−2/3 as n →∞.

27

Proof. Using Lemmas 7.3 and 7.4 we have

‖Vz‖2 = ‖V 2z ‖ = ‖K−1

z ‖ =‖Kz‖det Kz

, (7.5)

and an application of Lemmas 7.6 and 7.7 completes the proof.

We also need to estimate the so-called Lyapunov coefficient

Lz := ‖K−1z ‖3/2

∑x∈X

|x|3Erz |ν0(x)3| ≡ ‖Vz‖3

∑x∈X

|x|3µ3(x), (7.6)

where ν0(x) := ν(x)− Erz [ν(x)] and µ3(x) := Er

z |ν0(x)3| (see Section 6.2).

Lemma 7.9. As n →∞, one has Lz � |n|−1/3.

Proof. The result follows from (7.6) using Lemmas 6.6 and 7.8.

7.4. Estimates of the characteristic functions

Recall from Section 2 that the random variables {ν(x)}x∈X , with respect to the distributionQr

z, are independent and have negative binomial distribution with parameters r and zx. Inparticular, ν(x) has the characteristic function (see (6.13))

ϕν(x)(s) =

(1− zx

1− zxeis

)r

, (7.7)

hence the characteristic function of the vector ξ =∑

x∈X xν(x) is given by

ϕξ(λ) =∏x∈X

ϕν(x)(〈λ, x〉) =∏x∈X

(1− zx)r

(1− zx ei〈λ,x〉)r, λ ∈ R2.

Lemma 7.10. Let ϕν0(x)(s) := Erz(e

itν0(x)) be the characteristic function of the random vari-able ν0(x) := ν(x)− Er

z [ν(x)]. Then for all s ∈ R

|ϕν0(x)(s)| ≤ exp{−µ2(x)

2s2 +

µ3(x)

3|s|3}

. (7.8)

Proof. Let a random variable ν ′0(x) be independent of ν0(x) and have the same distribution.Then the characteristic function of ν(x) := ν0(x)− ν ′0(x) is given by

ϕν(x)(s) = ϕν0(x)(s) · ϕν0(x)(−s) = |ϕν0(x)(s)|2. (7.9)

Note that Erz [ν(x)] = 0 and Varr

z[ν(x)] = 2Varrz[ν(x)] = 2µ2(x). We also have

Erz |ν(x)3| ≤ 4Er

z |ν0(x)3| = 4µ3(x)

(see [4, § 8, Lemma 8.8, pp. 66–67]). Hence, by Taylor’s formula,

ϕν(x)(s) = 1− µ2(x)s2 +2θµ3(x)

3s3, (7.10)

where |θ| ≤ 1. Now, using (7.9) and the inequality |y| ≤ e(y2−1)/2 we get

|ϕν0(x)(s)| ≤ exp

{1

2

(|ϕν0(x)(s)|2 − 1

)}= exp

{1

2

(ϕν(x)(s)− 1

)},

and the lemma follows by (7.10).

28

The characteristic function of the vector ξ0 := ξ − az is given by

ϕξ0(λ) = Erz(e

i〈λ, ξ−a〉) = Erz

∏x∈X

ei〈λ, x〉ν0(x) =∏x∈X

ϕν0(x)(〈λ, x〉). (7.11)

Lemma 7.11. For all λ ∈ R2,

|ϕξ0(λVz)| ≤ exp{−1

2|λ|2 +

1

3Lz|λ|3

}. (7.12)

Proof. Using (7.11) and (7.8), we obtain

|ϕξ0(λVz)| ≤ exp

{−1

2

∑x∈X

〈λVz, x〉2µ2(x) +1

3

∑x∈X

|〈λVz, x〉|3µ3(x)

}. (7.13)

Recalling that Kz = V −2z (see (7.1)), we note that∑

x∈X

〈λVz, x〉2µ2(x) = Varrz〈λVz, ξ〉 = λVzCovr

z(ξ, ξ) (λVz)>

= λVzKzVzλ> = |λ|2.

(7.14)

Further, by the Cauchy–Schwarz inequality and on account of (7.6) we have∑x∈X

|〈λVz, x〉|3µ3(x) ≤ |λ|3 ‖Vz‖3∑x∈X

|x|3µ3(x)

= |λ|3 ‖Vz‖3M3 = |λ|3Lz.

(7.15)

Substituting (7.14), (7.15) into (7.13), we get (7.12).

Lemma 7.12. Let |λ| ≤ L−1z , then∣∣∣ϕξ0(λVz)− e−|λ|

2/2∣∣∣ ≤ 16Lz|λ|3e−|λ|

2/6. (7.16)

Proof. Let us first suppose that 12L−1/3z ≤ |λ| ≤ L−1

z . Then 18≤ Lz|λ|3 ≤ |λ|2, so (7.12)

implies |ϕξ0(λVz)| ≤ e−|λ|2/6. Hence,∣∣ϕξ0(λVz)− e−|λ|2/2∣∣ ≤ |ϕξ0(λVz)|+ e−|λ|

2/2 ≤ 2e−|λ|2/6 ≤ 16Lz|λ|3e−|λ|

2/6,

in accord with (7.16).Let now |λ| < 1

2L−1/3z . Using again Taylor’s formula, we have

ϕν0(x)(s)− 1 = −µ2(x)

2s2 +

θxµ3(x)

6s3, (7.17)

where |θx| ≤ 1. By Lyapunov’s inequality, µ2(x) ≤ µ3(x)2/3, so

|ϕν0(x)(s)− 1| ≤ 1

2|s|2µ3(x)2/3 +

1

6|s|3µ3(x). (7.18)

29

Putting here s = 〈λVz, x〉 we have (cf. (7.15))

|〈λVz, x〉|µ3(x)1/3 ≤ L1/3z |λ| ≤ 1

2, (7.19)

and so (7.18) yields

|ϕν0(x)(〈λVz, x〉)− 1| ≤ 1

2L2/3

z |λ|2 +1

6Lz|λ|3 ≤

1

2· 1

4+

1

6· 1

8<

1

2. (7.20)

Similarly, using the elementary inequality (a + b)2 ≤ 2(a2 + b2), from (7.18) we obtain

|ϕν0(x)(s)− 1|2 ≤ 1

2

(|s|µ3(x)1/3 +

1

9|s|3µ3(x)

)|s|3µ3(x),

whence in view of (7.19)

|ϕν0(x)(〈λVz, x〉)− 1|2 ≤ 1

2

(1

2+

1

9· 1

8

)|λ|3‖Vz‖3|x|3µ3(x)

≤ 1

3|λ|3‖Vz‖3|x|3µ3(x).

(7.21)

Consider the function ln(1 + u) of complex variable u, choosing the principal branch ofthe logarithm (i.e., such that ln 1 = 0). Taylor’s expansion implies ln(1 + u) = u + θu2 for|u| ≤ 1/2, where |θ| ≤ 1. By (7.17), (7.20) and (7.21) this yields

ln ϕν0(x)(〈λVz, x〉) =− 1

2〈λVz, x〉2µ2(x) +

1

2θx|λ|3‖Vz‖3|x|3µ3(x),

where |θx| ≤ 1. Substituting this into (7.11), due to (7.14) and (7.15) we obtain

ln ϕξ0(λVz) =∑x∈X

ln ϕν0(〈λVz, x〉) = −1

2|λ|2 +

1

2θ1Lz|λ|3 (|θ1| ≤ 1).

Using the known inequality |ey − 1| ≤ |y|e|y|, which holds for any complex y, we have∣∣ϕξ0(λVz)− e−|λ|2/2∣∣ = e−|λ|

2/2∣∣eθ1Lz |λ|3/2 − 1

∣∣≤ e−|λ|

2/2 · 1

2Lz|λ|3eLz |λ|3/2 ≤ e−|λ|

2/2Lz|λ|3,

and the proof is completed.

Lemma 7.13. For all λ ∈ R2,

|ϕξ0(λ)| ≤ exp{−r

4Jα(λ)

}, (7.22)

whereJα(λ) :=

∑x∈X

e−〈α, x〉(1− cos〈λ, x〉). (7.23)

30

Proof. According to (7.11) we have

|ϕξ0(λ)| = |ϕξ(λ)| = exp

{∑x∈X

ln |ϕν(x)(〈λ, x〉)|}

. (7.24)

Using (7.7) we note

ln |ϕν(x)(s)| =r

2ln

(1− zx)2

|1− zxeis|2=

r

2ln

1− 2zx + z2x

1− 2zx cos s + z2x

≤ −rzx(1− cos s)

|1− zxeis|2≤ −rzx(1− cos s)

4,

where we used the inequalities ln y ≤ y−1 and |1− zxeis| ≤ 2. Substituting this estimate into(7.24) and recalling (3.6) we obtain (7.22).

7.5. Proof of Theorem 7.1 and Corollary 7.2

Before proceeding to the proof of Theorem 7.1, let us derive Corollary 7.2. According toTheorem 5.1, az = Er

z(ξ) = n + o(|n|2/3). Together with Lemma 7.8 this implies

|(n− az)Vz| ≤ |n− az| · ‖Vz‖ = o(|n|2/3) O(|n|−2/3) = o(1).

Hence, by Lemma 7.6 we get

faz ,Kz(n) =1

2π(det Kz)

−1/2e−|(n−az)Vz |2/2 =r1/3κ

2π√

3(n1n2)

−4/3(1 + o(1)),

and (7.4) follows from (7.3).

Proof of Theorem 7.1. By the Fourier inversion formula, we have

Qrz{ξ = m} =

1

(2π)2

∫T

e−i〈λ,m−az〉ϕξ0(λ) dλ, (7.25)

where T := {λ = (λ1, λ2) : |λ1| ≤ π, |λ2| ≤ π}. On the other hand, the characteristicfunction corresponding to the probability density faz ,Kz(y) is given by

ϕaz ,Kz(λ) = ei〈λ,az〉−|λV −1z |2/2, λ ∈ R2.

Hence, by the inversion formula we obtain

faz ,Kz(m) =1

(2π)2

∫R2

e−i〈λ,m−az〉−|λV −1z |2/2 dλ. (7.26)

Note that if |λV −1z | ≤ L−1

z then, according to Lemmas 7.8 and 7.9,

|λ| ≤ |λV −1z | · ‖Vz‖ ≤ L−1

z ‖Vz‖ = O(|n|−1/3) = o(1), (7.27)

which implies that λ ∈ T . Using this observation, from (7.25) and (7.26) we get∣∣Qrz{ξ = m} − faz ,Kz(m)

∣∣ ≤ I1 + I2 + I3, (7.28)

31

where

I1 :=1

(2π)2

∫|λV −1

z |≤L−1z

∣∣ϕξ0(λ)− e−|λV −1z |2/2

∣∣ dλ,

I2 :=1

(2π)2

∫|λV −1

z |>L−1z

e−|λV −1z |2/2 dλ,

I3 :=1

(2π)2

∫T∩{|λV −1

z |>L−1z }

|ϕξ0(λ)| dλ.

By the substitution λ = λVz, the integral I1 is reduced to

I1 =(det Kz)

−1/2

(2π)2

∫|λ|≤L−1

z

∣∣ϕξ0(λVz)− e−|λ|2/2∣∣ dλ

= O(|n|−4/3) Lz

∫R2

|λ|3e−|λ|2/6 dλ = O(|n|−5/3), (7.29)

on account of Lemmas 7.6, 7.9, and 7.12.For I2, we again set λ = λVz and pass to the polar coordinates to get

I2 =1

2π

∫ ∞

L−1z

ρ e−ρ2/2 dρ =1

2πe−L−2

z /2 = o(|n|−5/3), (7.30)

as follows from Lemma 7.9.Estimation of I3 is the main part of the proof. Using Lemma 7.13, we obtain

I3 = O(1)

∫T∩{|λV −1

z |>L−1z }

e−Jα(λ) dλ, (7.31)

where Jα(λ) is given by (7.23). The condition |λV −1z | > L−1

z implies that |λ| > c|α| for asuitable (small enough) constant c > 0 and hence

max{|λ1|/α1, |λ2|/α2} > c,

for otherwise from Lemmas 7.7, 7.9 and 7.12 it would follow

1 < Lz|λV −1z | ≤ c|α|Lz ‖K‖1/2 = O(c) → 0 as c ↓ 0.

Hence, the estimate (7.31) is reduced to

I3 = O(1)

(∫|λ1|>cα1

+

∫|λ2|>cα2

)e−Jα(λ) dλ. (7.32)

To estimate the first integral in (7.32), by keeping in the sum (7.23) only pairs of the formx = (x1, 1), x1 ∈ Z+, we obtain

eα2Jα(λ) ≥∞∑

x1=0

e−α1x1(1− Re ei(λ1x1+λ2)

)=

1

1− e−α1− Re

eiλ2

1− e−α1+iλ1

≥ 1

1− e−α1− 1

|1− e−α1+iλ1|, (7.33)

32

because Re s ≤ |s| for any s ∈ C. Since cα1 ≤ |λ1| ≤ π, we have

|1− e−α1+iλ1| ≥ |1− e−α1+icα1| = α1

√1 + c2(1 + O(α1)) (α1 → 0).

Substituting this estimate into (7.33), we conclude that Jα(λ) is asymptotically bounded frombelow by (1− 1/

√1 + c2)α−1

1 � |n|1/3, uniformly in cα1 ≤ |λ1| ≤ π. Thus, the first integralin (7.32) is estimated by O(1) exp(−const · |n|1/3) = o(|n|−5/3).

The second integral in (7.32) where |λ2| > cα2, is estimated similarly by reducing sum-mation in (7.23) to that over x = (1, x2) only. As a result, we have shown that I3 = o(|n|−5/3).Substituting this estimate, together with (7.29) and (7.30), into (7.28), we get (7.3). The proofof Theorem 7.1 is completed.

8. Proof of the limit shape results

8.1. Limit shape under Qrz

Theorem 8.1. For each ε > 0,

limn→∞

Qrz

{sup

0≤t≤∞|n−1

i ξi(t)− g∗i (t)| ≤ ε

}= 1 (i = 1, 2),

where g∗1(t), g∗2(t) are defined in (4.4).

Proof. By Theorems 4.1 and 4.2, the expectation of the random process n−1i ξi(t) uniformly

converges to g∗i (t) as n →∞. Therefore, we only need to check that for each ε > 0

limn→∞

Qrz

{sup

0≤t≤∞n−1

i

∣∣ξi(t)− Erz [ξi(t)]

∣∣ > ε

}= 0. (8.1)

Note that the random process ξi(t) := ξi(t)−Erz [ξi(t)] has independent increments and zero

mean, hence it is a martingale with respect to the natural filtration Ft := σ{ν(x), x ∈ Xn(t)},t ∈ [0,∞]. From the definition of ξi(t) (see (4.2)), it is also clear that ξi(t) is a cadlag process,that is, its paths are everywhere right-continuous and have left limits. Therefore, applying theKolmogorov–Doob submartingale inequality (see, e.g., [31, § 2.2, Corollary 2.1, p. 14]) andusing Theorem 6.1, we obtain

Qrz

{sup

0≤t≤∞|ξi(t)| > εni

}≤ Varr

z[ξi]

(εni)2= O(n

−2/3i ) → 0.

The theorem is proved.

8.2. Limit shape under P rn

We are finally ready to prove our main result about universality of the limit shape under themeasures P r

n (cf. Theorem 1.1).

Theorem 8.2. For any ε > 0,

limn→∞

P rn

{sup

0≤t≤∞|n−1

i ξi(t)− g∗i (t)| ≤ ε

}= 1, i = 1, 2.

33

Proof. Like in the proof of Theorem 8.1, the claim of the theorem is reduced to the limit

limn→∞

P rn

{sup

0≤t≤∞|ξi(t)| > εni

}= 0,

where ξi(t) = ξi(t)− Erz [ξi(t)]. Using (2.9) we get

P rn

{sup

0≤t≤∞|ξi(t)| > εni

}≤

Qrz

{sup0≤t≤∞ |ξi(t)| > εni

}Qr

z{ξ = n}. (8.2)

Using the Kolmogorov–Doob submartingale inequality and Lemma 6.7 (with k = 3), weobtain

Qrz

{sup

0≤t≤∞|ξi(t)| > niε

}≤

Erz

(|ξi|6

)(εni)6

= O(|n|−2).

On the other hand, by Corollary 7.2

Qrz{ξ = n} ∼ r1/3κ

2√

3 π(n1n2)

−4/3 � |n|−4/3.

In view of these estimates, the right-hand side of (8.2) is dominated by a quantity of order ofO(|n|−2/3) → 0, and the theorem is proved.

9. Appendix

9.1. Tangential distance between convex paths

Let G0 = {γ} be the space of paths in R2+ starting from the origin and such that each path

γ ∈ G0 is continuous, piecewise C1-smooth (i.e., everywhere except a finite set), bounded andconvex, and furthermore, its tangent slope (where it exists) is non-negative and does not exceedπ/2. (Note that the value π/2 corresponding to vertical tangent is not ruled out; in particular,a path may finish off with a vertical edge.) Convexity implies that the slope is non-decreasingas a function of natural parameter (i.e., the length along the curve measured from the origin).

For γ ∈ G0, let gγ(t) = (g1(t), g2(t)) denote the right endpoint of the (closure of the) partof γ where the tangent slope does not exceed t ∈ [0,∞]. Note that the functions

u1 = g1(t), u2 = g2(t) (0 ≤ t ≤ ∞)

are cadlag (i.e., right-continuous with left limits), and

du2

du1

=g′2(t)

g′1(t)= t. (9.1)

More precisely, equation (9.1) holds at points where the tangent exists and its slope is strictlygrowing; cusp points of γ correspond to intervals where both functions g1 and g2 are constant,while flat (straight-line) pieces on γ lead to simultaneous jumps of g1 and g2.

The canonical limit shape curve γ∗ (see (1.3)) is determined by the parametric equations

u1 = g∗1(t) :=t2 + 2t

(1 + t)2, u2 = g∗2(t) :=

t2

(1 + t)2(0 ≤ t ≤ ∞). (9.2)

34

Indeed, it can be readily seen that the functions (9.2) satisfy the Cartesian equation (1.3) forγ∗; moreover, it is easy to check t in equations (9.2) is a tangential parameter:

dg∗1(t)

dt=

2

(1 + t)3,

dg∗2dt

=2t

(1 + t)3,

and hence (cf. (9.1))du2

du1

=dg∗2/dt

dg∗1/dt= t.

The tangential distance dT between paths in G0 is defined as follows:

dT (γ1, γ2) := sup0≤t≤∞

|gγ1(t)− gγ2(t)|, γ1, γ2 ∈ G0. (9.3)

Recall that the Hausdorff distance dH is defined in (1.2).

Lemma 9.1. The Hausdorff distance dH is dominated by the tangential distance dT :

dH(γ1, γ2) ≤ dT (γ1, γ2), γ1, γ2 ∈ G0. (9.4)

Proof. First of all, note that any path γ ∈ G0 can be approximated, simultaneously in metricsdH and dT , by polygonal lines γm (for instance, by inscribing polygons with refined edges inthe arc γ), so that

limm→∞

dH(γ, γm) = 0, limm→∞

dT (γ, γm) = 0.

This reduces inequality (9.4) to the case where γ1, γ2 are polygons. Moreover, by symmetry itsuffices to show that

maxx∈γ1

miny∈γ2

|x− y| ≤ dT (γ1, γ2). (9.5)

Note that if a point x ∈ γ1 can be represented as x = gγ1(t0) with some t0 (i.e., x is avertex of γ1), then

miny∈γ2

|x− y| = miny∈γ2

|gγ1(t0)− y| ≤ |gγ1(t0)− gγ2(t0)| ≤ dT (γ1, γ2),

and inequality (9.5) follows.Suppose now that x ∈ γ1 lies on an edge, say of slope t∗, between two consecutive vertices

gγ1(t∗) and gγ1(t∗), then x = sgγ1(t∗) + (1− s)gγ1(t

∗) with some s ∈ (0, 1) and

miny∈γ2

|x− y| = miny∈γ2

|sgγ1(t∗) + (1− s)gγ1(t∗)− y|

≤ |sgγ1(t∗) + (1− s)gγ1(t∗)− gγ2(t

∗)|≤ s|gγ1(t∗)− gγ2(t

∗)|+ (1− s)|gγ1(t∗)− gγ2(t

∗)|≤ s|gγ1(t∗)− gγ2(t

∗)|+ (1− s)dT (γ1, γ2).

(9.6)

Note that for all t ∈ [t∗, t∗) we have gγ1(t) = gγ1(t∗), hence

|gγ1(t∗)− gγ2(t)| = |gγ1(t)− gγ2(t)| ≤ dT (γ1, γ2) (t∗ ≤ t < t∗). (9.7)

If gγ2(t) is continuous at t = t∗, then inequality (9.7) extends to t = t∗:

|gγ1(t∗)− gγ2(t∗)| ≤ dT (γ1, γ2).

35

Substituting this into the right-hand side of (9.6), we get

miny∈γ2

|x− y| ≤ dT (γ1, γ2),

which implies (9.5).If gγ2(t

∗−0) 6= gγ2(t∗) then t∗ coincides with the slope of some edge on γ2 (with endpoints,

say, gγ2(t∗) and gγ2(t∗)), which is thus parallel to the edge on γ1 where the point x lies (i.e.,

with endpoints gγ1(t∗), gγ1(t∗)). Setting s∗ := max{t∗, t∗} < t∗, we have gγ1(t∗) = gγ1(s∗),

gγ2(t∗) = gγ2(s∗). To complete the proof, it remains to observe that the shortest distance froma point on a base of a trapezoid to the opposite base does not exceed the maximum length ofthe two lateral sides. Hence,

miny∈γ2

|x− y| ≤ min{|x− y|, y ∈ [gγ2(s∗), gγ2(t∗)]}

≤ max{|gγ1(s∗)− gγ2(s∗)|, |gγ1(t∗)− gγ2(t

∗)|}≤ dT (γ1, γ2),

and the bound (9.5) follows.

Remark 9.1. Note, however, that the metrics dH and dT are not equivalent. For instance, ifγ ∈ G0 is a smooth strictly convex curve with the curvature bounded below by a constantκ0 > 0, then for an inscribed polygon Γε with edges of length no more than ε > 0, itstangential distance from γ is of order of ε, but the Hausdorff distance is of order of ε2:

dT (Γε, γ) � ε, dH(Γε, γ) � κ0 ε2 (ε → 0).

Moreover, in the degenerate case where the curvature may vanish, the difference between thetwo metrics may be more dramatic. For instance, it is possible that two polygonal lines areclose to each other in the Hausdorff distance while their tangential distance is quite large. Foran example, consider two congruent straight line segments Γ1, Γ2 in R2

+, both starting fromthe origin and with very close slopes; then dH(Γ1, Γ2) is bounded by the Euclidean distancebetween their right endpoints, while dT (Γ1, Γ2) equals the Euclidean distance from the lowerendpoint to the origin.

9.2. Total variation distance between P rn and P 1

n

Note that if probability measures Pn and Pn on Πn are asymptotically close to each other intotal variation, that is, ‖Pn − Pn‖TV → 0 as n →∞, where

‖Pn − Pn‖TV := supA⊂Πn

|Pn(A)− Pn(A)|,

then the problem of universality of the limit shape is resolved in a trivial way, in that if a limitshape γ0 exists under Pn then the same curve γ0 provides the limit shape under Pn. Indeed,assuming that the event Aε = {d(Γn, γ0) > ε} satisfies Pn(Aε) → 0 as n →∞, we have

Pn(Aε) ≤ Pn(Aε) +∣∣Pn(Aε)− Pn(Aε)

∣∣≤ Pn(Aε) + ‖Pn − Pn‖TV → 0 (n →∞).

However, the family {P rn}, defined by formula (2.10) with the coefficients (2.11), is not

close to P 1n in total variation, at least uniformly in r ∈ (0,∞).

36

Theorem 9.2. For every fixed n, the limiting distance in total variation between P rn and P 1

n ,as r → 0 or r →∞, is given by

limr→0,∞

‖P rn − P 1

n‖TV = 1− 1

#(Πn). (9.8)

Proof. To obtain a lower bound for ‖P rn − P 1

n‖TV in the case r → ∞, consider the polygonΓ∗ ∈ Πn consisting of two edges, a horizontal one, from the origin to the point (n1, 0), anda vertical one, from (n1, 0) to the point n = (n1, n2). The corresponding configuration νΓ∗ isdetermined by the conditions νΓ∗(1, 0) = n1, νΓ∗(0, 1) = n2 and νΓ∗(x) = 0 otherwise. Notethat br

k ∼ rk/k! as r → ∞ (see (2.11)), hence br(Γ) = O(rNΓ), where NΓ :=

∑x∈X νΓ(x)

is the total number of integer points on Γ (excluding the origin). We have NΓ∗ = n1 + n2, sobr(Γ∗) = br

n1brn2∼ rn1+n2/(n1!n2!) as r → ∞. Observe that for any other polygon Γ ∈ Πn

(Γ 6= Γ∗), one has br(Γ) = o(rn1+n2). Indeed, for x ∈ X we have x1 + x2 ≥ 1, and moreover,x1 + x2 > 1 unless x = (1, 0) or x = (0, 1). Hence, for any Γ ∈ Πn (Γ 6= Γ∗),

NΓ =∑x∈X

νΓ(x) <∑x∈X

(x1 + x2)νΓ(x) = n1 + n2,

so that NΓ < n1 + n2 and br(Γ) = O(rNΓ) = o(rn1+n2) as r →∞. Therefore, from (2.10) weget

P rn(Γ∗) =

br(Γ∗)

br(Γ∗) +∑

Γ6=Γ∗ br(Γ)=

rn1+n2

rn1+n2 + o(rn1+n2)→ 1 (r →∞), (9.9)

and it follows

‖P rn − P 1

n‖TV ≥∣∣P r

n(Γ∗)− P 1n(Γ∗)

∣∣→ 1− 1

#(Πn)(r →∞). (9.10)

For the case r → 0, consider the polygon Γ∗ ∈ Πn consisting of one edge leading fromthe origin to n = (n1, n2). That is, νΓ∗(n/kn) = kn and ν(x) = 0 otherwise, where kn =GCD(n1, n2). Clearly, br(Γ∗) = br

kn= r/kn, while for any other polygon Γ ∈ Πn (i.e., with

more than one edge), by (2.11) we have br(Γ) = O(r2) as r → 0. Therefore, according to(2.10),

P rn(Γ∗) =

br(Γ∗)

br(Γ∗) +∑

Γ6=Γ∗br(Γ)

=r

r + O(r2)→ 1 (r → 0),

and, similarly to (9.10), we obtain

‖P rn − P 1

n‖TV ≥∣∣P r

n(Γ∗)− P 1n(Γ∗)

∣∣→ 1− 1

#(Πn)(r → 0). (9.11)

The upper bound for ‖P rn − P 1

n‖TV (uniform in r) follows from the known fact (see [7,p. 472]) that the total variation distance can be expressed in terms of a certain Vasershtein(Kantorovich–Rubinstein, cf. [28]) distance:

‖P rn − P 1

n‖TV = infX,Y

E[%(X, Y )].

37

Here the infimum is taken over all pairs of random elements X and Y defined on a commonprobability space with values in Πn and marginal distributions P r

n and P 1n , respectively, and

the function %(·, ·) on Πn × Πn is such that %(Γ, Γ′) = 1 if Γ 6= Γ′ and %(Γ, Γ′) = 0 if Γ = Γ′

(therefore defining a discrete metric in Πn). Choosing X and Y so that they are independentof each other, we obtain

‖P rn − P 1

n‖TV ≤ E[%(X, Y )] = 1− P(X = Y )

= 1−∑Γ∈Πn

P(X = Γ, Y = Γ)

= 1−∑Γ∈Πn

P(X = Γ) · P(Y = Γ)

= 1−∑Γ∈Πn

P rn(Γ) · P 1

n(Γ)

= 1− 1

#(Πn)

∑Γ∈Πn

P rn(Γ)

= 1− 1

#(Πn).

Combining this estimate with (9.10) and (9.11), we obtain (9.8).

In the limit n →∞, Theorem 9.2 yields

limn→∞

limr→0,∞

‖P rn − P 1

n‖TV = 1.

That is to say, in the successive limit r → 0 (∞), n → ∞, the measures P rn and P 1

n becomesingular with respect to each other.

Moreover, one can show that the distance ‖P rn − P 1

n‖TV is not small even for a fixed r. Tothis end, it suffices to find a function on the space Πn possessing a limit distribution (possiblydegenerate) under each P r

n , with the limit depending on the parameter r. Recalling Remark2.3, it is natural to seek such a function as one referring to integer points on the polygonΓ ∈ Πn. Indeed, for the statistic NΓ defined above the following law of large numbers holds(see Bogachev and Zarbaliev [5, Theorem 3]; Zarbaliev [32, § 1.10]):

Lemma 9.3. Under Assumption 3.1, for each r ∈ (0,∞) and any ε > 0

limn→∞

P rn(Ar

ε) = 1, where Arε :=

{∣∣∣∣ NΓ

(n1n2)1/3− r1/3

κ2

∣∣∣∣ ≤ ε

}. (9.12)

From (9.12) it follows that for any given r 6= 1 and for all ε > 0 small enough,

‖P rn − P 1

n‖TV ≥∣∣P r

n(Arε)− P 1

n(Arε)∣∣→ |1− 0| = 1 (n →∞),

and we arrive at the following result.

Theorem 9.4. For every fixed r 6= 1, the measures P rn and P 1

n are asymptotically singular asn →∞, that is,

limn→∞

‖P rn − P 1

n‖TV = 1.

38

Acknowledgments

We would like to express our deep gratitude to Ya. G. Sinai for introducing us to the beautifularea of random polygons and for stimulating discussions over the years. Our sincere thanks arealso due to Yu. A. Prokhorov and A. M. Vershik for useful remarks and discussions at variousstages of this work. The second named author gratefully acknowledges support by a LondonMathematical Society Scheme 5 Grant during his visit to the University of Leeds in March2003 when part of this research was done.

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