Size of the giant component in a random geometric graph

www.imstat.org/aihp 2013, Vol. 49, No. 4, 1130–1140

DOI:10.1214/12-AIHP498

© Association des Publications de l’Institut Henri Poincaré, 2013

Size of the giant component in a random geometric graph

Ghurumuruhan Ganesan

Theoretical Statistics and Mathematics Unit, Indian Statistical Institute, New Delhi, 7 S. J. S. Sansanwal Marg, New Delhi 110016, India.

E-mail:guru9r@isid.ac.in

Received 16 April 2012; revised 1 June 2012; accepted 5 June 2012

Abstract. In this paper, we study the size of the giant componentC_Gin the random geometric graphG=G(n, rn, f )ofnnodes independently distributed each according to a certain densityf (·)in[0,1]²satisfying inf_x∈[0,1]2f (x) >0. If^c_n¹ ≤r_n²≤c₂^logn_n for some positive constantsc₁, c₂andnr_n²−→ ∞asn→ ∞, we show that the giant component ofGcontains at leastn−o(n) nodes with probability at least 1−e^−βnrn2for allnand for some positive constantβ. We also obtain estimates on the diameter and number of the non-giant components ofG.

Résumé. Dans cet article nous étudions la composante principaleC_Gdans le graphe géométrique aléatoireG=G(n, r_n, f )avecn nœuds indépendants, chacun étant distribué selon une densitéf (·)dans[0,1]²telle que inf_x∈[0,1]2f (x) >0. Si ^c_n¹ ≤r_n²≤c₂^logn_n pour des constantes positivesc₁, c₂etnr_n²−→ ∞quandn→ ∞, nous montrons que la composante principale deGcontient au moinsn−o(n)nœuds avec probabilité minorée par 1−e^−βnrn2pour toutnet pour une constante positiveβ. Nous obtenons aussi des estimations sur les diamètres et sur le nombre des plus petites composantes deG.

MSC:Primary 60D05; secondary 60C05

Keywords:Random geometric graphs; Size of giant component; Number of components

1. Introduction

Considernnodes independently distributed inS= [0,1]²each according to a certain densityf (·)and say two nodes u=(x_u, y_u), v=(x_v, y_v)∈R²are connected to each other if the Euclidean distanced(u, v)between them is less than r_n. We denote the resulting random geometric graph (RGG) asG=G(n, r_n, f ). Throughout the paper we assume the densityf on[0,1]²satisfies

0< inf

x∈[0,1]²f (x)≤ sup

x∈[0,1]²

f (x) <∞. (1)

Random graphs as described above are important in many applications and properties like emergence of giant com- ponent, connectivity and area coverage have been studied before [2,4–6] for a variety of random graphs.

For the case of RGGs, we recollect the pertinent results below for convenience.

Theorem [4,6]. Ifr_n²=^c_n¹ for some constantc₁>0sufficiently large and the densityf (·)satisfies(1),then:

(a) There exists a constantε=ε(c₁) >0so that

P(Gcontains a componentCGsuch that#CG≥εn)−→1

and

#CG

n −→2εin probability asn→ ∞.Ifr_n²=c2logn

n for some constantc2>0and the densityf (·)satisfies(1),we have:

(b) Ifc₂is sufficiently large,thenP(Gis connected)−→1asn→ ∞. (c) Ifc₂is sufficiently small,thenlim infnP(Gis not connected) >0.

Here and henceforth any constant will always be independent of n and #CG denotes the number of nodes in C_G. Part (a) of the above result describes the size of the giant componentC_G of G. Parts (b) and (c) describe the behaviour ofGin the densely connected regime. Indeed when the densityf is uniform, parts (b) and (c) are proved in Corollary 3.1 and Corollary 2.1, respectively, of [4]. The proof for non-uniformf satisfying (1) is analogous. Part (a) and related results are discussed in Chapter 11 of [6]. Also, it is known thatε(c1)−→¹₂asc1→ ∞(see Chapters 9 and 11 of [6]).

Not much is known about the graph for intermediate values ofr_n. The size of the giant component is not known as a function of r_n. Our main contribution in this paper is developing techniques to analyze the structure of giant component in the intermediate range i.e., when ^c_n¹ ≤r_n²≤c₂^logn_n for sufficiently large positive constantsc₁, c₂and obtain estimates on the size and diameter of non-giant components (Theorem1). The advantage of our approach is that it can also be used to study related problems in RGGs.

Before we state the main result, we define the diameter of a graph. The diameter of any subgraphHofGis defined as

diam(H )=sup

u,v

d_H(u, v),

whered_H(u, v)represents the graph distance between the nodes uandv inH and the supremum is taken over all pairsu, vbelonging to the vertex set ofH. We state the main result of the paper below. LetTGdenote the collection of all components ofG. For a fixedβ >0 we define the following events: Let

Un=Un(β)=

#TG≤ 1 r_n²e^−βnrn2

denote the event that the number of components ofGis less than _r¹2 ne^−βnrn2, V_n=V_n(β)=

there existsC₀∈TGsuch that #C0≥n−ne^−βnrn2

denote the event that there exists a (giant) componentC₀inTGwhose size is at leastn−ne^−βnr2nand Wn=Wn(β)=Vn∩

sup

C∈TG\C₀

diam(C)≤ 1 β

logn nr_n²

2

denote the event that the diameter of every component ofGother than the giant componentC₀is less than_β¹(^logn

nr_n² )². Theorem 1. Consider the graphG=G(n, r_n, f ),where the densityf (x)satisfies(1)and the radiusr_nsatisfies

c₁

n ≤r_n²≤c₂logn

n (2)

for some fixed positive constantsc₁andc₂.LetU_nandW_nbe events as defined above and fixδ >1.Ifnr_n²−→ ∞as n→ ∞,there exists a positive constantβ=β(δ)sufficiently small so that:

(i) P(U_n)≥1−e^{−βn1−1/δ} and (ii) P(Wn)≥1−e^−βnrn2,for alln≥1.

The above result essentially says wheneverr_nis in the intermediate range as in (2), a giant component ofGexists with very high probability and moreover it contains nearly all the nodes.

2. Proof of Theorem1

Divide the unit squareSinto small ^r_Δⁿ×^r_Δⁿ closed squares{S_i}i≥1and chooseΔ=Δ_n∈ [4,5]so that ^Δ_r

n is an integer.

We choose such aΔso that nodes in adjacent squares can be joined by an edge inG. DefineSi to be occupied if it has at least one node and vacant otherwise.

2.1. Proof of(i)

We first count the number of vacant squares in the set {Si}i. We then use the fact that for each vacant squareSj, the ^8r_Δⁿ ×^8r_Δⁿ square with same centre asS_j intersects at most 81 distinct components ofGto prove (i). The choice of 8 is not crucial and any integer larger than 2 suffices since we only need to estimate the number of components

“associated” withS_j. The total number of squares ist=(_r^Δ

n)². To obtain an estimate on the total number of vacant squares, we let{Z_i}1≤i≤t be Bernoulli random variables taking values either zero or one. We setZ_i=1 if and only if the squareSi is vacant which happens if and only if none of thennodes are inSi.

We note that the sum

iZ_i equalskif and only if there are exactlykvacant squares. Since the random variables {Zi}i are not independent, we cannot evaluate the probability that

iZi =kusing standard binomial estimates. We therefore proceed as follows. The number of ways of choosingksquares from a total oftsquares is ^t_k

. The total area of theksquares isk^r2n

Δ² ≥ ^kr₂₅ⁿ² sinceΔ≤5. All theksquares chosen are empty with probability at mostp_kⁿ, where p_k=1−kinf

i

Si

f (x)dx≤1−β₀kr_n²≤e^−β0krn2 (3)

andβ0=₂₅¹ inf_x∈[0,1]²f (x) >0. Thus using the inequality ⁿ_k

≤(^ne_k)^k, we have

P _t

i=1

Z_i≥k

≤ t

j=k

t j

p_jⁿ≤

t

j=k

te j

j

p_jⁿ

≤ t

j=k

te j

j

e^−jβ0nr2n≤ t

j=k

te k

j

e^−jβ0nrn2.

Settingk=ete^{−θ nrn2}for some constantθ < β₀to be determined later and lettingβ₁=β₀−θ, we get for all sufficiently largenthat

P _t

i=1

Z_i≥ete^{−θ nr2n}

≤ t

j=k

e^−jβ1nrn2

≤ e^−kβ1nrn2

1−e^−β1nrn2 ≤2e^−kβ1nrn2

=2 exp −ete^{−θ nrn2}β₁nr_n²

=2 exp −β₁eΔ²ne^{−θ nrn2}

≤2 exp −16eβ1ne^{−θ nrn2} ,

where we uset=Δ²r_n⁻²andΔ≥4, respectively, in obtaining the last two expressions. We use the fact thatnr_n²−→

∞to obtain the third inequality. In what follows, the constants{β_i}i≥1are not necessarily same in each occurrence.

Letδ >1 be any constant. Sincer_n²≤c₂^logn_n for somec₂>0 (see (2)), we chooseθsufficiently small so that θ nr_n²≤θ c2logn≤1

δ logn.

This implies that P

_t

i=1

Z_i≥ete^{−θ nrn2}

≤2 exp −16eβ1n^1−1/δ .

Also, for each vacant squareS_j, the ^8r_Δⁿ×^8r_Δⁿ square with same centre asS_jintersects at most 81 distinct components ofG. Sincet=^Δ_r2²

n ≤²⁵_r2

n, we get from the above equation that P #TG≥2025er_n⁻²e^{−θ nrn2}

≤2 exp −16eβ1n^1−1/δ and (i) follows.

The rest of the proof is devoted to establishing (ii). The idea is to tileShorizontally and vertically into rectangles and show that each rectangle contains a crossing of edges in the longer direction with high probability. We then join together these crossings to form a “backbone” and show that it forms a part of the giant component. Throughout, we defineKn=^logn_nr2

n and allowKn to be an integer. (Later, we show that the tiling is (slightly) modified ifKnis not an integer without any change in the argument.)

From (2), we have thatK_n≥_c¹

2. For positive integersm₁andm₂, letRbe any^m_Δ^2rn×^m1K_Δ^nrn rectangle contained inSwhich contains exactlym1m2Knof the squares from{Si}i. We define a left–right crossing inRto be any set of distinct squaresL=(S_j1, . . . , S_jl)such that:

(a) For everyi, the squaresS_jiandS_ji+1 share an edge.

(b) Sj₁ intersects the left face ofRandSj_l intersects the right face.

If every square inLis occupied, we say thatLis an occupied left–right crossing. We define analogously occupied top–bottom and vacant crossings of R. The only difference in the definition of vacant crossings is that “edge” in condition (a) above is replaced by “corner”. Figure1illustrates an occupied left–right crossing in a ^m_Δ^2rn ×^m1K_Δ^nrn rectangleR. The nodes in the rectangle are illustrated as dark dots and the sequence of grey squares form an occupied left–right crossing inR. We need the following estimate on the probability of occurrence of an occupied left–right crossing inR.

Fig. 1. Occupied left–right crossing in the rectangleRfor someΔ≥4.

Fig. 2. Vacant top–bottom crossing of a 4×9 rectangle inZ²from the sitex. Circled sites are occupied.

Lemma 2. Forn≥N₀ (independent of the choices ofm₁ andm₂),the event that an occupied left–right crossing occurs inRhas probability at least

1− m₂

n^m1δ1 (4)

for some constantδ₁>0 (independent of the choices ofm₁andm₂).

We use the above estimate to construct a “backbone” ofGand thus prove (ii). Before we do so, we prove Lemma2.

The proof is independent of the rest of the proof of Theorem1.

Proof of Lemma2. To prove (4), we identify the centre of each squareS_i contained inRwith a vertex inZ²in the natural way. Thus the rectangleRhas an equivalent rectangleR˜ consisting of sites inZ². Say that a site is occupied if the corresponding squareSi is occupied and vacant otherwise. Analogous to crossings inR, define occupied and vacant crossings inR.˜

We now use the fact that either a left–right occupied crossing or a top–bottom vacant crossing must always occur inR˜but not both (see e.g., [1] or [3]). To evaluate the probability of a vacant top–bottom crossing, we fix a pointx in the top face ofR˜and consider a vacant crossing of lengthkstarting fromx (see Fig.2for illustration). The area enclosed by the corresponding vacant top–bottom crossingΠ₁inR⊂R²is ^kr_Δ2²ⁿ ≥^kr₂₅ⁿ², sinceΔ≤5. The probability that a particular node is present inΠ1is (see (3))

Π₁

f (x)dx≥kβ₀r_n²,

whereβ0=₂₅¹ inf_x∈[0,1]²f (x) >0. Therefore the probability thatΠ1is vacant is less than 1−kβ₀r_n²n

≤e^−knβ0r2n.

Since the number of vacant top–bottom crossings of lengthkstarting fromxis less than 8^k(at each step no more than eight choices are possible), the probability that there exists a vacant top–bottom crossing ofksquares starting from the square S_x with centrex and contained in R is bounded above by 8^ke^−knβ0r2n. Any top–bottom crossing starting fromS_xmust necessarily contain at leastm₁K_nand no more thanm₁m₂K_nsquares. Therefore the probability that there exists a vacant top–bottom crossing starting fromSxand contained inRis bounded above by

m1m2Kn

k=m1Kn

8^ke^−kβ0nrn2≤ e^−β1nrn2m1Kn

for a fixed constant 0< β₁< β₀and alln≥N₀, for some constantN₀independent of the choices ofm₁andm₂. In the above, we use the fact thatnr_n²−→ ∞and therefore that 8e^−β0nrn2<e^−β1nr2n for all nsufficiently large. Since

(a) (b)

Fig. 3. Construction of the backbone. (a) The eventEnin the unit square. Each horizontal wavy line is an occupied left–right crossing of^r_Δⁿ×^r_Δⁿ squares as in Fig.1. (b) Start horizontal tiling from below. The two topmost 1×^MK_Δ^nrn rectangles in the tiling overlap. Perform a vertical tiling analogously.

there arem2possibilities forSx, the probability that there exists a vacant top–bottom crossing ofRis bounded above by

m2 e^−β1nrn2m1Kn=m2e^−β1m1logn

=m2

1 n^β1

m₁

sinceK_n=^logn_nr2

n .

2.2. Proof of(ii)

Tile the squareShorizontally into a set of rectanglesRH each of size 1×^Mr_Δ^nKn and also vertically into rectangles each of size ^Mr_Δ^nKn ×1 for some fixed integer constantM≥1 to be determined later. The argument below is for a perfect tiling as in Fig.3(a). Otherwise we perform an analogous argument with tiling as in Fig.3(b). LetRbe a fixed 1×^MK_Δ^nrn rectangle in the tilingRH and letδ >1 be a fixed constant. From (4), we know thatRcontains an occupied left–right crossing with probability at least

1−Δ r_n

1

n^Mδ1 ≥1− Δ

√c₁

√n

n^Mδ1 ≥1− 1 n^δ+2

ifMis sufficiently large. Fix such anM. The first inequality above is becauser_n²≥^c_n¹ for some constantc₁(see (2)).

LetE_n^H denote the event that every rectangle inRH contains an occupied left–right crossing inGsatisfying (a)–(b) described above. The number of rectangles inRH is less than

Δ

Mr_nK_n ≤ Δ Mr_n

1 c₂≤ Δ

Mc₂

√n

√c₁≤D1

√n

for some constantD1>0. In evaluating the above we again use (2). The first inequality is becauseKn=^logn_nr2

n ≥_c¹₂ by our choice ofr_nin (2) and the second inequality follows becauser_n²≥^c_n¹. It follows that

P E_n^H

≥1−D1√ n

n^δ+2 ≥1− 1 n^δ+1

for alln sufficiently large. Following an analogous analysis for the vertically tiled rectangles described in the first paragraph of the proof and defining an analogous event E_n^V with occupied top–bottom crossings, we have that P(E_n^V)≥1−_nδ¹+1. Thus ifE_n=E^H_n ∩E_n^V, we have that

P(E_n)≥1− 2

n^δ+1. (5)

In Fig.3(a), we depict the occurrence of the eventEn. We see that the eventEnresults in a connected set of ^r_Δⁿ×^r_Δⁿ squaresB⊆ {S_i}iforming a “backbone” of crossings inS. LetC₀denote the component ofGcontaining nodes inB.

In the above, we have assumed thatKn=^logn_nr2

n is an integer. If not, we setKn= ^logn_nr2

nand starting from the base of the squareS, we perform an analogous horizontal tiling as above. The only difference is that the two topmost rectangles could overlap as seen in Fig.3(b). A similar situation occurs in the vertical tiling. Following an analogous analysis as above, we obtain (5) and a corresponding backbone. The rest of the argument below remains unchanged.

We note that the tiling ofSinto vertical and horizontal rectangles induces a tiling ofS into ^Mr_Δ^nKn ×^Mr_Δ^nKn size squares{S_i}i. If the eventE_n occurs, then the resulting backboneB(and hence the componentC₀) intersects each squareS_i“vertically” and “horizontally” as shown in Fig.3(a). Therefore, if there exists a connected componentCof Gdistinct fromC₀, it must necessarily be contained in a^2MK_Δ ⁿ×^2MK_Δ ⁿ square with centre at some^r_Δⁿ×^r_Δⁿ squareS_i. In Fig.4, the squareA1A2A3A4of Fig.3(a) is magnified and a componentCdistinct fromC0is shown. The centre of the hatched^r_Δⁿ×^r_Δⁿ square is also the centre ofA₁A₂A₃A₄.

Clearly in such a componentC, the minimum number of edges traversed in going from any nodeuto any other nodevis at most(^2MK_Δⁿ)²< (2MKn)²and therefore diam(C) < (2MKn)². To summarize, so far we have proved that if eventE_noccurs, then a backboneBand hence the componentC₀containing all the nodes in squares comprising the backbone and possibly other nodes exist. Moreover, any component ofGdistinct fromC₀has diameter less than (2MKn)². Recall thatTGis the set of all components ofGand forθ >0 let

F_n=F_n(θ )=

C∈TG: diam(C)<(2MKn)²

#C < ne^{−θ nrn2}

denote the event that the sum of sizes of components whose diameter does not exceed(2MKn)²is less thanne^{−θ nrn2}. We have the following estimate on probability of occurrence of the eventFn.

Fig. 4. The squareA₁A₂A₃A₄in Fig.3(a) is magnified to show a component not attached to the backbone.

Lemma 3. We have

P(F_n)≥1−e^−θ1nrn2 (6)

for some positive constantsθandθ1.

Before we prove the above result, we complete the proof of (ii). WheneverEn∩Fnoccurs, the componentC0con- tains at leastn−ne^{−θ nrn2} nodes and is therefore the giant component. Also, the diameter of any non-giant component is less than(2MKn)². Choosingθ1>0 smaller if necessary, we have from (5) and (6) that the eventEn∩Fnoccurs with probability

P(En∩Fn)≥1−e^−θ1nrn2− 2

n^δ+1≥1−2e^−θ1nrn2

for all n sufficiently large. In the above estimate, we have used the fact (2) that nr_n²≤c2logn for some positive constantc₂. This proves (ii) and hence Theorem1. The proof of Lemma3is independent of the proof of Theorem1 and is provided below.

Proof of Lemma3. Say that a set of squaresC⊆ {Si}i is a cluster if they form a connected set inR². We say that the clusterCis occupied if every square in the cluster is occupied.

Fixiand consider the squareS_i. IfS_iis occupied, denoteCito be the maximal occupied cluster containingS_i. Set X_ito be the number of nodes inCiifCi is contained in the 2(2MKn)²r_n×2(2MKn)²r_nsquareS_iⁱⁿwith same centre asS_i. Otherwise setX_i to be zero. Thus,

iX_i is an upper bound on the sum of size of components whose diameter is less than 2(2MKn)². In the beginning of the proof of (ii), we recall that to obtain the estimate(2MKn)² on the diameter of a component not attached to the backbone, we had considered a 2MKn×2MKn square appropriately centred (like A₁A₂A₃A₄ in Fig.4). In this subsection, however, we are not given any information regarding the backbone. Therefore, to obtain a bound on the size of a component whose diameter is less than(2MKn)²the only information we have is that the component is enclosed in a (slightly bigger) 2(2MKn)²×2(2MKn)²square.

We first estimate P({#Ci =k} ∩ {X_i =0}) for k≥1. Suppose thatX_i =0 and therefore that the clusterCi is contained in the squareS_iⁱⁿ. Our aim now is to obtain a sufficiently large number of vacant squares “attached to”Ci. ConsiderCi as a set inR²and let∂₁, . . . , ∂_T be its disjoint boundaries. Each∂_i is a circuit of edges(e_i,1, . . . , e_i,Li) (not necessarily self-avoiding) such thatei,1andei,L_i touch each other. SinceCi is connected, one of the boundaries, say∂₁, contains all squares ofCi and all the other boundaries in its interior. Also, any squareS_j that has an edge e1,j∈∂1and not contained inCi is necessarily vacant.

Letπ₁denote the set of distinct vacant squares that contain some edge in∂₁. The path∂₁containsL₁≥2 edges of which at least ^L₂¹ of them have an endvertex in the interior of the unit squareS. (Here we use the fact that the clusterCi is contained inS_iⁱⁿ. If we did not have such a bounding box for the clusterCi, the above statement will not hold; e.g. consider the event that each square in{S_k}kcontains at least one node.) From the discussion in the previous paragraph, each such “interior” edge has a vacant square “attached” to it. Since each vacant square is counted at most four times (once for each of its four edges), this implies that #π1≥ ^L₈¹. In Fig.5, the dark grey square isSi and the grey squares form Ci. The set of vacant squaresπ₁is shown by the squares markedΠ and the curve of thick lines represents∂1.

To compute the probability that such a vacant set of squares occurs, we set the centre ofS_i to be the origin and drawX- andY-axes parallel to the sides ofSi. Lete1,lastbe the “last” edge in∂1that intersects theX-axis at(xlast,0).

In other words, if an edgee_1,j in∂₁intersects theX-axis at(x_j,0), thenx_last> x_j. In Fig.5, the edgee_1,lastis also shown. Clearly, there are at mostL1possibilities for the location of edgee1,last. Also, the number of choices for∂1

starting frome_1,lastis less than 4^L1.

Now, the total area of squares inπ1is at least^L₈¹_Δ^rn22 ≥^L₈^1r₂₅ⁿ² sinceΔ≤5. Given∂1, with probability at least^L₈¹β0r_n² a particular node is present inπ1whereβ0=₂₅¹ inf_x∈[0,1]²f (x) >0 is as in (3). Therefore with probability at most

1−1

8β₀L₁r_n² n

≤e^{−β0L1nrn2/8} none of thennodes are present inπ1.

Fig. 5. The occupied clusterCiand the set of vacant squaresπ1(marked by the symbolΠ) are shown for the squareSithat is denoted by the dark square.

IfCi containsksquares, then the number of edgesL₁in∂₁satisfies

√k

4 ≤L₁≤4k. The upper bound is clear. To see why the lower bound is true, suppose that∂₁has less than

√k

4 edges. It is then necessary that∂₁is contained in the

√k 2

r_n

Δ×^√₂^kr_Δⁿ squareS_pkwith the same centre asS_i. The squareS_pkcontains at most ^k₄ squares from{S_j}j. This is a contradiction since the path∂₁containsCi in its interior andCi containsksquares. Thus fork≥1 we have from the above discussion that

P {#Ci=k} ∩ {X_i=0}

≤

√k/4≤l≤4k

e^{−lβ0nrn2/8}l4^l

≤4k

√k/4≤l≤4k

4e^−β0nrn2/8l

≤ke^−θ0nrn2

√k

(7) for a fixed positive constantθ₀<^β₄₀⁰ and alln≥N₀, whereN₀is a constant that does not depend onk. Here we use the fact thatnr_n²−→ ∞and hence that 4e^−β0nrn2/8<e^−5θ0nrn2for some constantθ₀>0 and for all sufficiently largen.

LettingN (A)denote the number of nodes in the setA, we therefore have that EXi =E

C0

S_j∈C0

N (Sj)1(Ci=C0)1(Xi=0)

=I1+I2,

where the summation in the first line is over all clustersC0that contain the squareS_i and are contained inS_iⁱⁿ. In the above equation,

I1=E

C0

S_j∈C0

N (Sj)1(Ci=C0)1 N (C0)≥2ekδ0nr_n²

1(Xi=0),

I₂=EX_i−I₁andδ₀=₁₆¹ sup_x∈Sf (x).

To evaluateI₁andI₂, we need a couple of preliminary estimates. For a fixedC0containingksquares, we estimate P(N (C0)≥2ekδ0nr_n²)first. Indeed since a particular node is present inC0with probability at mostqk=kδ0r_n², we

have that

P N (C0)≥2enqk

≤

2enqk≤j≤n

n j

q_k^j

≤

2enqk≤j≤n

ne j

j

q_k^j

≤

2enqk≤j≤n

ne 2enqk

j

q_k^j

≤

j≥2enq_k

1 2

j

≤e^−2β2knrn2 (8)

for some positive constantβ₂independent ofk, iand the choice ofC0. In the third inequality above, we have used the estimate ⁿ_k

≤(^ne_k)^k. Also, the expected number of nodes in any squareSi is bounded above by sup

j

EN (Sj)=nsup

j

Sj

f (x)dx≤n sup

x∈[0,1]²

f (x)r_n²

Δ²≤D1nr_n² (9)

for some positive constantD₁since sup_x∈[0,1]2f (x) <∞(see (1)) andΔ≥4. Analogously, sup

j

EN (S_j)²≤D₂ nr_n²2

(10) for some positive constantD2.

To evaluateI₁, we now use Cauchy–Schwarz inequality to obtain that I₁≤

k≥1

#C0=k

Sj∈C0

EN (S_j)1 N (C0)≥2ekδ0nr_n²

≤

k≥1

#C0=k

S_j∈C0

EN²(S_j)1/2

P N (C0)≥2ekδ0nr_n²1/2

≤D₃nr_n²

k≥1

#C0=k

Sj∈C0

e^−kβ2nrn2

for some positive constantD₃independent ofi. In obtaining the final estimate, we use (8) and (10) and the notation

#C0=k refers to the sum over all clustersC0containingksquares of which one of them isS_i. Since the number of such clusters is less than 8^k, we get

I1≤D3nr_n²

k≥1

k8^ke^−kβ2nrn2≤D4nr_n²e^−β3nrn2 for some positive constantsD₄andβ₃, independent ofi.

To evaluateI₂we write I₂=E

k≥1

#C0=k

Sj∈C0

N (S_j)1(Ci=C0)1 N (C0)≤2ekδ0nr_n²

1(Xi=0)

≤2eδ0nr_n²E

k≥1

k

#C0=k

S_j∈C0

1(Ci=C0)1(Xi=0)

=2eδ0nr_n²E

k≥1

k²

#C0=k

1(Ci=C0)1(Xi=0)

=2eδ0nr_n²

k≥1

k²P {#Ci=k} ∩ {Xi=0}

≤2eδ0nr_n²

k≥1

k³e^−θ0nrn2

√k≤D5nr_n²e^−β5nrn2

for some positive constantsD₅andβ₅independent ofi, where the second inequality follows from the estimate (7).

From the estimates ofI1andI2, we therefore have that

EX_i≤D₆nr_n²e^−β6nrn2 (11)

for some positive constantsD6andβ6independent ofi.

The number of squares in{Si}i isΔ²r_n⁻². By Markov inequality, we therefore have forθ >0 that P

_Δ²_r−2

n

i=1

X_i≥ne^{−θ nrn2}

≤

iEX_i n e^{θ nrn2}

≤ Δ²r_n⁻²D₆nr_n²e^−β6nrn2

n e^{θ nrn2}≤D7e^−θ1nrn2

for some positive constants θ1 andD7, if θ is sufficiently small. Since Fn= {

iXi < ne^{−θ nrn2}}, this proves the

lemma.

Acknowledgments

I thank Professor Rahul Roy for a careful reading of the manuscript and a referee for comments which led to an improvement of the paper. The support from a National Board for Higher Mathematics scholarship and a Sandwich Ph.D. programme scholarship is gratefully acknowledged. I thank Kristina Mattson for help with the typesetting.

References

[1] B. Bollobas and O. Riordan.Percolation. Cambridge Univ. Press, New York, 2006.MR2283880

[2] M. Franceschetti, O. Dousse, D. N. C. Tse and P. Thiran. Closing gap in the capacity of wireless networks via percolation theory.IEEE Trans.

Inform. Theory53(2007) 1009–1018.MR2302808

[3] G. Grimmett.Percolation, 2nd edition.Grundlehren der Mathematischen Wissenschaften321. Springer, Berlin, 1999.MR1707339

[4] P. Gupta and P. R. Kumar. Critical power for asymptotic connectivity in wireless networks. InStochastic Analysis, Control, Optimization and Applications547–566.Systems Control Found. Appl.Birkhäuser, Boston, MA, 1999.

[5] S. Muthukrishnan and G. Pandurangan. The bin-covering technique for thresholding random geometric graph properties. InProc. SODA 989–998. ACM, New York, 2005.MR2298358

[6] M. Penrose.Random Geometric Graphs. Oxford Studies in Probability5. Oxford Univ. Press, Oxford, 2003.MR1986198