Substituting (n)andr into Eqn.(4-8).and for all sufficient large n,we further have Then wen bound boud P() log(L.H.S of Eqn.(4-7)) PUF)≤∑PF) 2≥osm-kr(et+P+(xr+P）) ≥1ogm-km(ar+R2+8（a2rjP)） 26-- (414) =-5-kre(n+28)-2vk(n logn+5) ≤me-r(r-R2kna 40(y logn +)2 1 e2cvk(n log n) ekTe(na+28) e-ξ-1 -n(e-1) (4-10) where the second equality is due to Lemma 1. Since a+28<0 and a>0,u can be arbitrarily close to Since >0 and a+28 <0,there exists some constant 0 when n is large.Taking the exponent of both sides and let c>1 such that Eqn.(4-13)holds. 6=e#i<1,Lemma 2 follows. Then,we have the following lemma. V.CRITICAL TRANSMISSION RANGE FOR Lemma 3:If r =logmis(n)where a+28<0,0< CLUSTER-DENSE STATE kπ a≤1,0<y≤1 and lim(n)=ξ<+o,we have n-oo In this state,.we have a+23≥元andπR2=w(a): Different from the cluster-sparse state focusing on the cluster lim inf Peas(n,a,B,Y,r0p）≥e-(1-e-). (4-11) connectivity perspective,we investigate cluster-dense state mainly from the perspective of cluster-member connectivity, Proof:Denote K.as the event that Ci is the only since cluster members behave more like independent nodes disconnected cluster during the k time slots. now.Let Peds(n,a,B,,rp)denote the probability that g(n,,B,,rP)has some cluster members disconnected. Pess(n,a,B,Y:rp) ≥∑P(K) Proposition 2:In a correlated mobile k-hop clustered net- work g(n,B,Y,r)for the cluster-dense state,the crit- 2-Gn) ical transmission range is where ,0<a≤1,0<y≤1. ≠ ≥∑P)-(∑PF)月 A.Necessary Condition of Proposition 2 The probability that a cluster member is not connected ≥m1-r+2)n-(m(1-r-2))》 during k timeslots is (5-1) >0e--m2e-2xkn"(r-R)2 P(fis)=(1-πr2)km =0e-5-e-2e2kr8(na+2a0)+4vKme(n2V√iogn+） The following lemma will show the probability that there exists one cluster member not connected,if we regard each 0e-f-(1+2)e-2 cluster member as independent nodes. (4-12) Lemma4:fr=Va志，a+28≥，0<a≤l,for for sufficient large n and u2>0.Note that the fourth any fixed 0<0<1 and sufficiently large n,we have inequality is due to Lemma I and the fifth inequality is due to Lemma 2.Recall that 6 can be arbitrarily close to 1 and u2 e-f≤n1-r2)n ≤e-】 (5-2) is arbitrarily close to 0 for sufficient large n,we can obtain Lemma 3. ◆ Proof:For the right hand side of Eqn.(5-2). B.Sufficient Condition of Proposition I nl-r2)r≤ne-tamr-e- Recall that f is the event that cluster member is dis- Then employing similar technique in the proof of Lemma connected during.Therefore,for the sufficient condition of 2.for all sufficient large n we have Proposition 1,it suffices to show that when r =crep.(c >1), the following equality holds. sa-r2))2osm-an（(am2+8car2) ▣(g心0)=▣U)= =--50ogn+)2 6kna 会-ξ-43 (4-13) (5-3)
6 Substituting δ(n) and r = Èγ log n+ξ kπnα into Eqn. (4-8), and for all sufficient large n, we further have log L.H.S of Eqn. (4-7) ≥log m − knα π(r + R) 2 + 5 6 π(r + R) 2 2 ≥ log m − knα π(r + R) 2 + 5 6 π(2r) 2 2 = − ξ − kπΘ(n α+2β ) − 2 √ kπΘ(n α+2β 2 È γ log n + ξ) − 40(γ log n + ξ) 2 3knα , − ξ − µ1. (4-10) Since α + 2β < 0 and α > 0, µ1 can be arbitrarily close to 0 when n is large. Taking the exponent of both sides and let θ = e −µ1 < 1, Lemma 2 follows. Then, we have the following lemma. Lemma 3: If r = Èγ log n+ξ(n) kπnα where α + 2β < 0, 0 < α ≤ 1, 0 < γ ≤ 1 and limn→∞ ξ(n) = ξ < +∞, we have lim inf n→∞ Pcss(n, α, β, γ, rw.p. c ) ≥ e −ξ (1 − e −ξ ). (4-11) Proof: Denote Kc i as the event that Ci is the only disconnected cluster during the k time slots. Pcss(n, α, β, γ, rw.p. c ) ≥ Xm i=1 P(K c i ) ≥ Xm i=1 P(Fi) − X j̸=i P(Fi ∩ Fj ) ≥ Xm i=1 P(Fi) − Xm i=1 P(Fi) 2 ≥m 1 − π(r + R) 2 knα − m 1 − π(r − R) 2 knα 2 ≥θe−ξ − m2 e −2πknα(r−R) 2 =θe−ξ − e −2ξ e −2kπΘ(n α+2β )+4√ kπΘ n α+2β 2 √ γ log n+ξ ,θe−ξ − (1 + µ2)e −2ξ (4-12) for sufficient large n and µ2 > 0. Note that the fourth inequality is due to Lemma 1 and the fifth inequality is due to Lemma 2. Recall that θ can be arbitrarily close to 1 and µ2 is arbitrarily close to 0 for sufficient large n, we can obtain Lemma 3. B. Sufficient Condition of Proposition 1 Recall that f λ jκ is the event that cluster member ψ λ jκ is disconnected during λ. Therefore, for the sufficient condition of Proposition 1, it suffices to show that when r = crw.p. c (c > 1), the following equality holds. limn→∞ P [m j=1 [ϖ κ=1 ( \ k λ=1 f λ jκ) = limn→∞ P( [m j=1 Fj ) = 0. (4-13) Then we use union bound to bound P( Sm j=1 Fj ): P( [m j=1 Fj ) ≤ Xm j=1 P(Fj ) ≤ Xm j=1 1 − π(r − R) 2 knα ≤me−π(r−R) 2knα = 1 n(c 2−1)γ · e 2c √ kπγΘ(n α+2β 2 log n) e kπΘ(nα+2β) , (4-14) where the second equality is due to Lemma 1. Since γ > 0 and α + 2β < 0, there exists some constant c > 1 such that Eqn. (4-13) holds. V. CRITICAL TRANSMISSION RANGE FOR CLUSTER-DENSE STATE In this state, we have α + 2β ≥ ϵ k and πR2 = ω( 1 nα ). Different from the cluster-sparse state focusing on the cluster connectivity perspective, we investigate cluster-dense state mainly from the perspective of cluster-member connectivity, since cluster members behave more like independent nodes now. Let Pcds(n, α, β, γ, rw.p. c ) denote the probability that G(n, α, β, γ, rw.p. c ) has some cluster members disconnected. Proposition 2: In a correlated mobile k-hop clustered network G(n, α, β, γ, rw.p. c ) for the cluster-dense state, the critical transmission range is r w.p. c = È log n kπnα , where α + 2β ≥ ϵ k , 0 < α ≤ 1, 0 < γ ≤ 1. A. Necessary Condition of Proposition 2 The probability that a cluster member is not connected during k timeslots is P(fjκ) = (1 − πr2 ) knα . (5-1) The following lemma will show the probability that there exists one cluster member not connected, if we regard each cluster member as independent nodes. Lemma 4: If r = Èlog n+ξ kπnα , α + 2β ≥ ϵ k , 0 < α ≤ 1, for any fixed 0 < θ < 1 and sufficiently large n, we have θe−ξ ≤ n 1 − πr2 knα ≤ e −ξ . (5-2) Proof: For the right hand side of Eqn. (5-2), n 1 − πr2 knα ≤ ne−kπnαr 2 = e −ξ . Then employing similar technique in the proof of Lemma 2, for all sufficient large n, we have log n 1 − πr2 knα ≥ log n − knα πr2 + 5 6 (πr2 ) 2 = − ξ − 5(log n + ξ) 2 6knα , − ξ − µ3. (5-3)
As n goes to infinity,H3 will go to 0.Let =e#<1,and we have this lemma. Lemma 5 will be used to calculate the probability that at least two members in a same cluster fail to connect. Lemma 5:For a sufficiently large n,we have (5-4) 0 sin29.e-岩osm0=o( log n Fig.5.1:Illustrations of P(ff). Proof:Rearranging the equation,we have logn n sim20.e2-尝21ognd0 0 For the second term,let P1,P2l denote the distance =logn sin 20.e()logn do between two positions P and P2.By considering all the /o possible positions of two cluster members,we have =logn sin2t·e-(岩2)logn dt ()=(() ≤logn 2t.e-器1 ogn dt =(P吟吟l=PU庆n会1g,=) 0 (5-9) =2logn(-2logn kπ -24-2原。 4log2n When2r,we have kn2 (n--1), P(f六nf六1x,‖>2r）≤(1-2mr2)n 2logn (5-10) (5-5) ≤n~e-装 where in the second equality we let t=-0.Let noo where the second inequality is due to Lemma 4. and the lemma follows. Lemma6:fr=√s+d,where a+28≥元，0< When2r,the related positions of and krna a≤l,0<y≤1 and lim(m)=ξ<+oo,we have are shown in Figure 5.1.Therefore we have lim inf Ped(m,a,B,,rp）≥e(1-e). (5-6) P(I,e‖=x)P(会nfe|le,el=x)d ≤ 2”红1-22士Acs茎2Yr2上 f2r2rxd江 Proof:For a fixed E,let Km denote the event that fir is the only failed session during k time slots and we have 42 sin 20(1-2wr2+20r2-r2 sin 20)mdo m四 Pc(n,a,B,x,rp)≥∑∑P(c) 4r2 j=1k=1 ≤R0 sin 20e(2si 20)ndo 空-三.n加) =e-2mnor2 4p2 sin 20e 2do 台台 (5-7) 2 -∑∑∑∑(f, ≤n-是e-等e善4(logn+) sin 20e odo krna R2 Jo j=1K=1≠jk'=1 =n-e-等e是4logn+1n走 The second term to compute the probability that two members kmno Ra o(1ogn〉 (5-11) in a same cluster failed.The third term is to compute the probability that two memebrs in different clusters failed. where we let t=2r cos6 in the second equality,and the last equality is due to Lemma 5. For the first and third terms,we have By substituting Eqn.(5-10),Eqn.(5-11)and r into Eqn. ∑PU-∑∑∑三PUxn5) (5-9),we can bound Peds(n,a,B,r-p.)as follows. =1=】 j=1k=1i≠jk'=1 Pf_cds(n,a,B,Y,rP) ≥2U)-(∑PU (5-8) ≥9e-(1-e-)-e-2cn-y1+4e是logn+o kinaR20(oo// =n(1-πr2)km(1-n(1-r2m） =e-f1-e-)-e2x(n-2+4e是1+5 ≥0e-f(1-e-)】 色0e-f-(1+4e-2E knnoian-go(1) where the third inequality is due to Lemma 4. (5-12)
7 As n goes to infinity, µ3 will go to 0. Let θ = e −µ < 1, and we have this lemma. Lemma 5 will be used to calculate the probability that at least two members in a same cluster fail to connect. Lemma 5: For a sufficiently large n, we have Z π 2 0 sin 2θ · e 2θ−sin 2θ kπ log n dθ = o( n 1 k log n ). (5-4) Proof: Rearranging the equation, we have log n n 1 k · Z π 2 0 sin 2θ · e 2θ−sin 2θ kπ log n dθ = log n Z π 2 0 sin 2θ · e ( 2θ−sin 2θ kπ − 1 k ) log n dθ = log n Z π 2 0 sin 2t · e −( 2t+sin 2t kπ ) log n dt ≤ log n Z π 2 0 2t · e − 2t kπ log n dt =2 log n − kπ 2 log n te− 2 log n kπ t − k 2π 2 4 log2 n e − 2 log n kπ t | π 2 t=0 = − kπ2 2 n − 1 k − k 2π 2 2 log n (n − 1 k − 1), (5-5) where in the second equality we let t = π 2 − θ. Let n → ∞ and the lemma follows. Lemma 6: If r = Èlog n+ξ(n) kπnα , where α + 2β ≥ ϵ k , 0 < α ≤ 1, 0 < γ ≤ 1 and limn→∞ ξ(n) = ξ < +∞, we have lim inf n→∞ Pcds(n, α, β, γ, rw.p. c ) ≥ e −ξ (1 − e −ξ ). (5-6) Proof: For a fixed ξ, let Km jκ denote the event that fjκ is the only failed session during k time slots and we have Pcds(n, α, β, γ, rw.p. c ) ≥ Xm j=1 Xϖ κ=1 P(K m jκ) ≥ Xm j=1 Xϖ κ=1 P(fjκ) − Xm j=1 Xϖ κ=1 X κ′̸=κ P(fjκ ∩ fjκ′ ) − Xm j=1 Xϖ κ=1 X i̸=j Xϖ κ′=1 P(fiκ ∩ fjκ′ ). (5-7) The second term to compute the probability that two members in a same cluster failed. The third term is to compute the probability that two memebrs in different clusters failed. For the first and third terms, we have Xm j=1 Xϖ κ=1 P(fjκ) − Xm j=1 Xϖ κ=1 X i̸=j Xϖ κ′=1 P(fiκ ∩ fjκ′ ) ≥ Xm j=1 Xϖ κ=1 P(fjκ) − Xm j=1 Xϖ κ=1 P(fjκ) 2 = n(1 − πr2 ) knα 1 − n(1 − πr2 ) knα ≥ θe−ξ (1 − e −ξ ), (5-8) where the third inequality is due to Lemma 4. Fig. 5.1: Illustrations of P(f λ jκ ∩ f λ jκ′ ). For the second term, let ∥P1,P2∥ denote the distance between two positions P1 and P2. By considering all the possible positions of two cluster members, we have P(fjκ ∩ fjκ′ ) = P(f λ jκ ∩ f λ jκ′ ) k = Z x P(∥ψ λ jκ, ψλ jκ′∥ = x)P(f λ jκ ∩ f λ jκ′ | ∥ψ λ jκ, ψλ jκ′∥ = x)dxk . (5-9) When ∥ψ λ jκ, ψλ jκ′∥ > 2r, we have P f λ jκ ∩ f λ jκ′ | ∥ψ λ jκ, ψλ jκ′∥ > 2r ≤ (1 − 2πr2 ) n α ≤ n − 2 k e − 2ξ k , (5-10) where the second inequality is due to Lemma 4. When ∥ψ λ jκ, ψλ jκ′∥ ≤ 2r, the related positions of ψ λ jκ and ψ λ jκ′ are shown in Figure 5.1. Therefore we have Z 2r x=0 P ∥ψ λ jκ, ψλ jκ′∥ = x P f λ jκ ∩ f λ jκ′ | ∥ψ λ jκ, ψλ jκ′∥ = x dx ≤ Z 2r 0 2πx dx πR2 1 − 2πr 2 + 4 arccos x 2 r 2π · πr 2 − x É r 2 − x2 4 n α = 4r 2 R2 Z π 2 0 sin 2θ(1 − 2πr 2 + 2θr2 − r 2 sin 2θ) n α dθ ≤ 4r 2 R2 Z π 2 0 sin 2θe(−2πr2+2θr2−r 2 sin 2θ)n α dθ = e −2πnαr 2 4r 2 R2 Z π 2 0 sin 2θe(2θr2−r 2 sin 2θ)n α dθ ≤ n − 2 k e − 2ξ k e ξ k 4(log n + ξ) kπnαR2 Z π 2 0 sin 2θe 2θ−sin 2θ kπ log n dθ = n − 2 k e − 2ξ k e ξ k 4(log n + ξ) kπnαR2 o( n 1 k log n ). (5-11) where we let x = 2r cos θ in the second equality, and the last equality is due to Lemma 5. By substituting Eqn. (5-10), Eqn. (5-11) and r into Eqn. (5-9), we can bound Pcds(n, α, β, γ, rw.p. c ) as follows. Pf cds(n, α, β, γ, rw.p. c ) ≥ θe−ξ (1 − e −ξ ) − e −2ξn −γ 1 + 4e ξ k log n + ξ kπnαR2 o( n 1 k log n ) k = θe−ξ (1 − e −ξ ) − e −2ξ n − γ k + 4e ξ k 1 + ξ log n kπnα+2β− ϵ k o(1)k , θe−ξ − (1 + µ4)e −2ξ (5-12)