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 Š | 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 Š ≥ 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′ ) Šk = € Z x P(∥ψ λ jκ, ψλ jκ′∥ = x)P(f λ jκ ∩ f λ jκ′ | ∥ψ λ jκ, ψλ jκ′∥ = x)dxŠk . (5-9) When ∥ψ λ jκ, ψλ jκ′∥ > 2r, we have P € f λ jκ ∩ f λ jκ′ | ∥ψ λ jκ, ψλ jκ′∥ > 2r Š ≤ (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 ) Šk = 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)