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Fundamentals of Continuum Percolation

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Fundamentals of Continuum Percolation Fundamentals of continuum percolation Mathew D. Penrose University of Bath, UK Tutorial lecture Information Theory and Applications Workshop University of California San Diego February 3, 2010 1 CONTINUUM PERCOLATION (see Meester and Roy 1996) Let be a homogeneous Poisson point process in Rd with intensity Pλ λ, i.e. (A) Poisson( A ) Pλ ∼ | | and (A ) are independent variables for A , A ,... disjoint. Pλ i 1 2 Gilbert Graph. Form a graph := G( ) on by connecting two Gλ Pλ Pλ Poisson points x, y iff x y 1. | − |≤ Form graph 0 := G( 0 ) similarly on 0 . Gλ Pλ ∪ { } Pλ ∪ { } Let p (λ) be the prob. that the component of 0 containing 0 has k k Gλ vertices (also depends on d). ∞ p∞(λ):=1 p (λ) be the prob. that this component is infinite. − k=0 k P 2 RANDOM GEOMETRIC GRAPHS (finite analog) (Penrose 2003) Let U1,...,Un be independently uniformly randomly scattered in a cube of volume n/λ in d-space. Form a graph on 1, 2,...,n by Gn,λ { } i j iff U U 1 ∼ | i − j|≤ is a geometric alternative to the classical random graph model. Gn,λ Define C to be the order of the component of containing i. It | i| Gn,λ can be shown that n−1 n 1 C = k P p (λ) as n , i.e. i=1 {| i| } −→ k → ∞ for large n, P n P n−1 1 C = k p (λ) 1 {| i| }≈ k ≈ " i=1 # X That is, the proportionate number of vertices of Gn lying in components of order k, converges to pk(λ) in probability. 3 A FORMULA FOR pk(λ). k pk+1(λ)=(k + 1)λ h(x1,...,xk) Rd k Z( ) exp( λA(0, x ,...,x ))dx ...dx × − 1 k 1 k where h(x ,...,x ) is 1 if G( 0, x ,...,x ) is connected and 1 k { 1 k} 0 x x lexicographically, otherwise zero; ≺ 1 ≺···≺ k and A(0, x1,...,xk) is the volume of the union of 1-balls centred at 0, x1,...,xk. Not tractable for large k. 4 THE PHASE TRANSITION If p∞(λ) = 0, then has no infinite component, almost surely. Gλ If p∞(λ) > 0, then has a unique infinite component, almost surely. Gλ Also, p∞(λ) is nondecreasing in λ. Fundamental theorem: If d 2 then ≥ λ (d) := sup λ : p∞(λ) = 0 (0, ). c { }∈ ∞ If d = 1 then λ (d)= . From now on, assume d 2. The value of c ∞ ≥ λc(d) is not known. 5 Sketch proof of fundamental theorem. Be wise, discretize! Divide Rd into boxes of side ε (ε small, fixed). If C = there is an infinite path through occupied boxes | 0| ∞ successively distance less than two from each other. There exists finite γ such that the expected number of such paths of d length n is at most γn(1 e−λε )n. For λ small enough, this tends to − zero as n . Hence the probability of an infinite path is zero. → ∞ Conversely, if C is finite, there must be a path of empty boxes | 0| surrounding the origin and by a similar argument, can show for λ large enough the probability of this happening is less than 1. 6 CONTINUITY OF p∞(λ) Clearly p∞(λ) = 0 for λ < λc Also p∞(λ) is increasing in λ on λ > λc. Less trivially, it is known that p∞(λ) is continuous in λ on λ (λ , ) and is right continuous at λ = λ , i.e. ∈ c ∞ c p∞(λc)= lim p∞(λ). λ↓λc So p∞( ) is continuous on (0, ) iff · ∞ p∞(λc) = 0. This is known to hold for d = 2 (Alexander 1996) and for large d (Tanemura 1996). It is conjectured to hold for all d. 7 HIGH-INTENSITY ASYMPTOTICS Let θ denote the volume of the unit ball in Rd. As λ , → ∞ p∞(λ) 1 and in fact (Penrose 1991) → 1 p∞(λ) p (λ) = exp( λθ) − ∼ 1 − This says that for large λ, given the unlikely event that the component of 0 containing 0 is finite, it is likely to consist of an Gλ isolated vertex. Also, for k 1, ≥ λ log p (λ)= λθ +(d 1)k log + O(1) − k+1 − k as λ with k fixed (Alexander 1991). → ∞ 8 LARGE-k ASYMPTOTICS FOR pk(λ) Suppose λ < λc. Then there exists ζ(λ) > 0 such that −1 ζ(λ)= lim n log pn(λ) n→∞ − or more informally, p (λ) (e−ζ(λ))n. n ∼ n Proof uses subadditivity. With x1 := (x1,...,xn), recall n n n −λA(0,x1 ) n pn+1(λ)=(n + 1)λ h(x1 )e dx1 . Z Setting q := p /(n + 1), can show q q q − , so n n+1 n m ≤ n+m 1 log q /(n 1) inf ≥ ( log q /(n 1)) := ζ as n . − n − → n 1 − n − → ∞ That ζ(λ) > 0 is a deeper result. 9 LARGE-k ASYMPTOTICS: THE SUPERCRITICAL CASE Suppose λ > λc. Then −(d−1)/d lim sup n log pn(λ) < 0 n→∞ −(d−1)/d lim inf n log pn(λ) > n→∞ −∞ Loosely speaking, this says that in the supercritical case pn(λ) decays exponentially in n1−1/d, whereas in the subcritical case it decays exponentially in n. 10 HIGH DIMENSIONAL ASYMPTOTICS (Penrose 1996) d Let θd denote the volume of the unit ball in R . Suppose λ = λ(d) is chosen so that λ(d)θd = µ (this is the expected degree of the origin 0 in 0). Gλ Asymptotically as d with µ fixed, the structure of C converges → ∞ 0 to that of a branching process (Z0,Z1,...) with Poisson (µ) offspring distribution (Z0 = 1, Zn is nth generation size). So (Penrose 1996): p (λ(d)) p˜ (µ) := P [ ∞ Z = k]; k → k n=0 n ∞ p∞(λ(d)) ψ(µ) := P [P Zn = ]; → n=0 ∞ θdλc(d) 1. P → 11 DETAILS OF THE BRANCHING PROCESS THEORY: ∞ kk−2 p˜ (µ)= P [ Z = k]= µk−1e−kµ. k n (k 1)! n=0 X − t = 1 ψ(µ) is the smallest positive solution to t = exp(µ(t 1)). In − − particular, ψ(µ) = 0, if µ 1 ≤ ψ(µ) > 0, if µ> 1 12 LARGE COMPONENTS FOR THE RGG Consider again the random geometric graph (on n uniform Gn,λ random points in a cube of volume n/λ in d-space) Let L ( ) be the size of the largest component, and L ( ) the 1 Gn,λ 2 Gn,λ size of the second largest component (‘size’ measured by number of vertices). As n with λ fixed: → ∞ −1 P if λ > λ then n L ( ) p∞(λ) > 0 c 1 Gn,λ −→ if λ < λ then (log n)−1L ( ) P 1/ζ(λ) c 1 Gn,λ −→ and for the Poissonized RGG (N Poisson (n)), GNn,λ n ∼ L ( )= O(log n)d/(d−1) in probability if λ > λ 2 GNn,λ c 13 CENTRAL LIMIT THEOREMS. Let K( ) be the number of Gn,λ components of . As n with λ fixed, Gn,λ → ∞ t K( ) EK( ) 2 P Gn,λ − Gn,λ t Φ(t) := (2π)−1/2 e−x /2dx σ√n ≤ → Z−∞ and if λ > λc, L ( ) EL ( ) P 1 Gn,λ − 1 Gn,λ t Φ(t). τ√n ≤ → Here σ and τ are positive constants, dependent on λ. 14 ISOLATED VERTICES. Suppose d = 2. Let N ( ) be the 0 Gn,λ number of isolated vertices. The expected number of isolated vertices satisfies EN ( ) n exp( πλ) 0 Gn,λ ≈ − so if we fix t and take λ(n) = (log n + t)/π, then as n , → ∞ EN ( ) e−t. 0 Gn,λ(n) → Also, N0 is approximately Poisson distributed so P [N ( )=0] exp( e−t). 0 Gn,λ(n) → − 15 CONNECTIVITY. Note is connected iff K( )=1. Gn,λ Gn,λ Clearly P [K( )=1] P [N ( ) = 0]. Again taking Gn,λ ≤ 0 Gn,λ(n) λ(n) = (log n + t)/π with t fixed, it turns out (Penrose 1997) that −t lim P [K( n,λ(n))=1]= lim P [N0( n,λ(n)) = 0] = exp( e ) n→∞ G n→∞ G − or in other words, lim (P [K( n,λ(n)) > 1] P [N0( n,λ(n)) > 0]) = 0. n→∞ G − G This is related to the earlier result that at high intensity, the most likely way for a vertex to be in a finite component is if it is isolated. 16 THE CONNECTIVITY THRESHOLD Let V1,...,Vn be independently uniformly randomly scattered in [0, 1]d. Form a graph r on 1, 2,...,n by Gn { } i j iff V V r. ∼ | i − j |≤ Given the values of V1,...,Vn, define the connectivity threshold ρn(K = 1), and the no-isolated-vertex threshold ρn(N0 = 0), by ρ (K = 1) = min r : K( r ) = 1 ; n { Gn } ρ (N = 0) = min r : N ( r ) = 1 . n 0 { 0 Gn } The preceding result can be interpreted as giving the limiting distributions of these thresholds (suitably scaled and centred) as n : they have the same limiting behaviour. → ∞ 17 Taking λ(n) = (log n + t)/π with t fixed, we have √λ(n)/n P [K( )=1]= P [K( n )=1] Gn,λ(n) G = P ρ (K = 1) (log n + t)/(πn) n ≤ h i so the earlier result p −t lim P [K( n,λ(n))=1]= lim P [N0( n,λ(n)) = 0] = exp( e ) n→∞ G n→∞ G − implies 2 −t lim P [nπ(ρn(K = 1)) log n t] = exp(e ) n→∞ − ≤ and likewise for ρn(N0) = 0.
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