Fundamentals of Continuum Percolation

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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