Large Deviations and Random Graphs
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- Sophie Gilmore
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1 Large Deviations and Random Graphs S.R.S. Varadhan Courant Institute, NYU UC Irvine March 25, 2011 Large DeviationsandRandom Graphs p.1/39
2 Joint work with Sourav Chatterjee. Large DeviationsandRandom Graphs p.2/39
3 Joint work with Sourav Chatterjee. G is a graph Large DeviationsandRandom Graphs p.2/39
4 Joint work with Sourav Chatterjee. G is a graph V(G) is the set of its vertices Large DeviationsandRandom Graphs p.2/39
5 Joint work with Sourav Chatterjee. G is a graph V(G) is the set of its vertices E(G) is the set of its edges. Large DeviationsandRandom Graphs p.2/39
6 Joint work with Sourav Chatterjee. G is a graph V(G) is the set of its vertices E(G) is the set of its edges. Not oriented. Large DeviationsandRandom Graphs p.2/39
7 Joint work with Sourav Chatterjee. G is a graph V(G) is the set of its vertices E(G) is the set of its edges. Not oriented. V(G) is the number of vertices in G Large DeviationsandRandom Graphs p.2/39
8 Joint work with Sourav Chatterjee. G is a graph V(G) is the set of its vertices E(G) is the set of its edges. Not oriented. V(G) is the number of vertices in G E(G) are the number of edges in G. Large DeviationsandRandom Graphs p.2/39
9 G N is the set of all graphsgwithn vertices. Large DeviationsandRandom Graphs p.3/39
10 G N is the set of all graphsgwithn vertices. We think ofv(g) as {1,2,...,N} Large DeviationsandRandom Graphs p.3/39
11 G N is the set of all graphsgwithn vertices. We think ofv(g) as {1,2,...,N} The set E is all unordered pairs(i,j), i.e. the full set of edges. Large DeviationsandRandom Graphs p.3/39
12 G N is the set of all graphsgwithn vertices. We think ofv(g) as {1,2,...,N} The set E is all unordered pairs(i,j), i.e. the full set of edges. E(G) E Large DeviationsandRandom Graphs p.3/39
13 G is determined by its edge set E(G) E Large DeviationsandRandom Graphs p.4/39
14 G is determined by its edge set E(G) E G N = 2 (N 2) Large DeviationsandRandom Graphs p.4/39
15 G is determined by its edge set E(G) E G N = 2 (N 2) A random graph withn vertices is just a probability measure on G N, i.e. a collection of weights {p(g)} with G G N p(g) = 1 Large DeviationsandRandom Graphs p.4/39
16 What types of random graphs do we consider? Large DeviationsandRandom Graphs p.5/39
17 What types of random graphs do we consider? A pair of vertices are connected by an edge with probabilityp Large DeviationsandRandom Graphs p.5/39
18 What types of random graphs do we consider? A pair of vertices are connected by an edge with probabilityp Different edges are independent. Large DeviationsandRandom Graphs p.5/39
19 What types of random graphs do we consider? A pair of vertices are connected by an edge with probabilityp Different edges are independent. Th probabilityp N,p (G) of a graph G G N is given by Large DeviationsandRandom Graphs p.5/39
20 What types of random graphs do we consider? A pair of vertices are connected by an edge with probabilityp Different edges are independent. Th probabilityp N,p (G) of a graph G G N is given by p E(G) (1 p) (N 2) E(G) Large DeviationsandRandom Graphs p.5/39
21 What types of random graphs do we consider? A pair of vertices are connected by an edge with probabilityp Different edges are independent. Th probabilityp N,p (G) of a graph G G N is given by p E(G) (1 p) (N 2) E(G) Ifp = 1 2, the distribution is uniform and we are essentially counting the number of graphs. Large DeviationsandRandom Graphs p.5/39
22 We want to calculate the probabilities of the following types of events. Large DeviationsandRandom Graphs p.6/39
23 We want to calculate the probabilities of the following types of events. Let Γ be a finite graph. Large DeviationsandRandom Graphs p.6/39
24 We want to calculate the probabilities of the following types of events. Let Γ be a finite graph. The number of different occurrences ofγin G is # G (Γ) Large DeviationsandRandom Graphs p.6/39
25 We want to calculate the probabilities of the following types of events. Let Γ be a finite graph. The number of different occurrences ofγin G is # G (Γ) The number of different occurrences ofγin a complete graph with N vertices # N (Γ) Large DeviationsandRandom Graphs p.6/39
26 We want to calculate the probabilities of the following types of events. Let Γ be a finite graph. The number of different occurrences ofγin G is # G (Γ) The number of different occurrences ofγin a complete graph with N vertices # N (Γ) The ratio r G (Γ) = # G(Γ) # N (Γ) Large DeviationsandRandom Graphs p.6/39
27 We consider mapsv(γ) V(G) that are one to one. Large DeviationsandRandom Graphs p.7/39
28 We consider mapsv(γ) V(G) that are one to one. If V(G) = N and V(Γ) = k there are p(n,k) = N(N 1) (N k +1) of them. Large DeviationsandRandom Graphs p.7/39
29 We consider mapsv(γ) V(G) that are one to one. If V(G) = N and V(Γ) = k there are p(n,k) = N(N 1) (N k +1) of them. Of these a certain number p(g,n,k) will map a connected pair of vertices in Γ to connected ones in G. Large DeviationsandRandom Graphs p.7/39
30 We consider mapsv(γ) V(G) that are one to one. If V(G) = N and V(Γ) = k there are p(n,k) = N(N 1) (N k +1) of them. Of these a certain number p(g,n,k) will map a connected pair of vertices in Γ to connected ones in G. r G (Γ) = p(g,n,k) p(n,k). Large DeviationsandRandom Graphs p.7/39
31 There is some ambiguity here due to possible multiple counting. Large DeviationsandRandom Graphs p.8/39
32 There is some ambiguity here due to possible multiple counting. It is a multiple that depends only on Γ and will cancel out when we take the ratio. Large DeviationsandRandom Graphs p.8/39
33 There is some ambiguity here due to possible multiple counting. It is a multiple that depends only on Γ and will cancel out when we take the ratio. What probabilities do we want to estimate? Large DeviationsandRandom Graphs p.8/39
34 There is some ambiguity here due to possible multiple counting. It is a multiple that depends only on Γ and will cancel out when we take the ratio. What probabilities do we want to estimate? For each j {1,2,...,k} we are given a finite graphγ j and a number r j [0,1]. Large DeviationsandRandom Graphs p.8/39
35 There is some ambiguity here due to possible multiple counting. It is a multiple that depends only on Γ and will cancel out when we take the ratio. What probabilities do we want to estimate? For each j {1,2,...,k} we are given a finite graphγ j and a number r j [0,1]. We are interested in estimating the probability P N,p [ j, rg (Γ j ) r j ǫ ] Large DeviationsandRandom Graphs p.8/39
36 More precisely we are interested in calculating the function Large DeviationsandRandom Graphs p.9/39
37 More precisely we are interested in calculating the function ψ p ({Γ j,r j }) given by lim ǫ 0 lim N 1 [ ) logp N,p j, rg (Γ j ) r j ǫ ] ( N 2 Large DeviationsandRandom Graphs p.9/39
38 More precisely we are interested in calculating the function ψ p ({Γ j,r j }) given by lim ǫ 0 lim N 1 [ ) logp N,p j, rg (Γ j ) r j ǫ ] ( N 2 ψ p ({Γ j,r j }) = 0 if and only if r j = p E(Γ j) for j = 1,2,...,k. Large DeviationsandRandom Graphs p.9/39
39 Let us consider the set K = {f : [0,1] 2 [0,1];f(x,y) = f(y,x)} Large DeviationsandRandom Graphs p.10/39
40 Let us consider the set Define K = {f : [0,1] 2 [0,1];f(x,y) = f(y,x)} H p (f) = h p (f(x,y))dxdy Large DeviationsandRandom Graphs p.10/39
41 Let us consider the set Define K = {f : [0,1] 2 [0,1];f(x,y) = f(y,x)} Where H p (f) = 1 0 h p (f) = f log f p 1 0 h p (f(x,y))dxdy +(1 f)log 1 f 1 p Large DeviationsandRandom Graphs p.10/39
42 For any f K, finite graph Γ with vertices {1,2,...,k}, and edge set E(Γ) we define r Γ (f) = Π (i,j) E(Γ) f(x i,x j )Π k i=1dx i [0,1] V(Γ) Large DeviationsandRandom Graphs p.11/39
43 For any f K, finite graph Γ with vertices {1,2,...,k}, and edge set E(Γ) we define r Γ (f) = Π (i,j) E(Γ) f(x i,x j )Π k i=1dx i [0,1] V(Γ) For example ifγis the triangle, then r (f) = f(x 1,x 2 )f(x 2,x 3 )f(x 3,x 1 )dx 1 dx 2 dx 3 [0,1] 3 Large DeviationsandRandom Graphs p.11/39
44 For ak cycle it is [0,1] k f(x 1,x 2 )f(x 2,x 3 ) f(x k,x 1 )dx 1 dx 2 dx k Large DeviationsandRandom Graphs p.12/39
45 For ak cycle it is [0,1] k f(x 1,x 2 )f(x 2,x 3 ) f(x k,x 1 )dx 1 dx 2 dx k The main result is. ψ p ({Γ j,r j }) = inf {f : j, r Γ j(f)=r j } H p (f) Large DeviationsandRandom Graphs p.12/39
46 What is it good for? Large DeviationsandRandom Graphs p.13/39
47 What is it good for? Let us analyze ψ p (Γ,c). Large DeviationsandRandom Graphs p.13/39
48 What is it good for? Let us analyze ψ p (Γ,c). Calculus of variations. The infimum is attained. Proof later. Compactness and continuity. Large DeviationsandRandom Graphs p.13/39
49 What is it good for? Let us analyze ψ p (Γ,c). Calculus of variations. The infimum is attained. Proof later. Compactness and continuity. Euler equation log f(x,y) 1 f(x,y) log p 1 p = β 1 0 f(x,z)f(y,z)dx Large DeviationsandRandom Graphs p.13/39
50 Subject to [0,1] 3 f(x,y)f(y,z)f(z,x)dxdydz = c Large DeviationsandRandom Graphs p.14/39
51 Subject to [0,1] 3 f(x,y)f(y,z)f(z,x)dxdydz = c If c p 3 << 1, then f(x,y) = c 1 3 is the only solution and so is optimal. Large DeviationsandRandom Graphs p.14/39
52 Subject to [0,1] 3 f(x,y)f(y,z)f(z,x)dxdydz = c If c p 3 << 1, then f(x,y) = c 1 3 is the only solution and so is optimal. For any c, ifp << 1, then a clique f = 1 [0,c 1 3 ] (x)1 [0,c 1 3] (y) is a better option than f c 1 3. Large DeviationsandRandom Graphs p.14/39
53 A slight increase or decrease fromp 3 in the proportion of triangles is explained by a corresponding deviation in the number of edges frompto c 1 3. Large DeviationsandRandom Graphs p.15/39
54 A slight increase or decrease fromp 3 in the proportion of triangles is explained by a corresponding deviation in the number of edges frompto c 1 3. This is not the case if p is small but c is not. Large DeviationsandRandom Graphs p.15/39
55 A slight increase or decrease fromp 3 in the proportion of triangles is explained by a corresponding deviation in the number of edges frompto c 1 3. This is not the case if p is small but c is not. A similar story when c is small but p is not. Large DeviationsandRandom Graphs p.15/39
56 A slight increase or decrease fromp 3 in the proportion of triangles is explained by a corresponding deviation in the number of edges frompto c 1 3. This is not the case if p is small but c is not. A similar story when c is small but p is not. A bipartite graph is a better option. Large DeviationsandRandom Graphs p.15/39
57 What is the general Large Deviations setup and how do we apply it here? Large DeviationsandRandom Graphs p.16/39
58 What is the general Large Deviations setup and how do we apply it here? A metric space X and a sequence P n of probability distributions. Large DeviationsandRandom Graphs p.16/39
59 What is the general Large Deviations setup and how do we apply it here? A metric space X and a sequence P n of probability distributions. P n δ x0. Large DeviationsandRandom Graphs p.16/39
60 What is the general Large Deviations setup and how do we apply it here? A metric space X and a sequence P n of probability distributions. P n δ x0. IfAis such thatd(x 0,A) > 0 P n (A) 0 as n. Large DeviationsandRandom Graphs p.16/39
61 Want a lower semi continuos functioni(x) such that 1 n logp n[s(x,ǫ)] = I(x)+o(ǫ)+o ǫ (n) Large DeviationsandRandom Graphs p.17/39
62 Want a lower semi continuos functioni(x) such that 1 n logp n[s(x,ǫ)] = I(x)+o(ǫ)+o ǫ (n) lim sup ǫ 0 lim sup n 1 n logp n(s(x,ǫ)) I(x) Large DeviationsandRandom Graphs p.17/39
63 Want a lower semi continuos functioni(x) such that 1 n logp n[s(x,ǫ)] = I(x)+o(ǫ)+o ǫ (n) lim sup ǫ 0 lim sup n 1 n logp n(s(x,ǫ)) I(x) lim inf ǫ 0 lim inf n 1 n logp n(s(x,ǫ)) I(x) Large DeviationsandRandom Graphs p.17/39
64 IfK is compact lim sup n 1 n logp n(k) inf x K I(x) Large DeviationsandRandom Graphs p.18/39
65 IfK is compact lim sup n 1 n logp n(k) inf x K I(x) IfGis open lim inf n 1 n logp n(g) inf x G I(x) Large DeviationsandRandom Graphs p.18/39
66 For any l the set {x : I(x) l} is compact. Large DeviationsandRandom Graphs p.19/39
67 For any l the set {x : I(x) l} is compact. For any l <, there is a set K l such that C K l = implies lim sup n 1 n logp n(c) l Large DeviationsandRandom Graphs p.19/39
68 For any l the set {x : I(x) l} is compact. For any l <, there is a set K l such that C K l = implies lim sup n 1 n logp n(c) l It now follows that for any closed set C lim sup n 1 n logp n[c] inf x C I(x) Large DeviationsandRandom Graphs p.19/39
69 Contraction Principle. Large DeviationsandRandom Graphs p.20/39
70 Contraction Principle. {P n } satisfies LDP with ratei(x) on X, Large DeviationsandRandom Graphs p.20/39
71 Contraction Principle. {P n } satisfies LDP with ratei(x) on X, F : X Y is a continuous map. Large DeviationsandRandom Graphs p.20/39
72 Contraction Principle. {P n } satisfies LDP with ratei(x) on X, F : X Y is a continuous map. Q n = P n F 1 satisfies an LDP on Y Large DeviationsandRandom Graphs p.20/39
73 Contraction Principle. {P n } satisfies LDP with ratei(x) on X, F : X Y is a continuous map. Q n = P n F 1 satisfies an LDP on Y With rate function J(y) = inf x:f(x)=y I(x) Large DeviationsandRandom Graphs p.20/39
74 Let us turn to our case. The probability measures are on graphs withn vertices. Large DeviationsandRandom Graphs p.21/39
75 Let us turn to our case. The probability measures are on graphs withn vertices. The space keeps changing. Large DeviationsandRandom Graphs p.21/39
76 Let us turn to our case. The probability measures are on graphs withn vertices. The space keeps changing. Need to put them all on the same space. Large DeviationsandRandom Graphs p.21/39
77 Let us turn to our case. The probability measures are on graphs withn vertices. The space keeps changing. Need to put them all on the same space. Every graph is an adjacency matrix. Large DeviationsandRandom Graphs p.21/39
78 Let us turn to our case. The probability measures are on graphs withn vertices. The space keeps changing. Need to put them all on the same space. Every graph is an adjacency matrix. Random graph is a random symmetric matrix. X = {x i,j }, x i,i = 0,x i,j {0,1} Large DeviationsandRandom Graphs p.21/39
79 0 x 1,2 x 1,N x 2,1 0 x 2,N... x N,1 x N,2 0 Large DeviationsandRandom Graphs p.22/39
80 Imbed in K. Simple functions constant on small squares. 0 x 1,2 x 1,N x 2,1 0 x 2,N x N,1 x N,2 0 Large DeviationsandRandom Graphs p.23/39
81 Measures {Q N,p } onk. Large DeviationsandRandom Graphs p.24/39
82 Measures {Q N,p } onk. The space K needs a topology. Weak is good. Nice compact space. Large DeviationsandRandom Graphs p.24/39
83 Measures {Q N,p } onk. The space K needs a topology. Weak is good. Nice compact space. Q N,p δ p Large DeviationsandRandom Graphs p.24/39
84 Measures {Q N,p } onk. The space K needs a topology. Weak is good. Nice compact space. Q N,p δ p Lower Bound. Let f be a nice function in K. Large DeviationsandRandom Graphs p.24/39
85 Measures {Q N,p } onk. The space K needs a topology. Weak is good. Nice compact space. Q N,p δ p Lower Bound. Let f be a nice function in K. Create a random graph with probabilityf( i N, j N ) of connecting i and j Large DeviationsandRandom Graphs p.24/39
86 By law of large numbers Q f N δ f in the weak topology on K. Large DeviationsandRandom Graphs p.25/39
87 By law of large numbers Q f N δ f in the weak topology on K. The new measureq f N on K has entropy ( ) N H(Q f N,Q N,p) H p (f) 2 Large DeviationsandRandom Graphs p.25/39
88 Standard tilting argument dp P(A) = dq dq A 1 = Q(A) Q(A) Q(A)exp[ 1 dq log e dp dq A log dq Q(A) dp dq] = exp[ H(Q;P)+o(H(Q,P))] A Large DeviationsandRandom Graphs p.26/39
89 Upper Bound. Cramér. 2 N 2 logeq N,p [ N J(x, y)f(x, y)dxdy] log[pe J(x,y) +(1 p)]dxdy Large DeviationsandRandom Graphs p.27/39
90 Upper Bound. Cramér. 2 N 2 logeq N,p [ N J(x, y)f(x, y)dxdy] log[pe J(x,y) +(1 p)]dxdy Tchebychev. Half-plane. For small balls, optimize. Large DeviationsandRandom Graphs p.27/39
91 Upper Bound. Cramér. 2 N 2 logeq N,p [ N J(x, y)f(x, y)dxdy] log[pe J(x,y) +(1 p)]dxdy Tchebychev. Half-plane. For small balls, optimize. I(f) = H p (f) Large DeviationsandRandom Graphs p.27/39
92 Upper Bound. Cramér. 2 N 2 logeq N,p [ N J(x, y)f(x, y)dxdy] log[pe J(x,y) +(1 p)]dxdy Tchebychev. Half-plane. For small balls, optimize. I(f) = H p (f) Are we done! Large DeviationsandRandom Graphs p.27/39
93 NO!, Why? Large DeviationsandRandom Graphs p.28/39
94 NO!, Why? The object of interest is the map F = {r Γ j (f)}; K [0,1] k Large DeviationsandRandom Graphs p.28/39
95 NO!, Why? The object of interest is the map F = {r Γ j (f)}; K [0,1] k They are not continuous unless no two edges in consists Γ share a common vertex. Large DeviationsandRandom Graphs p.28/39
96 NO!, Why? The object of interest is the map F = {r Γ j (f)}; K [0,1] k They are not continuous unless no two edges in consists Γ share a common vertex. Well. Change the topology to L 1 Large DeviationsandRandom Graphs p.28/39
97 NO!, Why? The object of interest is the map F = {r Γ j (f)}; K [0,1] k They are not continuous unless no two edges in consists Γ share a common vertex. Well. Change the topology to L 1 No chance. Even the Law of large numbers fails. Large DeviationsandRandom Graphs p.28/39
98 NO!, Why? The object of interest is the map F = {r Γ j (f)}; K [0,1] k They are not continuous unless no two edges in consists Γ share a common vertex. Well. Change the topology to L 1 No chance. Even the Law of large numbers fails. In between topology! Cut topology. Large DeviationsandRandom Graphs p.28/39
99 d(f,g) = sup a 1 b 1 = sup A,B A B [f(x, y) g(x, y]a(x)b(y)dxdy [f(x,y) g(x,y]dxdy Large DeviationsandRandom Graphs p.29/39
100 d(f,g) = sup a 1 b 1 = sup A,B A B [f(x, y) g(x, y]a(x)b(y)dxdy [f(x,y) g(x,y]dxdy In the cut topology F is continuous. Half the battle! Large DeviationsandRandom Graphs p.29/39
101 d(f,g) = sup a 1 b 1 = sup A,B A B [f(x, y) g(x, y]a(x)b(y)dxdy [f(x,y) g(x,y]dxdy In the cut topology F is continuous. Half the battle! Law of large numbers? Large DeviationsandRandom Graphs p.29/39
102 Enough to takeaand B as unions of intervals of the form[ j N, j+1 N ] Large DeviationsandRandom Graphs p.30/39
103 Enough to takeaand B as unions of intervals of the form[ j N, j+1 N ] For each A B it is only the ordinary LLN for independent random variables. Large DeviationsandRandom Graphs p.30/39
104 Enough to takeaand B as unions of intervals of the form[ j N, j+1 N ] For each A B it is only the ordinary LLN for independent random variables. Error Boundse cn2 Large DeviationsandRandom Graphs p.30/39
105 Enough to takeaand B as unions of intervals of the form[ j N, j+1 N ] For each A B it is only the ordinary LLN for independent random variables. Error Boundse cn2 Number of rectangles2 n 2 n. That is good. LLN Holds. Large DeviationsandRandom Graphs p.30/39
106 Enough to takeaand B as unions of intervals of the form[ j N, j+1 N ] For each A B it is only the ordinary LLN for independent random variables. Error Boundse cn2 Number of rectangles2 n 2 n. That is good. LLN Holds. Three fourths of the battle! Large DeviationsandRandom Graphs p.30/39
107 IfKwere compact in the cut topology we would be done. Large DeviationsandRandom Graphs p.31/39
108 IfKwere compact in the cut topology we would be done. But it is not. The projection f f(x,y)dy is continuous. The image is L 1 and not compact. Large DeviationsandRandom Graphs p.31/39
109 IfKwere compact in the cut topology we would be done. But it is not. The projection f f(x,y)dy is continuous. The image is L 1 and not compact. The problem is invariant under a huge group. The permutation groupπ N. Large DeviationsandRandom Graphs p.31/39
110 IfKwere compact in the cut topology we would be done. But it is not. The projection f f(x,y)dy is continuous. The image is L 1 and not compact. The problem is invariant under a huge group. The permutation groupπ N. The function H p (f), r Γ (f) are invariant under the groupσ Σ of measure preserving transformations of[0,1]. Large DeviationsandRandom Graphs p.31/39
111 Go tok/σ. Large DeviationsandRandom Graphs p.32/39
112 Go tok/σ. It is compact! ( Lovász-Szegedy) Large DeviationsandRandom Graphs p.32/39
113 Go tok/σ. It is compact! ( Lovász-Szegedy) But what is it? Large DeviationsandRandom Graphs p.32/39
114 Go tok/σ. It is compact! ( Lovász-Szegedy) But what is it? It is the space of "graphons" Large DeviationsandRandom Graphs p.32/39
115 Go tok/σ. It is compact! ( Lovász-Szegedy) But what is it? It is the space of "graphons" What is a graphon? Large DeviationsandRandom Graphs p.32/39
116 G n a sequence of graphs. Becoming infinite. Large DeviationsandRandom Graphs p.33/39
117 G n a sequence of graphs. Becoming infinite. We say G n has limit if lim r G n n (Γ) = r(γ) exists for every finite graph Γ Large DeviationsandRandom Graphs p.33/39
118 G n a sequence of graphs. Becoming infinite. We say G n has limit if lim r G n n (Γ) = r(γ) exists for every finite graph Γ Graphon is the map Γ r(γ) Large DeviationsandRandom Graphs p.33/39
119 G n a sequence of graphs. Becoming infinite. We say G n has limit if lim r G n n (Γ) = r(γ) exists for every finite graph Γ Graphon is the map Γ r(γ) It has a representation as r Γ (f) for some f in K Large DeviationsandRandom Graphs p.33/39
120 f is not unique. But r Γ (f) r Γ (g) if and only if f(x,y) = g(σx,σy) for someσ Σ Large DeviationsandRandom Graphs p.34/39
121 f is not unique. But r Γ (f) r Γ (g) if and only if f(x,y) = g(σx,σy) for someσ Σ In other wordsr( ) K/Σ Large DeviationsandRandom Graphs p.34/39
122 SinceK/Σ is compact it is enough to prove the upper bound for ballsb( f,ǫ) in K/Σ Large DeviationsandRandom Graphs p.35/39
123 SinceK/Σ is compact it is enough to prove the upper bound for ballsb( f,ǫ) in K/Σ This means estimating Q N,p [ σ Σ B(σf,ǫ)] Large DeviationsandRandom Graphs p.35/39
124 SinceK/Σ is compact it is enough to prove the upper bound for ballsb( f,ǫ) in K/Σ This means estimating Q N,p [ σ Σ B(σf,ǫ)] Szemerédi s regularity lemma. Large DeviationsandRandom Graphs p.35/39
125 SinceK/Σ is compact it is enough to prove the upper bound for ballsb( f,ǫ) in K/Σ This means estimating Q N,p [ σ Σ B(σf,ǫ)] Szemerédi s regularity lemma. The permutation group Π N Σ by permuting intervals of length 1 N. Large DeviationsandRandom Graphs p.35/39
126 SinceK/Σ is compact it is enough to prove the upper bound for ballsb( f,ǫ) in K/Σ This means estimating Q N,p [ σ Σ B(σf,ǫ)] Szemerédi s regularity lemma. The permutation group Π N Σ by permuting intervals of length 1 N. Given ǫ > 0, there is a finite set {g j } K such that for sufficiently largen, K N = j σ ΠN B(σg j,ǫ) Large DeviationsandRandom Graphs p.35/39
127 It is therefore enough to estimate the probability Q N,p [ j σ ΠN B(σg j,ǫ)) [ σ Σ B(σf,ǫ)]] Large DeviationsandRandom Graphs p.36/39
128 It is therefore enough to estimate the probability Q N,p [ j σ ΠN B(σg j,ǫ)) [ σ Σ B(σf,ǫ)]] Sincej only varies over a finite set, it is enough to estimate for any g Q N,p [ [ σ π(n) B(σg,ǫ)] [ σ Σ B(σf,ǫ)] ] Large DeviationsandRandom Graphs p.36/39
129 It is therefore enough to estimate the probability Q N,p [ j σ ΠN B(σg j,ǫ)) [ σ Σ B(σf,ǫ)]] Sincej only varies over a finite set, it is enough to estimate for any g Q N,p [ [ σ π(n) B(σg,ǫ)] [ σ Σ B(σf,ǫ)] ] This is the same as Q N,p [ σ π(n) [B(σg,ǫ) σ Σ B(σf,ǫ)] ] Large DeviationsandRandom Graphs p.36/39
130 Q N,p is Π N invariant. N! << e cn2. Q N,p [ B(g,ǫ) σ Σ B(σf,ǫ) ] Large DeviationsandRandom Graphs p.37/39
131 Q N,p is Π N invariant. N! << e cn2. Q N,p [ B(g,ǫ) σ Σ B(σf,ǫ) ] If the intersection is nonempty, then it is contained in B(σf, 3ǫ) for some σ Σ. Large DeviationsandRandom Graphs p.37/39
132 Q N,p is Π N invariant. N! << e cn2. Q N,p [ B(g,ǫ) σ Σ B(σf,ǫ) ] If the intersection is nonempty, then it is contained in B(σf, 3ǫ) for some σ Σ. The choice of σ does not depend on N. Only on ǫ. Large DeviationsandRandom Graphs p.37/39
133 Q N,p is Π N invariant. N! << e cn2. Q N,p [ B(g,ǫ) σ Σ B(σf,ǫ) ] If the intersection is nonempty, then it is contained in B(σf, 3ǫ) for some σ Σ. The choice of σ does not depend on N. Only on ǫ. Balls are weakly closed. Large DeviationsandRandom Graphs p.37/39
134 Q N,p is Π N invariant. N! << e cn2. Q N,p [ B(g,ǫ) σ Σ B(σf,ǫ) ] If the intersection is nonempty, then it is contained in B(σf, 3ǫ) for some σ Σ. The choice of σ does not depend on N. Only on ǫ. Balls are weakly closed. We have upper bounds. H p (σf) = H p (f) Large DeviationsandRandom Graphs p.37/39
135 Q N,p is Π N invariant. N! << e cn2. Q N,p [ B(g,ǫ) σ Σ B(σf,ǫ) ] If the intersection is nonempty, then it is contained in B(σf, 3ǫ) for some σ Σ. The choice of σ does not depend on N. Only on ǫ. Balls are weakly closed. We have upper bounds. H p (σf) = H p (f) Done! Large DeviationsandRandom Graphs p.37/39
136 With p = 1 2, we have done the counting. Large DeviationsandRandom Graphs p.38/39
137 With p = 1 2, we have done the counting. The quantity D(N,ǫ) = # {G : r G (Γ j ) r j ǫforj = 1,2,...,k} Large DeviationsandRandom Graphs p.38/39
138 With p = 1 2, we have done the counting. The quantity D(N,ǫ) = # {G : r G (Γ j ) r j ǫforj = 1,2,...,k} Satisfies lim lim ǫ 0 N 2 N 2 logd(n,ǫ) = log2 ψ1 2 ({Γ j,r j }) Large DeviationsandRandom Graphs p.38/39
139 Thank You. Large DeviationsandRandom Graphs p.39/39
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MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is