Stein s Method for concentration inequalities

Transcription

1 Number of Triangles in Erdős-Rényi random graph, UC Berkeley joint work with Sourav Chatterjee, UC Berkeley Cornell Probability Summer School July 7, 2009

2 Number of triangles in Erdős-Rényi random graph Let T n be the number of triangles in G(n, p). Behavior of the upper tail of subgraph counts in G(n, p) is a problem of great interest in the theory of random graphs. Best known bounds due to Kim and Vu (2004) for triangles and Janson, Oleszkiewicz, and Ruciński (2004) for general subgraphs. For triangles the result of Kim and Vu essentially state that for any fixed ε > 0, e Θ(n2 p 2 log 1/p) P(T n (1 + ε) E[T n ]) e Θ(n2 p 2). We are interested in finding the function f (p, ε) such that P(T n (1 + ε) E[T n ]) = e n2 f (p,ε) (1+o(1)).

3 Large deviation result for number of triangles For r, p (0, 1), define I (r, p) := r log r p + (1 r) log 1 r 1 p. Theorem (Chatterjee & D.) Let T n be the number of triangles in G(n, p), where p > p 0 is fixed with p 0 = 2/(2 + e 3/2 ) Then for any r > p, ( P T n ( ) ] n )r 3 = exp [ n2 3 2 I (r, p) (1 + O(n 1/2 )). Moreover, if p p 0, there exist p, p (p, 1) such that the same result holds for all r (p, p ) (p, 1].

4 Figure: The set of (p, r) where our large deviation result holds.

5 Stein s method for concentration inequalities Let (X, X ) be an exchangeable pair of random variables. (X 0, X 1 ) is an exchangeable pair for a stationary reversible Markov chain X 0, X 1, X 2,... Let F (, ) be an antisymmetric function F (x, y) = F (y, x), such that E[F (X, X ) X ] = f (X ) with E[f (X )] = 0. By exchangeability and antisymmetry, for all g, we have E[g(X )f (X )] = E[g(X )F (X, X )] = 1 2 E[g(X )F (X, X ) + g(x )F (X, X )] = 1 2 E[(g(X ) g(x ))F (X, X )].

6 Concentration inequality using exchangeable pair In particular we have Var(f (X )) = 1 2 E[(f (X ) f (X ))F (X, X )]. Define (X ) := 1 2 E( f (X ) f (X ) F (X, X ) X ). Theorem (Chatterjee 05) Suppose (X ) Bf (X ) + C a.s., then for any t 0, ( P( f (X ) t) 2 exp t 2 2C + 2Bt ).

7 Concentration inequality using exchangeable pair In particular we have Var(f (X )) = 1 2 E[(f (X ) f (X ))F (X, X )]. Define (X ) := 1 2 E( f (X ) f (X ) F (X, X ) X ). Theorem (Chatterjee & D.) Suppose (X ) B f (X ) α + C a.s. for some real number α [0, 2), then for any t 0, ( P( f (X ) > t) 2 exp t 2 α 16 max{b, C 1 α/2 } ).

8 Towards the large deviation result... Any undirected graph on n vertices can be represented by x {0, 1} (n 2) where x ij = 1{edge (i, j) is present} for i < j and this correspondence is a bijection. We need to consider a Gibbs measure on the space of graphs on n vertices with Hamiltonian where T (x) = H(x) = β T (x) n 2 + he(x) i<j<kx ij x jk x ik and E(x) = i<j x ij. denote the number of triangles and edges in x respectively.

9 A new model? Let ψ n (β, h) := log x eh(x) denote log of the partition function. Note that for any β 0 we have ( ) n log P 0,h (T )r 3 β ( ) n r 3 + log E 0,h [e β 3 n 2 3 ( ) n r 3 = ψ n(β, h) ψ n (0, h). If β = 0 we get back G(n, p) with p = ψ n (0, h) = For β 0, ψ n (β, h) =? eh and 1+e h ( ) n log(1 + e h ). 2 n 2 T ]

10 Solution in the high temperature regime Define ϕ : [0, 1] [0, 1] by ϕ(x) = eβx+h 1 + e βx+h. Theorem (Chatterjee and D.) If (β, h) is in a high temperature regime, there is a unique p = p (β, h) [0, 1] such that p = ϕ(p 2 ) and ψ n (β, h) lim ( n n = 2) βp3 3 + hp p log p (1 p ) log(1 p ).

11 Main Step Define L ij := 1 n 2 the fraction of wedges at (i, j). X ik X jk k i,j Easy fact: E[ X ij rest of the edge configuration] = ϕ(l ij ). First step: For all β 0, h R we have ( ) L ij β ϕ(l ik )ϕ(l jk ) + O n 2 n with high probability. k i,j Proof uses suitable exchangeable pair using Glauber dynamics, suitable antisymmetric function and Stein s method for concentration inequalities theorem.

12 Other temperature free results Using similar techniques we have E(X) ( n ) ( 1 n ) ( ) 1 + β ϕ(l ij ) + O n 2 2 i<j T (X) ( n ) ( 1 n ) ( ) 1 + β ϕ(l ij )ϕ(l jk )ϕ(l ik ) + O. n 3 3 i<j<k

13 Rigorous result in high temperature regime When (β, h) is in high temperature regime, the system of equations a ij = 1 ϕ(a ik )ϕ(a jk ) n 2 k i,j has a unique solution at (a ij = u for all i < j). Lemma (High temperature regime) Let β, h be such that the equation u = ϕ(u) 2 has a unique solution at u. Assume that 2ϕ(u )ϕ (u ) < 1. Then for each i < j, we have E L ij u K(β, h) n where K(β, h) is a constant depending only on β, h.

14 High temperature regime Figure: The shaded region corresponds to the high temperature regime. This boundary was predicted to be the phase transition curve using physical arguments by Park and Newman (2005).

15 Partition function in high temperature regime Temperature free results then imply that ( ) E n E(X) p K(β, h)n 3/2 2 ( ) E n T (X) p 3 3 K(β, h)n5/2 where p = ϕ(u ). We have H(x) = β T (x) n 2 +he(x) where h is such that p = Hence ψ n (β, h) ( ) ( ) n βp (h h )p +h E(x) eh 1+e. h ( ) ( ) n βp (h h )p + ψ n (0, h ).

16 Proof of the Large deviation result So given p, r find β such that r = p (β, h) where p = The upper bound holds. The lower bound can be proved by showing that ( ) n P( T n )r 3 C(p, r)n 5/2 3 ( ) = exp n2 I (r, p) (1 + O(n 1/2 )) 2 for r, p as above. eh. 1+e h

17 More Applications Similar results for general subgraph counts in G(n, p). Optimal tail bounds for magnetization in Curie-Weiss model at critical temperature. Tail bounds for magnetization in Ising model on Z d.

18 Open Questions What happens in the low temperature regime for the exponential random graph model? Partial results are available. Extend the large deviation result to the rest of the (p, r) region. A complete solution for the exponential random graph is not enough.

19 Thank you!