Law of the Iterated Logarithm in G(n, p)

Transcription

1 Law of the Iterated Logarithm in G(n, p) Austen James Matt Larson Andrew Salmon Yale University 4 August 2016 of the Iterated Logarithm in G(n, p) 4August2016 1/18

2 G(n, p) G(n, p) is a random graph model. of the Iterated Logarithm in G(n, p) 4August2016 2/18

3 G(n, p) G(n, p) is a random graph model. Connect any two vertices with probability p where 0 < p < 1. of the Iterated Logarithm in G(n, p) 4August2016 2/18

4 G(n, p) G(n, p) is a random graph model. Connect any two vertices with probability p where 0 < p < 1. G(n, p) exhibits a number of well-known properties relating to connectivity, coloring, the appearance of certain subgraphs, etc. of the Iterated Logarithm in G(n, p) 4August2016 2/18

5 The Law of Iterated Logarithm The Law of Iterated Logarithm (LIL) is a property of sequences of random variables, similar to the Central Limit Theorem (CLT). of the Iterated Logarithm in G(n, p) 4August2016 3/18

6 The Law of Iterated Logarithm The Law of Iterated Logarithm (LIL) is a property of sequences of random variables, similar to the Central Limit Theorem (CLT). While the CLT involves the convergence to a normal distribution, the LIL examines the greatest deviations from 0. of the Iterated Logarithm in G(n, p) 4August2016 3/18

7 The Law of Iterated Logarithm The Law of Iterated Logarithm (LIL) is a property of sequences of random variables, similar to the Central Limit Theorem (CLT). While the CLT involves the convergence to a normal distribution, the LIL examines the greatest deviations from 0. Theorem (Khinchin (1924) and Kolmogorov (1929)) nx Let S n = T i with T i iid random variables with 0 mean and 1 variance. i=1 apple S P lim sup p n =1 =1 n!1 2n log log n of the Iterated Logarithm in G(n, p) 4August2016 3/18

8 The Law of Iterated Logarithm The Law of Iterated Logarithm (LIL) is a property of sequences of random variables, similar to the Central Limit Theorem (CLT). While the CLT involves the convergence to a normal distribution, the LIL examines the greatest deviations from 0. Theorem (Khinchin (1924) and Kolmogorov (1929)) nx Let S n = T i with T i iid random variables with 0 mean and 1 variance. i=1 apple S P lim sup p n =1 =1 n!1 2n log log n of the Iterated Logarithm in G(n, p) 4August2016 3/18

9 History of the LIL Not true in general without independence. of the Iterated Logarithm in G(n, p) 4August2016 4/18

10 History of the LIL Not true in general without independence s: Random Walks with some conditions of the Iterated Logarithm in G(n, p) 4August2016 4/18

11 History of the LIL Not true in general without independence s: Random Walks with some conditions 1970 s: Martingales of the Iterated Logarithm in G(n, p) 4August2016 4/18

12 History of the LIL Not true in general without independence s: Random Walks with some conditions 1970 s: Martingales 1985: Random Walks with weaker conditions of the Iterated Logarithm in G(n, p) 4August2016 4/18

13 History of the LIL Not true in general without independence s: Random Walks with some conditions 1970 s: Martingales 1985: Random Walks with weaker conditions 2016: Random Variables coming from Random Graphs of the Iterated Logarithm in G(n, p) 4August2016 4/18

14 Central Limit Theorems in Random Graph Counts 1988: Ruciński proves graph counts of small subgraphs are asymptotically normal. of the Iterated Logarithm in G(n, p) 4August2016 5/18

15 Central Limit Theorems in Random Graph Counts 1988: Ruciński proves graph counts of small subgraphs are asymptotically normal. 1994: Janson proves that some spanning graphs, like spanning trees, Hamiltonian cycles, and perfect matchings, are asymptotically log-normal. Janson (1994) Let T n be the number of trees contained in a random graph in G(n, p). log T n n µ n! N(0, 1) of the Iterated Logarithm in G(n, p) 4August2016 5/18

16 Central Limit Theorems in Random Graph Counts 1988: Ruciński proves graph counts of small subgraphs are asymptotically normal. 1994: Janson proves that some spanning graphs, like spanning trees, Hamiltonian cycles, and perfect matchings, are asymptotically log-normal. Janson (1994) Let T n be the number of trees contained in a random graph in G(n, p). log T n n µ n! N(0, 1) µ n (the mean ) is log E[T n ] 1 p p (recall p is the edge probability). of the Iterated Logarithm in G(n, p) 4August2016 5/18

17 Central Limit Theorems in Random Graph Counts 1988: Ruciński proves graph counts of small subgraphs are asymptotically normal. 1994: Janson proves that some spanning graphs, like spanning trees, Hamiltonian cycles, and perfect matchings, are asymptotically log-normal. Janson (1994) Let T n be the number of trees contained in a random graph in G(n, p). log T n n µ n! N(0, 1) µ n (the mean ) is log E[T n ] n (the standard deviation ) is 1 p p q 2(1 p) p (recall p is the edge probability). of the Iterated Logarithm in G(n, p) 4August2016 5/18

18 Amoreprecisestatement Main result Let T n be the numberapple of trees contained in a random graph in G(n, p). log T n µ P lim sup p n =1 =1 n!1 2 log log n n of the Iterated Logarithm in G(n, p) 4August2016 6/18

19 Amoreprecisestatement Main result Let T n be the numberapple of trees contained in a random graph in G(n, p). log T n µ P lim sup p n =1 =1 n!1 2 log log n µ n and n are as in Janson (1994). n of the Iterated Logarithm in G(n, p) 4August2016 6/18

20 Proof Strategy Work with a di erent random graph model, G(n, m) where graphs are drawn uniformly on n vertices under the constraint that they have exactly m edges. of the Iterated Logarithm in G(n, p) 4August2016 7/18

21 Proof Strategy Work with a di erent random graph model, G(n, m) where graphs are drawn uniformly on n vertices under the constraint that they have exactly m edges. If X nm is the number of trees in a random graph from G(n, m), we need to show that, for some conditions: P[X nm KEX nm ] apple n 4 of the Iterated Logarithm in G(n, p) 4August2016 7/18

22 Proof Strategy Work with a di erent random graph model, G(n, m) where graphs are drawn uniformly on n vertices under the constraint that they have exactly m edges. If X nm is the number of trees in a random graph from G(n, m), we need to show that, for some conditions: P[X nm KEX nm ] apple n 4 To prove this concentration lemma, it is enough to show, by Markov s inequality, that EX k nm apple C k (EX nm ) k of the Iterated Logarithm in G(n, p) 4August2016 7/18

23 Proof Strategy Work with a di erent random graph model, G(n, m) where graphs are drawn uniformly on n vertices under the constraint that they have exactly m edges. If X nm is the number of trees in a random graph from G(n, m), we need to show that, for some conditions: P[X nm KEX nm ] apple n 4 To prove this concentration lemma, it is enough to show, by Markov s inequality, that EX k nm apple C k (EX nm ) k Given this, we can prove the upper and lower bound for the limsup term using the Borel-Cantelli lemma. of the Iterated Logarithm in G(n, p) 4August2016 7/18

24 About the Moment EX k nm In G(n, m), the appearance of a (labeled) subgraph only depends on the number of edges that subgraph contains. of the Iterated Logarithm in G(n, p) 4August2016 8/18

25 About the Moment EX k nm In G(n, m), the appearance of a (labeled) subgraph only depends on the number of edges that subgraph contains. So the probability that the union of trees S k i=1 T i is a subgraph of some graph from G(n, m) depends only on the size of the edge overlap. of the Iterated Logarithm in G(n, p) 4August2016 8/18

26 About the Moment EX k nm In G(n, m), the appearance of a (labeled) subgraph only depends on the number of edges that subgraph contains. So the probability that the union of trees S k i=1 T i is a subgraph of some graph from G(n, m) depends only on the size of the edge overlap. If X nm is the number of trees in a random graph with m edges, then the kth moment E X k nm is equal to: h i E Xnm k = X " # [ P T i (T 1,...,T k ) i of the Iterated Logarithm in G(n, p) 4August2016 8/18

27 The Quantity M(a) But it makes more sense to try to group tuples of trees by the number of edges they contain. P k(n 1) e=n 1 P [e (S i T i)] # {(T 1,...,T k ):e( S i T i)=e} of the Iterated Logarithm in G(n, p) 4August2016 9/18

28 The Quantity M(a) But it makes more sense to try to group tuples of trees by the number of edges they contain. P k(n 1) e=n 1 P [e (S i T i)] # {(T 1,...,T k ):e( S i T i)=e} Rewriting in terms of missing edges yields: n 2 k(n 1) + a (k 1)(n 1) X a=0 M(a) m k(n 1) + a n 2 m of the Iterated Logarithm in G(n, p) 4August2016 9/18

29 The Quantity M(a) But it makes more sense to try to group tuples of trees by the number of edges they contain. P k(n 1) e=n 1 P [e (S i T i)] # {(T 1,...,T k ):e( S i T i)=e} Rewriting in terms of missing edges yields: n 2 k(n 1) + a Definition of M(a) (k 1)(n 1) X a=0 M(a) m k(n 1) + a n 2 m M(a) is the number of k-tuples (T 1,...,T k ) of trees on a graph on n labeled vertices such that: Sk e i=1 T i = k(n 1) a of the Iterated Logarithm in G(n, p) 4August2016 9/18

30 M(a) andthecombinatoricsoftrees From Cayley s formula, we know that there are n n 2 total trees on n labeled vertices. Figure: Cayley s formula for n =2, 3, 4, image taken from Wikipedia of the Iterated Logarithm in G(n, p) 4August /18

31 M(a) andthecombinatoricsoftrees From Cayley s formula, we know that there are n n 2 total trees on n labeled vertices. For two random trees, what is the distribution of the size of the edge intersection? Figure: Cayley s formula for n =2, 3, 4, image taken from Wikipedia of the Iterated Logarithm in G(n, p) 4August /18

32 M(a) andthecombinatoricsoftrees From Cayley s formula, we know that there are n n 2 total trees on n labeled vertices. For two random trees, what is the distribution of the size of the edge intersection? For k (uniformly chosen) random trees, how many edges will overlap in total? Figure: Cayley s formula for n =2, 3, 4, image taken from Wikipedia of the Iterated Logarithm in G(n, p) 4August /18

33 Electrical Networks and Spanning Trees We can convert a graph to an electrical network by setting the resistance to 1. of the Iterated Logarithm in G(n, p) 4August /18

34 Electrical Networks and Spanning Trees We can convert a graph to an electrical network by setting the resistance to 1. Spanning tree edge probability (Kircho ) Let T be a spanning tree chosen uniformly from a graph H. Lete be any edge in H. P[e 2 E(T )] = i(e) That is, the probability that this edge is in T is equal to the current traveling along this edge when we inject one amp at one of the endpoints of e and remove one amp from the other endpoint: of the Iterated Logarithm in G(n, p) 4August /18

35 Electrical Networks are Graphs Many laws relating to electrical networks can be formalized more generally using graph theory. of the Iterated Logarithm in G(n, p) 4August /18

36 Electrical Networks are Graphs Many laws relating to electrical networks can be formalized more generally using graph theory. Ohm s law Let ab be an edge in graph G. LetV ab be the voltage di erence across ab and let R ab be the resistance of ab. Then: i(ab) = V ab R ab of the Iterated Logarithm in G(n, p) 4August /18

37 Electrical Networks are Graphs Many laws relating to electrical networks can be formalized more generally using graph theory. Ohm s law Let ab be an edge in graph G. LetV ab be the voltage di erence across ab and let R ab be the resistance of ab. Then: Rayleigh Monotonicity Law i(ab) = V ab R ab Let G, G 0 be graphs such that G G 0. Then for every edge e 2 E(G): i G (e) i G 0(e) of the Iterated Logarithm in G(n, p) 4August /18

38 An iterative construction We need to construct k-tuples random trees with total edge intersection a. of the Iterated Logarithm in G(n, p) 4August /18

39 An iterative construction We need to construct k-tuples random trees with total edge intersection a. Choose some partition (`2,`3,...,`k) suchthat P i `i = a. of the Iterated Logarithm in G(n, p) 4August /18

40 An iterative construction We need to construct k-tuples random trees with total edge intersection a. Choose some partition (`2,`3,...,`k) suchthat P i `i = a. Given a sequence T 1,...,T t 1 of trees, an additional tree T is `t-overlapping if T shares exactly `t edges with S t 1 i=1 T i. of the Iterated Logarithm in G(n, p) 4August /18

41 An iterative construction We need to construct k-tuples random trees with total edge intersection a. Choose some partition (`2,`3,...,`k) suchthat P i `i = a. Given a sequence T 1,...,T t 1 of trees, an additional tree T is `t-overlapping if T shares exactly `t edges with S t 1 i=1 T i. Counting the number of such trees gives us how often our construction can continue. of the Iterated Logarithm in G(n, p) 4August /18

42 Counting `t-overlapping Trees Reduction (Moon, 1967): the maximum degree of random trees is almost always smaller than log n. of the Iterated Logarithm in G(n, p) 4August /18

43 Counting `t-overlapping Trees Reduction (Moon, 1967): the maximum degree of random trees is almost always smaller than log n. Overlapping Tree Lemma Suppose we have chosen trees T 1,...,T t 1 with maximum degree less than log n. LetS be a fixed set of `t edges and call a tree T t `t-overlapping if it only uses precisely the edges in S as well as unused edges not in T 1,...,T t 1. Then the number of `t-overlapping trees is at most 2`t e 2 2t+O(log5 (n)/n) n n 3 `t of the Iterated Logarithm in G(n, p) 4August /18

44 Counting `t-overlapping Trees Reduction (Moon, 1967): the maximum degree of random trees is almost always smaller than log n. Overlapping Tree Lemma Suppose we have chosen trees T 1,...,T t 1 with maximum degree less than log n. LetS be a fixed set of `t edges and call a tree T t `t-overlapping if it only uses precisely the edges in S as well as unused edges not in T 1,...,T t 1. Then the number of `t-overlapping trees is at most 2`t e 2 2t+O(log5 (n)/n) n n 3 `t The results from electrical networks are well-behaved because of the high minimum degree in K n \ S i T i. of the Iterated Logarithm in G(n, p) 4August /18

45 AProperUpperBoundonM(a) foranyk We can get an upper bound on M(a) becauseanytreesthatsatisfy the condition for M(a) contain a procedure of selecting `t-overlapping trees. of the Iterated Logarithm in G(n, p) 4August /18

46 AProperUpperBoundonM(a) foranyk We can get an upper bound on M(a) becauseanytreesthatsatisfy the condition for M(a) contain a procedure of selecting `t-overlapping trees. By summing over all possible partitions P(a), we use the multinomial theorem. of the Iterated Logarithm in G(n, p) 4August /18

47 AProperUpperBoundonM(a) foranyk We can get an upper bound on M(a) becauseanytreesthatsatisfy the condition for M(a) contain a procedure of selecting `t-overlapping trees. By summing over all possible partitions P(a), we use the multinomial theorem. Upper Bound The total number of trees with a edge overlaps is at most: M(a) apple X Y `t-overlapping Trees `2P(a) n k(n 2) e 2(k 2)+O(log 5 (n)/n) 2 k 2 = a! a of the Iterated Logarithm in G(n, p) 4August /18

48 Some Remarks on the Upper Bound For k = o(n), M(a) N k apple e2( k 2) 2 k 2 a! a (1 + o(1)) of the Iterated Logarithm in G(n, p) 4August /18

49 Some Remarks on the Upper Bound For k = o(n), M(a) N k apple e2( k 2) 2 k 2 a! a (1 + o(1)) The upper bound is a Poisson distribution with parameter =2 k 2. of the Iterated Logarithm in G(n, p) 4August /18

50 Some Remarks on the Upper Bound For k = o(n), M(a) N k apple e2( k 2) 2 k 2 a! a (1 + o(1)) The upper bound is a Poisson distribution with parameter =2 k 2. We P can infer convergence to a Poisson distribution because 1 i=0 M(a) must equal the total number of k-tuples of trees. of the Iterated Logarithm in G(n, p) 4August /18

51 The Law of Iterated Logarithm This asymptotic estimate on M(a) is su concentration equality. cient to prove the of the Iterated Logarithm in G(n, p) 4August /18

52 The Law of Iterated Logarithm This asymptotic estimate on M(a) is su concentration equality. cient to prove the Using similar techniques, we also proved the Law of Iterated Logarithm for the appearance of perfect matchings in random graphs in G(n, p). of the Iterated Logarithm in G(n, p) 4August /18

53 Acknowledgements Thanks to MAA Mathfest for hosting. Thanks to Daniel Montealegre for serving as our advisor. Thanks to SUMRY for giving us the opportunity to work on this problem. of the Iterated Logarithm in G(n, p) 4August /18