Finite rooted tree. Each vertex v defines a subtree rooted at v. Picking v at random (uniformly) gives the random fringe subtree.

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1 Finite rooted tree. Each vertex v defines a subtree rooted at v. Picking v at random (uniformly) gives the random fringe subtree. 1

2 T = { finite rooted trees }, countable set. T n : random tree, size n. F n : random fringe subtree of T n. Empirical Observation. For most natural families of random trees, F n F (say), as n. Remarks. 1. Convergence in distribution on countable set T. 2. Maybe asymptotic cycling instead. 3. The point is P (F = ) = asymptotic prop. of leaves in T n. degree of root of F = asymptotic distribution of out-degrees in T n. Informally, F describes limits of all local properties of T n. 2

3 The Pessimist s View of Life 1. You re born; you have a random number of children at random times; you die. 2. Your children behave in the same way, independently of you. The mathematical model is alternatively called the Crump-Mode-Jagers general continuoustime supercritical branching process. 3

4 Model. B.p. where each individual has C children (EC > 1) at times (ξ 1, ξ 2,..., ξ C ) (arbitrary distribution) after own birth. Standard facts. 1. Number born before t Ze θt for a certain constant θ. 2. Pick individual at random from those born before T ; look at descendants born before T. As T this random family tree has the following limit. Start the b.p. with 1 individual and watch for an exponential (θ) time. Point. Limits often exist for models of random trees with no b.p. structure. 4

5 Example where bare-hands calculations easy. Trie on 5 strings A natural model for a random trie T m on m strings is to take all bits independent uniform. For a given string v, it s easy to calculate (Binomial dist!) P (v is vertex of random trie ) P ( fringe subtree at v has r leaves ). random trie has a self-reducibility property: 5

6 given the subtree at v has exactly r leaves, it is distributed as the random trie on r strings. There is an asymptotic fringe dist. F (as m and log 2 m modulo (1) constant) which must be of the form P (F = ) = r a r P (T r = ) for constants (a r ). 6

7 By analogy with (S n a n )/b n Y, ask: What distributions π on T occur as limits of random fringe subtrees F n? Answer: π occurs iff πq = π, where Q = Q(t, t ) is the matrix counting first-generation subtrees. Q(t, ) = 2. Q(t, ) = 1. Q(t, ) = 1. Call such π s fringe distributions. Given such a π, define a Markov transition matrix by P π (t, t ) = π(t )Q(t, t)/π(t). Run this chain with initial dist. π... 7

8 Analogy: weak limits (e.g. Brownian motion) may have lots more structure that the approximating processes do.] Abstract b.p. with arbitrary types. Type space S. Individual, type s, has random number/type of children. R(s, ) = E(# offspring with type ). Given a p.m. ν on S, define random tree F to be family tree of descendants of individual, type ν. Suppose finite. Fact. F has a fringe dist. iff νr = ν. Empirical observation. (N.B. selection bias!) In all examples where asymptotic fringe dist. is known, it has description as abstract b.p. with simple type-space, even though underlying trees are not b.p. s. 8

9 Harder example: greedy undirected tree. n vertices, one distinguished (root). Start with no edges. Repeat add edge, chosen at random (uniformly) from set of all edges whose addition would not create a cycle Fact. Asymptotic fringe dist. is abstract b.p. with type space (0, ) and invariant dist. ν. Type s individual has: Poisson(λ(s)) offspring of type s Poisson (rate ρ(s )) process of offspring of types s < s So asymptotic proportion of leaves is 0 exp( λ(s) s 0 ρ(s )ds )ν(ds)

10 More precisely, let L n = (random) proportion of vertices of T n which are leaves: then EL n Guess: really L n in probability. Not easy to use Chebyshev s inequality here, but following abstract result applies. 10

11 Random trees T n, random fringe subtrees F n. Suppose asymptotic fringe F exists. This says Eφ(F n ) Eφ(F) (all bounded φ). Theorem. In order that φ(f n ) p φ(f) (all bounded φ), it suffices that the ancestor chain s type of ancestor of s R (s, ds ) = ν(ds )R(s, ds)/ν(ds) has trivial invariant σ-field. Remarks. 1. For CMJBP, ancestor chain is renewal process, and Choquet-Deny theorem applies. 2. Otherwise, seek to verify by coupling. 3. Proof: show dist(f) is extreme amongst all fringe distributions. 11

Random trees and branching processes

Random trees and branching processes Svante Janson IMS Medallion Lecture 12 th Vilnius Conference and 2018 IMS Annual Meeting Vilnius, 5 July, 2018 Part I. Galton Watson trees Let ξ be a random variable