Distributing labels on infinite trees

called A n+1i . Now, each sub-tree A n+1i is composed of a root and d factors of size n, in the set{A n 1 , . . . A n p }. In turn, they are all prolonged into trees of size n in a unique way. Therefore,P(n + 2) = p. By a direct induction, P(k) = p for all k ≥ n.4 implies 1: If the number of factors of size n is smaller than B for all n, then this means thatthe number of equivalence classes for ≡ n is smaller than n for all n, this means that G(T ) hasless than B nodes.3.1 Density of rational treesLet T be a rational tree and let G(T ) be its minimal multi-graph. The nodes of G(T ) arenumbered v 1 · · · , v K , with v 1 corresponding to the root of T .G(T ) can be seen as the transition kernel of a Markov chain by considering each arc of G(T )as a transition with probability 1/d.If G(T ) is irreducible then the Markov chain admits a unique stationary measure π on itsnodes. The density of T and the stationary measure π are related by the following theorem.Theorem 3.2. Let T be an irreducible rational tree with a minimal multigraph G(T ) with Knodes. Let l = (l 1 , . . . l K ) be the <strong>labels</strong> of the nodes of G(T ) and let π = (π 1 , . . . , π K ) be thestationary measure over the nodes of G(T ).If T is aperiodic, then T admits a density α = πl t .If T is periodic with period p then T admits an average density α = πl t .Proof. Let V n be a Markov chain corresponding to G(T ). Since G(T ) is irreducible, V n admits aunique stationary measure , say π = (π 1 , . . . , π K ). Let us call P the kernel of this Markov chain:P i,j = a/d if there are a arcs in G(T ) from v i to v j .Now, let us consider all the paths of length n in T , starting from an arbitrary node v i . Byconstruction of G(T ), the number of paths that end up in the nodes v 1 , · · · , v K respectively, ofG(T ), is given by the vector e i d n P n , where e i is the vector with all its coordinates equal to 0except the ith coordinate, equal to 1.∑ n−1Now, the number of ones in the tree of height n starting in v i is h n (v i ) = e i k=0 dk P k l t .Let us first consider the case where P is aperiodic. We denote by Π the matrix with all itslines equal to the stationary measure, π and by D k the matrix P k − Π. When P is aperiodic,then lim k→∞ ||D k || 1 = 0. Therefore, for all k > n, ||D k || 1 < ɛ n → 0.Then the density of ones d 2n (v i ) =size 2n into a factor of size n at the root and d. factors of size n. One getsd−1d 2n −1 h 2n(v i ) can be estimated by splitting the factors ofd 2n (v i ) ==d − 1d 2n − 1 e id − 1d 2n − 1 e i(n∑k=1d k P k l t + d − 1d 2n − 1 e in∑d k P k +k=12n−1∑k=n+1d k D k +2n−1∑k=n+12n∑k=n+1d k P k l t ,d k Π)l t .∑ nwhen n goes to infinity, the first term goes to 0 because e i k=1 dk P k l t ≤ d n+1 . As for thesecond termd−1d 2n −1 e ∑ 2n−1i k=n+1 dk D k l t ≤ 1d 2n −1 d2n ɛ n . This goes to 0 when n goes to infinity.d−1As for the last term,d 2n −1 e ∑ 2n−1i k=n+1 dk Πl t = 1d 2n −1 (d2n −d n+2 )(e i Π)l t this goes to πl t whenn goes to infinity.The same holds by computing the density of trees of size 2n + 1 by splitting them into thefirst n + 1 levels and the last n levels.This shows that the rooted density of all the trees in T is the same, equal to πl t .8