A phase transition for large values of bifurcating autoregressive models
aa r X i v : . [ m a t h . P R ] D ec A PHASE TRANSITION FOR LARGE VALUES OFBIFURCATING AUTOREGRESSIVE MODELS
VINCENT BANSAYE AND S. VAL`ERE BITSEKI PENDA
Abstract.
We describe the asymptotic behavior of the number Z n [ a n , ∞ ) of individuals witha large value in a stable bifurcating autoregressive process. The study of the associated firstmoment E ( Z n [ a n , ∞ )) is equivalent to the annealed large deviation problem P ( Y n ≥ a n ), where Y is an autoregressive process in a random environment and a n → ∞ .The population with large values and the trajectorial behavior of Z n [ a n , ∞ ) is obtained fromthe ancestral paths associated to the large deviations of Y together with its environment. Thestudy of large deviations of autoregressive processes in random environment is of independentinterest and achieved first in this paper. The proofs of trajectorial estimates for bifurcatingautoregressive process involves then a law of large numbers for non-homogenous trees.Two regimes appear in the stable case, depending on the fact that one of the autoregressiveparameter is greater than one or not. It yields two different asymptotic behaviors for the largelocal densities and maximal value of the bifurcating autoregressive process. Keywords : Branching process, autoregressive process, random environment, large deviations.
Mathematics Subject Classification (2010) : 60J80, 60J20, 60K37, 60F10, 60J20, 60C05,92D25. 1.
Introduction
The bifurcating autoregressive (BAR) process X = ( X n ) n ≥ is a model for affine random trans-mission of a real value along a binary tree. To define this process, we consider a real value randomvariable X , independent of a sequence of i.i.d bivariate random variables (cid:0) ( η k , η k +1 ) , k ≥ (cid:1) with law N (0 , Γ), where Γ = (cid:18) ρρ (cid:19) , ρ ∈ ( − , . Then X is defined inductively for k ≥ X k = αX k + η k X k +1 = βX k + η k +1 , where α, β are non-negative real numbers. Informally, the value X k of individual k is randomlytransmitted to its two offsprings 2 k and 2 k + 1 following an autoregressive process.This model has been introduced in the symmetric case α = β by Cowan [20] and Cowan andStaudte [21] to analyze cell lineages data. It allows to study the evolution and the transmission ofa trait after division, in particular it size or it growth rate. In these works, Cowan and Staudtehave focused on the study of the bacteria Escherichia Coli. E. Coli is a rod-shaped bacteriumwhich reproduces by dividing in the middle, producing two cells. Several extensions of their modelhave been proposed and we refer e.g. to the works of Basawa and Higgins [6, 7] and Basawa andZhou [8, 9], where the model of Cowan and Staudte is studied for long memory and non-Gaussian noise.In 2005, Steward et al. [43] have designed an experimental protocol which brings evidence ofaging and asymetry in E. Coli. In order to study the dynamic of evolution of values in a populationof cells, taking into account such possible asymmetry, Guyon and al. [32] have considered and usedthe model of Cowan and Staudte with ( α, β ) ∈ (0 , , which means that α and β may be different.Since then, this model has been extended to more complex settings and studied from a statisticaland probabilistic point of view. Bercu et al. [11] consider an extension of BAR model with non-Gaussian noise and long memory. They use martingale approach in order to study the asymptoticbehavior of least- squares estimators of unknown parameters of their model. In the following ofthis work, Bitseki and Djellout [13] and Bitseki et al. [14] have studied the deviation inequalitiesand the moderate deviations principle of unknown parameters in the BAR models introduced byBercu et al. [11] and Guyon [31]. Several extension of this model were proposed where the missingdata are taking into account in the study of cell lineage data, see de Saporta et al. [24, 25, 27]and Delmas and Marsalle [22]. Furthermore Bercu and Blandin [10] and de Saporta et al. [26],have considered random coefficients, while Bitseki Penda and Olivier [15] have worked on nonlinearautoregressive models.More generally, polarisation of cells is a fundamental process in cell biology. Asymmetry at celldivision have been observed and studied in different contexts, see e.g. [37] for plasmids, [42] forextra chromosomal DNA, [35] for mitochondrias and [5] for parasites. Asymmetry at division is akey feature of aging, cell variability and differentiation [19].The BAR models turns out to be an interesting toy model to explore mathematically the effectof randomness and asymmetry in the long time behavior of characteristics of cells in divisionprocesses. In this paper, we focus on large values among the cell population. We can, describehere how lineages and stochasticity from the Gaussian additive term interplay to explain this partof the distribution of traits when time goes to infinity.1.1. Stability and random cell lineage.
To analyze the long time behavior of the bifurcatingautoregressive model, the genealogy of the population is involved. It is given here by the binarytree. Each vertex is seen as a positive integer different from 0 and it denotes one individual of thepopulation. The initial individual is thus denoted by 1 and for n ∈ N , G n = { n , n + 1 , · · · , n +1 − } and T n = n [ m =0 G m denote respectively the n -th generation and the first ( n + 1) generations of the population. Thecollection of values ( X i : i ∈ G n ) of the individuals in generation n is represented by the randompunctual measure Z n = X i ∈ G n δ X i . The process Z is a multitype branching process where types are real valued. The first moment ofmeasure Z n satisfies the following simple many-to-one formula:(2) E ( Z n ([ a, b ])) = 2 n P ( Y n ∈ [ a, b ]) , for any real numbers a ≤ b , where ( Y n ) n is the autoregressive process in random environmentcorresponding to a uniform random path in the binary tree. More precisely, Y is defined by(3) Y = X and ∀ n ≥ , Y n = θ n Y n − + ε n , ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 3 where Θ = ( θ n , n ≥
1) is a sequence of i.i.d. random variables such that P ( θ = α ) = P ( θ = β ) = 1 / ε n , n ≥
1) is a sequence of i.i.d. centered standard Gaussian random variables independentof ( θ n , n ≥ n ≥ Y n = n Y k =1 θ k ! Y + Y n , where Y n = n X k =1 n Y ℓ = k +1 θ l ! ε k , using the convention Π ℓ ∈ ∅ = 1. Conditionally on ( θ n , n ≥ Y n is a Gaussian random variablewith standard deviation(5) A n = A ( θ , ..., θ n ) = vuut n X k =1 n Y ℓ = k +1 θ ℓ . Moreover, for each n , A n is distributed as A ∗ n = A ( θ n , ..., θ ) and by monotonicity the latterconverges a.s. as n → ∞ to A ∗∞ = vuut ∞ X k =1 k − Y ℓ =1 θ ℓ ∈ (0 , ∞ ] . If αβ ≥ A ∗∞ = ∞ a.s. and the mean proportion of individuals whose value is in a compactset goes to 0 as n goes to infinity, i.e. Z n ([ a, b ]) / n → αβ < . Then ( Y n ) n converges in law to its unique stationary distribution π on R , which is a mixed centeredgaussian random variable with random standard deviation distributed as A ∗∞ . Guyon [31] hasshown in that case that a strong law of large numbers holds :12 n Z n ([ a, b ]) n →∞ −→ π ([ a, b ]) a.s.1.2. Large deviations and local densities in the stable case : main results.
We areinterested here in the number of individuals with larges values and the extremal values among thepopulation. More precisely we study Z n ([ a n , ∞ )) = { i ∈ G n : X i ≥ a n } when a n → ∞ . In particular we expect a law of large number (or concentration) effect, informally Z n ([ a n , ∞ )) ≈ E ( Z n ([ a n , ∞ ))) = 2 n P ( Y n ∈ [ a n , ∞ ))at the logarithm scale.Proving this law of large number effect and determining the asymptotic behavior of E ( Z n ([ a n , ∞ ))leads us to study the large deviation event { Y n ≥ a n } . More precisely, we describe the past trajec-tory ( Y i : i ≤ n ) conditionally on this event { Y n ≥ a n } together with the associated environment( θ i : i ≤ n ). It allows us both to estimate the mean behavior P ( Y n ≥ a n ) and to control the subtreeassociated to the local density Z n ([ a n , ∞ )) and thus correlations of the values between individuals.This large deviation issue on the autoregressive process Y is both crucial here for the analysis of Z n and of independent interest. In the case α = β < Y n is a gaussian random variable and P ( Y n ≥ a n ) is derived analytically. In the general stable case αβ <
1, two regimes appear, whichcorrespond to two different behaviors of the past trajectory and associated environments θ . Weobtain the following classification VINCENT BANSAYE AND S. VAL`ERE BITSEKI PENDA i ) Case α = max { α, β } <
1. Then, the deviation event { Y n ≥ a n } is achieved by selecting themost favorable environment α in the last generations. The deviation event { Y n ≥ a n } relieson extreme events in these last generations, which are quantified by the tail of Gaussianrandom variables with standard deviation α . We obtain that for any x > n → + ∞ xn log P (cid:0) Y n ≥ √ xn (cid:1) = − − α . Moreover we prove that conditionally on { Y n ≥ a n } , the past trajectory ( Y n − i : 0 ≤ i ≤ n )is approximated by a geometric progression ( a n α i : 0 ≤ i ≤ n ). ii ) Case β < < α and αβ <
1. Then there exists a unique κ > α κ + β κ = 2and now the deviation event { Y n ≥ a n } comes from the deviation of the environment { A n ≥ a n } . To estimate the probability of that latter, we use Kesten theorem, whichensures that P ( A ∞ ≥ a n ) ∼ ca κn as n → ∞ . We thus restrict the study here to the casewhen { A ∗ n ≥ a n } is comparable to { A ∗∞ ≥ a n } , which will correspond to the fact that thesupremum over all time of the associated random walk drifting to −∞ is reached beforetime n when it is larger than a n . That leads us to introduce γ ∈ (0 ,
1) defined by γ := inf s ≥ ( α s + β s ) / . Then for a sequence a n tending to infinity not too fast, we prove that P ( Y n ≥ a n ) isequivalent to a − κn . More precisely we need that a κn is negligible compared to n / (1 /γ ) n as n tends to infinity. In particular, for any ρ ≥ ρ κ ∈ (0 , /γ ), we getlim n → + ∞ n log( ρ ) log P ( Y n ≥ ρ n ) = − κ. We expect that the regimes max( α, β ) = 1 and a n = exp( nρ ) with ρ > /γ could be studied withan approach similar to ii ). As far as we seen, { A ∗ n ≥ a n } will not be comparable to { A ∗∞ ≥ a n } if ρ > /γ and describing the behavior of { Y n ≥ a n } should require more work. We also mentionthat non gaussian noise could lead to different behavior, in particular if the tail of ε is at the samescale as the deviation probability coming from the environment.We have thus obtained both the asymptotic behavior of E ( Z n ([ a n , ∞ ))) = 2 n P ( Y n ∈ [ a n , ∞ ))and the ancestral path of ( Y, Θ) on the event Y n ≥ a n . Roughly speaking, a trajectorial versionof the many-to-one formula (2) allows us to derive the genealogy and process contributing to thelocal density Z n ([ a n , ∞ )) from the ancestral path of ( Y, Θ) before time n . We obtain the followinglong time estimates i ) Case α = max { α, β } <
1. Then, for any x ∈ (0 , / (1 − α )),lim n → + ∞ n log Z n ([ x √ n ; + ∞ )) = log(2) − x (1 − α ) / . In that case, the proof follows ideas of Biggins for branching random walks and we con-sider a subpopulation of individuals which realizes the deviation and thus the large values[ x √ n, ∞ ). It is described by a non-homogeneous Galton-Watson process, inherited fromthe trajectory of values in last generation following the geometric paths obtained above forthe deviation { Y n ≥ a n } . The main difference with branching random walk is this timenon-homogeneity since the trajectory associated with the deviation of the value (or value)is not straightline but geometric and limited to the last generations. ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 5 ii ) Case β < < α and αβ <
1. For all ρ ≥ ρ κ < min(2 , /γ ), we havelim n → + ∞ n log( Z n ([ ρ n , ∞ )) = log(2) − κ log( ρ ) in probability . The subtree associated with the individuals in [ ρ n , ∞ ) is very different from the previouscase. Indeed, only some particular paths of the binary genealogical tree are involved inlocal densities, corresponding to the deviation event { A ∗ n ≥ ρ n } . The evolution of the chainon these paths does not deviate from its usual Gaussian autoregressive expected behavior.The proof is here more involved since a study of the process restricted to the subtree of T n corresponding to the environment Θ n = ( θ i : i ≤ n ) associated to the large deviationof Y n . This phenomenon is linked to the random environment structure inherited fromthe tree and can be compared to the results in the weakly subcritical case in Kimmel’sbranching model [5]. In that latter, an analogous phase transition occurs but only themean behavior had been obtained. The technique developed in this paper should help toobtain a convergence in probability. The issue of the a.s. convergence remains open. Wecould expect to prove it by estimating for instance the speed of convergence in probability.In words, two sources of randomness are combined here : the gaussian additive term of theautoregressive process and the choice of paths in the binary genealogy. In the first regime, thelocal densities is inherited from both randomness, in the very last generations. The picture is quiteclose to branching random walks, but the fact that deviation creating local densities occurs onlyat the end, not starting from the beginning. In the second regime, only the path in the tree isinvolved for large traits and the deviation starts much earlier to create these local densities. Thepicture is very different from branching random walks since a deterministic subtree, inherited fromfunctional A n support the local density. The study of the autoregressive model Y together withits environment θ yields a natural way to analyse this issue.1.3. Extremal value.
The maximal value in generation nM n = max { X i : i ∈ G n } has different behavior and speed. If α = max { α, β } <
1, thenlim n → + ∞ √ n M n = r − α a.s . If β < < α , then lim n → + ∞ n log( M n ) = log(2) κ in probability . The issue of maximal value has attracted lots of attention and has been extensively studied forbranching random walks from the works of Hammersley [33], Kingman [38], Biggins [12] andBramson [18]. Fine estimates have been obtained recently, which have shed light on the branchingstructure of the process, see in particular Hu and Shi [34], Faraud et al. [29], Addario-Berry andReed [1] and A¨ıd´ekon [2].The behavior of the maximal value is different in this model. Technics developed for branchingrandom walk are used and adapted here, in particular the use of many to one formula to exploitthe description of the path leading to extremal particules [41] and the construction of an imbeddedbranching process. The main novelty lies in the weak regime β < < α where a law of largenumbers of a well chosen subtree of the binary tree is involved. VINCENT BANSAYE AND S. VAL`ERE BITSEKI PENDA
Notation.
We will use the following notations. Let ( x n ) and ( y n ) be two sequences of realnumbers. • We recall that x n = O (1) means that x n → n → + ∞ . • We write x n = O ( y n ) if x n /y n = O (1) . • We write x n ∼ y n if x n /y n → n → + ∞ . • We write x n . y n if there is a positive constant c , independent of n , such that x n ≤ cy n . • We write x n ≍ y n if there are two positives constants c and c such that c y n ≤ x n ≤ c y n as n → + ∞ .2. Large deviations and ancestral paths in the stable case
This section has two goals. The first is to evaluate the probabilities of large deviations ofrandom coefficients autoregressive process ( Y n , n ∈ N ). The second is to understand the asymptoticbehavior of the paths of this process conditionally to the large deviations events. For simplicity andowing to the motivation of this paper (see the next section), we focus on the case where the randomcoefficients take only two values with identical probability. Then, let ( θ n , n ≥
1) be a sequence ofi.i.d. random variables which take their values in { α, β } with 0 < β ≤ α , and P ( θ = α ) = 1 / ε n , n ≥
1) be a sequence of i.i.d. centered standard Gaussian random variables independent of( θ n , n ≥ Y = 0 and consider the process ( Y n , n ∈ N )defined by Y = 0 and ∀ n ≥ , Y n = θ n Y n − + ε n , in such a way that from (4), we have Y n = n X k =1 n Y ℓ = k +1 θ ℓ ! ε k . Large deviations for strictly stable case.
We consider here the case β ≤ α < Y n , n ∈ N ) is given by a geometricgrowth in envrionment α in the last generations. Theorem 1.
Let ( a n ) n ∈ N be a sequence of non-decreasing real numbers which tends to infinity.We have (6) lim n → + ∞ a n log P ( Y n ≥ a n ) = − − α . Moreover, for all sequence ( ℓ n ) n of integers such that n − ℓ n → ∞ and ℓ n = O (log a n ) , (7) lim n → + ∞ P sup ℓ ∈{ , ··· ,ℓ n } (cid:12)(cid:12)(cid:12)(cid:12) Y n − ℓ a n − α ℓ (cid:12)(cid:12)(cid:12)(cid:12) ≤ ε, ( θ n − ℓ n , . . . , θ n ) = ( α, . . . , α ) (cid:12)(cid:12)(cid:12)(cid:12) Y n ≥ a n ! = 1 for any ε > .Remark . More generally, one can see the proof to check that admissible sequences ( ℓ n ) n justneed to satisfy n − ℓ n → ∞ and lim n →∞ ℓ n exp (cid:0) − c ( a n α ℓ n ) (cid:1) = 0 for some finite constant c . Proof.
Let F n = σ ( θ , · · · , θ n ) be the filtration defined by θ , · · · , θ n . We recall that L ( Y n |F n ) = N (0 , A n ), the centered normal law with variance A n , where A n is defined in (5). We observe that A n ≤ ∞ X k =0 α k = 11 − α since β ≤ α . First, we have P ( Y n ≥ a n ) = E (cid:2) P (cid:0) N (0 , A n ) ≥ a n (cid:12)(cid:12) F n (cid:1)(cid:3) ≤ P (cid:18) N (cid:18) , − α (cid:19) ≥ a n (cid:19) . a n exp (cid:18) − (1 − α ) a n (cid:19) , (8)where the last inequality follows from the classical upper tail inequality for standard normal dis-tribution, see for instance [16].Next, we introduce a sequence ( ℓ n ) n ∈ N which tends to infinity such that ℓ n ≤ n and ℓ n = O (cid:0) a n (cid:1) .We set B ℓ n = { θ n = α, · · · , θ n − ℓ n = α } , u ℓ n = ℓ n X k =0 α k = 1 − α ℓ n +2 − α and using again L ( Y n |F n ) = N (0 , A n ), we have P ( Y n ≥ a n , B n ) = 12 ℓ n +1 × E (cid:2) P (cid:0) N (cid:0) , α ℓ n +2 A ( θ , · · · , θ n − ℓ n − ) + u ℓ (cid:1) ≥ a n (cid:12)(cid:12) F n − ℓ n − (cid:1)(cid:3) ≥ ℓ n +1 × P (cid:0) N (cid:0) , u n (cid:1) ≥ a n (cid:1) & ℓ n × u ℓ n a n exp (cid:18) − a n u ℓ (cid:19) , (9)where the last inequality follows from the classical lower bounds for the tail of standard normaldistribution (see for e.g. [16]).Now, from (8) and (9) we get12 ℓ n u ℓ n a n exp − a n u ℓ n ! . P ( Y n ≥ a n ) . a n exp (cid:18) − (1 − α ) a n (cid:19) . Finally, letting n go to infinity in the previous inequalities leads to − − α ≤ lim inf n → + ∞ a n log P ( Y n ≥ a n ) ≤ lim sup n → + ∞ a n log P ( Y n ≥ a n ) ≤ − − α ℓ n = O (cid:0) a n (cid:1) and u ℓ n → / (1 − α ), which ends the proof of (6) .For the second part, we focus on the case β < α , while the case α = β is simpler. We write τ n = sup { i = 1 , . . . , n : θ i = β } with convention sup ∅ = 0. We first observe that P ( Y n ≥ a n ; τ n = n − i ) = P ( θ n = α, · · · , θ n − i +1 = α, θ n − i = β, Y n ≥ a n )and L ( Y n | θ n = α, · · · , θ n − i +1 = α, θ n − i = β ) = L α i βY n − i − + i X k =0 α k ε n − k ! . VINCENT BANSAYE AND S. VAL`ERE BITSEKI PENDA
We get P ( Y n ≥ a n ; τ n = n − i )= (cid:18) (cid:19) i +1 × P α i βY n − i − + i X k =0 α k ε n − k ≥ a n ! = (cid:18) (cid:19) i +1 × E (cid:18) P (cid:18) N (cid:18) , − ( α ) i +1 − α + (cid:0) α i β (cid:1) A n − i − (cid:19) ≥ a n (cid:12)(cid:12) F n − i (cid:19)(cid:19) ≤ (cid:18) (cid:19) i +1 × P (cid:0) N (cid:0) , w i (cid:1) ≥ a n (cid:1) , where using again A n − i − ≤ / (1 − α ), w i is a non-negative real number defined by w i = 1 − ( α ) i +1 − α + (cid:0) α i β (cid:1) − α = 1 − α i ( α − β )1 − α . Then P ( Y n ≥ a n ; τ n = n − i ) . (cid:18) (cid:19) i +1 a n exp (cid:18) − a n w i (cid:19) and summing over i , we get by monotonicity of w i , P ( Y n ≥ a n , B cn ) . ℓ n X i =0 (cid:18) (cid:19) i +1 a n exp (cid:18) − a n w i (cid:19) . a n e − a n /w ℓn . Let s ∈ N such that α s < α − β . Using now P ( Y n ≥ a n ) ≥ P ( Y n ≥ a n , B ℓ n + s ) and (9), we getlim sup n →∞ P ( Y n ≥ a n , B cn ) P ( Y n ≥ a n ) . lim sup n →∞ ℓ n + s e − a n (1 /w ℓn − /u ℓn + s ) = 0 , since 1 /w ℓ n − /u ℓ n + s ∼ cα ℓ n with c > w ℓ n /u ℓ n + s → ℓ n = O (log a n ). It ensures that(10) lim n → + ∞ P ( B ℓ n | Y n ≥ a n ) = 1 . This proves a part of (7).Besides, recalling that n − ℓ n → ∞ , we consider k n ≤ n such that k n − ℓ n → ∞ and k n = O (log a n ).Then, (10) ensures thatlim sup n →∞ P sup ℓ =0 , ··· ,ℓ n (cid:12)(cid:12)(cid:12)(cid:12) Y n − ℓ a n − α ℓ (cid:12)(cid:12)(cid:12)(cid:12) > ε (cid:12)(cid:12)(cid:12) Y n ≥ a n ! ≤ lim sup n →∞ P sup ℓ =0 , ··· ,ℓ n (cid:12)(cid:12)(cid:12)(cid:12) Y n − ℓ a n − α ℓ (cid:12)(cid:12)(cid:12)(cid:12) > ε (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n ! . Conditionally on B k n , we can write Y n − ℓ a n = α k n − ℓ +1 a n Y n − k n − + 1 a n k n − ℓ X k =0 α k ε n − k − ℓ where Y n − k n − and ( ε n − k − ℓ : k = 0 , . . . , k n − ℓ ) are still independent and ( ε n − k − ℓ : k = 0 , . . . , k n − ℓ )are distributed as standard gaussian random variables. Then,(11) Y n − ℓ a n − α ℓ = α k n − ℓ +1 a n Y n − ℓ n − + k n − ℓ X k =0 (cid:18) α k a n ε n − k − ℓ − α ℓ (1 − α ) α k (cid:19) + η n,ℓ , ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 9 where(12) η n,ℓ = α ℓ (1 − α ) k n − ℓ X k =0 α k − α ℓ , sup ℓ ∈{ ,...,ℓ n } | η n,ℓ | n →∞ −→ . Next, for all ℓ ∈ { , · · · , ℓ n } , (8) yields P (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) α k n − ℓ +1 a n Y n − k n − (cid:12)(cid:12)(cid:12)(cid:12) ≥ ε (cid:19) ≤ P (cid:16) | Y n − k n − | ≥ a n α k n − ℓ n ε (cid:17) . α k n − ℓ n a n exp (cid:18) − (1 − α ) ε α − k n − ℓ n +1) a n (cid:19) . Dividing now by P ( Y n ≥ a n ) and using again the lowerbound (9) with k n yieldssup ℓ ∈{ , ··· ,ℓ n } P (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) α k n − ℓ +1 a n Y n − k n − (cid:12)(cid:12)(cid:12)(cid:12) ≥ ε (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n (cid:19) . k n α k n − ℓ n exp (cid:18) − (1 − α ) a n (cid:18) ε α − k n − ℓ n +1) − − ( α ) k n +1 (cid:19)(cid:19) = O (1 /ℓ n )since k n − ℓ n → ∞ and k n = O (log a n ). Using P (sup i X i ≥ a ) ≤ P i P ( X i ≥ a ) and the fact that Y n − k n − is independent of B k n , it ensures that(13) lim n → + ∞ P sup ℓ ∈{ ,...,ℓ n } (cid:12)(cid:12)(cid:12)(cid:12) α k n − ℓ +1 a n Y n − k n − (cid:12)(cid:12)(cid:12)(cid:12) ≥ ε (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n ! = 0 . Finally, to control the remaining term in (11), we prove that(14) lim n → + ∞ P sup ℓ ∈{ ,...,ℓ n } (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k n − ℓ X k =0 (cid:18) α k a n ε n − k − ℓ − α ℓ (1 − α ) α k (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ ε (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n ! = 0 . It amounts to solve a minimisation problem for the cost of a trajectory ( ε n − k − ℓ : ℓ ∈ { , . . . , ℓ n } ),under the constraint Y n ≥ a n . This problem can be solved explicitly and we actually prove inAppendix that for any k ∈ { , . . . , k n } (15) P (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) α k a n ε n − k − (1 − α ) α k (cid:12)(cid:12)(cid:12)(cid:12) > ε (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n (cid:19) . exp (cid:18) − C a n (cid:19) , for some positive constant C , which yields directly (14).Combining the last estimates ensures thatlim n → + ∞ P sup ℓ ∈{ , ··· ,ℓ n } (cid:12)(cid:12)(cid:12)(cid:12) a n Y n − ℓ − α ℓ (cid:12)(cid:12)(cid:12)(cid:12) ≥ ε (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n ! = 0 , which ends the proof recalling (10). (cid:3) Large deviations in the weakly stable case.
In this regime, αβ < β ≤ < α. Then there exists a unique κ > α κ + β κ = 2 . We also set κ > γ ∈ (0 , α ) α κ + log( β ) β κ = 0 , γ := ( α κ + β κ ) / s ≥ { ( α s + β s ) / } . We introduce now the random walk ( S k ) k ≥ defined by S = 0 and S k = k − X ℓ =1 log θ ℓ ( k ≥ . Recalling from the introduction that A ∗ n = A ( θ n , . . . , θ ), we can write( A ∗ n ) = n X k =1 exp (2 S k ) . We restrict here the study to the case when { A ∗ n ≥ a n } is comparable to { A ∗∞ ≥ a n } . In the proofbelow, it corresponds to the fact that the supremum S = sup k ≥ S k is reached before time n when it is larger than a n . It forces a n to be not too large in the statementbelow. Theorem 2.
Let ( a n , n ∈ N ) be a sequence of real numbers such that a κn .n − / .γ n → as n → ∞ .Then, (16) P ( Y n ≥ a n ) ≍ P ( A n ≥ a n ) ≍ a − κn . When the supremum is reached latter, additional work is needed and we expect that otherequivalents can be proved using a change of probability.
Proof.
First we determine the asymptotic behavior of P ( A n ≥ a n ) using the result of Kesten [36]which guarantees that ( A ∗∞ ) = P ∞ k =1 exp (2 S k ) is comparable to exp(2 S ). More precisely, we firstobserve that E h ( θ ) κ/ i = 1 and E [( θ ) κ/ log + θ ] < ∞ . Besides the following equalities hold:2 S k = k − X ℓ =1 log θ ℓ and 2 S = sup k k − X ℓ =1 log θ ℓ . Note that for any a > P ( A ∗ n ≥ a ) ≤ P (cid:0) ( A ∗∞ ) ≥ a (cid:1) . Now, using the results of Section XI.6 in[30] and Section 1 in [36] (see also Section 2 in [28]), we get, for some positive constant C , P (cid:0) ( A ∗∞ ) ≥ a (cid:1) ∼ a →∞ C × P (cid:0) exp (cid:0) S (cid:1) ≥ a (cid:1) = C × P (cid:0) exp (cid:0) S (cid:1) ≥ a (cid:1) . Moreover, we know, see for e.g. Section XI.6 in [30], that P (cid:0) exp (cid:0) S (cid:1) ≥ a (cid:1) ∼ a →∞ c a − κ , where c is positive constant. We conclude that P ( A ∗ n ≥ a n ) . P (cid:0) exp (cid:0) S (cid:1) ≥ a n (cid:1) ∼ c a − κn when n → ∞ . Besides, for any x ∈ (0 , P ( A ∗ n ≥ a n x ) ≥ P n X k =1 exp (2 S k ) ≥ a n x , τ S ≤ n ! ≥ P (cid:0) exp (cid:0) S (cid:1) ≥ a n x, τ S ≤ n (cid:1) , where τ S := inf { k > S k = S } is the first hitting time of the maximum of the random walk( S k , k > P ( τ S > n ) . n − / γ n , ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 11 and using n − / γ n = O ( a − κn ), we get that P ( τ S > n ) is negligible compared to P (cid:0) exp (cid:0) S (cid:1) ≥ a n (cid:1) and P ( A ∗ n ≥ a n x ) & P (cid:0) exp (cid:0) S (cid:1) ≥ a n x (cid:1) uniformly for n ≥ x ∈ (0 , θ , . . . , θ n ) ensures that A n is distributed as A ∗ n , we get(17) P ( A n ≥ a n x ) = P ( A ∗ n ≥ a n x ) ≍ P (cid:0) exp (cid:0) S (cid:1) ≥ a n x (cid:1) uniformly for x ∈ (0 ,
1] and(18) ∀ x ∈ (0 , , P (cid:0) exp (cid:0) S (cid:1) ≥ a n x (cid:1) ∼ n →∞ c ( xa n ) − κ ; sup x ∈ (0 , ,n ≥ P (cid:0) exp (cid:0) S (cid:1) ≥ a n x (cid:1) ( xa n ) − κ < ∞ . It ensures in particular that P ( A n ≥ a n ) ≍ a − κn and ends the proof of the second part of thestatement.We prove now that P ( Y n ≥ a n ) ≍ a − κn . Setting W n = A n /a n and recalling that L ( Y n |F n − ) = N (0 , a n W n ), we first consider P ( Y n ≥ a n , W n < ∼ E (cid:20) { W n < } W n exp (cid:18) − W n (cid:19)(cid:21) = Z (cid:18) x (cid:19) exp (cid:18) − x (cid:19) P ( A n ≥ a n x ) dx. Besides, (18) and the fact that (cid:0) x (cid:1) exp( − x ) x − κ is integrable on (0 ,
1] ensure by boundedconvergence that a κn Z (cid:18) x (cid:19) exp (cid:18) − x (cid:19) P (exp( S ) ≥ a n x ) dx n →∞ −→ c ∈ (0 , ∞ ) . Using the uniform estimate in (17) we obtain from the two previous displays that P ( Y n ≥ a n , W n < ≍ a − κn . Next, using again L ( Y n |F n − ) = N (0 , a n W n ) and the fact that P ( N (0 , a n x ) ≥ a n ) ∈ [ P ( N (0 , ≥ ,
1] for x ≥
1, we get P ( Y n ≥ a n , W n ≥ ≍ P ( W n ≥
1) = P ( A n ≥ a n ) ≍ a − κn , which ends the proof of the theorem. (cid:3) Local densities for bifurcating autoregressive processes
We can now study the autoregressive process Z n = X i ∈ G n δ X i defined in Introduction. We recall that T is the regular binary tree describing the underlyingpopulation and G n = { n , n + 1 , · · · , n +1 − } is the generation n . The strictly stable case.Theorem 3. If β ≤ α < , then for any x ∈ [0 , p / (1 − α )) , lim n → + ∞ n log Z n ([ x √ n ; + ∞ )) = log 2 − x (1 − α ) / a.s . The proof relies on the large deviation results of the previous section and a law of large numberprinciple, which is in the same spirit as proofs for branching random walk. The results of the pre-vious section also tells us that the bulk of Z n ([ x √ n ; + ∞ )) corresponds to individuals coming fromthe first daughter in the last divisions and that their value has undergone a geometric deviationfrom their stable distribution in these last divisions.The proof of the upper bound is a classical Borel Cantelli argument using the first moment. Proof of the upper bound of Theorem 3.
Let ε > α ( x ) = log 2 − x (1 − α ) /
2. ByMarkov inequality we have P (cid:18) n log Z n (cid:0) [ x √ n ; + ∞ ) (cid:1) ≥ α ( x ) + ε (cid:19) ≤ exp ( − n ( α ( x ) + ε )) × E (cid:2) Z n (cid:0) [ x √ n ; + ∞ ) (cid:1)(cid:3) . Besides, the many-to-one formula (2) and (8) yield E (cid:2) Z n (cid:0) [ x √ n ; + ∞ ) (cid:1)(cid:3) = 2 n P (cid:0) Y n ≥ x √ n (cid:1) . n exp (cid:18) − (1 − α ) x n (cid:19) . We obtain P (cid:18) n log Z n (cid:0) [ √ n ; + ∞ ) (cid:1) ≥ α ( x ) + ε (cid:19) ≤ exp ( − nε )and Borel Cantelli Lemma allows us to conclude thatlim sup n → + ∞ n log Z n (cid:0) [ √ n ; + ∞ ) (cid:1) ≤ α ( x ) a.s.which ends the proof of the upper bound of Theorem 3. (cid:3) For the lower bound, we need now to prove a law of large number result on the subpopulationproducing the local density [ x √ n, ∞ [ in generation n . Using the results of the previous section, itis achieved by following individuals who undergo a deviation in the last generation in environment α . We first derive from the previous section the following result on the large deviations of Y andthen proceed with the proof of the lower bound. Lemma 1.
For any x ≥ p →∞ lim inf n →∞ n + p log P ( Y p ≥ x √ n + p ) = − x (1 − α ) / . Proof.
We first observe that by monotonicity of Y with respect to the initial condition and Markovproperty, P ( Y n + p ≥ x √ n + p ) ≥ P ( Y n ≥ P ( Y p ≥ x √ n + p ) = P ( Y p ≥ x √ n + p ) / . Using (6), weget the upper boundlim inf p →∞ lim inf n →∞ n + p log P ( Y p ≥ x √ n + p ) ≤ − x (1 − α ) / . ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 13
We prove now the converse inequality. We know from Theorem 1 that conditionally on { Y n ≥ a n } ,the process favors the best environment, that is the coefficient α , at least in the last time. We thenhave P ( Y p ≥ x √ n + p ) ≥ P (cid:0) Y p ≥ x √ n + p, θ p = α, . . . , θ = α (cid:1) = 12 p × P p − X ℓ =0 α ℓ ε p − ℓ ≥ x √ n + p ! = 12 p × P N (0 , ≥ r − α − α p x √ n + p ! & p × x √ n + p exp (cid:18) − (1 − α ) x ( n + p )2(1 − α p ) (cid:19) . Now, applying the log function in both sides of the last inequality, dividing by n + p and letting n and p go to infinity gives the expected lower bound and ends the proof. (cid:3) Proof of the the lower bound of Theorem 3.
We first observe that for any n, p, a ≥ Z n + p ([ a, ∞ )) ≥ X u ∈ G n : X u ≥ { v ∈ G n + p : u v, X v ≥ a } a.s . Using the monotonicity of the autoregressive process with respect to its initial value and thebranching property, Z n + p ([ a, ∞ )) ≥ X u ∈ G n : X u ≥ X ( u ) p,a a.s . where ( X ( u ) p,a ) u ∈ G n are i.i.d. r.v. with the same law µ p,a defined by µ p,a = P δ ( Z p ([ a ; + ∞ )) ∈ · ) . Besides, Proposition 28 in [31] ensures that12 n { u ∈ G n : X u ≥ } n →∞ −→ Z R + µ ( dx ) a.s. , where µ is the law of Y ∗∞ = P ∞ k =1 ( Q k − ℓ =1 θ ℓ ) ε k . Actually, note that Z R + µ ( dx ) = lim n →∞ P ( Y n ≥
0) = 12 . Setting by now a = x √ n + p , the many-to-one formula yields u n,p = µ p,a = E [ X ( u ) p,a ] = E [ Z p [ x √ n + p, ∞ )] = 2 p P ( Y p ≥ x √ n + p ) . Besides, the large deviation estimate proved above in Lemma 1 ensures that(20) lim inf p →∞ lim inf n →∞ n + p log u n,p = − x (1 − α ) / . Recalling that x ∈ [0 , p / (1 − α )), we consider by now p large enough such thatlim inf n →∞ n log ( { u ∈ G n : X u ≥ } ) + lim inf n →∞ n log u n,p = log(2) + lim inf n →∞ n log u n,p > . Writing ( X n,i : i = 1 . . . , N n ) = ( X ( u ) p,a : u ∈ G n , X u ≥ n →∞ { u ∈ G n : X u ≥ } u n,p X u ∈ G n : X u ≥ X ( u ) p,a = 1 a.s . Then, lim inf n →∞ n − u n,p Z n + p ([ x √ n + p, ∞ )) ≥ n →∞ n log Z n ([ x √ n ; + ∞ )) ≥ log(2) − x (1 − α ) / (cid:3) We derive now the asymptotic behavior of the highest value M n = max { X i : i ∈ G n } . Corollary 1. lim n → + ∞ √ n M n = r − α a.s . Proof.
Setting v = 2 log(2) / (1 − α ) + ε for some ε >
0, we use the classical estimate P ( M n / √ n ≥ v + ) ≤ E ( Z n [ √ nv + , ∞ )) = 2 n P ( Y n ≥ √ nv + )and recalling (6), we get P n P ( M n / √ n ≥ v + ) < ∞ . Then lim sup n →∞ M n / √ n ≤ v + a.s. by BorelCantelli lemma, which proves the upperbound letting ε → v − = 2 log(2) / (1 − α ) − ε , we observe that { Z ([ √ nv − , ∞ )) > } ⊂ { M n ≥√ nv − } . So lim inf n →∞ M n / √ n ≥ v − a.s.is a direct consequence of Theorem 3. It ends the proof. (cid:3) The weakly stable case.
We assume now that α >
1. We recall that αβ <
1, so β < κ ∈ (0 , ∞ ) is defined by α κ + β κ = 2 . Theorem 4.
For all c ∈ [0 , ∞ ) such that c κ < min(2 , /γ ) and c > , we have lim n → + ∞ n log { i ∈ G n : X i ≥ c n } = log(2 /c κ ) in probability . The proof of the upperbound is achieved by a classical Borel Cantelli argument using the firstmoment evaluated in the previous section, as for the strictly stable case. The proof of the lower-bound is different and more involved. We need to focus on a subtree producing the large values[ c n , ∞ ) at time n and characterized by A ( θ , . . . , θ n ) ≥ c n . We then control correlations throughthe common ancestor of nodes and perform L estimates. Proof of the upperbound of Theorem 4.
Let ε >
0. By Markov inequality we have P (cid:18) n log { i ∈ G n : X i ≥ c n } ≥ log(2 /c κ ) + ε (cid:19) ≤ ( c κ / n exp( − nε ) × E [ { i ∈ G n : X i ≥ c n } ] . Besides, using the many-to-one formula and (16), E [ { i ∈ G n : X i ≥ c n } ] = 2 n P ( Y n ≥ c n ) . n c − nκ . ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 15
From the foregoing we are led to P (cid:18) n log { i ∈ G n : X i ≥ c n } ≥ log(2 /c κ ) + ε (cid:19) . exp ( − nε ) . Finally, the Borel Cantelli Lemma allows us to conclude thatlim sup n → + ∞ n log { i ∈ G n : X i ≥ c n } ≤ log(2 /c κ ) a.s . which ends the proof of the upper bound. (cid:3) Let us turn to the the proof of the lower bound and first give the outline. Recall that ( θ n , n ≥ θ and A n = A ( θ , . . . , θ n ). Forany i ∈ G n , we write ( i , · · · , i n ) ∈ { , } n its binary decomposition. This decompositions yieldsthe unique path in the tree from the root 1 to the vertex i . Setting ¯0 = α and ¯1 = β , we define A [ i ] = vuut n X k =1 n Y ℓ = k +1 i l and we consider the subset of G n defined by T n = { i ∈ G n : A [ i ] ≥ c n } . Note that we have(21) P ( A n ≥ c n ) = T n n and { i ∈ G n : X i ≥ c n } ≥ { i ∈ T n : X i ≥ c n } a.s . We set F n = { i ∈ T n : X i ≥ c n } T n . The subtree T n provides the bulk of Z n [ c n , ∞ ). Roughly the proof will follow from the fact thatlim inf n → + ∞ F n > . Indeed, using Theorem 2, it will ensure that(22) { i ∈ G n : X i ≥ c n } n ≥ { i ∈ T n : X i ≥ c n } n = F n × P ( A n ≥ c n ) & ( c n ) − κ . We actually prove using L computations that F n is close to the non degenerated sequence f n = E [ F n ] = 1 T n E " X i ∈ T n { X i ≥ c n } = 1 T n E " X i ∈ G n { X i ≥ c n } { i ∈ T n } = 2 n T n × P ( Y n ≥ c n , A n ≥ c n ) = P ( Y n ≥ c n | A n ≥ c n ) ≥ P ( N (0 , ≥ > , (23)where ( Y n , n ≥
0) is the autoregressive process with random coefficients defined in (3) and we usethe many to one formula coupling the process and its environment.For that purpose, for each i ∈ G n , we set Z n ( i ) = { X i ≥ c n } − P ( X i ≥ c n ) . Thus F n − f n = T n P i ∈ T n Z n ( i ) and E (cid:2) ( F n − f n ) (cid:3) = 1 T n X i ∈ T n E (cid:2) Z n ( i ) (cid:3) + 1 T n X ( i,j ) ∈ T n i = j E [ Z ( i ) Z ( j )] . (24) For the first term of the right hand side of (24), we have1 T n X i ∈ T n E (cid:2) Z n ( i ) (cid:3) ≤ T n X i ∈ T n E (cid:2) { X i ≥ c n } (cid:3) = P ( Y n ≥ c n | A n ≥ c n ) T n −−→ n → + ∞ , since T n → ∞ as n → ∞ . Let us deal with the second term of (24). For all p ∈ { , · · · , n − } ,we denote by T ( n ) p the set of all the individuals of the generation p who are ancestors of at leastone individual in the sub-population T n . For each i ∈ G p , we denote by T n ( i ) the set of individualsbelonging to T n who are descendants of i . Writing i j when i is an ancestor of j , it means that T ( n ) p = { i ∈ G p : ∃ j ∈ T n such that i j } and T n ( i ) = { j ∈ T n : i j } . Writing i ∧ j the most recent common ancestor of two individuals i and j and gathering the couplesin function of their most recent common ancestor, we obtain1 T n X ( i,j ) ∈ T n i = j E [ Z n ( i ) Z n ( j )]= 1 T n n − X p =0 X u ∈T ( n ) p X ( i,j ) ∈ T n i ∧ j = u E [ Z n ( i ) Z n ( j )]= 1 T n n − X p =0 X u ∈T ( n ) p E E X i ∈ T n (2 u ) Z n ( i ) (cid:12)(cid:12)(cid:12) X u × E X j ∈ T n (2 u +1) Z n ( j ) (cid:12)(cid:12)(cid:12) X u +1 . (25)For all p fixed in { , · · · , n − } and u ∈ G p , we set A p,un = A ( u , . . . , u p , θ p +1 , . . . , θ n ) and for any x ∈ R , P pn ( x, u ) = P (cid:16) Y n ≥ c n (cid:12)(cid:12)(cid:12) Y p = x, A p,un ≥ c n (cid:17) and by convention P pn ( x, u ) = 0 if P ( A p,un ≥ c n ) = 0. Besides the many-to-one formula ensuresthat(26) E X i ∈ T n ( u ) Z n ( i ) (cid:12)(cid:12)(cid:12) X u = T n ( u ) ( P pn ( X u , u ) − E [ P pn ( X u , u )]) . The rest of the proof of the theorem relies on the two following lemmas. The first one ensures thatan ergodic property holds on lineages, which allows to forget the beginning of the trajectory.
Lemma 2.
For any p ∈ N and any u ∈ G p , we have P pn ( X u , u ) − E [ P pn ( X u , u )] −−→ n → + ∞ in probability . Proof of Lemma 2.
Let p be a fixed natural integer and let u ∈ G p . First we will prove thatconditionally on { A n ≥ c n } , n Q k =1 θ k /c n converges to 0 in probability. For that purpose we will showthat for all ε >
0, we have(27) P n Y k =1 θ k ≥ εc n ! = O ( P ( A n ≥ c n )) . ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 17
We set S k = k X ℓ =1 log( θ ℓ ); S = sup { S k , k ≥ } and τ S = inf (cid:8) k ≥ , S k = S (cid:9) . Let ξ ∈ (0 ,
1) and δ > β ) ≤ − δ < E [log( θ )] <
0. We have P n Y k =1 θ k ≥ εc n ! = P n Y k =1 θ k ≥ εc n , τ S > ξn ! + P n Y k =1 θ k ≥ εc n , τ S ≤ ξn ! ≤ P ( τ S > ξn ) + P (cid:0) S > n (log( c ) + δ ) (cid:1) + ξn X j =1 P n Y k =1 θ k ≥ εc n , τ S = j, S j ≤ n (log( c ) + δ ) ! . (28)Using the asymptotic of τ S given in Theorem 15, Chapter 4 of [17], we obtain(29) P ( τ S > ξn ) . ( ξn ) − / γ ξn . Next, from Kesten’s results, see e.g. [36], Section 1 or [28] Section 2, we have(30) P (cid:0) S ≥ n (log( c ) + δ ) (cid:1) ∼ ( c n ) − κ exp( − κδn ) . Finally, for the last term of the inequality (28) we have ξn X j =1 P n Y k =1 θ k ≥ εc n , τ S = j, S j ≤ n (log( c ) + δ ) ! ≤ ξn X j =1 P (exp( S n − S j ) ≥ ε exp ( − nδ ) , τ S = j, S j ≥ S n ≥ n log( c ) + log( ε ))= ξn X j =1 P τ S = j, S j > n log( c ) + log( ε ) ! P (cid:16) exp( S n − S j ) ≥ ε exp( − nδ ) (cid:12)(cid:12)(cid:12) τ S = j (cid:17) . We first observe that conditioning on τ S = j make stochastically decrease the random after time j and P (cid:16) exp( S n − S j ) ≥ ε exp( − nδ ) (cid:12)(cid:12)(cid:12) τ S = j (cid:17) ≤ P ( S n − S j ≥ − nδ + log( ε )) . P ( S n − j ≥ − nδ ) ≤ exp (cid:18) − ( n − j ) ψ (cid:18) − nn − j δ (cid:19)(cid:19) , where ψ is the rate function associated to ( S k , k ≥ ψ is nonincreasing in ( −∞ , E [log( θ )]], we have for j ∈ { , · · · , ξn } ,exp (cid:18) − ( n − j ) ψ (cid:18) − nn − j δ (cid:19)(cid:19) . exp ( − (1 − ξ ) nψ ( − δ )) . From the foregoing we obtain(31) ξn X j =1 P n Y k =1 θ k ≥ εc n , τ S = j, S j ≤ n (log( c ) + δ ) ! . P (cid:0) S τ S > n log( c ) (cid:1) × exp ( − (1 − ξ ) nψ ( − δ )) . ( c n ) − κ × exp ( − (1 − ξ ) nψ ( − δ )) . From (29)-(31) we conclude that (27) holds and then that P n Y k =1 θ k ≥ εc n (cid:12)(cid:12)(cid:12) A n ≥ c n ! −−→ n → + ∞ . Similarly, it is easy to see that for p ∈ N and u ∈ G p and x ∈ R , we also have(32) P (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x Q nk = p +1 θ k c n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ ε (cid:12)(cid:12)(cid:12) A p,un ≥ c n ! −−→ n → + ∞ . We observe now that(33) P pn ( x, u ) = P x Q nk = p +1 θ k c n + e Y n c n ≥ (cid:12)(cid:12)(cid:12) A p,un ≥ c n ! , where e Y n = P nk = p +1 (cid:0)Q nℓ = k +1 θ l (cid:1) ε k . Combining (33), (32) and the fact that e Y n /c n is a mixed ofGaussian random variable and the fact that X u is stochastically bounded (since gaussian), we get P pn ( X u , u ) − P (cid:16) e Y n ≥ c n (cid:12)(cid:12)(cid:12) A p,un ≥ c n (cid:17) → . By dominated convergence theorem, we obtain that E [ P pn ( X u , u )] − P (cid:16) e Y n ≥ c n (cid:12)(cid:12)(cid:12) A p,un ≥ c n (cid:17) goesto 0 as n tends to infinity. The proof follows by combining these two limits. (cid:3) The next lemma proves that the common ancestor of lineages of the bulk Z n [ c n , ∞ ) are at thebeginning of the tree, thus providing the decorrelation needed for a law of large number. Lemma 3. sup n>p n − P k = p P u ∈T ( n ) k T n (2 u ) T n (2 u + 1) T n −−→ p → + ∞ . Proof of Lemma 3.
Let I n and J n be two independent indices uniformly drawn from T n andindependent of ( X i , i ∈ T ). Let p < n . We have P ( p ≤ | I n ∧ J n | < n ) = 1 T n X ( i,j ) ∈ T n ,i = j {| i ∧ j |≥ p } . Next, gathering pairs in function of their most recent common ancestor we obtain P ( p ≤ | I n ∧ J n | < n ) = 1 T n X ( i,j ) ∈ T n ,i = j {| i ∧ j |≥ p } = 1 T n n − X k = p X u ∈T ( n ) k T n (2 u ) . T n (2 u + 1) . Since for any u ∈ T ( n ) p , P ( p ≤ | I n ∧ J n | < n | u I n ) = T n ( u ) T n , then P ( p ≤ | I n ∧ J n | < n ) ≤ max n T n ( u ) : u ∈ T ( n ) p o T n . ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 19
Moreover, since α > β , we easily observe thatmax n T n ( u ) : u ∈ T ( n ) p o = T n (2 p ) . Thus we need to check that(34) sup n>p { T n (2 p ) } T n −−→ p → + ∞ . For that purpose, we set for p < n , N ( i ) = { k : i k = 1 } , B pn = { i ∈ T n (2 p ) : N ( i ) < ε ( n − p ) } , where ε > i ∈ G n , ( i , · · · , i n ) ∈ { , } n is the binary decompo-sition providing the unique path in the binary tree from the root to i . In word, B pn is the set ofindividuals of T n (2 p ) whose binary decomposition has less than ε ( n − p ) “1”. We first check thatthere exists ε > B pn T n −−→ p → + ∞ . It means that the number of extremal individuals created by the prolific individual “2 p ” is negligiblewith respect to the total number of extremal individuals when p becomes large. Now let ψ be ratefunction associated to the simple random walk (with step 0 and 1 with probability 1/2). For n > p large enough, we have B pn n − p = { i ∈ G n − p : N ( i ) < ε ( n − p ) } n − p ≤ exp ( − ( n − p ) ψ ( ε )) . Recalling that T n ∼ (2 /c κ ) n , there exists c > B pn T n ≤ c (cid:18) exp ( ψ ( ε ))2 (cid:19) p (cid:18) c κ exp ( ψ ( ε )) (cid:19) n . Now we recall that c κ < ψ (0) = log(2). So we can choose ε > c κ < exp ( ψ ( ε )).Then B pn T n ≤ c (cid:18) exp ( ψ ( ε ))2 (cid:19) p . (cid:18) c κ exp ( ψ ( ε )) (cid:19) p ≤ (cid:18) c κ (cid:19) p for n > p . The last term tends to 0 as p tends to infinity, which yields (35).Now using (35), a sufficient condition for (34) is(36) T n (2 p ) \ B pn T n −−→ p → + ∞ , where T n (2 p ) \ B pn is the set of individuals of T n (2 p ) whose the number of components which areequal to 1 is greater than ε ( n − p ). We prove (36) as follows. For i ∈ G n and 1 ≤ a < b ≤ n ,we define τ ab i ∈ G n as the label obtained from i by permutation of i a and i b in the binarydecomposition (or path in the binary tree). We also introduce I kn = { u ∈ T n (2 p ) : N ( u ) = k } the set of individuals in T n , whose ancestor in generation p is 2 p , and which contains exactly k components equal to 1.Let us first fix a with 1 ≤ a ≤ p and consider the set I kn ( a ) = { τ ab u : u ∈ I kn , p + 1 ≤ b ≤ n, u b = 1 } . First I kn ( a ) ∩ T n (2 p ) = ∅ and I kn ( a ) ⊂ T n since by definition of A n and using β < α , A n ( τ ab u ) ≥ A n ( u ) for any u such that u a = 0 and u b = 1. Moreover for any a = a ′ and k = k ′ , I kn ( a ) ∩ I k ′ n ( a ′ ) = ∅ , so T n \ T n (2 p ) ≥ [ k ≥ ( n − p ) ε ≤ a ≤ p I kn ( a ) = X k ≥ ( n − p ) ε, ≤ a ≤ p I kn ( a ) . Besides, I kn ( a ) = kn − p − k I kn , since • for any u ∈ I kn , we have k choices for b such that u b = 1; • for any v ∈ I kn ( a ), we have n − p − k choices for 0 in the n − p last components of v thatyou can change for 1 : (cid:8) ( u, b ) : u ∈ I kn , p + 1 ≤ b ≤ n, u b = 1 , τ ab u = v (cid:9) = ( n − p − k ) . Putting the three last facts together, we get T n \ T n (2 p ) ≥ X ( n − p ) ε ≤ k 1) corresponding to k = n − p , T n (2 p ) \ B pn is the union of thedisjoint sets I kn for ( n − p ) ε ≤ k < n − p . Thus, we get T n \ T n (2 p ) ≥ εp × T n (2 p ) \ B pn and lim p →∞ sup n ≥ p T n (2 p ) \ B pn T n \ T n (2 p ) = 0 . This proves (36) and ends the proof of Lemma 3. (cid:3) We can now proceed with the proof of Theorem 4. Proof of the lower bound of Theorem 4. Now let ε > 0. From Lemma 3 let p ε such that1 T n n − X k = p ε X u ∈T ( n ) k T n (2 u ) T n (2 u + 1) ≤ ε. Adding that | Z n ( i ) | ≤ p larger than p ε , we get1 T n X ( i,j ) ∈ T n i = j E [ Z n ( i ) Z n ( j )] ≤ R n + ε, where R n = 1 T n p ε − X p =0 X u ∈T ( n ) p X ( i,j ) ∈ T n i ∧ j = u E [ Z n ( i ) Z n ( j )] . ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 21 Besides, recalling (25)-(26), R n = 1 T n p ε − X p =0 X u ∈T ( n ) p . T n (2 u ) . T n (2 u + 1) × [ P p +1 n ( X u , u ) − E ( P p +1 n ( X u , u ))][ P p +1 n ( X u +1 , u + 1) − E ( P p +1 n ( X u +1 , u + 1))] . Next, P pn is bounded and by dominated convergence theorem, Lemma 2 yields R n −−→ n → + ∞ . Weconclude from the foregoing that 1 T n X ( i,j ) ∈ T n i = j E [ Z n ( i ) Z n ( j )] −−→ n → + ∞ E (cid:2) ( F n − f n ) (cid:3) −−→ n → + ∞ . Combining this limit with (22) and (23) ends the proof of the lowerbound and Theorem 4. (cid:3) Finally, we derive the asymptotic value of the maximal value M n in the weakly stable case. Corollary 2. lim n → + ∞ n log( M n ) = log(2) κ in probability . The proof follows the proof of Corollary 1 in the strictly stable case and is left to the reader.One need now to use the estimate of Theorem 2 for the upper bound and check inequalities inprobability for the lower bound. 4. Appendix Proof of (15) : minimization problem for large values in the strictly stable case. We first focus on the minimization problem and then use it to get the estimates of probabilitiesgiven in (15). We set ϑ = 1 /α > H = { y ∈ R N : X k ≥ y k ϑ k < ∞} . It is easy to see that H is an Hilbert space endowed with the inner product < x, y > = P k ≥ x k y k ϑ k .We write k y k = P k ≥ y k ϑ k the associated norm. Thanks to Cauchy-Schwarz inequality, one canobserve that y → P k ≥ y k is well defined and continuous on H . For D ⊂ H , we define(37) I ( D ) = inf k y k : X j ≥ y j ≥ , y ∈ D . and we have the following minimization result for a quadratic form with a linear constraint. Lemma 4. Writing v = ((1 − α ) α j ) j ≥ , for all k ≤ n and for all ε > , we have i ) I ( H ) = k v k = 1 − α ; ii ) I ( { y ∈ H : | y k − v k | ≥ ε } ) > I ( H ); iii ) I ( { y ∈ H : | y k − v k | ≥ ε, y i = 0 for i > n } ) ≥ I ( { y ∈ H : | y k − v k | ≥ ε } ) . Proof. First, since k y k is invariant by change of sign of a coordinate, we get I ( H ) = inf k y k : X j ≥ y j = 1 , y ∈ H . We observe that for all y ∈ H such that P j ≥ y j = 1 , the vector z = y − v belongs to H andsatisfies P j ≥ z j = 0 . Then y = v + z and < v, z > = 0 and k y k = k v k +2 < v, z > + k z k = k v k + k z k which proves i ) and ii ). Next, iii ) is a direct consequence of { y ∈ H : | y − v k | ≥ ε, and y i = 0 for i > n } ⊂ { y ∈ H : | y − v k | ≥ ε } . which ends the proof. (cid:3) We can now deal with the proof of (15). We recall that ( k n , n ∈ N ) is a sequence of integerssmaller than n and tending to infinity such that k n = O (log( a n )) and B k n = { θ n = α, · · · , θ n − k n = α } . For each n ≥ 1, we subdivise [ − , 1] as a collection of N n successive disjoint intervals ( I ni : 1 ≤ i ≤ N n ) of length (at most) 1 /n / :[ − , 1] = ∪ N n i =1 I ni , | I ni | ≤ /n / , I ni ∩ I nj = ∅ for i = j, N n ≤ n / + 1)and either I ni ⊂ R + or I ni ⊂ R − . We set I n = [0 , /n / ], I n = [ − /n / , 0) and I n = ( −∞ , − ∪ (1 , + ∞ ). Then we can write R = ∪ N n i =0 I ni . For k = 0 , . . . , k n and ε > 0, we write I n ( k ) the set ofpaths i ∈ { , . . . , N n } n such that the k th interval, I i k , does not intersect [ v k − ε, v k + ε ] : I n ( k ) = { i ∈ { , . . . , N n } n : I n i k ∩ [ v k − ε, v k + ε ] = ∅ } , where we recall that v k = (1 − α ) α k . We defineΞ j,n = α k n +1 a n θ n − k n − . . . θ n − j +1 ε n − j , for j ∈ { k n + 2 , . . . , n } and Ξ j,n = α kn +1 a n ε n − j , for j ∈ { , . . . , k n + 1 } . We set E n ( i ) = n \ j =0 n Ξ j,n ∈ I n i j o . For all ε > 0, we have(38) lim sup n →∞ P (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) α k a n ε n − k − v k (cid:12)(cid:12)(cid:12)(cid:12) > ε (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n (cid:19) ≤ lim sup n →∞ X i ∈I n ( k ) P ( E n ( i ) | Y n ≥ a n , B k n ) . Let r ∈ (0 , 1) fixed. We set I n ( k ) = { i ∈ I n ( k ) : ( i n r , · · · , i n − ) = { , } n − n r } and δ n = 1 − ( n − n r ) n / and D n,k = ( y ∈ R n ; n r − X ℓ =0 y ℓ ≥ − δ n ; | y k − v k | ≥ ε ) . Then, we have the following bounds. Lemma 5. For all i ∈ I n ( k ) c , (39) P ( E n ( i ) | Y n ≥ a n , B n ) . α n r n − / u − n exp (cid:18) − a n (cid:18)(cid:16) α − n r n − / (cid:17) − u − k n (cid:19)(cid:19) . ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 23 For i ∈ I n ( k ) , (40) P ( E n ( i ) | Y n ≥ a n , B k n ) . (1 / n − n r ( n / ) n r R n , where R n = exp − a n inf y ∈ D n,k n r − X j =0 α − j y j − − α − α k n +2 . Proof of (39) . First, note that for all j ≥ n r , we have P (cid:18) Ξ j,n ≥ n / (cid:12)(cid:12) Y n ≥ a n , B k n (cid:19) ≤ P (cid:0) Ξ n r ,n ≥ n / (cid:1) P ( B k n ) P ( Y n ≥ a n , B k n ) ≤ P (cid:0) ε n − n r ≥ a n α − n r n − / (cid:1) P ( B k n ) P ( Y n ≥ a n , B k n ) . α n r n − / u − k n exp (cid:18) − a n (cid:18)(cid:16) α − n r n − / (cid:17) − u − k n (cid:19)(cid:19) . In the same way, we have P (cid:18) Ξ j,n ≤ − n / (cid:12)(cid:12) Y n ≥ a n , B k n (cid:19) . α n r n − / u − k n exp (cid:18) − a n (cid:18)(cid:16) α − n r n − / (cid:17) − u − k n (cid:19)(cid:19) , which implies that(41) ∀ j ≥ n r , P (cid:18) | Ξ j,n | ≥ n / | Y n ≥ a n , B k n (cid:19) . α n r n − / u − k n exp (cid:18) − a n (cid:18)(cid:16) α − n r n − / (cid:17) − u − k n (cid:19)(cid:19) . Now, for i ∈ I n ( k ) c , let j n ≥ n r such that i j n = − . Using (41), we get P ( E n ( i ) | Y n ≥ a n , B k n ) ≤ P n r − \ j =0 n Ξ j,n ∈ I n i j o \ n Ξ j n ,n ∈ I n i jn o (cid:12)(cid:12)(cid:12) Y n ≥ a n , B k n ≤ P (cid:18) | Ξ j n ,n | ≥ n / | Y n ≥ a n , B k n (cid:19) . α n r n − / u − k n exp (cid:18) − a n (cid:18)(cid:16) α − n r n − / (cid:17) − u − k n (cid:19)(cid:19) and this ends the proof of (39). (cid:3) Proof of (40) . We set y i the lower bound of I i if I i ⊂ R + and the upper bound of I i if I i ⊂ R − .We note that for all j ∈ { n r , · · · , n − } , we have { Ξ j,n ∈ I n } ⊂ { ε n − j ≥ } and { Ξ j,n ∈ I n } ⊂{ ε n − j < } . We also note that, for all j ∈ { n r , . . . , n − } , we have − /n / ≤ Ξ j,n ≤ /n / andthen that 1 − ( n − n r ) n / ≤ − n − X j = n r Ξ j,n ≤ n − n r ) n / . Now since Y n /a n = P n r − j =0 Ξ j,n + P n − j = n r Ξ j,n , we get(42) P ( E n ( i ) , Y n ≥ a n , B k n ) ≤ (1 / n − n r P ( D n ( i )) , where D n ( i ) = n r − \ j =0 n Ξ j,n ∈ I n i j o \ n r − X j =0 Ξ j,n ≥ − δ n \ B k n . Using the convention 0 / D n ( i ) ⊂ n r − \ j =0 n Ξ j,n ∈ I n i j o \ n r − X j =0 ( y i j + n − / ) ≥ − δ n \ B k n , we get P ( D n ( i )) ≤ P ( B k n ) n r − Y j =0 (cid:18) α j a n | y i j | { y i j =0 } + 12 { y i j =0 } (cid:19) exp − a n n r − X j =0 α − j y i j ≤ ( n / ) n r exp − a n y ∈ D n,k n r − X j =0 α − j y j P ( B k n ) , where we used the fact that 1 / | y i j | ≤ n / if y i j = 0 for the last inequality. From (42) we obtain P ( E n ( i ) | Y n ≥ a n , B k n ) ≤ (1 / n − n r ( n / ) n r exp − a n y ∈ D nr,k n r − X j =0 α − j y j × P ( B k n ) P ( Y n ≥ a n , B k n ) − . Finally, (40) follows using (9). (cid:3) As a direct consequence of (39) and (40), we have, for some positive constant C ,(43) X i ∈I n ( k ) P ( E n ( i ) | Y n ≥ a n , B k n ) . (2 n / ) n e − C a n ( α − nr n − / ) + (2 n / ) n r (1 / n − n r R n . Recalling (37), we obtain from Lemma 4 thatinf y ∈ D n,k n r − X j =0 α − j y j ≥ I ( { y ∈ R n ; | y k − v k | ≥ ε } ) > I ( H ) = 1 − α . Since 1 − α k n +2 → n → ∞ , we get R n ≤ exp (cid:18) − C a n (cid:19) , for some positive constant C . From the latter inequality and (43), we get, since n → ∞ , X i ∈I n ( k ) P ( E n ( i ) | Y n ≥ a n , B k n ) . e − C a n ( α − nr n − / ) + e − C a n . e − C a n and using (38) this ends the proof of (15). ARGE DEVIATIONS AND LARGE LOCAL DENSITIES 25 A law of large numbers. We state here a strong law of large numbers useful for the proofof local densities in the strictly stable case. Proposition 1. Let {F n } ∞ be a filtration. Let { X n,i : n, i ≥ } be non-negative real valuedrandom variables such that for each n , conditionally on F n , { X n,i : i ≥ } are independent andidentically r.v. distributed as X n .Let { N n : n ≥ } be non-negative integer valued r.v. such that for each n , N n is F n measurableand lim inf n →∞ N n +1 /N n > . We assume that X n,i are uniformly bounded and lim sup n →∞ E ( X n ) / E ( X n +1 ) < lim inf n →∞ N n +1 /N n . Then N n E ( X n ) N n X i =1 X n,i n →∞ −→ a.s. . This result is a consequence of a classical law of large numbers where the variables X n,i dependboth on n and i , see Kurtz [39] and Athreya Kang [3]. We prove here that gathering terms in asuitable way allows to deal with the case when E ( X n ) → Proof. We write u n = ⌊ / E ( X n ) ⌋ ∨ N n X i =1 X n,i = V n − X k =0 Y n,k + R n , where V n = ⌊ N n /u n ⌋ goes to infinity and Y n,k = ( k +1) u n − X i = ku n X n,i , R n = N n X i = V n u n X n,i . Moreover, Markov inequality and boundedness of X n yields for any t ≥ n, k ≥ P ( Y n,k > t |F n ) ≤ e − t E ( e X n ) u n ≤ e − t (1 + C E ( X n )) u n ≤ C ′ e − t for some constants C, C ′ ≥ 0, since E ( X n ) is bounded. The right hand side is integrable on [0 , ∞ )and E ( Y n,k ) is bounded andlim inf n →∞ V n +1 /V n ≥ lim inf n →∞ N n +1 /N n . lim inf n →∞ E ( X n +1 ) / E ( X n ) > . We can apply a law of large numbers of Athreya and Kang (Lemma 1 in [3]) to the family of centeredand independent variables { Y n,k − E ( Y n, ) : n ≥ , k ≥ } which are stochastically dominated andget 1 V n V n − X k =0 ( Y n,k − E ( Y n, )) n →∞ −→ E ( Y n, ) = u n E ( X n ) and V n ∼ N n /u n as n → ∞ and(44) R n N n = P N n i = V n u n X n,i N n n →∞ −→ u n goes geometrically to infinity,since otherwise the boundedness of the r.v. X n,i allows to use a direct domination and Lemma 1in [3]. When u n grows geometrically, using again the law of large numbers in [3] yields P N n i = V n u n X n,i u n n →∞ −→ u n /N n → (cid:3) Acknowledgment. This work have been supported by the Chair “Mod´elisation Math´ematiqueet Biodiversit´e” of VEOLIA Environnement-´Ecole Polytechnique-MNHN-F.X and ANR ABIM(ANR-16-CE40-0001). References [1] Addario-Berry, L., and Reed, B. (2009). Minima in branching random walks , The Annals ofProbability, 37(3), 1044–1079.[2] A¨ıd´ekon, E. (2013). 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Vincent Bansaye, CMAP, ´Ecole Polytechnique, Route de Saclay, 91128 Palaiseau, France. E-mail address : [email protected] S. Val`ere Bitseki Penda, Institut de Math´ematiques de Bourgogne, UMR 5584, CNRS, Universit´e deBourgogne Franche- Comt´e, F-21000 Dijon, France. E-mail address ::