Extinction probability and total progeny of predator-prey dynamics on infinite trees

We consider the spreading dynamics of two nested invasion clusters on an infinite tree. This model was defined as the chase-escape model by Kordzakhia and it admits a limit process, the birth-and-assassination process, previously introduced by Aldous and Krebs. On both models, we prove an asymptotic equivalent of the extinction probability near criticality. In the subcritical regime, we give a tail bound on the total progeny of the preys before extinction.


Introduction
The chase-escape process is a stochastic predator-prey dynamics which was studied by Kordzakhia [15] on a regular tree. In an earlier paper, Aldous and Krebs [4] had introduced the birth-and-assassination (BA) process. The latter model can be seen as a natural limit of the chase-escape model. In [8] the two models were merged into the rumor scotching process. The original motivation of Aldous and Krebs was then to analyze a scaling limit of a queueing process with blocking which appeared in database processing, see Tsitsiklis, Papadimitriou and Humblet [24]. As pointed in [8], the BA process is also the scaling limit of a rumor spreading model which is motivated by network epidemics and dynamic data dissemination (see for example, [19], [5], [20]).
We may conveniently define the chase-escape processes as a SIR dynamics (see for example [19] or [5] for some background on standard SIR dynamics). This process represents the dynamics of a rumor/epidemic spreading on the vertices of a graph along its edges. A vertex may be unaware of the rumor/susceptible (S), aware of the rumor and spreading it as true/infected (I), or aware of the rumor and trying to scotch it/recovered (R).
We fix a locally finite connected graph G = (V, E). The chase-escape process is described by a Markov process on X = {S, I, R} V . If {u, v} ∈ E, we write u ∼ v. For Predator-prey dynamics on infinite trees Obviously, if T is finite then q T (λ) = 1 for any λ ≥ 0. Before stating our results, we first need to introduce some extra terminology.
There is a canonical way to represent the vertex set V as a subset of N f = ∪ ∞ k=0 N k with N 0 = ø and N = {1, 2 · · · }. If k ≥ 1 and v ∈ V is at distance k from the root, then v = (i 1 , · · · , i k ) ∈ V ∩ N k . The genitor of v is (i 1 , · · · , i k−1 ): it is the first vertex on the path from v to the root ø of length k. The offsprings of v are set of vertices who have genitor v. They are indexed by (i 1 , · · · , i k , 1), · · · , (i 1 , · · · , i k , n v ), where n v is the number of offsprings of v. The ancestors of v is the set of vertices (i 1 , · · · , i ), 0 ≤ ≤ k − 1 with the convention i 0 = . Similarly, the n-th generation offsprings of v are the vertices in V ∩ N k+n of the form (v, i k+1 , · · · , i k+n ).
Recall that the upper growth rate d ∈ [1, ∞] of a rooted infinite tree T is defined as where V k = V ∩ N k is the set of vertices at distance k from the root and | · | denotes the cardinal of a finite set. The lower growth rate is defined similarly with a lim inf. When the lim inf and the lim sup coincide, this defines the growth rate of the tree. For example, for integer d ≥ 1, we define the d-ary tree as the tree where all vertices have exactly d offsprings 1 . Obviously, the d-ary tree has growth rate d. More generally, consider a realization T of a Galton-Watson tree with mean number of offsprings d ∈ (1, ∞). Then, the Seneta-Heyde Theorem [23,14] implies that, conditioned on T infinite, the growth rate of T is a.s. equal to d. For background on random trees and branching processes, we refer to [6,22].
For integer n ≥ 1, we define T * n as the rooted tree on V obtained from T by putting an edge between all vertices and their n-th generation offsprings. For real d > 1, we say that T is a lower d-ary if for any 1 < δ < d, there exist an integer n ≥ 1 and v ∈ V such that the subtree of the descendants of v in T * n contains a δ n -ary tree. Note that if T is lower d-ary then its lower growth rate is at least d. Also, if T is the realization of a Galton-Watson tree with mean number of offsprings d ∈ (1, ∞) then, conditioned on T infinite, T is a.s. lower d-ary (for a proof see Lemma 5.5 in appendix).
The first result is an extension of [15,Theorem 1] where it is proved for d-ary trees.
It describes the phase transition of the event of survival.
Note that in the classical SIR dynamics, it is easy to check that the critical value of λ is λ = 1/(d − 1). Also, for any d > 1, λ 1 < 1 and, (1.1) The proof of Theorem 1.1 will follow a strategy parallel to [15,4]. We employ techniques akin to the study the infection process in the Richardson model. They will be based on large deviation estimates on the probability that a single vertex is I at time t.
To our knowledge, there is no known closed form expression for the extinction probability q T (λ). Our next result determines an asymptotic equivalent for the probability of survival for λ close to λ 1 . Our method does not seem to work on the sole assumption 1 It would be more proper to call this tree the complete infinite d-ary tree. that T has growth rate d > 1 and is lower d-ary. We shall assume that T is a realization of a Galton-Watson tree with offspring distribution P and d = ∞ k=1 kP (k) > 1.
We consider the annealed probability of extinction: where the expectation E (·) is with respect to the randomness of the tree and P λ (·) = E (P λ (·)) is the probability measure with respect to the joint randomness of T and X. Note that in the specific case d integer and P (d) = 1, T is the d-ary tree and the measures P λ and P λ coincide. Theorem 1.2. Assume further that the offspring distribution has finite second moment.
There exist constants c 0 , c 1 > 0 such that for all λ 1 < λ < 1, Note that the behavior depicted in Theorem 1.2 contrasts with the classical SIR dynamics, where 1 − q(λ) is of order (λ(d − 1) − 1) + . This result should however be compared to similar results in the Brunet-Derrida model of branching random walk killed below a linear barrier, see Gantert, Hu and Shi [11] and also Bérard and Gouéré [7]. As in this last reference, our approach is purely analytic. We will first check that q(λ) is related to a second order non-linear differential equation. Then, we will rely on comparisons with linear differential equations. A similar technique was already used by Brunet and Derrida [9], and notably also in Mueller, Mytnik and Quastel [18, section 2].
A possible parallel with the Brunet-Derrida model of branching random walk killed below a linear barrier is the following. Consider a branching random walk on Z started from a single particle at site 0 where the particles may only move by one step on the right. If we are only concerned by the extinction, we can think of this process as some branching process without walks where a particle at site k gives birth to particles at site k + 1. We can in turn represent this process by a growing random tree where the set of vertices at depth k is the set of particles at site k. Hence (I)-vertices play the role of the particles, the branching mechanism is the spreading of the (I)-vertices over the (S)-vertices and the set of (R)-vertices is a randomly growing barrier which absorbs the particles/(I)-vertices. Kortchemski [17] has recently built an explicit coupling of a branching random walk with the chase-escape process on a tree.
In the case 0 < λ < λ 1 , the process X stops a.s. evolving after some finite τ . We define Z as the total progeny of the root, i.e. the total number of recovered vertices (excluding the vertex o of T ↓ ) at time τ . It is the number of vertices which will have been infected before the process reaches its absorbing state. We define the annealed parameter: The scalar γ(λ) can be though as a power-tail exponent of the variable Z under the annealed measure P λ . In particular, for any 0 < γ < γ(λ), from Markov Inequality, there exists a constant c > 0 such that for all t ≥ 1, P λ (Z ≥ t) ≤ ct −γ . Conversely, if there exist c, γ > 0 such that for all t ≥ 1, P λ (Z ≥ t) ≤ ct −γ , then γ(λ) ≥ γ. We define (i) For any 0 < λ < λ 1 , .
It is straightforward to check that (i) is equivalent to (ii). Also, for u = 1, λ u coincides with λ 1 defined in Theorem 1.1. It follows that γ(λ) ≥ 1 for all 0 < λ < λ 1 . Theorem 1.3 contrasts with classical SIR dynamics. For example, if T is the d-ary tree, is the total progeny in the classical SIR dynamics. Here, the heavy-tail phenomenon is an interesting feature of the chase-escape process. Intuition suggest that large values of Z come from a (I)-vertex which is not recovered before an exceptionally long time. Indeed, in the chase escape process, a (I)-vertex which is not recovered by time t will typically have a progeny which is exponentially large in t (this is not the case in the classical sub-critical SIR dynamics, the progeny of such vertex will typically be of order 1) . A similar phenomenon appears also in the Brunet-Derrida model, see Addario-Berry and Broutin [1], Aïdékon [2] and Aïdékon Hu and Zindy [3]. Note finally that γ(λ) ∼ λ↓0 min 1 (d − 1)λ , γ P and γ(λ) ∼ λ↑λ1 1.
By recursion, we will also compute the moments of Z. The computation of the first moment gives Theorem 1.4 implies a surprising right discontinuity of the function λ → E λ Z at the critical intensity λ = λ 1 : Predator-prey dynamics on infinite trees The birth-and-assassination process We now turn to the BA process. It is a scaling limit in d → ∞ of the chase-escape process on the d-ary tree when λ is rescaled in λ/d.
Informally, the process can be described as follows. We start from a root vertex that produces offsprings according to a Poisson process of rate λ. Each offspring in turn produces children according to independent Poisson processes and so on. The children of the root are said to belong to the first generation and their children to the second generation and so forth. Independently, the root vertex is at risk at time 0 and dies after a random time D ø that is exponentially distributed with mean 1. Its offsprings become at risk after time D ø and the process continues in the next generations. We now make precise the above description.
As above, N f = ∪ ∞ k=0 N k denotes the set of finite k−tuples of positive integers (with N 0 = ø). Elements from this set are used to index the offspring in the BA process. Let {Ξ v }, v ∈ N f , be a family of independent Poisson processes with common arrival rate λ; these will be used to define the offsprings. Let {D v }, v ∈ N f , be a family of independent, identically distributed (iid) exponential random variables with mean 1; we use them to assign the lifetime for the appropriate offspring. The families {Ξ v } and {D v } are independent. The process starts at time 0 with only the root, indexed by ø. This produces offspring at the arrival times determined by Ξ ø that enter the system with indices (1), (2), · · · according to their birth order. Each new vertex v, immediately begins producing offspring determined by the arrival times of Ξ v . The offspring of v are indexed (v, 1), (v, 2), · · · also according to birth order. The root is at risk at time 0. It continues to produce offspring until time T ø = D ø , when it dies. Let k > 0 and let v = (n 1 , · · · , n k−1 , n k ), v = (n 1 , ..., n k−1 ) denote a vertex and its genitor. When a particle v dies (at time T v ), the particle v then becomes at risk; it in turn continues to produce offspring until time T v = T v + D v , when it dies (see figure 2).
The BA process can be equivalently described as a Markov process X(t) on {S, I, R} N f , where a particle/vertex in state S is not yet born, a particle in state I is alive and a particle in state R is dead. A particle is at risk if it is in state I and its genitor is in state R. We use the same notation as above : under P λ , the process X(t) has infection rate λ > 0, q(λ) is the probability of extinction and so on. The following result from [4] describes the phase transition on the probability of survival as a function of λ. Theorem 1.6. Consider the BA process with rate λ > 0. If λ ∈ [0, 1/4], then q(λ) = 1, while if λ > 1 4 , 0 < q(λ) < 1.
The critical case λ = 1/4 was established in [8]. Note also that the threshold λ = 1/4 is consistent with (1.1). Our final result is the analog of Theorem 1.2.
Theorem 1.7. Consider the BA process and assume that λ > 1/4. There exist constants c 0 , c 1 > 0 such that for all 1/4 < λ < 1, Note that the analog of Theorems 1.3-1.4 was already performed in [8]. The remainder of the paper is organized as follows. In section 2, we start with the study of the BA process and prove Theorem 1.7. Proofs on the BA process are simpler and this section is independent of the rest of the paper. We then come back to the chase escape process: in section 3, we prove Theorem 1.1, in section 4, we prove Theorem 1.2. Finally, in section 5, we prove Theorems 1.3-1.4.
2 Proof of Theorem 1.7

Differential equation for the survival probability
We first determine the differential equation associated to the probability of extinction for the BA process. Define Q λ (t) to be the extinction probability given that the root dies at time t > 0 so that (2.1) and Q λ (0) = 1. Let {ξ i } i≥1 be the arrival times of Ξ ø with 0 ≤ ξ 1 ≤ ξ 2 ≤ · · · . For integer i with 1 ≤ ξ i ≤ D ø , we define B i as the subprocess on particles iN f with ancestors i. For the process B to get extinct, all the processes B i must get extinct. Conditioned on Ξ ø , and on the root to die at time D ø = t, the evolutions of the (B i ) then become independent. Moreover, on this conditioning, B i is a birth-and-assassination process conditioned on their root to be at risk at time t − ξ i . Hence, we get where {ξ i } i≥1 is a Poisson point process of intensity λ and (D i ), i ≥ 1, independent exponential variables with parameter 1. Using Lévy-Khinchin formula, we deduce So finally, for any t ≥ 0, We perform the change of variable We find that for any t ≥ 0, .
ϕ(x(s))e −s ds, (2.4) Now, multiplying the above expression by e −t and differentiating once again, we find that x(t) satisfies the differential equation x − x + ϕ(x) = 0. (2.5) This non-linear ordinary differential equation is not a priori easy to solve. However, in the neighborhood of λ = 1/4 it is possible to obtain an asymptotic expansion as explained below. The idea will be to linearize the ODE near (x(0), x (0)) = (0, 0) and look at the long time behavior of the solutions of the linearized ODE. The critical value λ = 1/4 appears to be the threshold for oscillating solutions of the linearized ODE. From a priori knowledge on the long time behavior of the solution of (2.5) of interest (studied in §2.2), we will obtain an asymptotic equivalent for (x(0), x (0)) as λ ↓ 1/4 (in §2.4).

A fixed point equation
Let H be the set of measurable functions f : R + → R + such that f (0) = 0 and for any a > 0, lim s→∞ e −as f (s) = 0.
We define the map A : H → H defined by ϕ(y(s))e −s dsdx. (2.6) Consider the function x defined by (2.2). Using (2.3) we find that x ∈ H and satisfies also the fixed point x = A(x). If λ > 1/4, we know from Theorem 1.6 that x ≡ 0.
In the sequel, we are going to study any non-trivial fixed point of A. To this end, let x ∈ H such that x = A(x) and x ≡ 0. By induction, it follows easily that t → x(t) is twice continuously differentiable. In particular, since x(s) ≥ 0, x (t) ≥ 0 and the function x : R + → R + is non-decreasing. Moreover, by assumption there exists t 0 > 0 such that x(t 0 ) > 0. Since x is non-decreasing, we deduce that x(t) > 0 for all t > t 0 . Then, using again (2.4), we find that for all t ≥ 0, 0 < x (t) < λ. From the argument leading to (2.5), x satisfies (2.5) and we are looking for a specific non-negative solution of (2.5) which satisfies x(0) = 0. To characterize completely this solution, it would be enough to compute x (0) (which is necessary positive since x(0) = x (0) = 0 corresponds to the trivial solution x ≡ 0). We first give some basic properties of the phase portrait, see figure 3. We define and for all t ≥ 0, Proof: We have already checked that x satisfies (2.5) and x (0) > 0. Let us now prove that (2.9) holds. Define the trajectory Φ = {X(t) ∈ R 2 + : t ≥ 0}. Since for all t ≥ 0, (2.10) The graph of the function ϕ is the curve L = {(s, ϕ(s)) : s ∈ R + } and the set is the set of points below L. Assume that (2.9) does not hold. Then by (2.7) and the intermediate value Theorem, the curves L and Φ intersect. Then the exists s 0 > 0 such that f (s 0 ) = ϕ(s 0 ).
We have proved so far that for all t ≥ 0, X(t) ∈ ∆. This implies that x (t) is increasing. In particular lim t→∞ x(t) = ∞ and S = ∞. Since lim s→∞ ϕ(s) = λ, by (2.4), we readily deduce that x (t) converges to λ as t → ∞.

Comparison of second order differential equations
It is possible to compare the trajectories of solutions of second order ODE by using the phase diagram. Let D be the set of increasing Lipschitz-continuous functions ψ on Proof: Let us start with the hypothesis of (i). The proof is by contradiction : we also assume that y (0) < x (0). We set Y (t) = (y(t), y (t)) and G(y 1 , y 2 ) = (y 2 , y 2 − ψ(y 1 )). Define the trajectories Φ = {X(t) ∈ R 2 + : t ≥ 0}, and for τ > 0, Ψ(τ ) = {Y (t) ∈ R 2 + : 0 ≤ t ≤ τ }. By Lemma 2.1, Φ is the graph of an increasing function f : Hence, by the intermediate value Theorem, there exists a first time 0 < s 1 < s 0 such that the curves intersect: g(s 1 ) = f (s 1 ) and g(s) < f (s) on [0, s 1 ). However, it follows from (2.10) and ϕ ≤ ψ that for s ∈ [0, s 1 ), Hence, integrating over [0, s 1 ] the above inequality gives However, by construction, f (s 1 ) = g(s 1 ). Thus, the above inequality contradicts g(0) < f (0) and we have proved (i). The proof of (ii) is identical and is omitted.
In particular ϕ ≥ ψ. (2.16) We shall now consider the linear differential equation The solutions of (2.17) are A careful choice of a, κ will lead to the following lower bound.
Step three : End of proof. We now complete the proof of Theorem 1.7. We should consider the function x(t) defined by (2.2). We have seen that x = A(x) and x ≡ 0 if λ > 1/4. We start with the left hand side inequality. From (2.9) in Lemma 2.1, x (t) is increasing and we have .
It remains to use Lemma 2.4 and we obtain the left hand side of Theorem 1.7.
We now turn the right hand side inequality.
(see e.g. [13,Exercise 4.6]). Looking at the solution of y − y = 0 such that y(0) = 0 and y (0) = x (0), we get that We deduce from (2.1) that, for any T > 0, Now, we notice that in order to prove Theorem 1.7, by Lemma 2.3, we may choose λ close enough to 1/4 so that x (0) < 1. We finally take T = − ln(x (0)) and apply Lemma 2.3. This concludes the proof of Theorem 1.7.

Proof of Theorem 1.1
We define the set recovered and infected vertices as R(t) = {v ∈ V : X v (t) = R} and I(t) = {v ∈ V : X v (t) = I}. The set R(t) being non-decreasing, we may define R(∞) = ∪ t>0 R(t) and Z = |R(∞)| ∈ N ∪ {∞}. Note that also a.s. R(∞) = {v ∈ V : ∃t > 0, X v (t) = I}, in words, R(∞) is the set of vertices which have been infected at some time.
Throughout this section, the chase-escape process is constructed thanks to i.i.d.
is the time by which v ∈ V will be infected (resp. recovered) once its ancestor is infected (resp. recovered).

Subcritical regime
We fix 0 < λ < λ 1 . In this paragraph we prove that q T (λ) = 1 if T has upper growth rate at most d. It is sufficient to prove that E λ Z. To this end, we will upper bound the probability that v ∈ R(∞) for any v ∈ V . Let V k be the set of vertices of V which are at distance k from the root . Let v ∈ V n and v 0 , · · · , v n be the ancestor line of v: v 0 = and v n = v. The vertex v will have been infected if and only if for all 1 ≤ m ≤ n, v m−1 has infected v m before being recovered. We thus find The Chernov bound gives for any 0 < θ < 1, where we have used the independence of all variables at the last line. Now, the above expression is minimized for θ = (1 − λ)/2 > 0 (since λ < λ 1 < 1). We find Also, from the growth-rate assumption, there exists a sequence ε n → 0, It follows that It is now straightforward to check that 4dλ (λ + 1) 2 < 1, if λ < λ 1 . This concludes the first part of the proof.

Supercritical regime
We now fix λ > λ 1 . We should prove that q T (λ) < 1 under the assumption that T is lower d-ary. We are going to construct a random subtree of T whose vertices are elements of R(∞) and which is a supercritical Galton-Watson tree.
First observe that we can couple two chase-escape processes with intensities λ > λ on the same probability space in such a way that they share the same variables (D v ) v∈V and for all v ∈ V , ξ . The event of non-extinction is easily seen to be non-increasing in the variables (ξ v ) v∈V for the partial order on R V + of component-wise comparison. It follows that the function λ → q T (λ) is non-increasing. We may thus assume without generality that λ 1 < λ < 1. For δ > 0, we define the function g δ by, for all x > 0, Taking derivative, the minimum of g δ is reached at c = (1 + λ)/2. We deduce easily the following property of the function g d .
In the remainder of the proof, we fix such pair (c, δ).

Construction of a nested branching process.
We fix an integer m ≥ 1 that we will be completely specified later on. We assume that m is large enough such that T * m contains a δ m -ary subtree. We denote by T this subtree and by ρ ∈ V its root. For integer k ≥ 0, we define V k as the set of vertices of generation k in T . Note that by assumption |V k | = δ m k .
We may assume that the generation of ρ in T is larger than m. We denote a(ρ) ∈ V the m-th ancestor of ρ in T . For z ∈ V k and k ≥ 1, we denote by a(z) ∈ V k−1 its ancestor in T . For example, if z ∈ V 1 , a(z) = ρ. We now start a branching process as follows. We set ρ to be the root of the process, S 0 = {ρ}. For integer k ≥ 1, we define recursively the offsprings of the k-th generation as the set S k of vertices z ∈ V k satisfying the following three conditions : Thus for z ∈ V k , such that its ancestor a(z) ∈ S k−1 , we have that Notice that by construction, the number of offsprings of z = z in S k−1 are identically distributed and are independent. It follows that the process forms a Galton-Watson branching process. In the next paragraph, we will check that this branching process is supercritical, i.e. we will prove that M = v∈V 1 P λ (z ∈ S 1 ) > 1. It implies that with positive probability, the branching process does not die out (see Athreya is the set of the vertices on the path from a(ρ) to ρ. Note that the existence of such finite time t > 0 and such sequence (D vi ) 0≤i≤m−1 has positive probability. Let us denote by E such event. We set t 0 = t and, for integer k ≥ 1, By construction, if E holds and z ∈ S k then, at time t k , z and a(z) are both infected.
Hence on the events of E and of non-extinction of the nested branching process, the chase-escape process does not get extinct. It thus remains to prove that (3.2) holds.
The nested branching process is supercritical. We need a standard large deviation estimate. We define while, for any a < 1/λ, Note that the bounds of Lemma 3.2 hold for all a > 0 (even if they are sharp only for the above ranges). We may now estimate the terms in (3.1). We have from Lemma 3.2 that Thus we obtain a lower bound on the mean number of offspring in the first generation to be where g δ (.) is as defined in (3.1). If m was chosen large enough, we have that M > 1 and hence that the branching process is supercritical. Therefore with positive probability, this branching process does not die out. This proves the theorem.

Proof of Theorem 1.2
The proof of Theorem 1.2 parallels the proof of Theorem 1.7. Even if the strategy is the same, we will meet some extra difficulties in the study of the phase diagram (notably in the forthcoming Lemma 4.3).

Differential equation for the survival probability
We first determine a differential equation associated to the probability of extinction.
Under P λ , define Q λ (t) to be the extinction probability given that the root ø is recovered at time t ≥ 0 so that and Q λ (0) = 1. Now, in T , the offsprings of the root are {1, · · · , N }, where N has distribution P . The root infects each of its offspring after an independent exponential variable with intensity λ. Let {ξ i } 1≤i≤N be the infection times. Note that in T , the subtrees generated by each of the offsprings of the root are iid copies of T . Hence, if for integer i with 1 ≤ ξ i ≤ D ø , we define X i as the subprocess on vertices (iN f ) ∩ V with ancestors i. Conditioned on D ø = t, on N and (ξ i ) 1≤i≤N , the processes (X i ) are independent chase-escape processes conditioned on the fact that root becomes at risk at time t − ξ i (where we say that a Ivertex is at risk if its genitor is in state R). For the process X to get extinct, all the processes X i must get extinct. So finally, where (D i ), i ≥ 1, are independent exponential variables with parameter 1. Consider the generating function of P Recall that ψ is strictly increasing and convex on [0, 1] and ψ (1) = EN = d. We find, for any t ≥ 0, Performing the change of variable leads to ψ(x(s))e −s dsdx. We multiply the above expression by e λt and differentiate once, it gives e λt (λx(t) + x (t)) = λe (λ+1)t ∞ t ψ(x(s))e −s ds, (4.4) Now, multiplying the above expression by e −(λ+1)t and differentiating once again, we find that x(t) satisfies the differential equation
Let H be the set of non-increasing functions f : R + → R + such that f (0) = 1, lim t→∞ f (t) = ρ. The next lemma is an easy consequence of the monotony of the process.
Proof: As in the previous section, we may construct the chase escape process conditioned on the root is recovered at time t thanks to i.i.d. Exp(λ) variables (ξ v ) v∈V and independent i.i.d. Exp(1) variables (D v ) v∈V =ø and set D ø = t. The variable ξ v (resp. D v ) is the time by which v ∈ V will be infected (resp. recovered) once its ancestor is infected (resp. recovered). The event of extinction is then non-increasing in t. It follows that the map t → Q λ (t) is non-increasing. From (4.2), it follows that x(t) is also non-increasing. We may thus define a = lim t→∞ x(t). Using the continuity of ψ leads to λe −λt From (4.3), we get that a = ψ(a) which implies that a ∈ {ρ, 1}. Note however that Theorem 1.1 and Lemma 5.5 imply that q(λ) < 1 for all λ > λ 1 . Then (4.1) and the monotony of t → Q λ (t) give that for all t ≥ t 0 large enough, Q λ (t) < 1. From (4.2) it implies in turn that for all t ≥ t 0 , x(t) < 1. So finally a ≤ x(t 0 ) < 1 and a = ρ.
From now on in this section, we fix a small u > 0 and we assume that  Since max x∈[0,1] |ψ(x)| = 1 it is indeed straightforward to check that A(y) is in L ∞ (R + , R + ): A(y)(t) is bounded by 1. Note also that y ≡ 1 is a solution of the fixed point equation y = A(y). By (4.3), we find that the function x defined by (4.2) satisfies also the fixed point x = A(x). In the sequel, we are going to analyze the non trivial fixed points of A. Let x ∈ H such that x = A(x). Then x ≡ 1. By induction, it follows easily that t → x(t) is twice differentiable. In particular, from the argument following (4.3), x satisfies (4.5) and we are looking for a specific non-negative solution of (4.5) with x(0) = 1. To characterize completely this solution, it would be enough to compute x (0) (which is necessarily negative since x(0) = 1, x (0) = 0 corresponds to the trivial solution x ≡ 1).
We will perform this in the next subsection in an asymptotic regime. We start with some properties obtained from the phase diagram of the ODE (4.5).
Proof: Let us prove (i). We first observe that since x(t) is non-increasing, x(0) = 1 and x (0) < 0, we have that for all t > 0, x(t) < 1. Also, if x(t) = ρ for some t > 0, then x(s) = ρ for all s ≥ t (since x is non-increasing and has limit ρ). However y ≡ ρ being a distinct solution of (4.5), x and y cannot coincide on an interval. We thus have for all t > 0, ρ < x(t) < 1.
The next lemma is proved as Lemma 2.2, we omit its proof.

Proof of Theorem 1.2
The proof of Theorem 1.2 follows now closely the proof of Theorem 1.7. We first linearize (4.5) in the neighborhood of λ 1 .
In particular ϕ ≤φ. (4.14) We define the linear differential equation The solutions of (4.15) are In the sequel, o(1) denotes a function which goes to 0 as ω goes to 0. Lemma 4.6. If P has finite second moment, then there exists a constant c 1 > 0 such that for all λ 1 < λ < 1, if x ∈ H satisfies A(x) = x then (1 + o(1)).
Step three : end of proof. We may now complete the proof of Theorem 1.2. We start with the left hand side inequality. We first note that, by Lemma 4.3, Moreover by Lemma 4.3, we find x(t 0 ) ≤ 1 − δ. However, by Lemma (4.2)(ii), we have Hence t 0 ≥ δ. Then, by construction, on the interval [0, t 0 ], Since ψ(x) ≤ x on [ρ, 1], it follows from (4.1) that the survival probability may be lower bounded as It remains to use Lemma 4.6 and we obtain the left hand side of Theorem 1.2.
We turn to the right hand side inequality. For X = ( (see e.g. [13,Exercise 4.6]). Looking at the solution of y − (1 − λ)y = 0 such that y(0) = 1 and y (0) = x (0), we get that We deduce from (4.1)-(4.2) and the convexity of ψ that, We are first going to find a recursive distributional equation (RDE) associated to the total progeny of the chase-escape process on a Galton-Watson tree.
As already pointed, we can build the chase escape process on the tree T ↓ thanks to i.i.d. Exp(λ) variables (ξ v ) v∈V and independent i.i.d. Exp(1) variables (D v ) v∈V . The variable ξ v (resp. D v ) is the time by which v ∈ V will be infected (resp. recovered) once its ancestor is infected (resp. recovered). For t ≥ 0, we define Y (t) as the total number of recovered vertices when the process reach its absorbing state (without counting o) when we replace D ø by t. The variable Y (t) is the conditional variable Z conditioned on the root is recovered at time t. By definition, if D is an independent exponential variable with mean 1, where the symbol d = stands for distributional equality. In T , we denote the offsprings of the root by {1, · · · , N }. The random variable N has distribution P . The root infects each of its offspring after an independent exponential variable with intensity λ. Note that in T , the subtrees generated by each of the offsprings of the root are iid copies of T . Hence, the recursive structure of the tree T leads to the following equality in distribution where (ξ i ) i∈N are iid exponential variables with intensity λ, (Y i ) 1≤i≤N and (D i ) 1≤i≤N are independent copies of Y and D respectively. Note that since all variables are nonnegative, there is no issue with the case Y (t) = +∞ in the above RDE.
The RDE (5.1) is the cornerstone of the argument. In the remainder of this subsection, we will use it to derive a linear second order ODE for the first moment of Y (t). In the following subsection §5.2, we will extend this exact computation to any integer moment. Finally, using convexity inequalities, we will push further the method and obtain sharp lower and upper bounds for any moment of Y (t) (in §5.3 and §5.4).
We start with a lemma Now, assume that E λ Z < ∞. We may then take expectation in (5.1): Multiplying by e λt and taking derivative, we get: Then, multiplying by e −(λ+1)t , taking the derivative a second time and then re-multiplying by e t , we obtain: f 1 (t) − (1 − λ)f 1 (t) − λf 1 = −λ − λdf 1 (t). So, finally, f 1 solves a linear ordinary differential equation of the second order where x(t) solves the ordinary differential equation If 0 < λ < λ 1 , the discriminant is positive. The roots of the polynomial are The solutions of (5.3) are (1 − a)e αt + ae βt It remains to prove that if 0 < λ ≤ λ 1 then EZ < ∞ and Indeed, we would get EZ = To this end, define T n as the tree T stopped at generation n. As above, we denote by Y (n) (t) the total number of recovered particles in T n when the root is recovered at time t (D ø is replaced by t). As n → ∞, Y n (t) is non-decreasing and converges to Y (t). We have Y (0) (t) = 1 and for all n ≥ 0, as in RDE (5.1), , and D i are independent copies of Y (n) and D respectively. Since EN < ∞, the expectation of the number of vertices in T n is finite. In particular, for all n ≥ 0, g n (t) = EY (n) (t) < ∞ is bounded uniformly in t. Also, taking expectation in (5.7), we have for all t ≥ 0, where Φ is the mapping It is easy to check that Φ is indeed a mapping from H 1 to H 1 , where H 1 is the set of non-decreasing functions g : [0, ∞) → [1, ∞) such that sup t≥0 g(t)e −αt < ∞. Now, from what precedes the function is a fixed point of Φ. Denote by ≤ the partial order on H 1 of point-wise domination: g ≤ f if for all t ≥ 0, g(t) ≤ f (t). The mapping Φ is non-decreasing on H 1 for this partial order. We notice that g 0 ≤ h and g 0 ≤ g 1 . Composing by Φ, we obtain: g 1 = Φ(g 0 ) ≤ Φ(h) = h and g 1 ≤ g 2 . By recursion, it follows for any n ≥ 1, g n ≤ h and (g n ) n≥0 is non-decreasing. By monotone convergence, for any t ≥ 0, the limit g(t) = lim n→∞ g n (t) exists and is bounded by h(t). Also, since g n ≤ h, by dominated convergence, for any t ≥ 0, lim n→∞ Φ(g n )(t) = Φ(g)(t). Therefore g solves the integral equation (5.2) and is equal to x − 1/(d − 1) where x is given by (5.5) (or (5.6) if λ = λ 1 ) for some a ≥ 0. However, from what precedes, we get x(t) ≤ h(t) + 1/(d − 1) and the only possibility is a = 0 and g(t) = h(t).
Finally, since Y (n) (t) is non-decreasing and converges to Y (t), by monotone convergence we have that f 1 (t) = EY (t) = lim n→∞ EY (n) (t) = g(t). This concludes the proof of Theorem 1.4.
where the summation is taken over n-tuples of integers that sum up to p. Taking power p in the above stochastic inequality and expanding brutally, we thus get where the summation is taken over N + 1-tuples of integers that sum up to p. Now we define f (κ) Taking expectation and using independence leads to Consider a n + 1-tuple that sums up to p such that for all i = 1, · · · , n, p i ≤ p − 1, n i=1 p i = q ≤ p and n i=1 1 1 pi≥1 = m ≤ p. From the recursive hypothesis and (5.9), with L = max 1≤k≤p−1 C k , we get Now recall that | ln(1 − y) + y| ≤ y 2 2(1−y) for y ∈ (0, 1). Since Then, grouping together all such n + 1-tuples, from (5.11) we deduce where we have used the hypothesis p < γ P . We may then apply Lemma 5.2: there exists a constant C p such that f (κ) The monotone convergence Theorem implies that f p (t) = lim κ→∞ f κ) p (t) exists and is bounded by C p e αpt . The recursion is complete.

Proof of Theorem 1.3 : lower bound on γ(λ)
To prove Theorem 1.3, we shall prove two statements If E λ [Z u ] < ∞ then u ≤ γ, (5.13) If 1 < u < min(γ, γ P ) then E λ [Z u ] < ∞. (5.14) In this paragraph, we prove (5.14). The argument is a refinement of the argument in §5.2. Let κ > 0 and let f Proof: The lemma is already proved if u is an integer in (5.12). The general case extends of the same argument. We write u = p + v with v ∈ (0, 1) and integer p ≥ 1. We use the inequality, for all y i ≥ 0, 1 ≤ i ≤ n, n i=1 y i u ≤ n i=1 p1,··· ,pn n p 1 · · · p n y pi+v i 1≤j≤n,j =i y pj j , (5.15) where the summation is taken over n-tuples of integers that sum up to p (which follows from the inequality ( y i ) v ≤ y v i and the multinomial formula). Then from (5.10), we get the stochastic domination j (t − ξ j + D i ) pj , (5.16) where the summation is taken over N + 1-tuples of integers that sum up to p. From Lemma 5.3, for all 1 ≤ q ≤ p, f q (t) ≤ C q e qαt . Note also, by Jensen inequality, that for all 1 ≤ q ≤ p − 1, f q+v (t) ≤ f p (t) q+v p ≤ C p e (q+v)αt . The same argument (with p replaced by u) which led to (5.12) in the proof of Lemma 5.3 leads to the result. Statement (5.14) is a consequence of Lemma 5.2 and Lemma 5.4. Indeed, by Lemma 5.2, for all t ≥ 0, f (κ) u (t) ≤ C u e uαt for some positive constant C u independent of κ. From the monotone convergence Theorem, we deduce that, for all t ≥ 0, f u (t) ≤ C u e uαt .
We have thus proven that with probability at least 1 − (q + ε) = p − 2ε, the root of T is never removed by the pruning algorithm. However, on the latter event, by construction T contains a (1 − q − 2ε)m -ary tree rooted at (note that 1 − q − 2ε = p − 3ε).