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Branching Processes Chapter 6 Branching processes Branching processes arise naturally in the study of stochastic processes on trees and locally tree-like graphs. After a review of the basic extinction theory of branching processes, we give a few classical examples of applications in discrete probability. 6.1 Background We begin with a review of the extinction theory of Galton-Watson branching pro- cesses. 6.1.1 Basic definitions Recall the definition of a Galton-Watson process. Definition 6.1. A Galton-Watson branching process is a Markov chain of the fol- Galton-Watson lowing form: process • Let Z0 := 1. • Let X(i, t), i 1, t 1, be an array of i.i.d. Z+-valued random variables ≥ ≥ with finite mean m = E[X(1, 1)] < + , and define inductively, 1 Zt := X(i, t). 1 i Zt 1 X − To avoid trivialities we assume P[X(1, 1) = i] < 1 for all i 0. ≥ Version: November 19, 2020 Modern Discrete Probability: An Essential Toolkit Copyright © 2020 Sebastien´ Roch 309 In words, Zt models the size of a population at time (or generation) t. The random variable X(i, t) corresponds to the number of offspring of the i-th individual (if there is one) in generation t 1. Generation t is formed of all offsprings of the − individuals in generation t 1. − X(1,1) We denote by pk k 0 the law of X(1, 1). We also let f(s):=E[s ] be { } ≥ the corresponding probability generating function. By tracking genealogical relationships, that is, who is whose child, we obtain a tree T rooted at the single individual in generation 0 with a vertex for each indi- vidual in the progeny and an edge for each parent-child relationship. We refer to T as a Galton-Watson tree. Galton-Watson A basic observation about Galton-Watson processes is that their growth is ex- tree ponential in t. t Lemma 6.2 (Exponential growth I). Let Mt := m− Zt. Then (Mt) is a nonneg- ative martingale with respect to the filtration t = σ(Z0,...,Zt). In particular, t F E[Zt]=m . Proof. Recall the following measure-theoretic lemma (see e.g. [Dur10, Exercise 5.1.1]). Lemma 6.3. Let (⌦, , P) be a probability space. If Y1 = Y2 a.s. on B then F 2F E[Y1 ]=E[Y2 ] a.s. on B. |F |F Returning to the proof, observe that on Zt 1 = k { − } E[Zt t 1]=E X(j, t) t 1 = mk = mZt 1. |F − 2 F − 3 − 1 j k X 4 5 This is true for all k. Rearranging shows that(Mt) is a martingale. For the second claim, note that E[Mt]=E[M0]=1. In fact, the martingale convergence theorem gives the following. Lemma 6.4 (Exponential growth II). We have Mt M < + a.s. for some ! 1 1 nonnegative random variable M σ( t t) with E[M ] 1. 1 2 [ F 1 Proof. This follows immediately from the martingale convergence theorem for nonnegative martingales (Corollary 3.36) and Fatou’s lemma. 310 6.1.2 Extinction Observe that 0 is a fixed point of the process. The event Z 0 = t : Z =0 , { t ! } {9 t } is called extinction. Establishing when extinction occurs is a central question in extinction branching process theory. We let ⌘ be the probability of extinction. Throughout, we assume that p0 > 0 and p1 < 1. Here is a first observation about extinction. Lemma 6.5. A.s. either Zt 0 or Zt + . ! ! 1 Proof. The process (Zt) is integer-valued and 0 is the only fixed point of the pro- cess under the assumption that p1 < 1. From any state k, the probability of never k coming back to k>0 is at least p0 > 0, so every state k>0 is transient. The claim follows. In the critical case, that immediately implies almost sure extinction. Theorem 6.6 (Extinction: critical case). Assume m =1. Then Zt 0 a.s., i.e., ! ⌘ =1. Proof. When m =1, (Zt) itself is a martingale. Hence (Zt) must converge to 0 by Lemma 6.4. Zt We address the general case using generating functions. Let ft(s)=E[s ]. Note that, by monotonicity, ⌘ = P[ t 0:Zt = 0] = lim P[Zt = 0] = lim ft(0), (6.1) t + t + 9 ≥ ! 1 ! 1 Moreover, by the Markov property, ft has a natural recursive form Zt ft(s)=E[s ] Zt = E[E[s t 1]] |F − Zt 1 = E[f(s) − ] (t) = ft 1(f(s)) = = f (s), (6.2) − ··· where f (t) is the t-th iterate of f. Theorem 6.7 (Extinction: subcriticial and supercritical cases). The probability of extinction ⌘ is given by the smallest fixed point of f in [0, 1]. Moreover: • (Subcritical regime) If m<1 then ⌘ =1. 311 • (Supercritical regime) If m>1 then ⌘<1. Proof. The case p0 + p1 =1is straightforward: the process dies almost surely after a geometrically distributed time. So we assume p0 + p1 < 1 for the rest of the proof. We first summarize some properties of f. Lemma 6.8. On [0, 1], the function f satisfies: (a) f(0) = p0, f(1) = 1; (b) f is indefinitely differentiable on [0, 1); (c) f is strictly convex and increasing; (d) lims 1 f 0(s)=m<+ . " 1 Proof. See e.g. [Rud76] for the relevant power series facts. Observe that (a) is clear by definition. The function f is a power series with radius of convergence R 1. This implies (b). In particular, ≥ i 1 i 2 f 0(s)= ip s − 0, and f 00(s)= i(i 1)p s − > 0, i ≥ − i i 1 i 2 X≥ X≥ because we must have pi > 0 for some i>1 by assumption. This proves (c). Since m<+ , f (1) = m is well defined and f is continuous on [0, 1], which 1 0 0 implies (d). We first characterize the fixed points of f. See Figure 6.1 for an illustration. Lemma 6.9. We have: • If m>1 then f has a unique fixed point ⌘ [0, 1). 0 2 • If m<1 then f(t) >tfor t [0, 1). Let ⌘ := 1 in that case. 2 0 Proof. Assume m>1. Since f (1) = m>1, there is δ>0 s.t. f(1 δ) < 1 δ. 0 − − On the other hand f(0) = p0 > 0 so by continuity of f there must be a fixed point in (0, 1 δ). Moreover, by strict convexity and the fact that f(1) = 1, if x (0, 1) − 2 is a fixed point then f(y) <yfor y (x, 1), proving uniqueness. 2 The second part follows by strict convexity and monotonicity. It remains to prove convergence of the iterates to the appropriate fixed point. See Figure 6.2 for an illustration. Lemma 6.10. We have: 312 Figure 6.1: Fixed points of f in subcritical (left) and supercritical (right) cases. Figure 6.2: Convergence of iterates to a fixed point. 313 • If x [0,⌘ ), then f (t)(x) ⌘ 2 0 " 0 • If x (⌘ , 1) then f (t)(x) ⌘ 2 0 # 0 Proof. We only prove 1. The argument for 2. is similar. By monotonicity, for x [0,⌘ ), we have x<f(x) <f(⌘ )=⌘ . Iterating 2 0 0 0 x<f(1)(x) < <f(t)(x) <f(t)(⌘ )=⌘ . ··· 0 0 So f (t)(x) L ⌘ . By continuity of f we can take the limit inside of " 0 (t) (t 1) f (x)=f(f − (x)), to get L = f(L). So by definition of ⌘0 we must have L = ⌘0. The result then follows from the above lemmas together with Equations (6.1) and (6.2). Example 6.11 (Poisson branching process). Consider the offspring distribution X(1, 1) Poi(λ) with λ>0. We refer to this case as the Poisson branching ⇠ process. Then i X(1,1) λ λ i λ(s 1) f(s)=E[s ]= e− s = e − . i! i 0 X≥ So the process goes extinct with probability 1 when λ 1. For λ>1, the probability of extinction ⌘λ is the smallest solution in [0, 1] to the equation λ(1 x) e− − = x. The survival probability ⇣ := 1 ⌘ satisfies 1 e ⇣λ = ⇣ . λ − λ − − λ J We can use the extinction results to obtain more information on the limit in Lemma 6.4. Of course, conditioned on extinction, M =0a.s. On the other 1 hand: Lemma 6.12 (Exponential growth III). Conditioned on nonextinction, either M = 1 0 a.s. or M > 0 a.s. In particular, P[M = 0] ⌘, 1 . 1 1 2{ } Proof. A property of rooted trees is said to be inherited if all finite trees satisfy this property and whenever a tree satisfies the property then so do all the descendant trees of the children of the root. The property M =0 is inherited. The result { 1 } then follows from the following 0-1 law. 314 Lemma 6.13 (0-1 law for inherited properties). For a Galton-Watson tree T , an inherited property A has, conditioned on nonextinction, probability 0 or 1. Proof. Let T (1),...,T(Z1) be the descendant subtrees of the children of the root. We use the notation T A to mean that the tree T satisfies A. Then, by indepen- 2 dence, (i) Z1 P[A]=E[P[T A Z1]] E[P[T A, i Z1 Z1]] = E[P[A] ]=f(P[A]), 2 | 2 8 | so P[A] [0,⌘] 1 . Also P[A] ⌘ because A holds for finite trees. 2 [{ } ≥ That concludes the proof.
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