An algorithmic approach to branching processes with countably - PowerPoint PPT Presentation

An algorithmic approach to branching processes with countably infinitely many types Peter Braunsteins Supervisors: Sophie Hautphenne and Peter Taylor 29 June, 2016 1

Material of the talk The material of this talk is taken from S. Hautphenne, G. Latouche and G. Nguyen. Extinction probabilities of branching processes with countably infinitely many types. Advances in Applied Probability , 45(4) : 1068-1082, 2013. and P. Braunsteins, G. Decrouez, and S. Hautphenne. A pathwise iterative approach to the extinction of branching processes with countably many types. arXiv preprint arXiv :1605.03069 , 2016. 2

Multi-type Galton-Watson process Each individual has a type i ∈ S ≡ N The process initially contains a single individual of type ϕ 0 Each individual lives for a single generation At death individuals of type i have children according to the progeny distribution : p i ( r ) : r = ( r 1 , r 2 , . . . ), where p i ( r ) = probability that a type i gives birth to r 1 children of type 1, r 2 children of type 2, etc. All individuals are independent 3

Multi-type Galton-Watson process Population size : Z n = ( Z n 1 , Z n 2 , . . . ) , n ∈ N , where Z ni : # of individuals of type i at the n th generation In this example Z 3 = (0 , 1 , 1 , 2 , 1 , 0 , 0 , . . . ). { Z n } : ∞ -dim Markov process with state space ( N 0 ) ∞ and an absorbing state 0 = (0 , 0 , . . . ). 4

Multi-type Galton-Watson process Progeny generating vector G ( s ) = ( G 1 ( s ) , G 2 ( s ) , G 3 ( s ) , . . . ), where G i ( s ) is the progeny generating function of an individual of type i ∞ � � � p i ( r ) s r = s r k s ∈ [0 , 1] ∞ G i ( s ) = p i ( r ) k , k =1 r ∈ ( N 0 ) ∞ r ∈ ( N 0 ) ∞ Mean progeny matrix M with elements � � M ij = ∂ G i ( s ) � � ∂ s j s = 1 = expected number of direct offspring of type j born to a parent of type i There is a path from type i to j ⇔ there exists ℓ such that ( M ℓ ) ij > 0. 5

Global extinction probability Global extinction probability vector q = ( q 1 , q 2 , q 3 , . . . ), with entries � � � � ϕ 0 = i q i = P n →∞ Z n = 0 lim The vector q is the (componentwise) minimal nonnegative solution of s ∈ [0 , 1] ∞ s = G ( s ) , 6

Partial extinction probability Partial extinction probability vector � q = ( � q 1 , � q 2 , � q 3 , . . . ) , with � � � � ϕ 0 = i q i = P � ∀ ℓ : lim n →∞ Z n ℓ = 0 We have 0 ≤ q ≤ � q ≤ 1 . The vector � q also satisfies the fixed point equation s ∈ [0 , 1] ∞ s = G ( s ) , 7

Example 1 Suppose   1 / 6 , r = 3 e 1  p 1 ( r ) = 1 / 6 , r = 3 e 2   2 / 3 , r = 0 and   1 / 75 , r = 3 e i − 1     1 / 6 , r = 3 e i p i ( r ) =  1 / 6 , r = 3 e i +1     49 / 75 , r = 0 for i ≥ 2. 8

Example 1 The mean progeny matrix has entries M 11 = M 12 = 1 / 2 and M i , i − 1 = 1 / 25 , M i , i = M i , i +1 = 1 / 2 for i ≥ 2. Figure : A graphical representation of the mean progeny matrix. 9

Example 1 450 type 10 400 type 20 type 30 350 type 40 type 50 type 60 300 Population size total size 250 200 150 100 50 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 Generation Question : How to compute q and � q ? 10

Example 1 The progeny generating vector, G ( s ), has the form G 1 ( s ) = s 3 6 + s 3 6 + 2 1 2 3 G 2 ( s ) = s 3 75 + s 3 6 + s 3 6 + 49 1 2 3 75 . . . G i ( s ) = s 3 75 + s 3 6 + s 3 + 49 i − 1 i +1 i 6 75 . . . 11

Example 1 The fixed point equation, s = G ( s ), is s 1 = s 3 6 + s 3 6 + 2 1 2 3 s 2 = s 3 75 + s 3 6 + s 3 6 + 49 1 2 3 75 . . . s i = s 3 75 + s 3 6 + s 3 + 49 i − 1 i +1 i 6 75 . . . 12

Example 1 Take the first k elements of G ( s ) s 1 = s 3 6 + s 3 6 + 2 1 2 3 s 2 = s 3 75 + s 3 6 + s 3 6 + 49 1 2 3 75 . . . s i = s 3 6 + s 3 75 + s 3 + 49 i − 1 i +1 i 6 75 . . . s k = s 3 6 + s 3 75 + s 3 + 49 k − 1 k +1 k 6 75 13

Computing � q ( k ) Define { � Z n } by modifying { Z n } such that all types > k are sterile 14

Computing � q ( k ) q ( k ) : the (global) extinction probability of { � Denote � Z n } q ( k ) ց � � as k → ∞ q The proof is an application of the monotone convergence theorem q ( k ) can be computed, for instance using For each k , � functional iteration 15

Computing � q ( k ) ( s ), is In Example 1 the progeny generating vector, ˜ G ( s ) = s 3 6 + s 3 6 + 2 G ( k ) ˜ 1 2 1 3 ( s ) = s 3 75 + s 3 6 + s 3 6 + 49 G ( k ) ˜ 1 2 3 2 75 . . . ( s ) = s 3 6 + s 3 75 + s 3 + 49 G ( k ) ˜ i − 1 i +1 i i 6 75 . . . ( s ) = s 3 75 + s 3 6 + 1 6 + 49 G ( k ) ˜ k − 1 k k 75 16

Computing q Define { Z ( k ) n } by modifying { Z n } such that all types > k are replaced by an immortal type ∆ 17

Computing q Denote q ( k ) : the (global) extinction probability of { Z ( k ) n } q ( k ) ր q as k → ∞ The proof is again an application of the monotone convergence theorem For each k , q ( k ) can be computed, for instance using functional iteration 18

Computing q In Example 1 the progeny generating vector, G ( k ) ( s ), is ( s ) = s 3 6 + s 3 6 + 2 G ( k ) 1 2 1 3 ( s ) = s 3 75 + s 3 6 + s 3 6 + 49 G ( k ) 1 2 3 2 75 . . . ( s ) = s 3 6 + s 3 75 + s 3 + 49 G ( k ) i − 1 i +1 i i 6 75 . . . ( s ) = s 3 75 + s 3 6 + 0 + 49 G ( k ) k − 1 k k 75 19

Random replacement Z ( k ) Define { ¯ n } by modifying { Z n } such that All types > k are replaced by a type in { 1 , 2 . . . , k } The types of the replaced individuals are selected independently using the probability distribution � � α ( k ) = α ( k ) 1 , α ( k ) 2 , . . . , α ( k ) 3 For example α ( k ) = e 1 : replacement by type 1 α ( k ) = 1 / k : replacement by a type uniformly distributed on { 1 , . . . , k } α ( k ) = e k : replacement by type k q ( k ) : the (global) extinction probability of { ¯ Z ( k ) Denote ¯ n } 20

Random replacement An illustration when α ( k ) = e 1 21

Random replacement ( k ) ( s ), is In Example 1 the progeny generating vector, ¯ G ( s ) = s 3 6 + s 3 6 + 2 G ( k ) ¯ 1 2 1 3 ( s ) = s 3 75 + s 3 6 + s 3 6 + 49 G ( k ) ¯ 1 2 3 2 75 . . . ( s ) = s 3 6 + s 3 75 + s 3 + 49 G ( k ) ¯ i − 1 i +1 i i 6 75 . . . �� k � 3 ℓ =1 α ( k ) ℓ s ℓ ( s ) = s 3 75 + s 3 + 49 G ( k ) ¯ k − 1 k 6 + k 6 75 22

Random replacement What conditions on { Z n } and { α ( k ) } are required for q ( k ) → q ¯ as k → ∞ ? 23

Assumptions Assumption (1) i ∈S q i > 0 inf Assumption (2) There exists constants N 1 , N 2 ≥ 1 and a > 0 , all independent of k, such that min { N 1 , k } � α ( k ) ≥ a i i =1 for all k ≥ N 2 . 24

Main result Theorem Suppose Assumptions 1 and 2 hold. In addition, assume that there exists N 1 such that either ˜ q j < 1 for all j ∈ { 1 , . . . , N 1 } , or ˜ q j = 1 for all j ∈ { 1 , . . . , N 1 } , and there is a path from any j ∈ { 1 , . . . , N 1 } to the initial type i. Then q ( k ) k →∞ ¯ lim → q i i for any initial type i. 25

Coupling of the branching processes ( k ) We place { Z n } , { Z ( k ) Z ( k ) n } , { � n } , and { ¯ Z n } on the same probability space, for all k ≥ 1. 26

Coupling of the branching processes ( k ) We place { Z n } , { Z ( k ) Z ( k ) n } , { � n } , and { ¯ Z n } on the same probability space, for all k ≥ 1. 27

Example 1 28

Example 2 Consider the branching process with progeny generating function G ( s ) such that a , c > 0, d > 1 and G 1 ( s ) = cd 2 + 1 − cd t s t t , and for i ≥ 2,  cd i +1 + ad i − 1 + 1 − d ( a + c )   u s u u s u when i is odd, u G i ( s ) = c i +1 + a i − 1 + 1 − ( a + c )   dv s v dv s v when i is even, dv where t = ⌈ dc ⌉ + 1, u = ⌈ d ( c + a ) ⌉ + 1 and v = ⌈ ( c + a ) / d ⌉ + 1. 29

Example 2 When i ≥ 2 the mean progeny matrix M has entries, M i , i − 1 = ad and M i , i +1 = cd for i odd and M i , i − 1 = a / d and M i , i +1 = c / d for i even. 30

Example 2 a = 1 / 6, c = 7 / 8 and d − 1 = 0 . 95 a = 1 / 6, c = 7 / 8 and d − 1 = 0 . 93 31

Concluding remarks Example 2 demonstrates that when α ( k ) = e k , lim k →∞ ¯ q ( k ) does not necessarily exist For this example we can prove that when α ( k ) = e k for any a , c > 0 and d > 1, q ( k ) = q . lim inf k →∞ ¯ Under Assumption 1 we believe this to be true in general. When α ( k ) = 1 / k , we can construct an example where q ( k ) = � q < lim k →∞ ¯ q . 32

Questions ? 33