This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Stat 150 Stochastic Processes Spring 2009 Lecture 14: Branching Processes and Random Walks Lecturer: Jim Pitman Common setting: p ,p 1 ,p 2 ,... probability distribution of X on { , 1 , 2 ,... } . Two surprisingly related problems: Problem 1 : Use X as offspring variable of a branching process Z ,Z 1 ,Z 2 ,... . Consider the random variable := n =0 Z n , the total progeny of the branching process. Describe the distribution of given Z = k . Problem 2 : Use X 1 as the increment variable of a random walk S ( ) n , n = , 1 , 2 ,... on integer states. Let T := inf { n : S ( ) n } = 0. Describe the distribu tion of T given S ( ) = k . Theorem: These two problems have the same solution. Moreover, the solution is P ( = n  Z = k ) = P ( T = n  S ( ) = k ) = k n P ( S n = n k ) where S n := X 1 + + X n is the sum of n independent copies X i d = X . Notes : P ( <  Z = k ) = P ( T <  S ( ) = k ); Here ( < ) is the event of extinction of the branching process, which given k = 1 has probability which is the least root s [0 , 1] of s = ( s ), and for general k is the k th power of the probability for k = 1. . The formula is very explicit for any distribution of X for which there is a simple for the distribution of S n . e.g. for X d = Poisson( ), S n d = Poisson( n ), so P ( S n = n k ) = e n ( n ) n k ( n k )! Approach both problems with the technique of probability generating func tions.. This will show both solutions are the same. Then we can pick which one 1 Lecture 14: Branching Processes and Random Walks 2 to work with to establish the formula. Branching process problem: Introduce G.F. G ( s ) := n =0 s n P ( = n  Z = 1). Notice that given Z = k is the sum of k independent copies of given Z = 1, [ G ( s )] k = X n =0 s n P ( = n  Z = k ) . First step analysis : Start from Z = 1, condition on Z 1 = k for k = 0 , 1 , 2 ,... to see that (  Z = 1 ,Z 1 = k ) d = (1 +  Z = k ) which gives G ( s ) = X n =0 p k s [ G ( s )] k where s [ G ( s )] k is the GF of (1 +  Z = k ) = s ( G ( s )) = G ( s ) ( G ( s )) = s Define F ( z ) := z ( z ) , then F ( G ( s )) = s . So G is the functional inverse of F ( z )....
View Full
Document
 Spring '09
 Pitman,Jim

Click to edit the document details