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Lec14 - Stat 150 Stochastic Processes Spring 2009 Lecture...

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Stat 150 Stochastic Processes Spring 2009 Lecture 14: Branching Processes and Random Walks Lecturer: Jim Pitman Common setting: p 0 ,p 1 ,p 2 ,... probability distribution of X on { 0 , 1 , 2 ,... } . Two surprisingly related problems: Problem 1 : Use X as oﬀspring variable of a branching process Z 0 ,Z 1 ,Z 2 ,... . Consider the random variable Σ := n =0 Z n , the total progeny of the branching process. Describe the distribution of Σ given Z 0 = k . Problem 2 : Use X - 1 as the increment variable of a random walk S ( - ) n , n = 0 , 1 , 2 ,... on integer states. Let T 0 := inf { n : S ( - ) n } = 0. Describe the distribu- tion of T 0 given S ( - ) 0 = k . Theorem: These two problems have the same solution. Moreover, the solution is P (Σ = n | Z 0 = k ) = P ( T 0 = n | S ( - ) 0 = 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 0 = k ) = P ( T 0 < ∞| S ( - ) 0 = k ); Here (Σ < ) 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 = φ ( 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( λ ), S n d = Poisson( ), so P ( S n = n - k ) = e - ( ) 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

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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 (Σ = n | Z 0 = 1). Notice that Σ given Z 0 = k is the sum of k independent copies of Σ given Z 0 = 1, [ G ( s )] k = X n =0 s n P (Σ = n | Z 0 = k ) . First step analysis : Start from Z 0 = 1, condition on Z 1 = k for k = 0 , 1 , 2 ,... to see that | Z 0 = 1 ,Z 1 = k ) d = (1 + Σ | Z 0 = k ) which gives G ( s ) = X k =0 p k s [ G ( s )] k where s [ G ( s )] k is the GF of (1 + Σ | Z 0 = k ) = ( G ( s )) = G ( s ) φ ( G ( s )) = s Deﬁne F ( z ) := z φ ( z ) , then F ( G ( s )) = s . So G is the functional inverse of F ( z ).
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Lec14 - Stat 150 Stochastic Processes Spring 2009 Lecture...

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