Lec14[1]

# Lec14[1] - Stat 150 Stochastic Processes Spring 2009...

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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 Σ := ∑ ∞ n =0 Z n , the total progeny of the branching process. Describe the distribution of Σ 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 (Σ = 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 (Σ < ∞| Z = k ) = P ( T < ∞| S (- ) = 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( nλ ), so P ( S n = n- k ) = e- nλ ( nλ ) 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 (Σ = n | Z = 1). Notice that Σ given Z = k is the sum of k independent copies of Σ given Z = 1, [ G ( s )] k = ∞ X n =0 s n P (Σ = n | Z = k ) . First step analysis : Start from Z = 1, condition on Z 1 = k for k = 0 , 1 , 2 ,... to see that (Σ | Z = 1 ,Z 1 = k ) d = (1 + Σ | 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 + Σ | Z = k ) = sφ ( 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[1] - Stat 150 Stochastic Processes Spring 2009...

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