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Unformatted text preview: Summary of the branching process The branching process is the first example of simple stochastic processes. Suppose we have cells that are capable of producing cells of like kind. a117 Z = 1 a16 a16 a16 a16 a16 a16 a16 a16 a16 a0 a0 a0 a64 a64 a64 a80 a80 a80 a80 a80 a80 a80 a80 a80 a117 a117 a117 a117 Z 1 = 4 a8 a8 a8 a8 a8 a8 a0 a0 a0 a64 a64 a64 a117 a117 a117 a117 a117 Z 2 = 5 a8 a8 a8 a8 a8 a8 a0 a0 a0 a0 a0 a0 a64 a64 a64 a72 a72 a72 a72 a72 a72 a0 a0 a0 a117 a117 a117 a117 a117 a117 a117 a117 Z 3 = 8 a114 a114 a114 a114 a114 a114 a114 a114 ♣ Assumptions (a) Each cell has the probability of p j of producing exactly j new cells, j = 0 , 1 , 2 , . . . , ∑ ∞ j =0 p j = 1. Note that { p , p 1 , . . ., } = { p j } ∞ j =0 is called the family size distribution. (b) All cells act independently. ♣ Notations (a) Let Z n =# of cells in the n th generation; Z = 1 for simplicity. (b) Let X ni =# of cells produced by ith cell in the n th generation. Then Z n +1 = Z n summationdisplay i =1 X ni . We assume i. X n 1 , X n 2 , ··· are iid r.v.s; ii. P ( X ni = j ) = p j , j = 0 , 1 , 2 , . . . and ∑ ∞ j =0 p j = 1 iii. Z n and X ′ ni s are independent. (c) Let θ = ∑ ∞ j =0 jp j and σ 2 = ∑ ∞ j =0 ( j − θ ) 2 p j , which are the mean and the variance of family size distribution. In general, we are interested in three things: (a) the expectation of population size ( μ n = E ( Z n )) in the n th generation; (b) the variance of the population size ( σ 2 n = V ar ( Z n )) in the n th generation; (c) the extinct probability by the n th generation ( q n = p ( X n = 0)) and the probabil ity that the population eventually become extinct q = lim n →∞ q n . 1 STAT333 ♣ Mean and Variance of Z n Note that Z n +1 = ∑ Z n i =1 X ni , which is the random sum of iid random variables. So using the formula for the expectation and the fact that E ( Z ) = 1 or μ = 1, E ( Z n +1 ) = E ( Z n ) E ( X n 1 ) ⇒ μ n +1 = μ n θ, n ≥ ⇒ μ n = θ n , n ≥ . Using the formula for the variance, V ar ( Z n +1 ) = E ( Z n ) V ar ( X n 1 ) + V ar ( Z n ) E 2 ( X n 1 ) ⇒ σ 2 n +1 = μ n σ 2 + σ 2 n θ 2 , n ≥ . Thus σ 2 n +1 = θ n σ 2 + θ 2 σ 2 n , n ≥ ....
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This note was uploaded on 01/19/2011 for the course STATISTICS STAT 333 taught by Professor Menzhongxian during the Fall '10 term at Waterloo.
 Fall '10
 MenZhongxian
 Probability

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