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branching_process - STAT333 Summary of the branching...

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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 0 = 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 0 , 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 0 = 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
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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 0 ) = 1 or μ 0 = 1, E ( Z n +1 ) = E ( Z n ) E ( X n 1 ) μ n +1 = μ n θ, n 0 μ n = θ n , n 0 . 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 0 . Thus σ 2 n +1 = θ n σ 2 + θ 2 σ 2 n , n 0 . Note σ 2 0 = 0, we find σ 2 n = braceleftBigg θ n - 1 (1 θ n ) 1 θ σ 2 θ negationslash = 1 2 θ = 1 , n = 1 , 2 , . . . , which can be established by inductive argument.
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