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EE465-USC-Fall08-Assignment6_Solution

EE465-USC-Fall08-Assignment6_Solution - EE 465 Homework 6...

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EE 465 : Homework 6 Solutions 1. a) Since all the states can communicate due to irreducibility, they are all recurrent and hence probability of ever visting state j starting at any state i is 1. b) x 0 = 0 , x N = 1 x i = N - 1 j =1 P ij x j + P iN i = 1 , . . . , N - 1 c) Check if x i = i N is a solution, x i = N - 1 X j =1 P ij j N + P iN = N X j =0 P ij j N = 1 N N X j =0 jP ij = i N 2. a) Yes, the next state depends only on the present, and not on the past. b) The whole chain is one class. Their periods are 1. All states are recurrent. c) Note that each transition, X n will either increase by 1, decrease by 1 or stay the same, which implies P ij = 0, if | i - j |≥ 2. P i,i +1 = p N - i N , i = 0 , 1 , . . . , N - 1 P i,i - 1 = (1 - p ) i N , i = 1 , 2 , . . . , N P i,i = p i N + (1 - p ) N - i N , i = 0 , 1 , . . . , N d) For n = 2, we have the following probability transition matrix using the equations in part (c), P = 1 - p p 0 1 - p 2 1 2 p 2 0 1 - p p Using Π = Π P , we have Π 0 = (1 - p 0 + 1 - p 2 Π 1 Π 1 = 2 p 1 - p Π 0 Π 2 = p 2 Π 1 + p Π 2 Π 2 = p 2(1 - p ) Π 1 = p 2 (1 - p ) 2 Π 0 Π 0 + Π 1 + Π 2 = 1 Π 0 ˆ 1 + 2 p 1 - p + p 2 (1 - p ) 2 ! = 1 Π 0 = (1 - p ) 2 , Π 1 = 2 p (1 - p ) , Π 2 = p 2 . e) Based on the result of part (d), we can say that the stationary distribution is binomial, that is Π i = ˆ N i !
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