sol3 - STA3007 Applied probability (08-09) Chapter 3...

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Applied probability (08-09) Solution to Assignment 3 Chapter 3 Problem 1.1 The state space is { 0 , 1 , 2 , 3 , 4 , 5 } .It is obvious that P 00 = P 55 = 1. If there are k ( k = 1 , 2 , 3 , 4) people who are diseased, the number of pairs of individuals in which the disease transmission may take place(one of these persons is diseased and the other not) is C k 1 C 5 - k 1 = k (5 - k ), and the number of the total pairs of individuals is C 5 2 = 10. Since the probability that the disease is transmitted to the healthy people is α = 0 . 1, thus the ( k, k + 1)th entry of the transition matrix P is P k,k +1 = α k (5 - k ) 10 , and the ( k, k )th entry P k,k = 1 - P k,k +1 = 1 - α k (5 - k ) 10 . Other entries of P are 0. Therefore, the transition matrix is as follows: P = 1 0 0 0 0 0 0 0 . 96 0 . 04 0 0 0 0 0 0 . 94 0 . 06 0 0 0 0 0 0 . 94 0 . 06 0 0 0 0 0 0 . 96 0 . 04 0 0 0 0 0 1 . Problem 1.4 The state space is { 0 , 1 , 2 , 3 } . Since X n +1 = max { ξ 1 , · · · , ξ n +1 } = max { X n , ξ n +1 } , thus Pr( X n +1 = j | X n = i ) = 0 , if j < i, i k =0 Pr { ξ n +1 = k } , if j = i, Pr { ξ n +1 = j } , if j > i. 1
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This note was uploaded on 02/16/2011 for the course ECON 1224 taught by Professor Afdsa during the Spring '11 term at CUHK.

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sol3 - STA3007 Applied probability (08-09) Chapter 3...

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