Hw2-ORIE3510_S12_solns

Hw2-ORIE3510_S12_solns - ORIE 3510 Homework 2 Solutions...

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ORIE 3510 – Homework 2 Solutions Instructor: Mark E. Lewis due Wednesday February 8, 2012 (ORIE Hallway drop box) 1. Diagram for each Markov chain with the given transition matrix: 0.7 1/4 1/3 i) iii) 0.3 1 2 0.5 1 1/4 2 0.5 1/4 1/4 1/3 0.25 1/3 ii) 3 0.25 3 4 0.5 1/2 0.5 1/2 1 11 2 0.5 2. First, note that each X n can take value in { 0 , 1 , 2 , 3 , 4 } . Since the coins have equal probability of showing head, we have that X n +1 = ( X n + B (4 - X n , 0 . 4) if X n 3, B (4 , 0 . 4) if X n = 4, where B ( n,p ) denotes a binomial random variable with parameters n (number of trials) and p (probability of success). From this, we can see that the distribution of X n +1 given X 0 ,...,X n only depends on X n , i.e. P ( X n +1 = x n +1 | X 0 = x 0 n = x n ) = P ( X n +1 = x n +1 | X n = x n ) So { X n ,n 1 } is a Markov chain with state space { 0 , 1 , 2 , 3 , 4 } . Let b ( k ; ) = P ( B ( ) = k ) = ± n k ² p k (1 - p ) n - k , then the transition matrix has the folowing entries p ij = b ( j - i ; 4 - i, 0 . 4) if j i and i 3, b ( j ; 4 , 0 . 4) if i = 4, 0 otherwise , 1
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where 0 i,j 4. We have that the transition matrix is P = 0 . 1296 0 . 3456 0 . 3456 0 . 1536 0 . 0256 0 . 0000 0 . 2160 0 . 4320 0 . 2880 0 . 0640 0 . 0000 0 . 0000 0 . 3600 0 . 4800 0 . 1600 0 . 0000 0 . 0000 0 . 0000 0 . 6000 0 . 4000 0 . 1296 0 . 3456 0 . 3456 0 . 1536 0 . 0256 3. First, note that when
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This note was uploaded on 03/08/2012 for the course ORIE 3510 taught by Professor Resnik during the Spring '09 term at Cornell.

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Hw2-ORIE3510_S12_solns - ORIE 3510 Homework 2 Solutions...

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