HW 4 - P X n& 1 = i n& 1;X = i j X n = i 3 Let X = f X n g be a Markov chain with state space S and transition probability matrix P = P ij

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APMA 1200, 2010: Assignment #4 1. Assume that S = f S n g is a simple random walk starting at the origin. That is, S n = n X i =1 Y i ; S 0 = 0 where Y 1 ;Y 2 ;: :: are iid random variables with P ( Y i = 1) = p; P ( Y i = 1) = q = 1 p: (a) For every 0 ± k such that 2 k ± n , argue that P ( S 1 ² 0 ;S 2 ² 0 ; ³ ³ ³ ;S n 1 ² 0 j S n = n 2 k ) = n 2 k + 1 n k + 1 (b) Suppose n = 2 m is an even number. Argue that P ( S 1 6 = 0 ;S 2 6 = 0 ;: :: ;S 2 m 1 6 = 0 ;S 2 m = 0) = 2 m 2 m 2 m 1 ± p m q m : Hint: For part (a), imagine the random walk as if it started from ( 1 ; 1) in class. For part (b), use part (a) to compute the NUMBER of paths that satis±es S 1 6 = 0 ;: :: ;S n 1 6 = 0 ;S n = 0 . And note that each of these path has probability p m q m . 2. Argue that for any Markov chain X = f X n : n ² 0 g , P ( X n + k = j k ;: :: ;X n +1 = j 1 ;X n 1 = i n 1 ;: :: ;X 0 = i 0 j X n = i ) = P ( X n + k = j k ;: :: ;X n +1 = j 1 j X n = i ) ³
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Unformatted text preview: P ( X n & 1 = i n & 1 ;: :: ;X = i j X n = i ) 3. Let X = f X n g be a Markov chain with state space S and transition probability matrix P = [ P ij ] . Let Y be a Markov chain with state space W and transition probability matrix Q = [ Q kl ] . Furthermore, assume that X and Y are independent. Then the two-dimensional process Z = f Z n g with Z n : = ( X n ;Y n ) is also a Markov chain (you do not need to prove this, but at least try to understand it intuitively). (a) Identify the state space of Z . (b) Identify the transition probability matrix of Z . 1...
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This note was uploaded on 02/24/2010 for the course APMA 1200 taught by Professor Roz during the Spring '10 term at Brown.

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