# ass3sol - STAT455/855 Fall 2009 Applied Stochastic...

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Unformatted text preview: STAT455/855 Fall 2009 Applied Stochastic Processes Assignment #3, Solutions Total Marks: 44 for 455 and 50 for 855. 1. From Sheet. (7 marks) (a) (3 marks) Conditioning on X we have P ( X n = j ) = X i S P ( X n = j X = i ) P ( X = i ) = X i S p ij ( n ) P ( X = i ) . Therefore, lim n P ( X n = j ) = lim n X i S p ij ( n ) P ( X = i ) = X i S P ( X = i ) lim n p ij ( n ) = X i S P ( X = i ) j = j X i S P ( X = i ) = j , where taking the limit inside the sum in the second equality is justified by the bounded convergence theorem and the third equality follows from our result in class that lim n p ij ( n ) = j . The result follows since (also from class) j = 1 / j . (b) (4 marks) We first show that if a sequence { x n } n =1 has a limit x (i.e., lim n x n = x ) then lim n 1 n n X i =1 x i = x as well (i.e., the limit of partial averages converges to x ). Let &gt; 0 be given. Let N be such that | x n- x | &lt; / 2 for all n &gt; N , and let N 1 = 2 N X i =1 | x i- x | . STAT 455/855 -- Assignment 3 Solutions p. 2 Then for n &gt; N , max( N ,N 1 ), we get 1 n n X i =1 x i- x = 1 n n X i =1 ( x i- x ) 1 n n X i =1 | x i- x | = 1 n N X i =1 | x i- x | + 1 n n X i = N +1 | x i- x | &lt; 1 N 1 N X i =1 | x i- x | + 1 n n X i = N +1 2 = 2 + n- N n 2 &lt; 2 + 2 = . Therefore 1 n n i =1 x i x as n . Now letting x n = P ( X n = j ) and x = j we have that x n x as n from part(a) (since j = 1 / j ) and so the desired result follows directly. 2. From Sheet. (6 marks) (a) (3 marks) If X has period d &gt; 1 then suppose C ,...,C d- 1 are the cyclic classes. Let ( i,j ) be a state in the Z chain where i and j are in the same cyclic class. If ( k,` ) is a state of the Z chain such that k and ` are in different cyclic classes, then if Z starts in state ( i,j ) it will never reach state ( k,` ). Therefore ( i,j ) and ( k,` ) do not communicate and so the Z chain cannot be irreducible. (b) (3 marks) Consider state ( i,j ) in the Z chain. Let T ij denote the first time the Z chain returns to state ( i,j ), starting in ( i,j ), and let T i denote the first time the X chain returns to state i , starting in i . If the Z chain returns to ( i,j ) this implies that the X chain has returned to state i so the Z chain cannot return to ( i,j ) before the X chain returns to i . Thus T ij T i . If the X chain is null recurrent then E [ T i ] = and therefore E [ T ij ] = . Therefore, state ( i,j ) cannot be positive recurrent. STAT 455/855 -- Assignment 3 Solutions p. 3 3. Ross, Chapter 4, #74. (10 marks) (a) (4 marks) We set the state space S to be the set of all permutations of { 1 , 2 ,...,n } ....
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## This note was uploaded on 09/15/2010 for the course STAT 455/855 taught by Professor Glentakahara during the Fall '09 term at Queens University.

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ass3sol - STAT455/855 Fall 2009 Applied Stochastic...

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