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# HW3solns - March 8, 2007 HW3 Problem Solutions #2.6.4. What...

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Unformatted text preview: March 8, 2007 HW3 Problem Solutions #2.6.4. What we must show (since the finite constant factor 1 / ( A ) 6 = 0 does not affect the validity of the assertion) is: i, j A : i q ij = j q ji The equation holds trivially when i = j , and it follows immediately from the corresponding equation for P when i 6 = j . #3.1.5. Begin by defining G ( N ) = N n =0 P n , and then observing, using first-step analysis, that G ( N ) ij = I [ i = j ] + X k S P ik G ( N- 1) kj from which you can take the monotone nondecreasing extended-real-valued lim- its to prove the equation G = I + P G . By definition of G = lim N G ( N ) , it is easy to see that P G = GP , where the entries on both sides may be + . Therefore G = I + GP . However, if all entries of G are finite and the number | S | of elements in the state-space is finite, then this equation is equivalent to G ( I- P ) = I , which leads to a contradiction since post-multiplying by the vector 1 containing all 1s gives the zero-vector on the left-hand side but 1 on the right. Therefore the finite-by-finite matrix G has at least some infinite entries G ij , which implies that the corresponding state j is recurrent, and in the irreducible case this means that all states are recurrent. #3.2.1. As mentioned in the correction/hint, we do this verification only for n 1: by definition, for i 6 = 0, o p oi ( n + 1) = P ( k : k 6 =0 { X n = k, X n +1 = i } [ X 1 , . . . , X n- 1 6 = 0]) which by the Markov property is equal to = X k : k 6 =0 p k ( n ) P ki ....
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## This note was uploaded on 09/09/2009 for the course STAT 650 taught by Professor Slud during the Spring '09 term at Maryland.

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HW3solns - March 8, 2007 HW3 Problem Solutions #2.6.4. What...

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