Unformatted text preview: 3 ) 2 + ( 1 3 ) 3 + ... = 1 2 < ∞ Therefore, State 0 is a transient state. ∑ ∞ n =0 P ( n ) 11 = 1 3 < ∞ Therefore, State 1 is a transient state. ∑ ∞ n =0 P ( n ) 22 = 0 + 1 + 0 + 1 + ... = ∞ Therefore, State 2 and 4 are recurrent states since they are in the same class. ∑ ∞ n =0 P ( n ) 33 = 0 < ∞ Therefore, State 3 is a transient state. ∑ ∞ n =0 P ( n ) 55 = 1 + 1 + 1 + 1 + ... = ∞ Therefore, State 5 is a recurrent state. iv. Given P (0) 00 = 1 ,f 00 = 0 Finding out P (2) 00 = 0 . 25 ,P (3) 00 = 0 . 125 ,P (4) 00 = 0 . 375 ,P (5) 00 = 0 . 2188 By solving P ( n ) ii = ∑ n k =0 f ( k ) ii P ( nk ) ii , n = 1 ,..., 5 f (1) 00 = 0 ,f (2) 00 = 0 . 25 ,f 3) 00 = 0 . 125 ,f (4) 00 = 0 . 3125 ,f (5) 00 = 0 . 1563 1...
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 Spring '11
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 Markov Chains, Probability, Markov chain, Andrey Markov, transient state

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