midterm09sol - STAT 455/855 Midterm Solutions Fall 2009...

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STAT 455/855 – Midterm Solutions Fall, 2009 1. (15 marks) (a) (6 marks) The chain is irreducible (so only 1 equivalence class). It is recurrent and the period is 2. The cyclic classes are C 0 = { 0 , 5 } and C 1 = { 1 , 2 , 3 , 4 , 6 } . (b) (4 marks) Let M 01 denote the mean number of steps to get to state 1 starting in state 0. Conditioning on the first step we have M 01 = 1 4 (1) + 1 4 (2 + M 01 ) + 1 4 (2 + M 01 ) + 1 4 (4 + M 01 ) , where the four terms on the right above come from the first move being to state 1, 2, 3, or 4, respectively. Therefore, M 01 = 9 4 + 3 4 M 01 or M 01 = 9 . (c) (5 marks) The equations for π j , j = 1 , . . . , 6 from π = π P are π 1 = π 2 = π 3 = π 4 = 1 4 π 0 and π 6 = π 5 = π 4 . These give π j = 1 4 π 0 for j = 1 , . . . , 6. The normalization constraint then gives π 0 (1 + 6 / 4) = 1, or π 0 = 0 . 4. Then π 1 = π 2 = π 3 = π 4 = π 5 = π 6 = 0 . 1. 2. (15 marks) (a) (7 marks) We have p 00 ( n ) = 1 if n is even 0 if n is odd p 01 ( n ) = p if n is odd 0 if n is even p 02 ( n ) = q if n is odd 0 if n is even. This gives P 00 ( s ) = n =0 s 2 n = 1 1 - s 2 , (1)
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STAT 455/855 -- Solutions: Midterm p. 2 P 01 ( s ) = n =0 ps 2 n +1 = ps n =0 s 2 n = ps 1 - s 2 , (2) and similarly P
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