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Unformatted text preview: STAT455/855 Fall 2005 Applied Stochastic Processes Final Exam, Brief Solutions 1. (15 marks) (a) (5 marks) If there is one red and one blue ball in the urn, then with probability 1/2 one draw will be required and with probability 1/2 two draws will be required. Thus, T 1 , 1 = 1 2 (1) + 1 2 (2) = 3 2 . If there is one red ball and n > 1 blue balls in the urn then the red ball is equally likely to drawn k th, for k = 1 , . . . , n + 1. If it is drawn k th then k draws are required. Thus, T 1 ,n = n +1 X k =1 k 1 n + 1 = ( n + 1)( n + 2) 2( n + 1) = n + 2 2 . (b) (5 marks) If there are m > 1 red balls and one blue ball in the urn, then then m + 1 draws will be required unless the blue ball is drawn last, which occurs with probability 1 / ( m + 1), in which case m draws will be required. So T m, 1 = m m + 1 ( m + 1) + 1 m + 1 ( m ) = m ( m + 2) m + 1 . (c) (5 marks) Conditioning on the first draw, we have T m,n = m m + n (1 + T m- 1 ,n ) + n m + n (1 + T m,n- 1 ) . (1) Based on the answers from parts (a) and (b), a general form for T m,n that reduces to T 1 ,n and T m, 1 is T m,n = m ( m + n + 1) m + 1 . (2) We verify this guess by checking that is satisfies the recursion given in (1). Plug- ging (2) into the right hand side of (1) we get m m + n 1 + ( m- 1)( m + n ) m + n n + m 1 + m ( m + n ) m + 1 = 1 + m- 1 + nm m + 1 = m ( m + 1) + nm m + 1 = m ( m + n + 1) m + 1 , which is Eq.(2). STAT 455/855 -- Final Exam Solutions, 2005 Page 2 of 5 2. (15 marks) (a) (2 marks) The Markov chain is finite, therefore there is at least one positive recurrent state. It is irreducible, therefore all states are positive recurrent. This implies a stationary distribution exits (and it is unique).implies a stationary distribution exits (and it is unique)....
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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.
- Fall '09