midterm05sol - STAT455/855 Fall 2005 Applied Stochastic...

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Unformatted text preview: STAT455/855 Fall 2005 Applied Stochastic Processes Midterm, Brief Solutions 1. (15 marks) (a) (9 marks) (i) (1 mark) M (1) = 1, since the first step of the random walk will go to state 1 or state -1 with probability 1. (ii) (2 marks) Conditioning on the first move we obtain M ( N ) = 1 2 (1 + M ( N ) 1 ) + 1 2 (1 + M ( N )- 1 ) = 1 + 1 2 ( M ( N ) 1 + M ( N )- 1 ) . But M ( N )- 1 = M ( N ) 1 by spatial homogeneity so M ( N ) = 1 + M ( N ) 1 . (iii) (3 marks) For k = 1 ,...,N- 1 we condition on the first move to compute M ( N ) k , giving M ( N ) k = 1 + 1 2 ( M ( N ) k- 1 + M ( N ) k +1 ) 2 M ( N ) k = 2 + M ( N ) k- 1 + M ( N ) k +1 M ( N ) k- M ( N ) k- 1 = 2 + M ( N ) k +1- M ( N ) k This gives a simple recursion for M ( N ) k- M ( N ) k- 1 . Defining D ( N ) k = M ( N ) k- M ( N ) k- 1 and starting with k = 1, we have D ( N ) 1 = 2 + D ( N ) 2 = 2(2) + D ( N ) 3 = 2(3) + D ( N ) 4 . . . = 2( N- 1) + D ( N ) N , or M ( N ) 1- M ( N ) = 2( N- 1) + M ( N ) N- M ( N ) N- 1 . STAT 455/855 -- Midterm Exam Solutions, 2005 Page 2 of 4 Using M ( N ) 1- M ( N ) =- 1 from part(ii) and the boundary condition M ( N ) N = 0 this gives- 1 = 2( N- 1)- M ( N ) N- 1 or M ( N ) N- 1 = 1 + 2( N- 1) ....
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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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midterm05sol - STAT455/855 Fall 2005 Applied Stochastic...

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