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Unformatted text preview: Pstat160 Winter 2009 Homework #5 Solution Problem 1. 1. Let = 2 p 1. By conditioning on the first step, and using the fact that after the first step, the random walk is like a new random walk with one less step, we get E ( X n ) = E ( X n  up) p + E ( X n  down)(1 p ) = ( 1 + E ( X n 1 ) ) p + ( 1 + E ( X n 1 ) ) (1 p ) = 2 p 1 + E ( X n 1 ) = E ( X n 1 ) + Iterating this simple recursion for E ( X n 1 ) , E ( X n 2 ) , we get E ( X n ) = E ( X n 2 ) + + = = E ( X ) + n 2. a) A path is a succession of state that the random walk take. Path counting is the result of evaluating the number of such different paths with some specified property. In general, the properties contains the number of steps ( n ), the starting point, and the end point. More properties are often specified. b) Path properties: i) As stated in a) the minimal property requirement is specified starting point, end point, number of steps N n ( a,b ) = n n + b a 2 , n + b a even,  b a  n ii) paths that intersect the xaxis N n ( a,b ) = n n + b + a 2 , n + b + a even, b + a n iii) paths that start at 0 and never hit the xaxis afterwards N > n ( b ) = b n n n + b 2 (if n + b even, b n ) iv) paths that start at 0 hit the level a > 0 sometimes in the first n steps N a n ( b ) = N n (0 , 2 a b ) = ( n n +2 a b 2 )...
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This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.
 Spring '08
 EduardoMontoya

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