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Consider a symmetric random walk, starting at the origin. Let f) be the prob- ability of the first time of reaching to position r = 1 being at time n. i. Calculate f(2). ii. It can be shown that 1 f(2+1) (2n)! 2(2n+1 n(n+1)! Using without proof that the probability generating function is of the form Gi(s) = Σf{2n+1) g2n+1, n=0 show that G₁(s) = [1-(1-8²)]/s You may wish to use the equality I f(2n+1) 1 (2n)! 22n+1 n(n+1)! =(-1)" -1)" (+1). (n=0,1,2,...). iii. Calculate G₁(1). What does this tell us about the random walk's future behaviour? =

Consider a symmetric random walk, starting at the origin. Let f) be the prob- ability of the first time of reaching to position r = 1 being at time n. i. Calculate f(2). ii. It can be shown that 1 f(2+1) (2n)! 2(2n+1 n(n+1)! Using without proof that the probability generating function is of the form Gi(s) = Σf{2n+1) g2n+1, n=0 show that G₁(s) = [1-(1-8²)]/s You may wish to use the equality I f(2n+1) 1 (2n)! 22n+1 n(n+1)! =(-1)" -1)" (+1). (n=0,1,2,...). iii. Calculate G₁(1). What does this tell us about the random walk's future behaviour? =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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(c) Consider a symmetric random walk, starting at the origin. Let f(n) be the prob- ability of the first time of reaching to position r = 1 being at time n. i. Calculate f(2) ii. It can be shown that 1 (2n)! f(2n+1) 2(2n+1 n(n+1)! Using without proof that the probability generating function is of the form Gi(s) = Σf{2n+1) g2n+1, n=0 show that G₁(s) = [1-(1-8²)]/s You may wish to use the equality I 1 f(2n+1) (2n)! = (n = 0, 1, 2,...) 22n+1 230+ m²(n + 1)! = (-1)^(n+1). iii. Calculate G₁(1). What does this tell us about the random walk's future behaviour? =
Transcribed Image Text:(c) Consider a symmetric random walk, starting at the origin. Let f(n) be the prob- ability of the first time of reaching to position r = 1 being at time n. i. Calculate f(2) ii. It can be shown that 1 (2n)! f(2n+1) 2(2n+1 n(n+1)! Using without proof that the probability generating function is of the form Gi(s) = Σf{2n+1) g2n+1, n=0 show that G₁(s) = [1-(1-8²)]/s You may wish to use the equality I 1 f(2n+1) (2n)! = (n = 0, 1, 2,...) 22n+1 230+ m²(n + 1)! = (-1)^(n+1). iii. Calculate G₁(1). What does this tell us about the random walk's future behaviour? =
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