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Unformatted text preview: Math136/Stat219 Fall 2009 Sample Final Examination Write your name and sign the Honor code in the blue books provided. You have 3 hours to solve all questions, each worth points as marked (maximum of 100). Complete reasoning is required for full credit. You may cite lecture notes and homework sets, as needed, stating precisely the result you use, why and how it applies. Important note: If you wish to use a result that is contained in an Exercise in the course notes that was not assigned for homework or proved in lecture, you must prove the result yourself (i.e. you cannot just site the Exercise number.) 1. (5x5) Consider a simple, symmetric random walk. That is, let ξ 1 ,ξ 2 ,... be a sequence of independent random variables with P ( ξ k = 1) = P ( ξ k = 1) = 1 / 2 for all k . Let S = 0 and S n = ∑ n k =1 ξ k for n = 1 , 2 ,... . Let F n = σ ( ξ 1 ,...,ξ n ). For fixed positive integers a and b let τ = min { n ≥ 0 : S n / ∈ ( a,b ) } . You may assume that τ < ∞ a.s. a) Show that τ is an {F n }stopping time. ANS: { τ = n } = { S 1 ∈ ( a,b ) ,...,S n 1 ∈ ( a,b ) ,S n / ∈ ( a,b ) } ∈ F n since F n = σ ( S 1 ,...,S n ) (because S n is defined through an invertible transformation of ξ 1 ,...,ξ n ; see Corollary 1.2.17). This is enough to show that τ is a discrete time {F n }stopping time. b) Show that { S 2 n n } is an {F n }martingale. ANS: Clearly, {F n } is a filtration to which { S 2 n n } is adapted. Checking integrability, since ξ k ∈ { 1 , 1 } , we have  S n  ≤ n and so IE  S 2 n n  ≤ n 2 + n < ∞ for any fixed n . Finally using the independence and distributions of ξ n : IE [ S 2 n +1 ( n + 1) F n ] = IE [( S n + ξ n +1 ) 2 F n ] ( n + 1) = S 2 n + IE ( ξ 2 n +1 ) 2 S n IE ( ξ n +1 ) ( n + 1) = S 2 n + 1 + 0 ( n + 1) = S 2 n n c) Show that E ( τ ∧ N ) = E ( S 2 τ ∧ N ) for any fixed positive integer N . ANS: For any fixed N , τ ∧ N is an {F n }stopping time (see Exercise 4.3.3). Consider the stopped process { S 2 n ∧ ( τ ∧ N ) n ∧ ( τ ∧ N ) } . By same reasoning in part (b) we have that 1  S 2 n ∧ ( τ ∧ N ) n ∧ ( τ ∧ N )  ≤ N 2 + N for all n ; thus the stopped process is uniformly integrable. By the optional stopping theorem (Theorem 4.3.8), IE [ S 2 τ ∧ N τ ∧ N ] = IE [ S 2 0] = 0 , which yields the result. d) Verify why the equality E ( τ ) = E ( S 2 τ ) follows from the result in part (c). ANS: As N → ∞ , 0 ≤ τ ∧ N ↑ τ a.s. Thus by the monotone convergence theorem, IE [ τ ∧ N ] → IE [ τ ]. Using the definition of τ , we have that S 2 τ ∧ N ≤ max( a 2 ,b 2 ) for all N ....
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This document was uploaded on 04/12/2010.
 Spring '09
 Math

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