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Unformatted text preview: MATH 507b TAKEHOME FINAL EXAM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) Suppose a n % a . Then { T a n } is a bounded increasing sequence so T a n % S for some finite S . Then a n = B T an → B S since B t is continuous, so B S = a , and therefore S = T a . Thus T a n % T a which means T a is leftcontinuous in a . (b) Let a ≥ 0. By the Strong Markov Property and Chapter 7 Theorem 2.6, we have inf { t > B T a : B t > a } = B T a a.s. Therefore given > 0 there exists a 1 > a such that B t = a 1 for some t ∈ ( T a ,T a + ). This means T a ∈ ( T a ,T a + ) for all a ∈ ( a ,a 1 ). This shows that with probability one, T a is right continuous at a . (c,d) By the Strong Markov Property and and Chapter 7 Theorem 2.6, given > 0, with probability one the maximum of B t in [ T c ,T c + ] occurs at some s ∈ ( T c ,T c + ) with B s = a for some a > c . This means T h < s for all h < a , but T h > T c + for all h > a . This means T h is discontinuous at h = a , with a jump from at most s to at least T c + . Thus with probability one, T h is discontinuous at h = a for some a > c satisfying T a < T...
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 Spring '08
 Alexander
 Math, Probability, Trigraph, TA, tc, FN, strong Markov property

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