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m507a-finsoln-f07

m507a-finsoln-f07 - MATH 507a FINAL EXAM SOLUTIONS Fall...

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MATH 507a FINAL EXAM SOLUTIONS Fall 2007 Prof. Alexander (1) ϕ S n /n ( t ) = Ee itS n /n = ϕ S n t n = ϕ t n n . For fixed t , as n → ∞ , ϕ t n n = 1 + ϕ (0) t n + o 1 n n e ϕ (0) t = e iat , so S n /n a in distribution, hence also in probability. (2)(a) P ( X n > c (log n ) α ) = e - ac 2 (log n ) 2 α . (1) If α > 1 / 2 then 2 α > 1 so for large n , the exponent in (1) is > 2 log n , so P ( X n > c (log n ) α ) e - 2 log n = 1 n 2 , so n P ( X n > c (log n ) α ) < so P ( X n > c (log n ) α i.o.) = 0. If α < 1 / 2 then 2 α < 1 so for large n then exponent in (1) is < 1 2 log n and P ( X n > c (log n ) α ) e - 1 2 log n = 1 n , so n P ( X n > c (log n ) α ) = so P ( X n > c (log n ) α i.o.) = 1. If α = 1 / 2 then P ( X n > c (log n ) α ) = e - ac 2 log n = n - ac 2 , so n P ( X n > c (log n ) α ) < ∞ ⇐⇒ ac 2 > 1 ⇐⇒ c > 1 a . Hence P ( X n > c (log n ) α i . o . ) = 0 , if c > 1 / a, 1 , if c 1 / a. (b) From (a), b n = 1 a log n . (3) E ( S n +1 | F n ) = E X 1 + · · · + X n n + 1 + X n +1 n + 1 F n = n n + 1 A n + 1 n + 1 E ( X n +1 | F n ) = n n + 1 A n + 1 n + 1 A n = A n . (2) 1
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Clearly A n ∈ F n , so the proof is complete.
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