09831HW3solutions

09831HW3solutions - ω> 0 then Y n ω = 0 for all n such...

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HOMEWORK 3 SOLUTIONS Problem 1 A basic property of Poisson random variables is that if { X k } are i.i.d. Poisson( θ ), then S n is Poisson( ). Thus, X 0 k na e - ( ) k k ! = P ( S n na ) = P ( S n - n ( a - θ )) . By the WLLN, for all ± > 0 P ±² ² ² ² S n n - θ ² ² ² ² ± ³ 0 . Hence, if a < θ we have P ( S n - n ( a - θ )) P ( | S n - | ≥ n ( θ - a )) 0 , while for a > θ, P ( S n - n ( a - θ )) = 1 - P ( S n - nθ > n ( a - θ )) 1 . Problem 2 P ± S n E S n 2 ³ P ± | S n - E S n | ≥ E S n 2 ³ 4 V ar ( S n ) ( E ( S n )) 2 = 4 nV ar ( X n 1 ) ( E ( S n )) 2 4 nV ar ( X 2 n 1 ) ( E ( S n )) 2 4 Cn E ( X n 1 ) ( E ( S n )) 2 = 4 C E ( S n ) ( E ( S n )) 2 = 4 C E S n 0 . Consequently, if n is large enough so that E S n > 2 k , then P ( S n k ) P ± S n E S n 2 ³ 0 by the above. Problem 3 (a) Claim 1: X n 0 a.s. p k < . ( ) The Borel - Cantelli lemma implies P ( X n = 1 i.o. ) = 0 , therefore P ( n 0 ( ω ) s.t. n n 0 X n = 0) = 1 . ( ) p k = ∞ ⇒ P ( X n = 1 i.o. ) = 1 by the second B-C lemma, hence with probability 1, X n 0 fails. 1
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2 HOMEWORK 3 SOLUTIONS Claim 2: X n P 0 p n 0 . X n P 0 ⇔ ∀ ± > 0 , P ( | X n | > ± ) 0 P ( X n = 1) 0 p n 0 . So any sequence { p n } with p n 0 but p k = is such that X n P 0 but not a.s. For example p n = n - 1 . (b) Now for any p n 0 ,Y n 0 a.s. This is because if
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Unformatted text preview: ω > 0 then Y n ( ω ) = 0 for all n such that p n < ω. Thus Y n ( ω ) → 0 for all ω ∈ (0 , 1] , which is a probability 1 subset. The second Borel - Cantelli lemma fails because { Y n } are not independent: P { Y n = 1 ,Y n +1 = 1 } = p n +1 6 = p n p n +1 as long as 0 < p n ,p n +1 < 1 . Problem 4 Apply the second B-C lemma: For any C < ∞ , define j ( n ) = log 2 ([ Cn log 2 n ]) . Then P ( X n > Cn log 2 n ) = X j :2 j >Cn log 2 n 2-j = 2-j ( n ) ≥ 1 Cn log 2 n = ∞ Therefore, P ( X n > Cn log 2 n i.o. ) = 1 . Thus, S n n log 2 n ≥ X n n log 2 n > C i.o. and since C was arbitrary we can make it arbitrarily large and we have the conclusion....
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09831HW3solutions - ω> 0 then Y n ω = 0 for all n such...

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