hw6ans - Answers to Homework 6 AMS 570 4.2.1 Since a n and...

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Unformatted text preview: Answers to Homework 6 AMS 570 4.2.1 Since { a n } and a are all constant random variables, P ( | a n- a | < ) can only be 0 or 1. It is equal to 1 when | a n- a | < and 0 when | a n- a | ≥ . a n → a, as n → ∞ ⇐⇒ ∀ > , ∃ N ∈ N such that P ( | a n- a | < ) = 1 , when n > N ⇐⇒ ∀ < ε < 1 , > , ∃ N ∈ N s.t. P ( | a n- a | < ) > 1- ε > , when n > N ⇐⇒ ∀ < ε < 1 , > , ∃ N ∈ N s.t. P ( | a n- a | < )- 1 < ε , when n > N ⇐⇒ ∀ > , lim n →∞ P ( | a n- a | < ) = 1 ⇐⇒ a n P-→ a, as n → ∞ 4.2.2 (a) Y n ∼ b ( n,p ). Y n can be expressed as the sum of n independent Bernoulli(p) variables X 1 ,X 2 ,...,X n . That is Y n = n X i =1 X i ; P ( X i = 1) = p = 1- P ( X i = 0) , E( X i ) = p ∀ i. By the weak law of large numbers, X n = Y n n P-→ E( X i ) = p. (b) g 1 ( x ) = 1- x is continuous at p, < p < 1. By Theorem 4.2.4, g 1 Y n n = 1- Y n n P-→ 1- p....
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This note was uploaded on 03/24/2010 for the course AMS 54039 taught by Professor Hongshikahn during the Spring '10 term at SUNY Stony Brook.

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hw6ans - Answers to Homework 6 AMS 570 4.2.1 Since a n and...

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