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Unformatted text preview: HOMEWORK 5 SOLUTIONS Problem 1 a) We can show this immediately using Corollary 6.4 from Chapter 1 of Durret. b)Let X = c and for any > 0 define the open interval A = ( c ,c + ). The boundary A = { c ,c + } , therefore P { X A } = 0 . By the Portmanteau Theorem 1 = P { X A } = lim n P { X n A } = lim n P { X n c  < } Problem 2: Observe that f n is always nonnegative. To show that is a density function we need to compute the integral: Z 1 1 cos(2 nx ) dx = x sin(2 nx ) 2 n 1 x =0 = 1 . For x [0 , 1] , define F n ( x ) = x sin(2 nx ) 2 n . Observe that for all x [0 , 1], (1) lim n F n ( x ) = F ( x ) = x which is the distribution function of a uniform [0,1] r.v. We actually proved that for any point of continuity of F ( x ) the convergence (1) holds, therefore the n converge weakly to the Lebesque measure. Problem 3: Let 2 = V ar ( X 1 ) . Classical CLT implies that S n n = X N (0 , 1) . Let M > . The set (...
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This note was uploaded on 11/18/2009 for the course MATH 241 taught by Professor Reed during the Spring '09 term at Duke.
 Spring '09
 Reed

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