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Unformatted text preview: EE 351K PROBABILITY & RANDOM PROCESSES FALL 2011 Instructor: Sujay Sanghavi sanghavi@mail.utexas.edu Homework 7 Solution Problem 1 Let X 1 , X 2 , ...be independent random variables that are uniformly distributed over [ 1 . 1] . For each of the following cases, find whether Y 1 , Y 2 , ...convergence to some limit in probability. (a) Y n = X n /n . (b) Y n = ( X n ) n . (c) Y n = min { X 1 ,...,X n } . (d) Y n = max { X 1 ,...,X n } . Sol : Using the definition for convergence in probability, if Y n converge to a number c , then for any > , P (  Y n c  ) . (a) Claim: Y n converge to 0 in probability. Proof: For any > , P (  Y n  ) = P (  Y n  > ) = P ( X n > n ) + P ( X n < n ) . Note that for any > , we can always find N such that when n > N , n > 1 . So when n > N , using X n is uniformly distributed over [ 1 , 1] , we have P ( X n > n ) = 0 and P ( X n < n ) = 0 . Thus, P (  Y n  ) , which implies Y n converge to 0 in probability. (b) Claim: Y n converge to 0 in probability. Proof: For any > , P (  Y n  ) = P (  Y n  > ) = P ( X n > n ) + P ( X n < n ) . Note that for any > , n 1 as n , so P ( X n > n ) and P ( X n < n ) . Thus, P (  Y n  ) , which implies Y n converge to 0 in probability. (c) Claim: Y n converge to 1 in probability. Proof: For any > , P (  Y n ( 1)  ) = P (  Y n + 1  > ) = P ( Y n >  1) . Note that Y n = min { x 1 ,...,X n } , so Y n >  1 means X i >  1 for any 1 i n . Using X i are independent, we will have P ( Y n >  1) = n i =1 P ( X i >  1) . If < 2 , then P ( X i >  1) = 1 / 2 , so P (  Y n ( 1)  ) = (1 / 2) n ; if 2 , then P ( X i >  1) = 0 , so P (  Y n ( 1)  ) = 0 . Thus, P (  Y n ( 1)  ) , which implies Y n converge to 1 in probability....
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This note was uploaded on 02/20/2012 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.
 Spring '07
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