Homework 7 Solution - EE 351K PROBABILITY &...

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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.

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Homework 7 Solution - EE 351K PROBABILITY &...

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