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Homework 7 Solution - EE 351K PROBABILITY RANDOM PROCESSES...

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EE 351K PROBABILITY & RANDOM PROCESSES FALL 2011 Instructor: Sujay Sanghavi [email protected] 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 ϵ > 0 , P ( | Y n - c | ≥ ϵ ) 0 . (a) Claim: Y n converge to 0 in probability. Proof: For any ϵ > 0 , P ( | Y n - 0 | ≥ ϵ ) = P ( | Y n | > ϵ ) = P ( X n > nϵ ) + P ( X n < - ) . Note that for any ϵ > 0 , we can always find N ϵ such that when n > N ε , nϵ > 1 . So when n > N ϵ , using X n is uniformly distributed over [ - 1 , 1] , we have P ( X n > nϵ ) = 0 and P ( X n < - ) = 0 . Thus, P ( | Y n - 0 | ≥ ϵ ) 0 , which implies Y n converge to 0 in probability. (b) Claim: Y n converge to 0 in probability. Proof: For any ϵ > 0 , P ( | Y n - 0 | ≥ ϵ ) = P ( | Y n | > ϵ ) = P ( X n > n ϵ ) + P ( X n < - n ϵ ) . Note that for any ϵ > 0 , n ϵ 1 as n → ∞ , so P ( X n > n ϵ ) 0 and P ( X n < - n ϵ ) 0 . Thus, P ( | Y n - 0 | ≥ ϵ ) 0 , which implies Y n converge to 0 in probability. (c) Claim: Y n converge to - 1 in probability. Proof: For any ϵ > 0 , P ( | Y n - ( - 1) | ≥ ϵ ) = P ( | Y n + 1 | > ϵ ) = P ( Y n > ϵ - 1) . Note that Y n = min { x 1 , . . . , X n } , so Y n > ϵ - 1 means X i > ϵ - 1 for any 1 i n . Using X i are independent, we will have P ( Y n > ϵ - 1) = n i =1 P ( X i > ϵ - 1) . If ϵ < 2 , then P ( X i > ϵ - 1) = 1 - ϵ/ 2 , so P ( | Y n - ( - 1) | ≥ ϵ ) = (1 - ϵ/ 2) n 0 ; if ϵ 2 , then P ( X i > ϵ - 1) = 0 , so P ( | Y n - ( - 1) | ≥ ϵ ) = 0 . Thus, P ( | Y n - ( - 1) | ≥ ϵ ) 0 , which implies Y n converge to - 1 in probability. (d) Claim: Y n converge to 1 in probability. Proof: If define Z i = - X i , then Z i is also uniformly distributed over [ - 1 , 1] by the symmetry, and note that min { Z 1 , . . . , Z n } = - max { X 1 , . . . , X n } . So using the result from part (c), we will have Y n converge to 1 in probability.
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