hw06soln - ECE 310 Spring 2005 HW 6 solutions Problem E5.1...

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ECE 310, Spring 2005, HW 6 solutions Problem E5.1 a) Yes, it is a cdf for Ω = R (see example 5.4) b) No, because 5.6a fails: lim x →-∞ U ( x ) + [1 - U ( x )] 1 + e x 2 = 1 / 2 c) No, because 5.2 and 5.6b fail. d) Yes. e) No, it is not even a function since for x=0, it assumes two values: 0 and 1. Problem E5.2 Since K T must be non-negative integer: F K ( x ) = X i =0 P ( K T = i ) U ( x - i ) a) P ( K T = 1) = F K (1) - F K (1 - ) = F K (1) - F K (1 / 2) = 0 . 1 b) Assuming K T is distributed according Poisson, we can find λT using the result of the part a): P ( K T = 1) = 0 . 1 = e - λT λT . Solving numerically, we obtain two solutions, λ 1 T 0 . 1118 and λ 2 T 3 . 5772. Therefore, F K (1) = P ( K T = 0) + P ( K T = 1) = e - λT + 0 . 1 and substituting the two possible value of λ , we get: F 1 K (1) 0 . 9942 F 2 K (1) 0 . 1280 With the information given in the problem, it is not possible to decide which of these two values is the “true” one. Problem E5.8 First Plot: P ( X < - 3) 0 . 05 P ( X ≤ - 1) 0 . 27 P ( X = 0) = 0 P ( X > 2) = 1 - P ( X 2) 0 . 12 Second Plot: P ( X < - 3) = 0 P ( X ≤ - 1) = 0 . 2 P ( X = 0) = 0 P ( X > 2) = 1 - P ( X 2) = 0 1
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Problem E5.10 F X ( x ) = 3 X i =0 b ( i, 3 , 1 / 3) U ( x - i ) Problem E5.12 a) P ( Y ≤ - 1) = F Y ( - 1) = 0 . 1 P ( Y = 1) = F Y (1) - F Y (1 - ) = 0 . 2 P ( Y = 2) = F Y (2) - F Y (2 - ) = 0 Problem E5.21 a) P ( X 1) = F X (1) = 2 / 3 P ( X 2) = 1 - P ( X < 2) = 1 - F X (2 - ) = 1 - 2 / 3 = 1
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