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

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Unformatted text preview: 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 ≈ . 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) ≈ . 9942 F 2 K (1) ≈ . 1280 With the information given in the problem, it is not possible to decide which of these two values is the “true” one....
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hw06soln - ECE 310 Spring 2005 HW 6 solutions Problem E5.1...

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