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hw7_print - P X = n = 2-n n ≥ 1 E X = ∞ X n =1 n 2-n...

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STAT416: Probability — Spring 2011 Assignments 1 Homework #7 (Solution) March 29, 2011 1 Total possible points is 35 1. Ex. 4.35 (15 pts) . (a) There are infinitely many examples, no unique distribution. As long as they show μ > σ 2 they get 5 points. See exercise I.49 below. (b) There are infinitely many examples, no unique distribution. As long as they show μ = σ 2 they get 5 points. Let X be the number of tosses of a fair coin needed to obtain the first head. Then, E ( X ) = 1 p = 1 1 / 2 = 2, V ar ( X ) = 1 - p p 2 = 1 2 1 4 = 2. or, X 1 2 3 4 5 6 7 P ( X = x ) 1 7 1 7 1 7 1 7 1 7 1 7 1 7 Then, E ( X ) = V ar ( X ) = 4. (c) There are infinitely many examples, no unique distribution. As long as they show μ < σ 2 they get 5 points. Let P ( X = 0) = 0 . 99 and P ( X = 100) = 0 . 01 then E ( X ) = 1 and V ar ( X ) = 99. 2. Ex. I.35 (a) (5 pts)
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Unformatted text preview: . P ( X = n ) = 2-n , n ≥ 1. E ( X ) = ∞ X n =1 n · 2-n = ∞ X n =1 n 2 n = 1 2 + 2 4 + 3 8 + 4 16 + ··· = 2 . 3. Ex. I.42 (5 pts) . P ( X = 1) = P ( X = 2) = 2 P ( X = 3) = 3 P ( X = 4). Let P ( X = 1) = p 1 , then p 1 + p 1 + 1 2 p 1 + 1 3 p 1 = 1 ⇒ 17 6 p 1 = 1 ⇒ p 1 = 6 17 = 0 . 353 . X 1 2 3 4 P ( X = x ) 0.353 0.353 0.176 0.118 4. Ex. I.49 (5 pts) . Let P ( X = 0) = 1 10 4 +1 = 1 10001 , and P ( X = 100 . 01) = 10 4 10 4 +1 = 10000 10001 . Then, E ( X ) = 100, V ar ( X ) = 1. or, X 99 100 101 P ( X = x ) 0.5 0.5 Purdue University hw7˙print.tex; Last Modified: March 29, 2011 (Prof. Sharabati)...
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