Unformatted text preview: . P ( X = n ) = 2n , n ≥ 1. E ( X ) = ∞ X n =1 n · 2n = ∞ 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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 Spring '08
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 Probability, 5 pts, Purdue University, p1 + p1, Prof. Sharabati, inﬁnitely many examples, unique distribution

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