final_sol - IE 300 / GE 331 Spring 2009 Final Exam...

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IE 300 / GE 331 Spring 2009 Final Exam Solutions May 11th, 2009 Problem 1. [6 points] (a) T. f X | Y ( x | y ) = f X ( x ) = 1 2 , so X and Y are independent. (b) T. var( Y ) = var( kX ) = k 2 var( X ) k 2 = var( Y ) var( X ) = 9 25 k = 3 5 var( X - Y ) = var( 2 5 X ) = 4 25 var( X ) = 4. Problem 2. [18 points] (a) b. P ( E F G ) = P ( E )+ P ( F )+ P ( G ) - P ( E F ) - P ( F G ) - P ( E G ) = 1. Because E and G are mutually exclusive, P ( E G ) = 0. Therefore, P ( F ) = 1 - P ( E ) - P ( G ) + P ( E F ) + P ( F G ) = 1 - 0 . 4 - 0 . 4 + 0 . 2 + 0 . 2 = 0 . 6 . (b) b. P ( F | E ) = P ( E F ) P ( E ) = 0 . 2 0 . 4 = 0 . 5 . (c) c. P ( X 2) = 1 - P ( X < 2) = 1 - P ( X = 1) - P ( X = 0) = 1 - 3 e - 3 1! - e - 3 = 1 - 4 e - 3 . (d) d. For a, R -∞ f ( x )d x = R 1 - 1 1d x = 2. For b, R -∞ f ( x )d x = R 2 0 1 4 d x + R 4 2 x 8 d x = 5 4 . For c, the value of a valid pdf can not be greater than 1 at any point. For d, R -∞ f ( x )d x = R - 1 - 2 1 2 d x + R 3 1 1 4 d x = 1. 1
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2 (e) b. P ( - 1 < X 2) = P ( X 2) - P ( X ≤ - 1) = P ± X - 2 2 2 - 2 2 ² - P ± X - 2 2 - 1 - 2 2 ² = P ± X - 2 2 0 ² - P ± X - 2 2 ≤ - 3 2 ² = Φ(0) - Φ( - 3 2 ) = Φ(0) - 1 + Φ( 3 2 ) = Φ( 3 2 ) - 1. (f) a. λ = 30 60 = 1 2 , P ( X 10) E [ X ] a = 1 λa = 1 1 2 × 10 = 2 10 . Problem 3. [15 points] (a) [4 points] Z 1 - 1 c 0 x 2 d x = c 0 x 3 3 ³ ³ ³ ³ ³ 1 - 1 = 2 c 0 3 = 1 Z 3 - 3 c 1 (3 - | x | )d x = 2 Z 3 0 c 1 (3 - x )d x = 9 c 1 = 1 Then, we have c 0 = 3 2 , c 1 = 1 9 . (b) [11 points] Using the MAP rule, decision H 1 is given for the region Γ 1 such that (1 - π ) f 1 > πf 0 and decision H 0 is given for the complementary region of Γ 1 , Γ 0 , such that (1 - π ) f 1 < πf 0 . It is the same that [ - 3 , - 1) (1 , 3]
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This note was uploaded on 09/08/2009 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.

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final_sol - IE 300 / GE 331 Spring 2009 Final Exam...

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