# HW9 - Chapter 5 Multivariate Probability Distributions 97...

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Chapter 5: Multivariate Probability Distributions 97 Instructor’s Solutions Manual 5.23 a. 1 0 , 3 ) ( 2 2 2 2 3 2 3 1 1 1 2 2 2 = = y y dy y y f y . b. Defined over y 2 y 1 1, with the constant y 2 0. c. First, we have 1 0 , 3 3 ) ( 1 2 2 2 0 1 1 1 1 = = y y dy y y f y . Thus, 1 2 1 1 2 0 , / 1 ) | ( y y y y y f = . So, conditioned on Y 1 = y 1 , we see Y 2 has a uniform distribution on the interval (0, y 1 ). Therefore, the probability is simple: P ( Y 2 > 1/2 | Y 1 = 3/4) = (3/4 – 1/2)/(3/4) = 1/3. 5.24 a. 1 0 , 1 ) ( 1 1 1 = y y f , 1 0 , 1 ) ( 2 2 2 = y y f . b. Since both Y 1 and Y 2 are uniformly distributed over the interval (0, 1), the probabilities are the same: .2 c. 1 0 2 y . d. 1 0 , 1 ) ( ) | ( 1 1 2 1 = = y y f y y f e. P (.3 < Y 1 < .5 | Y 2 = .3) = .2 f. P (.3 < Y 2 < .5 | Y 2 = .5) = .2 g. The answers are the same. 5.25 a. 0 , ) ( 1 1 1 1 > = y e y f y , 0 , ) ( 2 2 2 2 > = y e y f y . These are both exponential density functions with β = 1. b. . 2858 . ) 5 . 2 1 ( ) 5 . 2 1 ( 5 . 2 1 2 1 = = < < = < < e e Y P Y P c. y 2 > 0. d. 0 , ) ( ) | ( 1 1 1 2 1 1 > = = y e y f y y f y . e. 0 , ) ( ) | ( 2 2 2 1 2 2 > = = y e y f y y f y . f. The answers are the same. g. The probabilities are the same. 5.26 a. 1 0 , 2 ) ( ; 1 0 , 2 4 ) ( 2 2 2 1 1 1 0 2 2 1 1 1 = = = y y y f y y dy y y y f . b. 4 / 1 2 2 4 ) 4 / 3 | 2 / 1 ( 2 / 1 0 1 1 1 4 / 3 2 2 2 / 1 0 1 4 / 3 2 1 2 1 2 1 = = = ∫∫ dy y dy y dy dy y y Y Y P . c. 1 0 , 2 ) ( ) | ( 1 1 1 1 2 1 = = y y y f y y f . d. 1 0 , 2 ) ( ) | ( 2 2 2 2 1 2 = = y y y f y y f . e. 16 / 9 2 ) 4 / 3 ( ) 2 / 1 | 4 / 3 ( 4 / 3 0 1 1 1 2 1 = = = = dy y Y P Y Y P .

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98 Chapter 5: Multivariate Probability Distributions Instructor’s Solutions Manual 5.27 a. ; 1 0 , ) 1 ( 3 ) 1 ( 6 ) ( 1 2 1 1 2 2 1 1 1 = = y y dy y y f y 1 0 ), 1 ( 6 ) 1 ( 6 ) ( 2 2 2 0 1 2 2 2 2 = = y y y dy y y f y . b. . 63 / 32 ) 1 ( 3 ) 1 ( 6 ) 4 / 3 | 2 / 1 ( 4 / 3 0 1 2 1 2 / 1 00 2 1 2 1 2 2 = = ∫∫ dy y dy dy y Y Y P y c. 1 0 , / 1 ) | ( 2 1 2 2 1 = y y y y y f . d. 1 0 , ) 1 /( ) 1 ( 2 ) | ( 2 1 2 1 2 1 2 = y y y y y y f . e. From part d , 1 2 / 1 ), 1 ( 8 ) 2 / 1 | ( 2 2 2 = y y y f . Thus, . 4 / 1 ) 2 / 1 | 4 / 3 ( 1 2 = = Y Y P 5.28 Referring to Ex. 5.10: a. First, find 1 0 ), 1 ( 2 1 ) ( 2 2 2 2 1 2 2 2 = = y y dy y f y . Then, 25 . ) 5 . ( 2 = Y P . b. First find . 2 2 , ) | ( 1 2 ) 1 ( 2 1 2 1 2 = y y y y f y Thus, 2 1 , 1 ) 5 . | ( 1 1 = y y f –– the conditional distribution is uniform on (1, 2). Therefore, 5 . ) 5 . | 5 . 1 ( 2 1 = = Y Y P 5.29 Referring to Ex. 5.11: a. 1 0 ), 1 ( 2 1 ) ( 2 2 1 1 1 2 2 2 2 = = y y dy y f y y . In order to find f 1 ( y 1 ), notice that the limits of integration are different for 0 y 1 1 and –1 y 1 0. For the first case: 1 1 0 2 1 1 1 1 ) ( 1 y dy y f y = = , for 0 y 1 1. For the second case, 1 1 0 2 1 1 1 1 ) ( 1 y dy y f y + = = + , for –1 y 1 0. This can be written as | | 1 ) ( 1 1 1 y y f = , for –1 y 1 1. b. The conditional distribution is | | 1 1 1 2 1 ) | ( y y y f = , for 0 y 1 1 – | y 1 |. Thus, 3 / 4 ) 4 / 1 | ( 2 = y f . Then, = = > 4 / 3 2 / 1 2 1 2 3 / 4 ) 4 / 1 | 2 / 1 ( dy Y Y P = 1/3.
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HW9 - Chapter 5 Multivariate Probability Distributions 97...

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