hw4s - MSPE PROGRAM ECON 506- FALL 2011 SOLUTION TO HW4...

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Unformatted text preview: MSPE PROGRAM ECON 506- FALL 2011 SOLUTION TO HW4 (The problem numbers refer to numbers in 7 th edition) 5.10 a. Geometrically, since Y 1 and Y 2 are distributed uniformly over the triangular region, using the area formula for a triangle k = 1. b. This probability can also be calculated using geometric considerations. The area of the triangle specified by Y 1 3 Y 2 is 2/3, so this is the probability. 5.12 a. P ( Y 1 3/4, Y 2 3/4) = 1 P( Y 1 > 3/4) P( Y 2 > 3/4) = 1 ( 29 ( 29 ( 29 ( 29 8 7 4 1 4 1 2 1 4 1 4 1 2 1 =- . b. . 2 / 1 2 ) 2 / 1 , 2 / 1 ( 2 / 1 2 / 1 2 1 2 1 = = dy dy Y Y P 5.28 Referring to Ex. 5.10: a. First, find 1 ), 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 5.34 a. Given Y 1 = y 1 , Y 2 has a uniform distribution on the interval (0, y 1 ). b. Since f 1 ( y 1 ) = 1, 0 y 1 1, f ( y 1 , y 2 ) = f ( y 2 | y 1 ) f 1 ( y 1 ) = 1/ y 1 , 0 y 2 y 1 1. c. - = = 1 2 2 1 1 2 2 2 1 ), ln( / 1 ) ( y y y dy y y f . 5.54 The ranges of y 1 and y 2 depend on each other so Y 1 and Y 2 cannot be independent. 5.58 1 ), 1 ( 12 6 ) ( 1 1 2 1 2 2 2 2 1 1 1 1 1 - =...
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hw4s - MSPE PROGRAM ECON 506- FALL 2011 SOLUTION TO HW4...

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