Homework11 - HOMEWORK PROBLEMS #11 SOLUTIONS TO EVEN...

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HOMEWORK PROBLEMS #11 SOLUTIONS TO EVEN PROBLEMS 10. (a) Since the area of the support is 2 × 1 2 = 1, then k = 1. (b) We just need to find the area of the triangle with vertices (0 , 0) , (2 , 0) and (2 , 2 / 3): P ( Y 1 3 Y 2 ) = 2 × 2 3 2 = 2 3 . 22. (a) y 1 0 1 p 1 ( y 1 ) .76 .24 and y 2 0 1 2 p 2 ( y 2 ) .55 .16 .29 (b) y 2 0 1 2 p 2 | 1 ( y 2 | Y 1 = 0) 19/38 7/38 12/38 (c) P ( Y 1 = 0 | Y 2 = 2) = p (0 , 2) p 2 (2) = . 24 . 29 0 . 83. 26. (a) f 1 ( y 1 ) = Z 1 0 4 y 1 y 2 dy 2 = 4 y 1 y 2 2 2 ± ± ± ± 1 0 = 2 y 1 for 0 y 1 1. Similarly, f 2 ( y 2 ) = 2 y 2 for 0 y 2 1. (b) Since f ( y 1 ,y 2 ) = f 1 ( y 1 ) · f 2 ( y 2 ), then Y 1 and Y 2 are independent. Therefore, P ( Y 1 1 / 2 | Y 2 3 / 4) = P ( Y 1 1 / 2) = Z 1 / 2 0 2 y 1 dy 1 = y 2 1 ± ± 1 / 2 0 = 1 4 . (c) and (d) Since they are independent, the conditional distributions are just the marginals. (e) P ( Y 1 3 / 4 | Y 2 = 1 / 2) = P ( Y 1 3 / 4) = 1 - P ( Y 1 < 3 / 4) = 1 - Z 3 / 4 0 2 y 1 dy 1 = 1 - y 2 1 ± ± 3 / 4
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Homework11 - HOMEWORK PROBLEMS #11 SOLUTIONS TO EVEN...

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