Discussion5

Discussion5 - Discussion 5 Emily Mower February 19, 2010...

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Discussion 5 Emily Mower February 19, 2010 Contents 1 Conditional Probability Distribution 2 1.1 Schaum’s 3.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Schaum’s 3.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Schaum’s 3.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2 Combinations 8 2.1 Textbook: 2.34 . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.2 Textbook: 2.119 . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3 Permutations 11 3.1 Textbook: 2.48 . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.2 Textbook: 2.136 . . . . . . . . . . . . . . . . . . . . . . . . . . 12 4 Conditional Probability and Bayes Law 13 4.1 Textbook: 2.89 . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4.2 Textbook: 2.128 . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.3 Textbook: 2.135 . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1
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1 Conditional Probability Distribution 1.1 Schaum’s 3.18 The joint pdf of a bivariate r.v. (X,Y) is given by f XY ( x,y ) = ( kxy, 0 < x < 1 , 0 < y < 1 0 , otherwise where k is a constant (a). Find the value of k . (b). Are X and Y independent? (c). Find P ( X + Y < 1) (d). Find f ( x | y ) Solutions: (a). Find the value of k . Z -∞ Z -∞ f ( x,y ) d x d y = 1 = Z 1 0 Z 1 0 kxy d x d y = 1 2 k Z 1 0 yx 2 ± ± ± ± 1 0 d y = 1 2 k Z 1 0 y d y = k 4 k = 4 2
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X and Y independent? f ( x ) = Z -∞ f ( x,y ) d y = Z 1 0 4 xy d y = 2 xy 2 ± ± ± ± 1 0 = 2 x f ( y ) = Z -∞ f ( x,y ) d x = Z 1 0 4 xy d x = 2 x 2 y ± ± ± ± 1 0 = 2 y Since f ( x,y ) = f ( x ) f ( y ) X and Y are independent (c). Find P ( X + Y < 1) P ( X + Y < 1) = Z 1 0 Z 1 - x 0 f ( x,y ) d y d x = 1 2 Z 1 0 4 xy 2 ± ± ± ± 1 - x 0 d x = 1 2 Z 1 0 4 x (1 - x ) 2 d x = 1 2 Z 1 0 4 x [1 - 2 x + x 2 ] d x = 1 2 Z 1 0 [4 x - 8 x 2 + 4 x 3 ]d x = 1 2 [2 x 2 - 8 3 x 3 + x 4 ] ± ± ± ± 1 0 P ( X + Y < 1) = 1 6 (d). Find f ( x | y ) Intuition: Since X and Y are statistically independent, f ( x | y ) = f ( x ). f
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Discussion5 - Discussion 5 Emily Mower February 19, 2010...

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