sample_quiz3_solution

# sample_quiz3_solution - X 1 X n So X = f Y Z = X ± Y = f Y...

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Math 187 Prof. Garsia (typed by Alex Brik) 04-20-2010 SAMPLE QUIZ 3 solution. 1. a. E [ Z ] = X z P ( Z = z ) z E [ Z ] = 1 P ( Z = 1) + 2 P ( Z = 2) + 3 P ( Z = 3) = = 1 1 3 + 2 1 3 + 3 1 3 = 2 : c. Generally for two random variables X; Y E ( X + Y ) = E ( X ) + E ( Y ) provided that E ( X ) and E ( Y ) E ( X + Y ) = E ( X ) + E ( Y ) E ( X ) = ± 3 1 2 + 2 1 4 + 5 1 4 = 1 4 E ( Y ) = ± 4 1 6 ± 5 1 6 ± 6 1 6 + 1 1 12 + 0 1 12 ± 1 1 12 + 4 1 12 + 3 1 12 + 2 1 12 = ± 7 4 Thus E ( X + Y ) = 1 4 ± 7 4 = ± 3 2 d. Generally if X; Z are independent random variables E ( XZ ) = E ( X ) E ( Z ) Are X; Z independent? Random variable X; Z are independent if for all x; z P ( X = x; Z = z ) = P ( X = x ) P ( Z = z ) : Probabilities: P ( X = x; Z = z ) # x ; z ! 1 2 3 -3 1/6 1/6 1/6 2 1/12 1/12 1/12 5 1/12 1/12 1/12 P ( Z = z; X = ± 3) = 1 6 = 1 2 1 3 = P ( X = ± 3) P ( Z = z ) 1

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P ( Z = z; X = 5) = 1 12 = 1 4 1 3 = P ( X = 5) P ( Z = z ) P ( Z = z; X = 2) = 1 12 = 1 4 1 3 = P ( X = 2) P ( Z = z ) : Thus X and Z are indpendent. Thus E ( XZ ) = E ( X ) E ( Z ) = 1 4 2 = 1 2 e. Dependence is not negation of independence. Dependence of variable X n on X 1 ; :::; X n 1 means X n is a deterministic function of
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Unformatted text preview: X 1 ; :::; X n . So X = f ( Y ) Z = X ± Y = f ( Y ) ± Y Y = g ( X; Z ) f. ( X; Z ) are indpendent (see ( d ) ) ( Z; Y ) are dependent: P ( Y = 0 ; Z = 1) = 0 6 = 1 12 & 1 3 = P ( Y = 0) P ( Z = 1) ( X; Y ) are dependent P ( Y = 0 ; X = ± 3) = 0 6 = 1 12 & 1 2 = P ( Y = 0) P ( X = ± 3) g. P ( Y = 4 j Z = 1) = P ( Y = 4 ; Z = 1) P ( Z = 1) = 1 = 12 1 = 3 = 1 4 h. E [ Y j Z = 1] = X y y & P ( Y = y j Z = 1) = X y y P ( Y = y; Z = 1) P ( Z = 1) = 1 P ( Z = 1) X y y & P ( Y = y; Z = 1) = = 3 & & 1 & 1 12 + 4 & 1 12 ± 4 & 1 6 ± = ± 3 4 2...
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sample_quiz3_solution - X 1 X n So X = f Y Z = X ± Y = f Y...

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