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quiz3_solution

# quiz3_solution - L throwing dice and losing T throwing dice...

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Quiz 3 solution Math 187, Prof. Garsia Alex Brik UC San Diego April 23, 2010 1. a. P ( X = 0 j Y = ° 1) = P ( X =0 ;Y = ° 1) P ( Y = ° 1) = 1 = 4 3 = 8 = 2 3 b. X,Y are not independent P ( X = 1 ; Y = 0) = 1 4 6 = 1 4 ± 1 4 = P ( X = 1) P ( Y = 0) X,Z are not independent P ( X = 3 ; Z = 2) = 1 8 6 = 1 8 ± 1 3 = P ( X = 3) P ( Z = 2) Y,Z are independent P ( Y = 0 ; Z = z ) = 1 12 = 1 4 ± 1 3 = P ( Y = 0) P ( Z = z ) P ( Y = 1 ; Z = z ) = 1 8 = 3 8 ± 1 3 = P ( Y = 1) P ( Z = z ) P ( Y = ° 1 ; Z = z ) = 1 8 = 3 8 ± 1 3 = P ( Y = ° 1) P ( Z = z ) c. X = f ( Y; Z ) Y = g ( X; Z ) X is not a function of Y X is not a function of Z Y is not a function of X Y is not a function of Z Z is not a function of X,Y because if X=1, Y=0 Z could be 1,2, or 3 d. E [ Y ] = 1 ± 3 8 ° 1 ± 3 8 = 0 e. E [ Z j X ² 0] = X z z ± P ( Z j X ² 0) = 1 P ( X ± 0) X z z ± P ( Z = z; X ² 0) = 4 3 ° 1 ± 5 24 + 2 ± 1 3 + 3 ± 5 24 ± = 2 f. E [ Y Z ] = E [ Y ] E [ Z ] = 0 (since Y; Z are independent and E [ Y ] = 0 ) 1

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g. E [ Y + Z ] = E [ Y ] + E [ Z ] = 0 + E [ Z ] E [ Z ] = 1 ± 1 3 + 2 ± 1 3 + 3 ± 1 3 = 2 Thus E [ Y + Z ] = 2 2. a. P ( h 2 ; 2 i ) = 1 36 P ("7") = 1 6 P ( h 1 ; 3 i ) = 1 18 P ( neither winning not losing ) = 27 36 Let W encode throwing dice and winning,
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Unformatted text preview: L- throwing dice and losing, T- throwing dice and neither winning nor losing. Then possible games can be encoded as follows: W , L , TW , TL , TTW , TTL , .... , T k W , T k L ,... where T k denotes T occuring k times. Then E = 1 X k =0 " 7 & 27 36 ± k 1 36 ± 1 & & 27 36 ± k 8 36 # = & 7 36 ± 8 36 ± 1 X k =0 & 27 36 ± k = = ± 1 36 & & 1 1 ± 27 = 36 ± = ± 1 36 & 36 9 = ± 1 9 b. E ( x ) = 0 E ( x ) = & x 1 36 ± 1 8 36 ± 1 X k =0 & 27 36 ± k Thus x = \$8 : Answer: \$8 BONUS. GO HANG A SALAMI IM A LASAGNA HOG 2...
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