STAT 400 Lecture AL1 Answers for 4.1, 4.4 (2) Spring 2015 Dalpiaz Independent Random Variables 1. Consider the following joint probability distribution p ( x, y ) of two random variables X and Y: x \ y 0 1 2 1 0.15 0.10 0 0.25 2 0.25 0.30 0.20 0.75 0.40 0.40 0.20 Recall: A and B are independent if and only if P ( A B ) = P ( A ) P ( B ). a) Are events {X = 1} and {Y = 1} independent? P( X = 1 Y = 1 ) = p ( 1, 1 ) = 0.10 = 0.25 0.40 = P( X = 1 ) P( Y = 1 ). {X = 1} and {Y = 1} are independent. Def Random variables X and Y are independent if and only if discrete p ( x, y ) = p X ( x ) p Y ( y ) for all x, y. continuous f ( x, y ) = f X ( x ) f Y ( y ) for all x, y. F ( x, y ) = P ( X x, Y y ). f ( x, y ) = 2 F ( x, y )/ x y . Def Random variables X and Y are independent if and only if F ( x, y ) = F X ( x ) F Y ( y ) for all x, y. b) Are random variables X and Y independent? p ( 1, 0 ) = 0.15 0.25 0.40 = p X ( 1 ) p Y ( 0 ). X and Y are NOT independent.2. Let the joint probability density function for ( X , Y ) be otherwise01 ,10 ,10 60 ,2 yxyxyxyxf Recall: f X ( x ) = 2 2 1 30 xx , 0 < x < 1, f Y ( y ) = 3 1 20 yy , 0 < y < 1. Are random variables X and Y independent? The support of ( X, Y ) is not a rectangle. X and Y are NOT independent. 3. Let the joint probability density function for ( X , Y ) be otherwise010 ,10 ,yxyxyxf Are X and Y independent? X and Y are NOT independent.4. Let the joint probability density function for ( X , Y ) be otherwise00 ,101 12 ,2 yxxxyxfye Are X and Y independent? f X ( x ) = 1 6 1 1202 xxdyxxye , 0 < x < 1. f Y ( y ) = yyee dxxx2102 2 1 12 , y > 0. Since f ( x, y ) = f X ( x ) f Y ( y ) for all x, y, X and Y are independent. If random variables X and Y are independent, then E ( g ( X ) h ( Y ) ) = E ( g ( X ) ) E ( h ( Y ) ). 5. Suppose the probability density functions of T 1 and T 2 are f T 1 ( x ) = e – x, x > 0, f T 2 ( y ) = e – y, y > 0, respectively. Suppose T 1 and T 2 are independent. Find P ( 2 T 1 > T 2 ). P ( 2 T 1 > T 2 ) = 02 02 αααββα ββdydxdydxyxyyyxeee = βαβ22 ββ02 02 αα ββ dydyyyyeee.6. Let X and Y be two independent random variables, X has a Geometric distribution with the probability of “success” p = 1/3 , Y has a Poisson distribution with mean 3. That is, p X ( x ) = 1 3231x, x = 1, 2, 3, … , p Y ( y ) = ! 3 3yey, y = 0, 1, 2, 3, … . a) Find P ( X = Y ). P ( X = Y ) = 1YX kkpkp = 131! 3 3231 kkkke = 113! 2 kkke = 1 2 203! kkke = 1 223ee = 231 ee 0.159. b) Find P ( X = 2 Y ). P ( X = 2 Y ) = 1Y X 2 kkpkp = 131 2! 3 3231 kkkke = 13! 1 34 2kkke = 1 2343 ee 0.069544. For fun: c) P ( X > Y ) = 0 131! 3 3231 y yxyxye = 03! 3 32 yyyye = 03! 2 yyye =
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