STAT 400 Lecture AL1 Answers for 4.1, 4.4 (Part 1) Spring 2015 Dalpiaz Multivariate Distributions Let X and Y be two discrete random variables. The joint probability mass function p ( x, y ) is defined for each pair of numbers ( x, y ) by p ( x, y ) = P( X = x and Y = y ). Let A be any set consisting of pairs of ( x, y ) values. Then P( ( X, Y ) A ) = yx Ayxp,, . Let X and Y be two continuous random variables. Then f ( x, y ) is the joint probability density function for X and Y if for any two-dimensional set A P( ( X, Y ) A ) = Adydxyxf ,. 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 2 0.25 0.30 0.20 a) Find P( X > Y ). P( X > Y ) = p ( 1, 0 ) + p ( 2, 0 ) + p ( 2, 1 ) = 0.15 + 0.25 + 0.30 = 0.70. b) Find P( X + Y = 2 ). P( X + Y = 2 ) = p ( 1, 1 ) + p ( 2, 0 ) = 0.10 + 0.25 = 0.35.The marginal probability mass functions of X and of Y are given by p X ( x ) = yyxpall,, p Y ( y ) = xyxpall,. The marginal probability density functions of X and of Y are given by f X ( x ) = , dyyxf, f Y ( y ) = , dxyxf. c) Find the (marginal) probability distributions p X ( x ) of X and p Y ( y ) of Y. y p Y ( y ) x p X ( x ) 0 0.40 1 0.25 1 0.40 2 0.75 2 0.2 If p ( x, y ) is the joint probability mass function of ( X, Y ) OR f ( x, y ) is the joint probability density function of ( X, Y ), then discrete continuous E ( g( X, Y ) ) = x y yxpyxg all all),(),( E ( g( X, Y ) ) = dydxyxfyxg ),(),( d) Find E ( X ), E ( Y ), E ( X + Y ), E ( X Y ). E ( X ) = 1 0.25 + 2 0.75 = 1.75. E ( Y ) = 0 0.40 + 1 0.40 + 2 0.20 = 0.8. E ( X + Y ) = 1 0.15 + 2 0.25 + 2 0.10 + 3 0.30 + 3 0 + 4 0.20 = 2.55. OR E ( X + Y ) = E ( X ) + E ( Y ) = 1.75 + 0.8 = 2.55. E ( X Y ) = 0 0.15 + 0 0.25 + 1 0.10 + 2 0.30 + 2 0 + 4 0.20 = 1.5.2. Alex is Nuts, Inc. markets cans of deluxe mixed nuts containing almonds, cashews, and peanuts. Suppose the net weight of each can is exactly 1 lb, but the weight contribution of each type of nut is random. Because the three weights sum to 1, a joint probability model for any two gives all necessary information about the weight of the third type. Let X = the weight of almonds in a selected can and Y = the weight of cashews. Then the region of positive density is D = { ( x , y ) : 0 x 1, 0 y 1, x + y 1 }. Let the joint probability density function for ( X , Y ) be otherwise01 ,10 ,10 60 ,2 yxyxyxyxf a) Verify that yxf , is a legitimate probability density function. 1. f ( x , y ) 0 for all ( x , y ). 2. dydxyxf , = 10102 60 dxdyyxx = 1022 1 30 dxxx = 10432 30 60 30 dxxxx = 0 1 6 15 10 543 xxx = 1. b) Find the probability that the two types of nuts together make up less than 50% of the can. That is, find the probability P( X + Y < 0.50 ). ( Find the probability that peanuts make up over 50% of the can.) P( X + Y < 0.50 ) = 5.005.002 60 xyyx dxd = 5.0022 5.0 30 xxx d 0 5.0 6 5.7 5.2 30 30 5.7 5435.00432 xxxdxxxx = 321 = 0.03125.c) Find the probability that there are more almonds than cashews in a can. That is, find the probability P( X > Y ). P( X > Y ) = 210 12 60 dydxyxyy = 210 12 3 20 dydxxyyy = 210 33 1 20 dyyyy = 210 3 2 2331 20 dxyyyy = 210 4 3 2 40606020 dxyyyy = 1611 021 5 4 3 2 8152010 yyyy = 0.6875. OR P( X > Y ) = 1 – 210 12 60 dxdyyxxx = 1 – 210 1 2 2 30 dxdyyxxx = 1 – 210 22 2 1 30 dxxxx = 1 – 210 2 21 30 dxxx = 1 – 210 3 2 6030 dxxx = 1 – 1611 021 4 3 1510 xx = 0.6875. OR P( X > Y ) = 210 02 60 dxdyyxx + 121 102 60 dxdyyxx = …d) Find the probability that there are at least twice as many cashews as there are almonds. That is, find the probability P( 2 X Y ). P( Y ≥ 2 X ) = 3101 2 2 60 dxdyyxxx = 3102 22 2 1 30 dxxxx = 3104 3 2 9060 30 dxxxx = 3104 3 2 9060 30 dxxxx = 031543 18 15 10 xxx = 2431881152710 = 91. e) Find the marginal probability density function for X. f X ( x ) = xdyyx10 2 60 = xdyyx10 2 2 30 = 2 2 1 30 xx , 0 < x < 1. f) Find the marginal probability density function for Y. f Y ( y ) = ydxyx10 2 60 = ydxxy10 2 3 20 = 3 1 20 yy , 0 < y < 1. g) Find E ( X ), E ( Y ), E ( X + Y ), E ( X Y ). E ( X ) = 10 2 2 1 30 dxxxx = 10543 30 60 30 dxxxx = 0 1 5 12 5.7 654 xxx = …
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