Dr. Hackney STA Solutions pg 45

# Dr. Hackney STA Solutions pg 45 - 5(6 5 = 12 36 = 1 3 P X =...

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Chapter 4 Multiple Random Variables 4.1 Since the distribution is uniform, the easiest way to calculate these probabilities is as the ratio of areas, the total area being 4. a. The circle x 2 + y 2 1 has area π , so P ( X 2 + Y 2 1) = π 4 . b. The area below the line y = 2 x is half of the area of the square, so P (2 X - Y > 0) = 2 4 . c. Clearly P ( | X + Y | < 2) = 1. 4.2 These are all fundamental properties of integrals. The proof is the same as for Theorem 2.2.5 with bivariate integrals replacing univariate integrals. 4.3 For the experiment of tossing two fair dice, each of the points in the 36-point sample space are equally likely. So the probability of an event is (number of points in the event)/36. The given probabilities are obtained by noting the following equivalences of events. P ( { X = 0 ,Y = 0 } ) = P ( { (1 , 1) , (2 , 1) , (1 , 3) , (2 , 3) , (1 , 5) , (2 , 5) } ) = 6 36 = 1 6 P ( { X = 0 ,Y = 1 } ) = P ( { (1 , 2) , (2 , 2) , (1 , 4) , (2 , 4) , (1 , 6) , (2 , 6) } ) = 6 36 = 1 6 P ( { X = 1 ,Y = 0 } ) = P ( { (3 , 1) , (4 , 1) , (5 , 1) , (6 , 1) , (3 , 3) , (4 , 3) , (5 , 3) , (6 , 3) , (3 , 5) , (4 , 5) , (5
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Unformatted text preview: , 5) , (6 , 5) } ) = 12 36 = 1 3 P ( { X = 1 ,Y = 1 } ) = P ( { (3 , 2) , (4 , 2) , (5 , 2) , (6 , 2) , (3 , 4) , (4 , 4) , (5 , 4) , (6 , 4) , (3 , 6) , (4 , 6) , (5 , 6) , (6 , 6) } ) = 12 36 = 1 3 4.4 a. R 1 R 2 C ( x + 2 y ) dxdy = 4 C = 1, thus C = 1 4 . b. f X ( x ) = ± R 1 1 4 ( x + 2 y ) dy = 1 4 ( x + 1) < x < 2 otherwise c. F XY ( x,y ) = P ( X ≤ x,Y ≤ y ) = R x-∞ R y-∞ f ( v,u ) dvdu . The way this integral is calculated depends on the values of x and y . For example, for 0 < x < 2 and 0 < y < 1, F XY ( x,y ) = Z x-∞ Z y-∞ f ( u,v ) dvdu = Z x Z y 1 4 ( u + 2 v ) dvdu = x 2 y 8 + y 2 x 4 . But for 0 < x < 2 and 1 ≤ y , F XY ( x,y ) = Z x-∞ Z y-∞ f ( u,v ) dvdu = Z x Z 1 1 4 ( u + 2 v ) dvdu = x 2 8 + x 4 ....
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