hw5vqn - Vinh Nguyen Stat 120A(Fall 2007): HW5 Solutions...

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Vinh Nguyen Stat 120A(Fall 2007): HW5 Solutions November 12, 2007 Text: Probabilty and Statistics , by Degroot and Schervish (3 rd ed) Problems: 3.4.1, 3.4.8, 3.5.8, 3.6.3, 4.6.12, 4.8.8, and 5.12.5 3.4.1 We have the following joint pdf for ( X, Y ): f X,Y ( x, y ) = ± c 0 x 2 , 0 y 1 , 0 otherwise. (a) In order for this joint pdf to be a valid pdf, the integration over all x and y must be equal 1. Thus, 1 Z -∞ Z -∞ f X,Y ( x, y ) dx dy = Z 1 0 Z 2 0 c dx dy = Z 1 0 [ cx ] 2 0 dy = Z 1 0 2 c dy = [2 cy ] 1 0 = 2 c which implies c = 1 / 2 . (b) The following is a graph of the domain (rectangle) for our joint distribution: -0.5 0.0 0.5 1.0 1.5 2.0 2.5 -0.5 0.0 0.5 1.0 1.5 x y y=x The shaded region is the area in the domain where x y . To get P ( X Y ), we must integrate over this region, or equivalently, integrate over the triangular region and subtract this probability from 1 (law of total probability): P ( X Y ) = 1 - P ( X < Y ) = 1 - Z 1 0 Z y 0 1 2 dx dy = 1 - Z 1 0 h x 2 i x = y x =0 dy = 1 - Z 1 0 y 2 dy = 1 - ² y 2 4 ³ 1 0 = 3 4 3.4.8 Suppose ( X, Y ) take on values in the set { ( x, y ) : 0 x 3 , 0 y 4 } and have the following joint distribution over that set: F X,Y ( x, y ) = 1 156 xy ( x 2 + y ) . 1
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(a) Recall equation 3.4.5 on p124: P ( a < X < b, c < Y < d ) = F X,Y ( b, d ) - F X,Y ( a, d ) - F X,Y ( b, c ) + F X,Y ( a, c ) . Thus, we have P (1 X 2 , 1 Y 2) = F X,Y (2 , 2) - F X,Y (1 , 2) - F X,Y (2 , 1) + F X,Y (1 , 1) . (b) We have P (2 X 4 , 2 Y 4) = P (2 X 3 , 2 Y 4) = F X,Y (3 , 4) - F X,Y (2 , 4) - F X,Y (3 , 2) + F X,Y (2 , 2) . Note that X is defined only in the interval [0 , 3]. (c) Note that the maximum value
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This note was uploaded on 01/21/2010 for the course STATS 37850 taught by Professor Phelan during the Spring '09 term at UC Irvine.

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hw5vqn - Vinh Nguyen Stat 120A(Fall 2007): HW5 Solutions...

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