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Problem Solutions – Chapter 4 Problem 4.1.1 Solution (a) The probability P [ X 2 , Y 3 ] can be found be evaluating the joint CDF F X , Y ( x , y ) at x = 2 and y = 3. This yields P [ X 2 , Y 3] = F X , Y ( 2 , 3 ) = ( 1 e 2 )( 1 e 3 ) (1) (b) To find the marginal CDF of X, F X ( x ) , we simply evaluate the joint CDF at y =∞ . F X ( x ) = F X , Y ( x , ) = ± 1 e x x 0 0 otherwise (2) (c) Likewise for the marginal CDF of Y , we evaluate the joint CDF at X . F Y ( y ) = F X , Y ( , y ) = ± 1 e y y 0 0 otherwise (3) Problem 4.1.2 Solution (a) Because the probability that any random variable is less than −∞ is zero, we have F X , Y ( x , −∞ ) = P [ X x , Y ≤−∞ ] P [ Y ] = 0 (1) (b) The probability that any random variable is less than infinity is always one. F X , Y ( x , ) = P [ X x , Y ≤∞ ] = P [ X x ] = F X ( x ) (2) (c) Although P [ Y ≤∞]= 1, P [ X ≤−∞]= 0. Therefore the following is true. F X , Y ( −∞ , ) = P [ X , Y ] P [ X ] = 0 (3) (d) Part (d) follows the same logic as that of part (a). F X , Y ( −∞ , y ) = P [ X , Y y ] P [ X ] = 0 (4) (e) Analogous to Part (b), we find that F X , Y ( , y ) = P [ X , Y y ] = P [ Y y ] = F Y ( y ) (5) 130
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Problem 4.1.3 Solution We wish to find P [ x 1 X x 2 ] or P [ y 1 Y y 2 ] . We define events A ={ y 1 Y y 2 } and B y 1 Y y 2 } so that P [ A B ] = P [ A ] + P [ B ] P [ AB ] (1) Keep in mind that the intersection of events A and B are all the outcomes such that both A and B occur, specifically, x 1 X x 2 , y 1 Y y 2 } . It follows that P [ A B ] = P [ x 1 X x 2 ] + P [ y 1 Y y 2 ] P [ x 1 X x 2 , y 1 Y y 2 ] . (2) By Theorem 4.5, P [ x 1 X x 2 , y 1 Y y 2 ] = F X , Y ( x 2 , y 2 ) F X , Y ( x 2 , y 1 ) F X , Y ( x 1 , y 2 ) + F X , Y ( x 1 , y 1 ) . (3) Expressed in terms of the marginal and joint CDFs, P [ A B ] = F X ( x 2 ) F X ( x 1 ) + F Y ( y 2 ) F Y ( y 1 ) (4) F X , Y ( x 2 , y 2 ) + F X , Y ( x 2 , y 1 ) + F X , Y ( x 1 , y 2 ) F X , Y ( x 1 , y 1 ) (5) Problem 4.1.4 Solution Its easy to show that the properties of Theorem 4.1 are satisfied. However, those properties are necessary but not sufficient to show F ( x , y ) is a CDF. To convince ourselves that F ( x , y ) is a valid CDF, we show that for all x 1 x 2 and y 1 y 2 , P [ x 1 < X 1 x 2 , y 1 < Y y 2 ] 0 (1) In this case, for x 1 x 2 and y 1 y 2 , Theorem 4.5 yields P [ x 1 < X x 2 , y 1 < Y y 2 ] = F ( x 2 , y 2 ) F ( x 1 , y 2 ) F ( x 2 , y 1 ) + F ( x 1 , y 1 ) (2) = F X ( x 2 ) F Y ( y 2 ) F X ( x 1 ) F Y ( y 2 ) (3) F X ( x 2 ) F Y ( y 1 ) + F X ( x 1 ) F Y ( y 1 ) (4) =[ F X ( x 2 ) F X ( x 1 ) ][ F Y ( y 2 ) F Y ( y 1 ) ] (5) 0 (6) Hence, F X ( x ) F Y ( y ) is a valid joint CDF. Problem 4.1.5 Solution In this problem, we prove Theorem 4.5 which states P [ x 1 < X x 2 , y 1 < Y y 2 ] = F X , Y ( x 2 , y 2 ) F X , Y ( x 2 , y 1 ) (1) F X , Y ( x 1 , y 2 ) + F X , Y ( x 1 , y 1 ) (2) 131
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(a) The events A , B , and C are Y X x 1 y 1 y 2 Y X x 1 x 2 y 1 y 2 Y X x 1 x 2 y 1 y 2 ABC (3) (b) In terms of the joint CDF F X , Y ( x , y ) , we can write P [ A ] = F X , Y ( x 1 , y 2 ) F X , Y ( x 1 , y 1 ) (4) P [ B ] = F X , Y ( x 2 , y 1 ) F X , Y ( x 1 , y 1 ) (5) P [ A B C ] = F X , Y ( x 2 , y 2 ) F X , Y ( x 1 , y 1 ) (6) (c) Since A , B , and C are mutually exclusive, P [ A B C ] = P [ A ] + P [ B ] + P [ C ] (7) However, since we want to express P [ C ] = P [ x 1 < X x 2 , y 1 < Y y 2 ] (8) in terms of the joint CDF F X , Y ( x , y ) , we write P [ C ] = P [ A B
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