p2soln10 - STAT 330 SOLUTIONS Part II 16. (a) f (x, y ) =...

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STAT 330 SOLUTIONS Part II 16 . ( a ) f ( x,y )= e 2 x !( y x )! ,x =0 , 1 ,...,y ; y , 1 ,... OR f ( e 2 x y x )! ,y = x,x +1 ; x , 1 f 1 ( x P y f ( P y = x e 2 x y x )! = e 2 x ! P y = x 1 ( y x )! let k = y x = e 2 x ! P k =0 1 k ! = e 2 e 1 x ! = e 1 x ! x , 1 by the Exponential Series Note that X v POI (1) . f 2 ( y P x f ( x, y y P x =0 e 2 x y x )! = e 2 y ! y P x =0 y ! x !( y x )! = e 2 y ! y P x =0 μ y x 1 x = e 2 y ! (1 + 1) y by the Binomial Series = 2 y e 2 y ! y , 1 Note that Y v POI (2) . 16 . ( b ) Since (for example) f (1 , 2) = e 2 6 = f 1 (1) f 2 (2) = e 1 2 2 e 2 2! =2 e 3 therefore X and Y are not independent random variables. OR The support set of X is A 1 = { x : x , 1 } , the support set of Y is A 2 = { y : y , 1 } and the support set of ( X,Y ) is A = { ( ): x , 1 ; y , 1 } . Since A 6 = A 1 × A 2 therefore by the Factorization Theorem for Independence (FTI) X and Y are not independent random variables. 1
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18 . ( a ) The support set of ( X,Y ) is pictured below: x x y y y=x 2 1 1 -1 0 (x,1) (x,x 2 ) (sqrt(y),y) (-sqrt(y),y) | | | Figure 1: 1 k = 1 Z x = 1 1 Z y = x 2 x 2 ydydx = 1 Z x = 1 x 2 1 2 y 2 | 1 x 2 ¸ dx = 1 2 1 Z x = 1 x 2 ¡ 1 x 4 ¢ dx = 1 2 1 3 x 3 1 7 x 7 | 1 1 ¸ = 1 2 1 3 1 7 1 3 ( 1) + 1 7 ( 1) ¸ = 1 3 1 7 = 4 21 Therefore k =21 / 4 and f ( x,y )= 21 4 x 2 y, x 2 <y< 1 , 1 <x< 1 or f ( 21 4 x 2 y<x< 0 1 . 2
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18 . ( b ) The marginal p.d.f. of X is f 1 ( x )= 21 4 1 Z x 2 x 2 ydy = 21 x 2 8 h y 2 | 1 x 2 i = 21 x 2 8 ¡ 1 x 4 ¢ , 1 <x< 1 . The marginal p.d.f. of Y is f 2 ( y 21 4 y Z x = y x 2 ydx = 21 2 Z y 0 x 2 ydx because of symmetry = 7 2 y h x 3 | y 0 i = 7 2 y ³ y 3 / 2 ´ = 7 2 y 5 / 2 , 0 <y< 1 . X and Y are not independent random variables since (for example) f μ 1 2 , 1 10 =0 6 = f 1 μ 1 2 · f 2 μ 1 10 > 0 . OR The support set of X is A 1 = { x : 1 1 } , the support set of Y is A 2 = { y :0 1 } and the support set of ( X,Y ) is A = © ( x,y ):0 <x 2 1 ª . Since A 6 = A 1 × A 2 therefore by the Factorization Theorem for Independence (FTI) X and Y are not independent random variables. 3
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18 . ( c ) x 0 y 1 -1 1 | | x | y=x 2 y=x (x,x 2 ) (x,x) Figure 2: P ( X Y )= 1 Z x =0 x Z y = x 2 21 4 x 2 ydydx = 21 8 1 Z 0 x 2 £ y 2 | x x 2 ¤ dx = 21 8 1 Z 0 x 2 ¡ x 2 x 4 ¢ dx = 21 8 1 Z 0 ¡ x 4 x 6 ¢ dx = 21 8 1 5 x 5 1 7 x 7 | 1 0 ¸ = 21 8 μ 7 5 35 = 3 20 4
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18 . ( d ) The conditional p.d.f. of X given Y = y is f 1 ( x | y )= f ( x,y ) f 2 ( y ) = 21 4 x 2 y 7 2 y 5 / 2 = 3 2 x 2 y 3 / 2 , y<x< y, 0 <y< 1 . Check: Z −∞ f 1 ( x | y ) dx = 1 2 y 3 / 2 y Z y 3 x 2 dx = y 3 / 2 h x 3 | y 0 i = y 3 / 2 y 3 / 2 =1 . The conditional p.d.f. of Y given X = x is f 2 ( y | x f ( ) f 1 ( x ) = 21 4 x 2 y 21 x 2 8 (1 x 4 ) = 2 y (1 x 4 ) ,x 2 1 , 1 <x< 1 . Check: Z −∞ f 2 ( y | x ) dy = 1 (1 x 4 ) 1 Z x 2 2 ydy = 1 (1 x 4 ) £ y 2 | 1 x 2 ¤ = 1 x 4 1 x 4 . 5
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20 . ( d )( i ) The support of ( X,Y ) is pictured below: x y y=x 0 1 1 (x,x) (y,y) x y Figure 3: 1= Z −∞ Z −∞ f ( x,y ) dxdy = 1 Z y =0 y Z x =0 k ( x + y ) dxdy = k 1 Z y =0 μ 1 2 x 2 + xy | y x =0 ¸ dy = k 1 Z 0 μ 1 2 y 2 + y 2 dx = k 1 Z 0 3 2 y 2 dy = k 2 £ y 3 | 1 0 ¤ = k 2 .
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p2soln10 - STAT 330 SOLUTIONS Part II 16. (a) f (x, y ) =...

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