p2soln09 - STAT 330 SOLUTIONS Part II 15. (a) 1 k = PP y =0...

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STAT 330 SOLUTIONS Part II 15 . ( a ) 1 k = P y =0 P x =0 q 2 p x + y = q 2 P y =0 p y μ P x =0 p x = q 2 P y =0 p y μ 1 1 p by the Geometric Series since 0 <p< 1 = q P y =0 p y since q =1 p = q μ 1 1 p by the Geometric Series Therefore k . 15 . ( b ) f 1 ( x )= P ( X = x P y f ( x,y P y =0 q 2 p x + y = q 2 p x à P y =0 p y ! = q 2 p x μ 1 1 p by the Geometric Series = qp x ,x =0 , 1 ,... By symmetry f 2 ( y y ,y , 1 The support set of ( X,Y ) is A = { ( ): x , 1 ; y , 1 } .S in ce f ( f 1 ( x ) · f 2 ( y ) for all ( ) A therefore X and Y are independent random variables. 15 . ( c ) P ( X = x | X + Y = t P ( X = x,X + Y = t ) P ( X + Y = t ) = P ( X = x, Y = t x ) P ( X + Y = t ) . Now P ( X + Y = t P ( x,y ): P x + y = t q 2 p x + y = q 2 t P x =0 p x +( t x ) = q 2 p t t P x =0 1 = q 2 p t ( t +1) ,t , 1 Therefore P ( X = x | X + Y = t q 2 p x +( t x ) q 2 p t ( t = 1 t +1 , 1 ,...,t. 1
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17 . ( a ) The support of ( X,Y ) is pictured below: x x y 0 1 1 -1 (x,1-x 2 ) (sqrt(1-y),y) (-sqrt(1-y),y) y=1-x 2 Figure 1: 1= Z −∞ Z −∞ f ( x,y ) dxdy = k 1 Z x = 1 1 x 2 Z y =0 ¡ x 2 + y ¢ dydx = k 1 Z 1 x 2 y + 1 2 y 2 | 1 x 2 0 ¸ dx = k 1 Z 1 x 2 ¡ 1 x 2 ¢ + 1 2 ¡ 1 x 2 ¢ 2 ¸ dx = k 1 Z 0 h 2 x 2 ¡ 1 x 2 ¢ + ¡ 1 x 2 ¢ 2 i dx by symmetry = k 1 Z 0 ¡ 1 x 4 ¢ dx = k x 1 5 x 5 | 1 0 ¸ = 4 5 k and thus k = 5 4 . Therefore f ( )= 5 4 ¡ x 2 + y ¢ , 0 <y< 1 x 2 , 1 <x< 1 or f ( 5 4 ¡ x 2 + y ¢ , p 1 y<x< p 1 y, 0 1 . 2
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17 . ( b ) The marginal p.d.f. of X is f 1 ( x )= Z −∞ f ( x,y ) dy = 5 4 1 x 2 Z 0 ¡ x 2 + y ¢ dy = 5 8 ¡ 1 x 4 ¢ , 1 <x< 1 The marginal p.d.f. of Y is f 2 ( y Z −∞ f ( ) dx = 5 4 1 y Z 1 y ¡ x 2 + y ¢ dx = 5 2 1 y Z 0 ¡ x 2 + y ¢ dx because of symmetry = 5 2 1 3 x 3 + yx | 1 y 0 ¸ = 5 2 1 3 (1 y ) 3 / 2 + y (1 y ) 1 / 2 ¸ = 5 6 (1 y ) 1 / 2 [(1 y )+3 y ] = 5 6 (1 y ) 1 / 2 (1 + 2 y ) , 0 <y< 1 X and Y are not independent random variables since (for example) f μ 3 4 , 1 2 =0 6 = f 1 μ 3 4 · f 2 μ 1 2 > 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 = © ( ):0 1 x 2 , 1 1 ª .S ince 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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17 . ( c ) x x y 1 1 -1 0 y=x+1 y=1-x 2 (x,x+1) (x,1-x 2 ) | Figure 2: P ( Y X +1) = 1 P ( Y X +1)=1 0 Z x = 1 1 x 2 Z y = x +1 5 4 ¡ x 2 + y ¢ dydx =1 5 4 0 Z x = 1 x 2 y + 1 2 y 2 | 1 x 2 x +1 ¸ dx 5 8 0 Z 1 n [2 x 2 (1 x 2 )+ ¡ 1 x 2 ¢ 2 ] [2 x 2 ( x +1)+( x +1) 2 ] o dx 5 8 0 Z 1 £ x 4 2 x 3 3 x 2 2 x ¤ dx =1+ 5 8 1 5 x 5 + 1 2 x 4 + x 3 + x 2 | 0 1 ¸ 5 8 1 5 ( 1) + 1 2 +( 1) + 1 ¸ 5 8 μ 2+5 10 3 16 = 13 16 4
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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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This note was uploaded on 02/22/2010 for the course STAT 330 taught by Professor Paulasmith during the Fall '08 term at Waterloo.

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p2soln09 - STAT 330 SOLUTIONS Part II 15. (a) 1 k = PP y =0...

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