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s330p2soln08 - STAT 330 SOLUTIONS Part II 13(a 1 k = XX X y...

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STAT 330 SOLUTIONS Part II 13 . ( a ) 1 k = X y =0 X x =0 q 2 p x + y = q 2 X y =0 p y à X x =0 p x ! = q 2 X y =0 p y μ 1 1 p by the Geometric Series since 0 < p < 1 = q X y =0 p y since q = 1 p = q μ 1 1 p by the Geometric Series = 1 Therefore k = 1 . 13 . ( b ) f 1 ( x ) = P ( X = x ) = X y f ( x, y ) = X y =0 q 2 p x + y = q 2 p x à X 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 ) = qp y , y = 0 , 1 , ... The support set of ( X, Y ) is A = { ( x, y ) : x = 0 , 1 , ... ; y = 0 , 1 , ... } . Since f ( x, y ) = f 1 ( x ) · f 2 ( y ) for all ( x, y ) A therefore X and Y are independent random variables. 13 . ( 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 ) = X ( x,y ): X x + y = t q 2 p x + y = q 2 t X x =0 p x +( t x ) = q 2 p t t X x =0 1 = q 2 p t ( t + 1) , t = 0 , 1 , ... Therefore P ( X = x | X + Y = t ) = q 2 p x +( t x ) q 2 p t ( t + 1) = 1 t + 1 , x = 0 , 1 , ..., t. 1
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14 . ( a ) f ( x, y ) = e 2 x ! ( y x )! , x = 0 , 1 , ..., y ; y = 0 , 1 , ... OR f ( x, y ) = e 2 x ! ( y x )! , y = x, x + 1 , ... ; x = 0 , 1 , ... f 1 ( x ) = X y f ( x, y ) = X y = x e 2 x ! ( y x )! = e 2 x ! X y = x 1 ( y x )! let k = y x = e 2 x ! X k =0 1 k ! = e 2 e 1 x ! = e 1 x ! x = 0 , 1 , ... by the Exponential Series Note that X v POI (1) . f 2 ( y ) = X x f ( x, y ) = y X x =0 e 2 x ! ( y x )! = e 2 y ! y X x =0 y ! x !( y x )! = e 2 y ! y X x =0 μ y x 1 x = e 2 y ! (1 + 1) y by the Binomial Series = 2 y e 2 y ! y = 0 , 1 , ... Note that Y v POI (2) . 14 . ( 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 = 0 , 1 , ... } , the support set of Y is A 2 = { y : y = 0 , 1 , ... } and the support set of ( X, Y ) is A = { ( x, y ) : x = 0 , 1 , ..., y ; y = 0 , 1 , ... } . Since A 6 = A 1 × A 2 therefore by the Factorization Theorem for Independence (FTI) X and Y are not independent random variables. 2
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15 . ( 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 Z 1 x = 1 Z 1 x 2 y =0 ¡ x 2 + y ¢ dydx = k Z 1 1 x 2 y + 1 2 y 2 | 1 x 2 0 ¸ dx = k Z 1 1 x 2 ¡ 1 x 2 ¢ + 1 2 ¡ 1 x 2 ¢ 2 ¸ dx = k Z 1 0 h 2 x 2 ¡ 1 x 2 ¢ + ¡ 1 x 2 ¢ 2 i dx by symmetry = k Z 1 0 ¡ 1 x 4 ¢ dx = k x 1 5 x 5 | 1 0 ¸ = 4 5 k Therefore k = 5 / 4 and f ( x, y ) = 5 4 ¡ x 2 + y ¢ , 0 < y < 1 x 2 , 1 < x < 1 or f ( x, y ) = 5 4 ¡ x 2 + y ¢ , p 1 y < x < p 1 y, 0 < y < 1 . 3
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15 . ( b ) The marginal p.d.f. of X is f 1 ( x ) = Z −∞ f ( x, y ) dy = 5 4 Z 1 x 2 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 ( x, y ) dx = 5 4 Z 1 y 1 y ¡ x 2 + y ¢ dx = 5 2 Z 1 y 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 < x < 1 } , the support set of Y is A 2 = { y : 0 < y < 1 } and the support set of ( X, Y ) is A = © ( x, y ) : 0 < y < 1 x 2 , 1 < x < 1 ª .
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