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solutions04 - Stat 351 Fall 2007 Solutions to Assignment#4...

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Stat 351 Fall 2007 Solutions to Assignment #4 Problem #3, page 55: Suppose that X + Y = 2. By definition of conditional density, f X | X + Y =2 ( x ) = f X,X + Y ( x, 2) f X + Y (2) . We now find the joint density f X,X + Y ( x, 2). Let U = X and V = X + Y so that X = U and Y = V - U . The Jacobian of this transformation is J = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v = 1 0 - 1 1 = 1 . Since X and Y are independent Γ(2 , a ), the joint density of ( X, Y ) is f X,Y ( x, y ) = f X ( x ) · f Y ( y ) = xy a 4 e - ( x + y ) /a , for x > 0 , y > 0 , 0 , otherwise , The joint density of ( U, V ) is therefore given by f U,V ( u, v ) = f X,Y ( u, v - u ) · | J | = u ( v - u ) a 4 e - v/a provided that u > 0 and v > u . The marginal density for V is therefore f V ( v ) = v 0 u ( v - u ) a 4 e - v/a du = a - 4 e - v/a v 0 u ( v - u ) du = v 3 6 a 4 e - v/a , u > 0 . Since V = X + Y , we can rewrite these densities as f X,X + Y ( x, 2) = x (2 - x ) a 4 e - 2 /a , 0 < x < 2, and f X + Y (2) = 2 3 6 a 4 e - 2 /a . Finally, we conclude f X | X + Y =2 ( x ) = f X,X + Y ( x, 2) f X + Y (2) = x (2 - x ) a 4 e - 2 /a 2 3 6 a 4 e - 2 /a = 3 x (2 - x ) 4 provided that 0 < x < 2. Problem #9, page 56 : (a) The density function for Y is given by f Y ( y ) = 0 x 2 2 y 3 · e - x y dx provided that 0 < y < 1. Let u = - x y so that du = - 1 y dx , from which it follows that f Y ( y ) = 0 x 2 2 y 3 · e - x y dx = 1 2 0 u 2 e - u du = 1 2 Γ(3) = 2! 2 = 1 . That is, Y U (0 , 1).
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(b) The conditional density of X given Y = y is therefore f X | Y = y ( x ) = f X,Y ( x, y ) f Y ( y ) = x 2 2 y 3 · e - x y 1 = x 2 2 y 3 · e - x y provided that x > 0. That is, X | Y = y Γ(3 , y ). (c) Since Y U (0 , 1), we know that E ( Y ) = 1 2 and Var( Y ) = 1 12 . We also use the fact from page 260 that the mean of a Γ( p, a ) random variable is pa and the variance is pa 2 . Thus, we find that the mean of X is E ( X ) = E ( E ( X | Y )) = E (3 Y ) = 3 E ( Y ) = 3 2 and the variance of X is Var( X ) = Var( E ( X | Y )) + E (Var( X | Y )) = Var(3 Y ) + E (3 Y 2 ) = 9 Var( Y ) + 3 E ( Y 2 ) = 9 Var( Y ) + 3 Var( Y ) + ( E ( Y )) 2 = 9 12 + 3 1 12 + 1 4 = 7
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