solutions2006 - Statistics 351Probability I Fall 2006...

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Unformatted text preview: Statistics 351Probability I Fall 2006 (200630) Final Exam Solutions Instructor: Michael Kozdron 1. (a) Solving for X and Y gives X = UV and Y = V- UV , so that the Jacobian of this transformation is J = x u x v y u y v = v u- v 1- u = v- vu + vu = v. By Theorem I.2.1, the joint density of ( U,V ) is therefore given by f U,V ( u,v ) = f X,Y ( uv,uv- uv ) | J | = - - ( )( ) ( uv ) - 1 ( v- uv ) - 1 exp n- v o v = - - ( )( ) u - 1 (1- u ) - 1 v + - 1 exp n- v o = ( + ) ( )( ) u - 1 (1- u ) - 1 - - ( + ) v + - 1 exp n- v o provided that 0 < u < 1 and 0 < v < . 1. (b) We recognize that the joint density for U and V can be factored as a product of the densities for U and V , respectively. Thus, f U ( u ) = ( + ) ( )( ) u - 1 (1- u ) - 1 , < u < 1 which we recognize as the density of a Beta( , ) random variable. 2. (a) We see that f X,Y ( x,y ) 0 for all x , y , and that Z - Z - f X,Y ( x,y ) dx dy = Z 1 Z x 12 y 2 dy dx = Z 1 4 x 3 dx = x 4 1 = 1 . Thus, f X,Y is a legitimate density. 2. (b) We compute f X ( x ) = Z - f X,Y ( x,y ) dy = Z x 12 y 2 dy = 4 x 3 , < x < 1 . 2. (c) We compute f Y ( y ) = Z - f X,Y ( x,y ) dx = Z 1 y 12 y 2 dx = 12 y 2 (1- y ) , < y < 1 . 2. (d) We compute f X | Y = y ( x ) = f X,Y ( x,y ) f Y ( y ) = 12 y 2 12 y 2 (1- y ) = 1 1- y , y < x < 1 . 2. (e) We compute f Y | X = x ( y ) = f X,Y ( x,y ) f X ( x ) = 12 y 2 4 x 3 = 3 y 2 x 3 , < y < x. 2. (f) We compute E ( X ) = Z - xf X ( x ) dx = Z 1 x 4 x 3 dx = 4 5 . 2. (g) We compute E ( Y | X = x ) = Z - yf Y | X = x ( y ) dy = Z x y 3 y 2 x 3 dy = 3 x 4 4 x 3 = 3 4 x. 2. (h) Using properties of conditional expectation (Theorem II.2.1), we compute E ( Y ) = E ( E ( Y | X )) = E 3 4 X = 3 4 E ( X ) = 3 4 4 5 = 3 5 . 2. (i) Solution 1: Using properties of conditional expectation (Theorem II.2.2) gives E ( XY ) = E ( E ( XY | X )) = E ( X E ( Y | X )) = E X 3 4 X = 3 4 E ( X 2 ) ....
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solutions2006 - Statistics 351Probability I Fall 2006...

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