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solutions2006 - Statistics 351Probability I Fall...

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Statistics 351–Probability 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 | = θ - α - β Γ( α )Γ( β ) ( uv ) α - 1 ( v - uv ) β - 1 exp - v θ v = θ - α - β Γ( α )Γ( β ) u α - 1 (1 - u ) β - 1 v α + β - 1 exp - v θ = Γ( α + β ) Γ( α )Γ( β ) u α - 1 (1 - u ) β - 1 · θ - α - β Γ( α + β ) v α + β - 1 exp - v θ 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 ) = Γ( α + β ) Γ( α )Γ( β ) u α - 1 (1 - u ) β - 1 , 0 < u < 1 which we recognize as the density of a Beta( α, β ) random variable. 2. (a) We see that f X,Y ( x, y ) 0 for all x , y , and that -∞ -∞ f X,Y ( x, y ) dx dy = 1 0 x 0 12 y 2 dy dx = 1 0 4 x 3 dx = x 4 1 0 = 1 . Thus, f X,Y is a legitimate density. 2. (b) We compute f X ( x ) = -∞ f X,Y ( x, y ) dy = x 0 12 y 2 dy = 4 x 3 , 0 < x < 1 . 2. (c) We compute f Y ( y ) = -∞ f X,Y ( x, y ) dx = 1 y 12 y 2 dx = 12 y 2 (1 - y ) , 0 < 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 .
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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 , 0 < y < x. 2. (f) We compute E ( X ) = -∞ xf X ( x ) dx = 1 0 x · 4 x 3 dx = 4 5 . 2. (g) We compute E ( Y | X = x ) = -∞ yf Y | X = x ( y ) dy = x 0 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 ) . Since E ( X 2 ) = -∞ x 2 f X ( x ) dx = 1 0 x 2 · 4 x 3 dx = 4 6 = 2 3 , we conclude E ( XY ) = 3 4 · 2 3 = 1 2 . Solution 2: By definition, E ( XY ) = -∞ -∞ xyf X,Y ( x, y ) dx dy = 1 0 x 0 xy · 12 y 2 dy dx = 12 1 0 x x 0 y 3 dy dx = 12 1 0 x · 1 4 x 4 dx = 3 1 0 x 5 dx = 3 6 = 1 2 . 3. (a) Let B = 1 - 2 1 0 1 1 so that Y = B X . By Theorem V.3.1, Y is MVN with mean = 1 0 1 0 2 0 0 0 0 = 0 0 and covariance matrix B Λ B = 1 - 2 1 0 1 1 1 0 - 1 0 2 - 1 - 1 - 1 1 1 0 - 2 1 1 1 = 12 - 3 - 3 1 .
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3. (b) Note that det 12 - 3 - 3 1 = 12 - 9 = 3
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