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solutions2007

# solutions2007 - Statistics 351Probability I Fall...

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Statistics 351–Probability I Fall 2007 (200730) Final Exam Solutions Instructor: Michael Kozdron 1. (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 y 0 8 xy dx dy = 1 0 4 y 3 dy = y 4 1 0 = 1 . Thus, f X,Y is a legitimate density. 1. (b) We compute f X ( x ) = -∞ f X,Y ( x, y ) dy = 1 x 8 xy dy = 4 x (1 - x 2 ) , 0 < x < 1 . 1. (c) We compute E ( X ) = -∞ xf X ( x ) dx = 1 0 x · 4 x (1 - x 2 ) dx = 4 3 - 4 5 = 8 15 . 1. (d) We compute f Y | X = x ( y ) = f X,Y ( x, y ) f X ( x ) = 8 xy 4 x (1 - x 2 ) = 2 y (1 - x 2 ) , x < y < 1 . 1. (e) We compute E ( Y | X = x ) = -∞ yf Y | X = x ( y ) dy = 1 x y · 2 y (1 - x 2 ) dy = 2(1 - x 3 ) 3(1 - x 2 ) . 1. (f) Using properties of conditional expectation (Theorem II.2.1), we compute E ( Y ) = E ( E ( Y | X ) ) = E 2(1 - X 3 ) 3(1 - X 2 ) = 1 0 2(1 - x 3 ) 3(1 - x 2 ) · 4 x (1 - x 2 ) dx = 8 3 1 0 x - x 4 dx = 4 5 . 2. (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 2 - 1 0 1 - 1 0 0 0 = 0 0 and covariance matrix B Λ B = 1 2 - 1 0 1 - 1 2 0 - 1 0 1 - 1 - 1 - 1 2 1 0 2 1 - 1 - 1 = 14 8 8 5 .

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2. (b) Note that det 14 8 8 5 = 70 - 64 = 6 so that 14 8 8 5 - 1 = 5 6 - 8 6 - 8 6 14 6 . Thus, we can conclude f Y 1 ,Y 2 ( y 1 , y 2 ) = 1 2 π · 1 6 exp - 1 2 5 6 y 2 1 - 8 3 y 1 y 2 + 7 3 y 2 2 . 2. (c) Since Λ = 14 8 8 5 we can immediately conclude that ϕ ( t 1 , t 2 ) = exp - 1 2 ( 14 t 2 1 + 16 t 1 t 2 + 5 t 2 2 ) . 3. (a) Using the results of Section V.6 (in particular, equation (6.2) on page 130) we know that X 2 | X 1 = x ∈ N μ 2 + ρ σ 2 σ 1 ( x - μ 1 ) , σ 2 2 (1 - ρ 2 ) . Since σ 2 1 = 1, σ 2 2 = 25, we conclude that ρ = Cov( X 1 , X 2 ) σ 1 σ 2 = α 5 . Therefore, 16 = Var( X 2 | X 1 ) = σ 2 2 (1 - ρ 2 ) = 25 1 - α 2 25 = 25 - α 2 implying that α 2 = 9. Hence, the two possible values of α are α = 3 and α = - 3. 3. (b) From (a) , we conclude that 1 = E ( X 2 | X 1 = 6) = μ 2 + ρ σ 2 σ 1 ( x - μ 1 ) = β + α 5 · 5 1 (6 - 5) = β + α. Therefore, if α = 3, then β = - 2 and if α = - 3, then β = 4. 4. (a) In order to find the eigenvalues of Λ, we must find those values of λ such that det(Λ - λI ) = 0. Therefore, det(Λ - λI ) = det 6 - λ 2 2 9 - λ = (6 - λ )(9 - λ ) - 4 = λ 2 - 15 λ +54 - 4 = λ 2 - 15 λ +50 = ( λ - 5)( λ - 10) so that the eigenvalues of Λ are λ 1 = 5 and λ 2 = 10.
4. (b) Since λ 1 = 5, - λ 1 I | 0) = 1 2 0 2 4 0 1 2 0 0 0 0 and since λ 2 = 10, - λ 2 I | 0) = - 4 2 0 2 - 1 0 2 - 1 0 0 0 0

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