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Unformatted text preview: 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 Z ∞∞ Z ∞∞ f X,Y ( x,y ) dx dy = Z 1 Z y 8 xy dx dy = Z 1 4 y 3 dy = y 4 1 = 1 . Thus, f X,Y is a legitimate density. 1. (b) We compute f X ( x ) = Z ∞∞ f X,Y ( x,y ) dy = Z 1 x 8 xy dy = 4 x (1 x 2 ) , < x < 1 . 1. (c) We compute E ( X ) = Z ∞∞ xf X ( x ) dx = Z 1 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 ) = Z ∞∞ yf Y  X = x ( y ) dy = Z 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 ) = Z 1 2(1 x 3 ) 3(1 x 2 ) · 4 x (1 x 2 ) dx = 8 3 Z 1 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 Bμ = 1 2 1 0 1 1 = and covariance matrix B Λ B = 1 2 1 0 1 1 2 1 1 1 1 1 2 1 2 1 1 1 = 14 8 8 5 . 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....
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 Fall '08
 MichaelKozdron
 Statistics, Probability

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