{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutions2006

# solutions2006 - Statistics 351Probability I Fall...

This preview shows pages 1–4. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
3. (b) Note that det 12 - 3 - 3 1 = 12 - 9 = 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern