h4soln - Homework 4 Solutions The George Washington...

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

View Full Document Right Arrow Icon
Homework # 4 Solutions The George Washington University ApSc 116 1 / 11
Image of page 1

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

View Full Document Right Arrow Icon
Problem (Page 136, # 7) Suppose that the joint p.d.f. of X and Y is as follows: f ( x , y ) = 2 xe - y for 0 x 1 and 0 < y < , 0 otherwise. Are X and Y independent? Solution: Since f ( x , y ) = 0 outside a rectangle and f ( x , y ) can be factored as in Eq. (3.5.7), i.e., X and Y are independent iff f ( x , y ) = g 1 ( X ) g 2 ( Y ) , for some non-negative functions g 1 and g 2 . Then inside the rectangle, using g 1 ( x ) = 2 x and g 2 ( y ) = exp ( - y ), it follows that X and Y are independent. 2 / 11
Image of page 2
Problem (Page 136, # 10) Suppose that a point ( X , Y ) is chosen at random from the circle S defined as follows S = { ( x , y ) : x 2 + y 2 1 } . (a) Determine the joint p.d.f. of X and Y , the marginal p.d.f. of X, and the marginal p.d.f. of Y . (b) Are X and Y independent? Solution: (a) f ( x , y ) is constant over the circle S . The area of S is π units, and it follows that f ( x , y ) = 1 inside S and f ( x , y ) = 0 outside S . Next, the possible values of x range from -1 to 1/ For any value x in this interval, f ( x , y ) > 0 only for values of y between - (1 - x 2 ) 1 / 2 and (1 - x 2 ) 2 . Hence for - 1 x 1, f 1 ( x ) = (1 - x 2 ) 1 / 2 - (1 - x 2 ) 1 / 2 1 π dy = 2 π (1 - x 2 ) 1 / 2 .
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.
  • Fall '08
  • TomMazzuchi
  • Probability distribution, Probability theory, probability density function, George Washington University, marginal p.d.f.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern