hw08soln - ECE 310 Spring 2005 HW 8 solutions Problem E7.1...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 310, Spring 2005, HW 8 solutions Problem E7.1 a) lim x,y →∞ F ( x, y ) = 0 . 5 + 0 . 5 = 1 b) F Y ( y ) = lim x →∞ F ( x, y ) = 0 . 5 F 2 ( y ) + 0 . 5 F 4 ( y ) Yes, by Theorem 5.3.1 c) f ( x, y ) = 2 F ( x, y ) ∂x∂y = 0 . 5 f 1 ( x ) f 2 ( y ) + 0 . 5 f 3 ( x ) f 4 ( y ) d) Yes, since f ( x, y ) 0 and R R 2 f ( x, y ) = 1 e) Yes. Problem E7.5 a) f Y ( y ) = Z -∞ f X,Y ( x, y ) dx = = ‰ R 1 0 3 x 2 dx, if 0 y 1, 0, otherwise. = 1 , if 0 y 1, 0, otherwise. = U ( y ) U (1 - y ) b) F Y ( y ) = Pr ( Y y ) = Z y -∞ f Y ( z ) dz = yU ( y )[1 - U ( y - 1)] + U ( y - 1) c) P ( X Y ) = ZZ y x f X,Y ( x, y ) dxdy = Z 1 x =0 Z x y =0 3 x 2 dydx = = Z 1 x =0 3 x 2 [ y | x 0 ] dx = Z 1 0 3 x 3 dx = 3 x 4 4 | 1 0 = 3 4 d) f X ( x ) = 3 x 2 U ( x ) U (1 - x ) f X,Y ( x, y ) = f X ( x ) f Y ( y ) product construction 1
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 E7.12 P ( Y X + 1 2 ) = Z Z x + 1 2 y f X,Y ( x, y ) dxdy = Z 1 2 x =0 Z 1 y = x + 1 2 ( x + y ) dydx = = Z 1 2 0 ( xy + y 2 2 ) | 1 x + 1 2 dx = Z 1 2 0 [( x + 1 2 ) - ( x 2 + x 2 + 1 2 ( x 2 + x + 1 4 ))] dx = = Z 1 2 0 ( 3 8 - 3 x 2 2 ) dx = ( 3 x 8 - x 3 2 ) | 1 2 0 = 3 16 - 1 16 = 1 8 Problem E7.20 b) By Lemma 7.4.2, C 1 , 3 = 6 2 2 4 and m 1 , 3 = 0 2 Then, we have: C - 1 1 , 3 = 0 . 2 - 0 . 1 - 0 . 1 0 . 3 And the corresponding pdf: f X 1 ,X 3 ( x 1 , x 3 ) = 1 4 π 5 exp( - 1 2 [0 . 2 x 2 1 - 0 . 2 x 1 ( x
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ 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