17s418hw6sol.pdf - Stat 418 HW#6 Problem 6.8 Ry(a fY(y =...

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

Stat 418 HW #6 Problem 6.8 (a) f Y ( y ) = c y R - y ( y 2 - x 2 ) e - y dx = 4 3 cy 3 e - y , 0 < y < Since R 0 f Y ( y ) dy = 1, c = 1 8 (b) f X ( x ) = R | x | 1 8 ( y 2 - x 2 ) e - y dy = 1 4 e -| x | (1 + | x | ) -∞ < x < ( by integration by parts, R y 2 e - y dy = - e - y ( y 2 + 2 y + 2) ) f Y ( y ) = 1 6 y 3 e - y 0 < y < (c) E[X]=0, since f X ( x ) is even function. Problem 6.19 1 R 0 x R 0 1 x dydx = 1 R 0 dx = 1 (a) 1 R y 1 x dx = - log y 0 < y < 1 (b) x R 0 1 x dy = 1 0 < x < 1 (c) E [ X ] = 1 2 (d) E [ y ] = 1 R 0 - y log ydy = 1 4 (integrating by parts, R y log ydy = 1 2 y 2 log y - 1 2 R ydy ) 1
Image of page 1

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

Problem 6.22 (a) No, since the joint density does not factor. ( f ( x, y ) 6 = f ( x ) f ( y )) (b) f X ( x ) = 1 R 0 ( x + y ) dy = x + 1 2 0 < x < 1 (c) P ( X + Y < 1) = R 1 0 R 1 - x 0 ( x + y ) dydx = 1 3 Problem 6.28 Let service time for A.J’s car = X and service time for M.J’s car = Y. Then, (a) P ( Y + t < X | X > t ) * P ( X > t ) = 1 2 * e - t (b) P ( X + Y < 2) = R 2 0 R 2 - x 0 e - ( x + y ) dydx = 1 - 3 e - 2 Problem 6.39 P ( Y | X = 1) = 1 y=1 P ( Y | X = 2) = ( 2 / 36 3 / 36 = 2 3 y=1 1 / 36 3 / 36 = 1 3 y=2 P ( Y | X = 3) = 2 / 36 5 / 36 = 2 5 y=1 2 / 36 5 / 36 = 2 5 y=2 1 / 36 5 / 36 = 1 5 y=3 P ( Y | X = 4) = 2 / 36 7 / 36 = 2 7 y=1 2 / 36 7 / 36 = 2 7 y=2 2 / 36 7 / 36 = 2 7 y=3 1 / 36 7 / 36 = 1 7 y=4 P ( Y | X = 5) = 2 / 36 9 / 36 = 2 9 y=1 2 / 36 9 / 36 = 2 9 y=2 2 / 36 9 / 36 = 2 9 y=3 2 / 36 9 / 36 = 2 9 y=4 1 / 36 9 / 36 = 1 9 y=5 P ( Y | X = 6) = 2 / 36 11 / 36 = 2 11 y=1 2 / 36 11 / 36 = 2 11 y=2 2 / 36 11 / 36 = 2 11 y=3 2 / 36 11 / 36 = 2 11 y=4 2 / 36 11 / 36 = 2 11 y=5 1 / 36 11 / 36 = 1 11 y=6 P(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