{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw4sol - AMS316 Homework 4 Solution 1 Since | < 1 we have...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
AMS316 Homework 4 Solution 1. Since | α | < 1, we have X t = Z t (1 - α ß) = Z t + αZ t - 1 + α 2 Z t - 2 + ... and V ar ( X t ) = σ 2 Z (1 - α 2 ) Furthermore, γ ( k ) = E [ X t X t + k ] = E [ X α i Z t - i X α j Z t + k - j ] = σ 2 Z X i =0 α i α ( k + i ) = α k σ 2 Z (1 - α 2 ) Hence ρ ( k ) = α | k | k = 0 , ± 1 , ± 2 , ... and ρ ( k ) = 0 . 7 | k | . 2. The roots of equation y 2 - 1 3 y - 2 9 = 0 are 2 / 3 and -1 / 3, then the ACFs are given by ρ ( k ) = A 1 ( 2 3 ) | k | + A 2 ( - 1 3 ) | k | . Since A 1 + A 2 = ρ (0) = 1 and ρ (1) = 2 3 A 1 - 1 3 A 2 and ρ (1) = 3 / 7, solving for A 1 and A 2 gives A 1 = 16 21 and A 2 = 5 21 . 3. The roots of equation y 2 - 1 12 y - 1 12 = 0 are 1 / 3 and - 1 / 4. Then the ACFs are given by ρ ( k ) = A 1 ( 1 3 ) | k | + A 2 ( - 1 4 ) | k | . 1
Image of page 1

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

View Full Document Right Arrow Icon
Since A 1 + A 2 = ρ (0) = 1 and ρ (1) = 1 3 A 1 - 1 4 A 2 and ρ (1) = 1 / 11, solving for A 1 and A 2 gives A 1 = 45 77 and A 2 = 32 77 . 4. (a) The root of 1 - 0 . 3 B = 0 is 1 / 0 . 3, which lies outside the unit circle, so X t is stationary. AR process is always invertible. X t = ( 1 1
Image of page 2
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