TimeSeriesBook.pdf

The second part of the equation system consists again

Info icon This preview shows pages 62–65. Sign up to view the full content.

View Full Document Right Arrow Icon
The second part of the equation system consists again of a linear homo- geneous difference equation in γ ( h ) whereas the first part can be used to determine the initial conditions. Note that the initial conditions depend ψ 1 , . . . , ψ q which have to be determined before hand. The general solution of the difference equation is: γ ( h ) = c 1 z - h 1 + . . . + c p z - h p (2.4) where z 1 , . . . , z p are the distinct roots of the polynomial Φ( z ) = 1 - φ 1 z - . . . - φ p z p = 0. 6 The constants c 1 , . . . , c p can be computed from the first p initial conditions after the ψ 1 , . . . ψ q have been calculated like in the first procedure. The form of the solution shows that the autocovariance and hence the autocorrelation function converges to zero exponentially fast. A numerical example We consider the same example as before. The second part of the above equation system delivers a difference equation for γ ( h ): γ ( h ) = φ 1 γ ( h - 1) + φ 2 γ ( h - 2) = 1 . 3 γ ( h - 1) - 0 . 4 γ ( h - 2), h 2. The general solution of this difference equation is (see Appendix B): γ ( h ) = c 1 (0 . 8) h + c 2 (0 . 5) h , h 2 where 0 . 8 and 0 . 5 are the inverses of the roots computed from the same polynomial Φ( z ) = 1 - 1 . 3 z - 0 . 4 z 2 = 0. 6 In case of multiple roots the formula has to be adapted accordingly. See equation (B.2) in the Appendix.
Image of page 62

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

View Full Document Right Arrow Icon
44 CHAPTER 2. ARMA MODELS The first part of the system delivers the initial conditions which determine the constants c 1 and c 2 : γ (0) - 1 . 3 γ ( - 1) + 0 . 4 γ ( - 2) = σ 2 (1 + 0 . 4 × 1 . 7) γ (1) - 1 . 3 γ (0) + 0 . 4 γ ( - 1) = σ 2 0 . 4 where the numbers on the right hand side are taken from the first procedure. Inserting the general solution in this equation system and bearing in mind that γ ( h ) = γ ( - h ) leads to: 0 . 216 c 1 + 0 . 450 c 2 = 1 . 68 σ 2 - 0 . 180 c 1 - 0 . 600 c 2 = 0 . 40 σ 2 Solving this equation system in the unknowns c 1 and c 2 one gets finally gets: c 1 = (220 / 9) σ 2 and c 2 = - 8 σ 2 . 2.4.3 Third procedure Whereas the first two procedures produce an analytical solution which relies on the solution of a linear difference equation, the third procedure is more suited for numerical computation using a computer. It rests on the same equation system as in the second procedure. The first step determines the values γ (0) , γ (1) , . . . , γ ( p ) from the first part of the equation system. The following γ ( h ) , h > p are then computed recursively using the second part of the equation system. A numerical example Using again the same example as before, the first of the equation delivers γ (2) , γ (1) and γ (0) from the equation system: γ (0) - 1 . 3 γ ( - 1) + 0 . 4 γ ( - 2) = σ 2 (1 + 0 . 4 × 1 . 7) γ (1) - 1 . 3 γ (0) + 0 . 4 γ ( - 1) = σ 2 0 . 4 γ (2) - 1 . 3 γ (1) + 0 . 4 γ (0) = 0 Bearing in mind that γ ( h ) = γ ( - h ), this system has three equations in three unknowns γ (0) , γ (1) and γ (2). The solution is: γ (0) = (148 / 9) σ 2 , γ (1) = (140 / 9) σ 2 , γ (2) = (614 / 45) σ 2 . This corresponds, of course, to the same numerical values as before. The subsequent values for γ ( h ) , h > 2 are then determined recursively from the difference equation γ ( h ) = 1 . 3 γ ( h - 1) - 0 . 4 γ ( h - 2).
Image of page 63
2.5. EXERCISES 45 2.5 Exercises Exercise 2.5.1. Consider the AR(1) process X t = 0 . 8 X t - 1 + Z t with Z WN(0 , σ 2 ) .
Image of page 64

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

View Full Document Right Arrow Icon
Image of page 65
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