TimeSeriesBook.pdf

# Remark 21 if the ar and the ma polynomial have common

• 435

This preview shows pages 53–56. Sign up to view the full content.

Remark 2.1. If the AR and the MA polynomial have common roots, there are two possibilities:

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

2.3. CAUSALITY AND INVERTIBILITY 35 No common roots lies on the unit circle. In this situation there exists a unique stationary solution which can be obtained by canceling the common factors of the polynomials. If at least one common root lies on the unit circle then more than one stationary solution may exist (see the last example below). Some examples We concretize the above Theorem and Remark by investigating some exam- ples starting from the ARMA model Φ(L) X t = Θ(L) Z t with Z t WN(0 , σ 2 ). Φ(L) = 1 - 0 . 05L - 0 . 6L 2 and Θ(L) = 1 : The roots of the polynomial Φ( z ) are z 1 = - 4 / 3 and z 2 = 5 / 4. Because both roots are absolutely greater than one, there exists a causal representation with respect to { Z t } . Φ(L) = 1 + 2L + 5 / 4L 2 and Θ(L) = 1 : The roots are conjugate complex and equal to z 1 = - 4 / 5 + 2 / 5 ı and z 2 = - 4 / 5 - 2 / 5 ı . The modulus or ab- solute value of z 1 and z 2 equals | z 1 | = | z 2 | = p 20 / 25. This number is smaller than one. Therefore there exists a stationary solution, but this solution is not causal with respect to { Z t } . Φ(L) = 1 - 0 . 05L - 0 . 6L 2 and Θ(L) = 1 + 0 . 75L : Φ( z ) and Θ( z ) have the common root z = - 4 / 3 6 = 1. Thus one can cancel both Φ(L) and Θ(L) by 1 + 3 4 L to obtain the polynomials ˜ Φ(L) = 1 - 0 . 8L and ˜ Θ(L) = 1. Because the root of ˜ Φ( z ) equals 5/4 which is greater than one, there exists a unique stationary and causal representation with respect to { Z t } . Φ(L) = 1 + 1 . 2L - 1 . 6L 2 and Θ(L) = 1 + 2L : The roots of Φ( z ) are z 1 = 5 / 4 and z 2 = - 0 . 5. Thus one root is outside the unit circle whereas one is inside. This would suggest that there is no causal solution. However, the root - 0 . 5 6 = 1 is shared by Φ( z ) and Θ( z ) and can therefore be canceled to obtain ˜ Φ(L) = 1 - 0 . 8L and ˜ Θ(L) = 1. Because the root of ˜ Φ( z ) equals 5 / 4 > 1, there exists a unique stationary and causal solution with respect to { Z t } . Φ(L) = 1 + L and Θ(L) = 1 + L : Φ( z ) and Θ( z ) have the common root - 1 which lies on the unit circle. As before one might cancel both poly- nomials by 1 + L to obtain the trivial stationary and causal solution { X t } = { Z t } . This is, however, not the only solution. Additional so- lutions are given by { Y t } = { Z t + A ( - 1) t } where A is an arbitrary
36 CHAPTER 2. ARMA MODELS random variable with mean zero and finite variance σ 2 A which is inde- pendent from both { X t } and { Z t } . The process { Y t } has a mean of zero and an autocovariance function γ Y ( h ) which is equal to γ Y ( h ) = ( σ 2 + σ 2 A , h = 0; ( - 1) h σ 2 A , h = ± 1 , ± 2 , . . . Thus this new process is therefore stationary and fulfills the difference equation. Remark 2.2. If the AR and the MA polynomial in the stochastic difference equation Φ(L) X t = Θ(L) Z t have no common root, but Φ( z ) = 0 for some z on the unit circle, there exists no stationary solution. In this sense the stochastic difference equation does no longer define an ARMA model. Models with this property are said to have a unit root and are treated in Chapter 7. If Φ( z ) has no root on the unit circle, there exists a unique stationary solution.

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

This is the end of the preview. Sign up to access the rest of the document.
• Spring '17
• Raffaelle Giacomini

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern