TimeSeriesBook.pdf

# The second part of the equation system consists again

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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.

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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).
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 ) .

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