TimeSeriesBook.pdf

# In order to make the point consider the following

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In order to make the point, consider the following given autocovariance function: γ ( h ) = γ 0 , h = 0; γ 1 , h = ± 1; 0 , | h | > 1 . The problem now consists of determining the parameters of the MA(1) model, θ and σ 2 from the values of the autocovariance function. For this purpose we

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1.5. PROPERTIES OF THE AUTOCOVARIANCE FUNCTION 21 equate γ 0 = (1 + θ 2 ) σ 2 and γ 1 = θσ 2 (see equation (1.1)). This leads to an equation system in the two unknowns θ und σ 2 . This system can be simplified by dividing the second equation by the first one to obtain: γ 1 0 = θ/ (1+ θ 2 ). Because γ 1 0 = ρ (1) = ρ 1 one gets a quadratic equation in θ : ρ 1 θ 2 - θ + ρ 1 = 0 . The two solutions of this equation are θ = 1 2 ρ 1 1 ± q 1 - 4 ρ 2 1 . The solutions are real if and only if the discriminant 1 - 4 ρ 2 1 is positive. This is the case if and only if ρ 2 1 1 / 4, respectively | ρ 1 | ≤ 1 / 2. Note that one root is the inverse of the other. The identification problem thus takes the following form: | ρ 1 | < 1 / 2 : there exists two observationally equivalent MA(1) processes. ρ 1 = ± 1 / 2 : there exists exactly one MA(1) process with θ = ± 1. | ρ 1 | > 1 / 2 : there exists no MA(1) process with this autocovariance function. The relation between the first order autocorrelation coefficient, ρ 1 = ρ (1), and the parameter θ of the MA(1) process is represented in Figure 1.10. As can be seen, there exists for each ρ (1) with | ρ (1) | < 1 2 two solutions. The two solutions are inverses of each other. Hence one solution is absolutely smaller than one whereas the other is bigger than one. In Section 2.3 we will argue in favor of the solution smaller than one. For ρ (1) = ± 1 / 2 there exists exactly one solution, namely θ = ± 1. For | ρ (1) | > 1 / 2 there is no solution. For | ρ 1 | > 1 / 2, ρ ( h ) actually does not represent a genuine autocorrelation function as the fourth condition in Theorem 1.1, respectively Theorem 1.2 is violated. Taking ρ 1 = 1 2 and setting a = (1 , - 1 , 1 , - 1 , . . . , 1 , - 1) 0 one gets: n X i,j =1 a i ρ ( i - j ) a j = n - 2( n - 1) ρ 1 < 0 , if n > 2 ρ 1 2 ρ 1 - 1 . For ρ 1 = - 1 2 one sets a = (1 , 1 , . . . , 1) 0 . Hence the fourth property is violated.
22 CHAPTER 1. INTRODUCTION -3 -2 -1 0 1 2 3 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 first order ‘‘moving average’’ parameter: θ first order autocorrelation coefficient: ρ (1) θ /(1+ θ 2 ) 0 0.5 1 2 Figure 1.10: Relation between the autocorrelation coefficient of order one, ρ (1), and the parameter θ of a MA(1) process

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1.6. EXERCISES 23 1.6 Exercises Exercise 1.6.1. Let the process { X t } be generated by a two-sided moving- average process X t = 0 . 5 Z t +1 + 0 . 5 Z t - 1 with Z t WN(0 , σ 2 ) . Determine the autocovariance and the autocorrelation function of { X t } . Exercise 1.6.2. Let { X t } be the MA(1) process X t = Z t + θZ t - 2 with Z t WN(0 , σ 2 ) . (i) Determine the autocovariance and the autocorrelation function of { X t } for θ = 0 . 9 . (ii) Determine the variance of the mean ( X 1 + X 2 + X 3 + X 4 ) / 4 .
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