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TIME SERIES

Solutions to Exercise Sheet 5

1. Following Example 5.5 in Section 5.2.2 of the lecture notes, to calculate 33, we needto find the function

f(2) = 1X2 + 2X1,

where 1 and 2 are chosen to minimise

E[{X3 f(2)}2] = E(X23 +

21X

22 +

22X

21 21X2X3 22X1X3 + 212X1X2)

= (0) + 21(0) + 22(0) 21(1) 22(2) + 212(1).

Taking the derivatives with respect to 1 and 2 and setting them equal to zero, weobtain

21(0) 2(1) + 22(1) = 0

and 22(0) 2(2) + 21(1) = 0.

Solving these yields 1 = and 2 = 0. Thus, we have f(2) = X2. It follows that

33 = corr(X3 X2, X0 X2) = corr(Z3, X0 X2) = 0.

2. (a) The model Xt = 0.3Xt1 + Zt is an ARMA(p,q) process with p = 1 and q = 0,that is, it is AR(1). In the backward shift operator form, it is (1 0.3B)Xt = Zt.It is a causal process, since the root of the associated polynomial (z) = 1 0.3z isz = 10/3 > 1, that is, it is outside the interval [1, 1].

(b) The model Xt = Zt 1.3Zt1 + 0.4Zt2 is an ARMA(p,q) process with p = 0 andq = 2, that is, it is MA(2). In the backward shift operator form, it can be writtenas Xt = (1 1.3B + 0.4B

2)Zt. To check invertibility, we examine the roots of theassociated polynomial (z) = 1 1.3z+ 0.4z2. Here, we have two real roots z1 = 1.25and z2 = 2. Since both roots are outside the interval [1, 1], the process is invertible.(c) The model Xt 0.5Xt1 = Zt 1.3Zt1 + 0.4Zt2 can be written in the backwardshift operator form as (10.5B)Xt = (11.3B + 0.4B

2)Zt. First, however, we need tocheck if there are common factors. The associated polynomial (z) = 1 1.3z+ 0.4z2

can be written as

(1 1.251z)(1 21z) = (1 0.8z)(1 0.5z).

Hence, there is a common factor (1 0.5z) and the model can be simplified. Dividingboth sides of the model by (1 0.5z), we obtain

Xt = (1 0.8B)Zt.

This is an ARMA(p,q) process with p = 0 and q = 1, that is, it is MA(1). This isan invertible process, since the root of the associated polynomial (z) = 1 0.8z isoutside the interval [1, 1].

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3. (a) The model Xt = Zt + 0.7Zt1 represents an invertible MA(1) process with = 0.7.For an MA(1) process, the autocorrelation and partial autocorrelation functions are

() =

1+2if = 1,

0 if > 1,

and

= ()(1 2)

1 2(+1)for 1.

When = 0.7, we obtain

() =

0.47 if = 1,

0 if > 1,

and11 = 0.47, 22 = 0.28, 33 = 0.19, 44 = 0.13, 55 = 0.09.

Thus, the autocorrelation function cuts off after lag 1, whereas the partial autocorre-lation function alternates sign and tails off.(b) The model Xt = 0.9Xt1 0.2Xt2 + Zt represents a causal AR(2) process with1 = 0.9 and 2 = 0.2. To obtain the autocorrelation function, we may use the dif-ference equations of order two, as in Example 5.4 in Section 5.2.1 of the lecture notes,which yield

() = c1z1 + c2z

2 ,

where the roots of the associated polynomial (z) = 1 0.9z+ 0.2z2 are z1 = 2 andz2 = 2.5. The constants c1 and c2 can be found from the initial conditions and areequal to c1 = 3.5 and c2 = 2.5. Hence, we obtain

() = 3.5 2 2.5 2.5 for = 1, 2, . . . .

The first five values of the autocorrelation function are

(1) = 0.75, (2) = 0.48, (3) = 0.28, (4) = 0.15, (5) = 0.08.

From Remark 5.11 in Section 5.3.1 of the lecture notes, the partial autocorrelationfunction of an AR(2) process is

=

112

= 0.75 if = 1,

2 = 0.2 if = 2,0 if > 2.

Thus, the autocorrelation function tails off, whereas the partial autocorrelation func-tion cuts off after lag 2.

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