Consider the following ARMA(1,1) process (1 0.4B)Xt = (1 + 0.3B)Zt where {Zi} is an i.i.d. sequence of random variables with E(Zi)

1. Consider the following ARMA(1,1) process (1 0.4B)Xt = (1 + 0.3B)Zt where {Zi} is an i.i.d. sequence of random variables with E(Zi) = 0 and V ar(Zi) = 2 < 1. [18](i) Show that the process has a causal stationary solution and present {Xt} as an MA(1) process. Is this process invertible? If yes write this inverse representation. Solution. Solve (1 + 0.3B)(1 0.4B)=0 gives B = 10/3 and B = 10/4 and both are outside unit circle. As a result the process has a causal stationary solution. Moreover the process is invertible. From page 91 of slides we know 0 = 1, j = ( + ✓)j1 = 0.7(0.4)j1 . Therefore Xt = Zt + 0.7 X1 j=1 0.4j1 From page 98 of slides write Zt = Xt 0.7 X1 j=0 (0.3)j1 Xtj . 2 [7](ii) Find the autocorrelation function for the process {Xt} and draw a graph manually for the acf[1:5]. You can use the formula we derived in class. Solution. From page 95 of slides, we have ⇢(0) = 1, ⇢(h) = ( + ✓)(1 + ✓) (1 + ✓2 + 2✓) |h| , |h| 1. This gives for |h| 1 we have ⇢(h) = (0.7)(1.12) 0.4(1 + 0.09 + 0.24(0.4)|h| . ● ● ● ● ● ● ● ● ● ● 2 4 6 8 10 0.0 0.2 0.4 0.6 0.8 1.0 Index r acf[1:10] Figure 1: acf for Question 1(ii) 3 [14] 2. Consider an AR(1) process Xt = Xt1 Xt2 + Zt where {Zt} is a sequence of i.i.d. random variables with E(Z1) = 0, X0 = X1 = 0 and V ar(Z1) = 2 < 1. Is there a stationary process ? Why? Find Cov(X3, X4). Solution. Solve 1 B + B2 = 0 to get B = 1/2 ± i p 3/2 which is on the boundary of unit circle. Therefore there is no stationary solution (causal or non-causa). We have X2 = Z2, X3 = Z2 + Z3, X4 = Z2 + Z3 Z2 + Z4 = Z3 + Z4. Therefore Cov(X3, X4) = Cov(Z2 + Z3, Z3 + Z4) = Cov(Z3, Z3) = 2 . 4 3. [14](i) Find V ar(X¯) for the process Xt = Zt 0.5Zt1. Solution. We have V ar(X¯) = Xn h=n (1 |h|/n)(h)/n Since (0) = 1.252 , (±1) = 0.52 we have V ar(X¯)=((0) + 2(1 1/n)(1))/n = 2 n (0.25 + 1/n). 5 [15](ii) For a stationary AR(1) process Xt µ = (Xt1 µ) + Zt where Zi ⇠ W N(0, 2). Find the asymptotic distribution of X¯ as n ! 1. Solution. From page 114 of lecture slides we know X¯ ⇠ N ✓ µ, 2 n(1 )2 ◆ 6 [12] 3. Let Xt and Yt be two uncorrelated AR(1) processes Xt = 1Xt1 + Z1t and Yt = 2Yt1 + Z2t where Z1t ⇠ W N(0, 2 1) and Z2t ⇠ W N(0, 2), 1 and 2 are two constants in (1, 1). Also let a, b be two constants. Show that Ut = aXt + bYt + c is stationary and find the autcovariance function for the process Ut. Solution. From page 53 of the textbook we know U (h) = a2 X(h) + b2 Y (h) = a2 2 1 |h| 1 1 2 1 + b2 2 2 |h| 2 1 2 2 . 7 [10] 4. Find the autocorrelation function for the following MA(2) process Xt = Zt 0.2Zt1 0.1Zt2 Solution. See page 69 of the lecture slides to see that the process is an MA(2) process. We have ⇢(0) = 1, ⇢(±1) = 0.2+0.02 1+0.04 + 0.01 = 0.171, ⇢(±2) = 0.1 1+0.04 + 0.01 = 0.095 and ⇢(h) = 0 when |h| > 2. 8 [10] 5. Prove that the random variable f(X1, X2) that minimizes E[(Y f(X1, X2))2] is f(X1, X2) = E(Y |X1, X2). Solution. We need to minimize E(Y f(X1, X2) 2 = E[E[(Y f(X1, X2))2 |X1, X2]]. In other words we need to minimize E[(Y f(X1, X2))2 |X1, X2]. Since both X1, X2 are given in the condition f(X1, X2) = c is a given constant. To minimize E(Y c)2 = g(c) we know fromthe question in assignment 1, that c = E(Y |X1, X2). 9 Formula Sheet. • For the stationary MA(1) process Xt = X1 j=0 jZtj we have V ar(Xt) = 2X1 j=0 2 j , Cov(Xt, Xt+h) = X(h) = 2X1 j=0 j j+h and X(h) = X(h) = X(|h|). • Backward and di↵erence operator BXt = Xt1,(1 B)Xt = Xt Xt1. • ARMA(p, q) process (B)Xt = ✓(B)Zt. The stationary solution is Xt = ✓(B) (B) Zt • When |r| < 1 we have X1 i=k ri = rk + rk+1 + rk+2 + ··· = rk 1 r , X1 k=1 krk = r (1 r)2 10 and X1 k=1 k2 rk = r2 + r (1 r)3 . • Cov(aX+bY +k, cZ+dT+l) = acCov(X, Z)+adCov(X, T)+bcCov(Y,Z)+bdCov(Y,T). where a, b, c, d, k and l are constants.

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