Time Series Solutions HT Let fx t g be the ARMA(1, 1) process, where jffij < 1 and j j < 1. Show that the autocorrelation function of

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1 ffi ; ρ(h) =ffih 1 ρ(1) for h > 1: Time Series Solutions HT Let fx t g be the ARMA(1, 1) process, X t ffix t 1 = ffl t + ffl t 1 ; fffl t gοwn(0;ff 2 ); where jffij < 1 and j j < 1. Show that the autocorrelation function of fx t g is given by (1 + ffi )(ffi + ) ρ(1) = 1

2 X t = ffix t 1 + ffl t + ffl t 1 = ffi[ffix t 2 + ffl t 1 + ffl t 2 ]+ffl t + ffl t 1 = ffi 2 X t 2 + ffiffl t 1 + ffi ffl t 2 + ffl t + ffl t 1 Solution. Taking E(X expectations )=ffie(x t 1 t ), and ffi<1 using and stationarity we E(X get )=E(X t 1 t )=0. k > For 2: multiplying t = ffix t 1 + ffl t + ffl t 1 X X by t k and taking expectations we fl get = ffifl k 1 k, and hence k = ffi k 1 fl 1 for k > 2. fl Multiplying the same equation X by t and taking expectations we get and fl 0 = ffifl 1 + E[X t (ffl t + ffl t 1 )] 2

3 fl 0 = ffifl 1 + E[(ffi 2 X t 2 + ffiffl t 1 + ffi ffl t 2 + ffl t + ffl t 1 )(ffl t + ffl t 1 )] fl 1 = E(X t X t+1 ) so Also = ffifl 1 + ff 2 [ffi ]: = E[X t (ffix t + ffl t+1 + ffl t )] = ffifl 0 + E[(ffiX t 1 + ffl t + ffl t 1 )(ffl t+1 + ffl t )] = ffifl 0 + ff 2 : We can now solve the two equations involving fl 0 ;fl 1, and then find fl k, and hence ρ k, as required. 3

4 2. Consider a process consisting of a linear trend plus an additive noise term, that is, X t = fi 0 + fi 1 t + ffl t where fi 0 and fi 1 are fixed constants, and where the ffl t are independent random variables with zero means and variances ff 2. Show that X t is non-stationary, but that the first difference series rx t = X t X t 1 is second-order stationary, and find the acf of rx t. 4

5 Y t = fi 0 + fi 1 t + ffl t ffi 0 + fi 1 (t 1) + ffl t 1 g cov(y t ;Y t+k )=cov(ffl t ffl t 1 ;ffl t+k ffl t+k 1 ) >< 2ff 2 k =0 Solution. E(X t )=E(fi 0 + fi 1 t + ffl t )=fi 0 + fi 1 t which depends on t, hence X t is non-stationary. Let Y t = rx t = X t X t 1. Then So = fi 1 + ffl t ffl t 1 : = E(ffl t ffl t+k ffl t 1 ffl t+k ffl t ffl t+k 1 + ffl t 1 ffl t+k 1 ) 8 = ff 2 k =1 >: 0 k > 2: 5

6 ρ k == >< Hence Y t is stationary and its acf is 8 1 k =0 1 2 k =1 >: 0 k > 2: 6

7 3. Let fs t ;t =0; 1; 2;:::g be the random walk with constant drift μ, defined by S 0 =0and S t = μ + S t 1 + ffl t ; t =1; 2;:::; where ffl 1 ;ffl 2 ;::: are independent and identically distributed random variables with mean 0 and variance ff 2. Compute the mean of S t and the autocovariance of the process fs t g. Show that frs t g is stationary and compute its mean and autocovariance function. 7

8 S t = ffl t + μ + S t 1 = ffl t + μ + ffl t 1 + μ + S t 2 = ffl t + ffl t 1 +2μ + S t 2 ffl t j + tμ + S 0 Solution. = ::: Xt 1 = So E(S t )=0+tμ +0=tμ. j=0 8

9 E[fS t tμgfs t+k (t + k)μg] =E( = tff 2 E(ffl t j ffl t j ) ffl t+k i ) For the autocovariance of S t, the autocovariance at lag k is Xt 1 t+k 1 X t j ffl j=0 i=0 Xt 1 = j=0 since, when moving from the first line to the second line of the above display, E(ffl t j ffl t+k i )=0unless i = j + k. t = rs t = S t S t 1 = μ + ffl t, which is clearly stationary. Y t )=μ. E(Y 9

10 For the autocovariance of Y t, note Y t μ = ffl t, and similarly 0 μ = ffl t t 0, and so t 6= t for each Y 0 t depends on a ffl different t, Y and therefore ;Y t t 0)=0for t 6= t all. So the autocovariance function cov(y 0 is at lag 0, and is zero at all other lags. 2 ff 10

11 cos ff cos fi = If X t = a cos( t) +ffl t where ffl t ο WN(0;ff 2 ), and where a and are constants, show that fx t g is not stationary. Now consider the process X t = a cos( t + ) where is uniformly distributed on (0; 2ß), and where a and are constants. Is this process stationary? Find the autocorrelations and the spectrum of X t. [To find the autocorrelations you may want to use the identity + fi) + cos(ff fi)g.] fcos(ff 11

12 t )=ae(cos( t + )) E(X 2ß Z a = 2ß a = [sin( t + )]2ß 0 2ß Solution. E(X t )=E(a cos( t) +ffl t )=a cos( t), which depends on t, so X t is not stationary. Now for X t = a cos( t + )we need to consider the joint distributions of (X(t 1 );:::;X(t k )) and of (X(t 1 + fi );:::;X(t k + fi )). Since shifting time by t is equivalent to shifting by t, and since is uniform on (0; 2ß), these two joint distributions are the same, and so X t is stationary. cos( t + ) d 0 =0 12

13 fl t = E(X t X 0 )=a 2 E(cos( ) cos( t + )) = a2» 1 Z 2ß 2 cos( t) = a 2 E 2 fcos( t + 2 ) + cos( t)g a2 = [ 1 2 cos( t +2 ) + cos( t) d ] 2ß 0 So ρ t =cos( t). R ß F The fl spectrum = ß is df where t (!). Try the discrete eit! distribution F ( ) =F ( ) for, =c, F (!) a constant, F =0otherwise. 13

14 Then fl t = e it c + e it c = c[cos(t ) +i sin(t ) + cos(t ) i sin(t )] =2c cos( t): So we want 2c = a 2 =2,orc = a 2 =4. SoF ( ) =F ( ) =a 2 =4. 14

15 k = ρ jkj 5. Find the Yule-Walker equations for the AR(2) process X t = 1 3 X t X t 2 + ffl t where ffl t ο WN(0;ff 2 ). Hence show that this process has autocorrelation function 1 : 3 jkj [To solve an equation of the aρ form + bρ k 1 + cρ k 2 k =0, try ρ k = A k for some constants A and : solve the resulting quadratic equation for and deduce that ρ k is of the form ρ k = A k 1 + B k 2 where A and B are constants.] 15

16 k = ( 2 3 )k ( 1 3 )k : ρ Solution. The Yule-Walker equations are k = 1 3 ρ k ρ k 2: ρ So as in the hint, to solve ρ k 1 3 ρ k ρ k 2 =0 try ρ k = A k. Substituting this into the above equation, and cancelling a factor of k 2, we get =0 which has roots = 2 3 and = 1 3,soρ k = A( 2 3 )k + B( 1 3 )k. We ρ also require ρ 0 = =1and ρ 1 1. Hence we can solve A for 3 A = and B: and.so B =

17 6. Let fy t g be a stationary process with mean zero and let a and b be constants. (a) If X t = a + bt + s t + Y t where s t is a seasonal component with period 12, show that rr 12 X t =(1 B)(1 B 12 )X t is stationary. (b) If X t =(a + bt)s t + Y t where s t is again a seasonal component with period 12, show that r 2 12X t =(1 B 12 )(1 B 12 )X t is stationary. 17

18 rx t = a + bt + s t + Y t [a + b(t 1) + s t 1 + Y t 1 ] = b + s t s t 1 + Y t Y t 1 rr 12 X t = b + s t s t 1 + Y t Y t 1 = Y t Y t 1 Y t 12 + Y t 13 Solution. (a) [b + s t 12 s t 13 + Y t 12 Y t 13 ] and this is a stationary process since Y t is stationary. (We have used the fact that s t = s t 12 for all t.) 18

19 = Y t +12bs t 12 Y t 12 r 2 12X t = Y t +12bs t 12 Y t 12 = Y t 2Y t 12 + Y t 24 (b) r 12 X t =(a + bt)s t + Y t [(a + b(t 12))s t 12 + Y t 12 ] [Y t bs t 24 Y t 24 ] and this is stationary since Y t is stationary (again using s t = s t 12 for all t.) 19

20 7. Consider the univariate state-space model given by state conditions X 0 = W 0, X t = X t 1 + W t, and observations Y t = X t + V t, =1; 2;:::, where t W and t are independent, Gaussian, white t noise V processes with and 2 V var(w t)=ff 2 W t. Show that the data )=ff var(v follow an ARIMA(0,1,1) ry model, that is, t follows an MA(1) model. Include in your answer an expression for the autocorrelation function of t in terms of and. 2 ff2 W V ff ry 20

21 ry t = Y t Y t 1 =(X t + V t ) (X t 1 + V t 1 ) = X t X t 1 + V t V t 1 = W t + V t V t 1 fl 0 = Var(rY t ) Solution. and so ry t is an MA(1). As V t, V t 1 and W t are independent, = ff 2 W +2ff2 V : 21

22 fl 1 = Cov(rY t ; ry t+1 ) ρ 1 = ff 2 V Furthermore, = Cov(W t + V t V t 1 ;W t+1 + V t+1 V t ) = ff 2 V ; and, from the independence, fl k =0for jkj 2. Hence the acf is ρ 0 =1, and ρ k =0for jkj 2. ff 2 W +2ff2 V ; 22