Solutions Guide Version 1.1

Spectral Analysis for

Univariate Time Series:

Solutions Guide (Version 1.1) Donald B. Percival Andrew T. Walden 1 / 4

Answer to Exercise [1.1]1 Answer to Exercise [1.1]Sincejaibj 2 =a 2 +b 2 for real-valuedaandb, we have k N 2

N1 X t=0 xte i2fjt

2 = k N 2

N1 X t=0 xtcos (2fjt) !i

N1 X t=0 xtsin (2fjt) !

2 = k N 2 2 4

N1 X t=0 xtcos (2fjt) !2 +

N1 X t=0 xtsin (2fjt) !2 3

5:

We obtain Equation (13a) by lettingk= 2 in the above. For evenN, Equation (13a) follows by lettingk= 1 because in that casefj=f N=2= 1=2 and hence N1 X t=0 xtsin (2fjt) = N1 X t=0

xtsin (t) = 0: 2 / 4

2Answer to Exercise [1.2] Answer to Exercise [1.2]To show part (a), note that, for any complex numberz, (1z) N1 X t=0 z t = N1 X t=0 (1z)z t = N1 X t=0 z t

N1 X t=0 z t+1 = N1 X t=0 z t

N X t=1 z t = 1z N

:

Ifz6= 1, we can divide through by 1zto get the stated result.To show part (b), note that Euler's relationship and the facts that cos(x) = cos(x) and sin(x) =sin(x) give e i2f = cos (2f) + i sin (2f) and e i2f

= cos (2f)i sin (2f):

Addition and subtraction yields e i2f

• e

i2f = 2 cos (2f) and e i2f e i2f = 2i sin (2f); which leads to the desired result.To show part (c), letz= e i2f

. Iff62Z, thenz6= 1, so we can apply part (a):

N1 X t=0 e i2f t = 1e i2N f 1e i2f = e iN f

e iN f e iN f

e if (e if e if ) = e i(N1)f e iN f e iN f e if e if

:

Part (b) says e iN f e iN f = 2i sin (Nf) and e if e if = 2i sin (f) { hence N1 X t=0 e i2f t = e i(N1)f sin (Nf) sin (f) =Ne i(N1)f DN(f); () as required. Iff2Z, thenz= 1; hencez t = 1 for all t, so N1 X t=0 z t = N1 X t=0 e i2f t =N=Ne i(N1)f DN(f) because e i(N1)f DN(f) = e i(N1)f (1) (N1)f = (1) (N1)f (1) (N1)f = 1; where we have made use of the fact that exp(i) = cos() + i sin() =1.To show part (d), note that, for anyf, we have N1 X t=0 e i2f t = N1 X t=0 cos (2ft) + i N1 X t=0

sin (2ft): ()

For 1j < N, we havefj def =j=N62Z, so () applies; however, sin (Nfj) = sin (j) = 0 for allj, so the left-hand side of () is 0, which in turns implies that we must have both N1 X t=0 cos (2fjt) = 0 and N1 X t=0 sin (2fjt) = 0; 3 / 4

Answer to Exercise [1.2]3 as required.To show part (e), we have N X t=1 e i2f t = N1 X t=0 e i2f(t+1) = e i2f N1 X t=0 e i2f t = e i2f Ne i(N1)f DN(f) = e i(N+1)f DN(f) and N1 X t=(N1) e i2f t = 2N2 X t=0 e i2f(tN+1) = e i2f(N1)

(2N1)1

X t=0 e i2f t = e i2f(N1) (2N1)e i(2N2)f D2N1(f) = (2N1)D2N1(f); where we have made use of () withNreplaced by 2N1.

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