Time Series Analysis. Solutions to problems in Chapter 7 IMM

Transcription

1 Time Series Analysis Solutions to problems in Chapter 7 I

2 Solution 7.1 Question 1. As we get by subtraction kx t = 1 + B + B +...B k 1 )ǫ t θb) = 1 + B B k 1 ), and BθB) = B + B B k ), 1 B)θB) = 1 B k ) θb) = 1 Bk 1 B Thus, the spectral density for {X t } can written as Question. f x ω) = σ ǫ 1 e iωk )1 e iωk ) k 1 e iω )1 e iω ) = σ ǫ coskω)) k cosω)) = σ ǫ sin 0.5kω) k sin 0.5ω) An expression for the window specified in frequency domain is found as the fourier transform of the window in the time domain, i.e. Gω) = 1 = a Tiω = ai = a π gt)e iωt dt = 1 e iωt e iωt) T T a T e iωt dt cosωt) i sinωt) cosωt) i sin ωt) Tω sinωt) ωt

3 Solution 7. Question 1. λ k 0 k 1 a + a cos 1 a a 0 a 0.5 ) πk 0 k Question. Using the general Tukey window the following spectral estimate is obtained ˆfω) = 1 )) ) πk C a + a cos C k cosωk) k=1 ) = 1 C a)c k cosωk) k=1 + 1 ac k cos ωk + π ) k + ac k cos ωk π )) ) k k=1 = a ˆf 1 ω π ) + 1 a) ˆf 1 + a ˆf 1 ω + π ) Question 3. 3

4 ν ˆfω) fω) χ ν) [ ν Var ˆfω) ] = ν fω) [ ] ˆfω) Var = fω) ν The degrees of freedom are ν = N/ k= λ k ), where k= )) πk λ k = 1 a a cos k= ) )) πk πk = 1 a) + 4a cos + 1 a)4a cos k= ) ) = 1 a) + 1) + 4a 1 ) k cos + 1 k=1 ) ) πk + 1 a)4a 1 + cos k=0 = 1 a) + 1) + 4a 1 + ) + 1 a)4a 1) = + 1a 8a) a 8a) Thus the variance relation is [ ] ˆfω) Var = fω) k= λ k /N = + 1a 8a) a 8a) ) 4

5 For the Tukey-Hanning window with a = 0.5 the variance relation becomes [ ] ˆfω) Var = 3 fω) 4N, which is in accordance with Table 7. on page 199. Question 4. For a Parzen window the variance relation is [ ] ˆfω) Var = fω) N. I.e. the variance relation is unchanged for p = 3 T N 4N P = 1.39 T = = 56 5

6 Solution 7.3 The spectral window is found as the Fourier transforme of the lag-window the window in the time domain), i.e. kω) = 1 = 1 = 1 π = 1 πω = λτ)e iτω dτ 1 τ )e iτω dτ 1 τ) cosτω)dτ cosτωdτω 1 πω 1 0 τω cosτω)dτω πω [sinτω)]1 0 1 πω [τω sinτω) + cosτω)]1 0 = 1 ) 1 1 cosω) = 1 cosω πω ω πω = sin ω/) πω 6

7 Solution 7.4 A 1001-α)% konfidence interval for fω) is given by [ ν ˆfω) ν, ˆfω) ] χ ν) 1 α/ χ ν) α/ Taking the logarithm we get [ ] νω) + log χ ν) ˆfω)), νω) + log 1 α/ χ ν) ˆfω)) α/ which means that for a logarithmic scale the width of the confidence interval is independent of ω. For a Parzen window with N=400 and =48 the degrees of freedom becomes ν Since the given figure for the spectrum is shown in decades it is suitable to use the confidence interval in logarithm to the base 10. ) 31 log 10 = 31 χ 31) ) 31 log 10 = 31 χ 31) On the figure the ordinate values are 1 log 10 1) = 0 log 10 ) = This distance is about the value we estimated for the confidence interval =0.9), and we can therefore draw in an approximate confidence interval for the smoothed spectrum. We find that the theoretical spectrum is not everywhere inside a 80% confidence interval. 7

8 Solution 7.5 Question 1. Cross-covariance function γ 1 τ) = Cov[X 1t, X t+τ ] = Cov[ǫ 1t, β 1 ǫ 1t+τ + β ǫ 1t+τ 1 + ǫ t ] β 1 σ1 τ = 0 = β σ1 τ = 1 0 otherwise Question. Cross-spectrum: Co-spectrum: Quadrature spectrum: f 1 ω) = 1 k= γ 1 τ)e iωτ = σ 1 β 1 + β cosω iβ sin ω) c 1 ω) = σ 1 β 1 + β cosω) q 1 ω) = σ 1 β sin ω) Cross-amplitude spectrum: Phase spectrum: Question 3. α 1 ω) = σ 1 β 1 + β cos ω + β 1 β cosω + β sin ω ) = σ 1 β 1 + β + β 1β cosω )1 ) β sin ω φ 1 ω) = arctan β 1 + β cosω 8

9 For β /β 1 = 1 β = β 1 holds α 1 ω) = σ 1 β 1 + cosω) 1 σ = 1 β 1 4 cos ω ) sin ω φ 1 ω) = arctan = arctan 1 + cosω sin ω ) = arctan cos ω = ω For β /β 1 = 1 β = β 1 holds α 1 ω) = σ 1 β 1 cosω) 1 σ = 1 β 1 4 sin ω ) sin ω φ 1 ω) = arctan = arctan 1 cosω cos ω ) = arctan sin ω = arctancot ω ) π = arctan tan ω )) = π ω ) 1 = σ 1 sin ω cos ω cos ω )1 = σ 1 sin ω cos ω sin ω π β 1 cos ω ) π β 1 sin ω ) The cross-amplitude and the phase spectrum is shown in Figure 1. The crossamplitude spectrum shows that the covariance between the two processes are dominated by low frequencies if β /β 1 = 1 and by high frequencies if β /β 1 = 1. In generel if the cross-correlation is of same sign the covariances will be dominated by low frequencies. The Phase spectrum shows that for β /β 1 = 1 the variation in X t follows the variation in X 1t, while the opposite case occurs when β /β 1 = 1. 9

10 pi/ β /β 1 =1 β /β 1 = 1 σ 1 /π)β1 β /β 1 =1 β /β 1 = 1 φ 1 ω) 0 α 1 ω) pi/ 0 \pi/ \pi ω 0 0 \pi/ \pi ω Figure 1: Left: Phase spectrum. Right: Cross-amplitude spectrum. 10