X random; interested in impact of X on Y. Time series analogue of regression.

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1 Multiple time series Given: two series Y and X. Relationship between series? Possible approaches: X deterministic: regress Y on X via generalized least squares: arima.mle in SPlus or arima in R. We have done this. X random; interested in impact of X on Y. Time series analogue of regression. X random; interested in joint behaviour. Multivariate ARIMA modelling. 310

2 Definition: Two processes X and Y are jointly (strictly) stationary if L(X t,..., X t+h, Y t,..., Y t+h ) for all t and h. = L(X 0,..., X h, Y 0,..., Y h ) Definition: X and Y are jointly second order stationary if each is second order stationary and also C XY (h) Cov(X t, Y t+h ) = Cov(X 0, Y h ) for all t and h. Notice that negative values of h give, in general, different covariances than positive values of h. 311

3 Regression: estimate impulse response Assume: E(X) = E(Y ) = 0. Example to illustrate range of models: Y t = ax t + bx t 2 + cx t+4 + ν t where ν is noise, not necessarily white noise. Crucial: ν independent of X. General model: Y t = a s X t s + ν t Problem: estimate impulse response function a. 312

4 Alternative description Find best predictor of Y from X? Minimize E [ {Y t f t (..., X t 1, X t,...)} 2] over all functions f t. Joint stationarity implies: f t+1 (BX) = f t (X) For Gaussian mean 0 series: f t (X) = a s X t s for some coefficients a s So: minimize E ( Y t s ) 2 a s X t s 313

5 Expand out: E(Y 2 t ) 2 a s E(Y t X t s )+ a r a s E(X t r X t s ) which is E(Y 2 t ) 2 a s C XY (s) + a r a s C X (r s) Derivative wrt a k : 2C XY (k) + 2 s a s C X (k s) Set to 0 to find minimum: C XY (k) = s a s C X (k s) Solve? With data? 314

6 Use spectral methods: Definition: : convolution of two sequences a t, b t is (a b) t = r a r b t r = b r a t r Fourier transform: f a (ω) = a r e 2πrωi f a b (ω) = = s r= r (a b) r e 2πωri a s b r s e 2πω(r s+s)i = ( as e 2πωsi) ( br e 2πωri) = f a (ω)f b (ω) So f CXY (ω) = f a (ω)f CX (ω) 315

7 Cross spectrum f XY (ω) = f CXY (ω) = C XY (r)e 2πrωi Property: f XY (ω) = f Y X ( ω) = f Y X (ω) because C XY (h) = C Y X ( h) Conclusion: Frequency response function, A, of filter a is: A(ω) = a r e 2πrωi = f XY (ω)/f X (ω) Estimate A from estimates of f XY and f X. Estimate f X by smoothing periodogram. 316

8 Estimating the cross-spectrum Use cross-periodogram: ˆf XY (ω) = Ŷ (ω) ˆX(ω) where Ŷ and ˆX are discrete Fourier transforms. Smooth to improve estimate. Estimating the impulse response Recall Fourier inversion 1 a t = 0 A(ω)e 2πtωi dω Replace A by estimate  to get â. Gain is sometimes used for A. Write A(ω) = A(ω) e iθ(ω) for some function θ with π < θ(ω) π Call θ the phase shift of the filter. 317

9 MSE of optimal filter? E ( Y t s a s X t s ) 2 = C Y (0) 2 a s C XY (s) + a r a s C X (r s) Recall Fourier inversion formulas like C Y (h) = f Y (ω)e 2πhωi dω Put h = 0 get Notice C Y (0) = f Y (ω)dω as C XY (s) = a s C Y X (0 s) But f a CY X = f a f CY X = f a f Y X so by Fourier inversion as C Y X (0 s) = f a (ω)f Y X (ω)dω 318

10 Also ar a s C X (r s) = = r a r f a CX (ω)e 2πrωi dω f a (ω)f X (ω)f a ( ω)dω But f a is just A and so ar a s C X (r s) = f X (ω) A(ω) 2 dω Assemble: MSE = {f Y (ω) 2A(ω)f Y X (ω) Factor out f Y, recall A = f XY /f X : MSE = { + f X (ω) A(ω) 2 }dω f Y (ω) 1 2 f XY (ω) 2 f X (ω)f Y (ω) + f XY (ω) 2 } dω f X (ω)f Y (ω) 319

11 Definition: Coherence between X and Y at frequency ω is γ XY (ω) = f XY (ω) f X (ω)f Y (ω) So MSE = f Y (ω) { 1 γ 2 XY (ω)} dω Interpretation: γ is a correlation coefficient. Note: suppose we use raw periodgrams to estimate cross-spectrum: Then ˆγ XY (ω) = Ŷ (ω) ˆX(ω) Ŷ (ω) ˆX(ω) = 1 Conclusion: must smooth to get useful estimates of γ. 320

12 A spectral analysis example Chandler wobble: Measure direction earth s axis of ratation points it wobbles. Data set 1: Monthly measurements of direction of North Pole. Two co-ordinates: X and Y. X direction toward Greenwich along Greenwich meridian. 321

13 Study X and Y motions of pole together. Plot of X and Y together: X and Y co ordinates of North Pole milli arcseconds Time in Years 322

14 Parametric plots of X and Y: X vs Y for first 19 years X vs Y for second 19 years Y Y X X X vs Y for third 19 years X vs Y for last 19 years Y Y X X 323

15 Re-examine plot of X: X co ordinate of North Pole milli arcseconds Time in Years Clear cycles, what else, what periods? 324

16 Look at spectrum of X: Power Frequency (cycles per month ) Notice lots of spikes interpret by plotting against periods over various ranges. 325

17 Power Period ( days ) Power Period ( days ) Power Period ( years ) 326

18 Peak at days: external forcing due to seasons. No surprise. Secondary peak at 730.5= Overtone due to imperfect annual sinusoid. Peaks at 427, 433 days, roughly: Chandler wobble. Peak at 30 Years? Compare amplitudes smaller in middle plot. 327

19 Modelling possibilities: 1) Fit multiviate Autogregression: Let and fit models like Z t µ = Z t = p j=1 [ Xt Y t ] A s (Z t s µ) + ǫ t where now µ,z t and ǫ t are vectors and each A s is a matrix. General theory exists. Fitted to Chandler in R picks AR(21). (Warning, not full maximum likelihood.) 2) Transfer function modelling. Try to predict say Y from X. (Done as an example not well motivated scientifically.) 328

20 Plot the ACF: x x & y ACF Lag Lag y & x y ACF Lag Lag 329

21 And the PACF x x & y Partial ACF Lag Lag y & x y Partial ACF Lag Lag 330

22 Two periodograms superimposed Raw Periodograms Chandler Wobble Data spectrum frequency bandwidth = 1.88e

23 Unsmoothed coherency Raw Coherency Chandler Wobble Data squared coherency frequency 332

24 Smoothed coherency Smoothed Coherency Chandler Wobble Data squared coherency frequency 333

25 plot(chandler) acf(chandler,lag.max=60) acf(chandler,lag.max=60,type="partial") ar.fit <- ar(chandler) plot(ar.fit$resid) spectrum(chandler, main="raw Periodograms\nChandler Wobble Data", log="no",pad=200,xlim=c(0,1.2)) spectrum(chandler, main="raw Coherency\nChandler Wobble Data", log="no",pad=200,plot.type="coherency",xlim=c(0,1.2)) spectrum(chandler, main="raw Phase\nChandler Wobble Data", log="no",pad=200,plot.type="phase",xlim=c(0,1.2)) spectrum(chandler, main="smoothed Coherency\nChandler Wobble Data", log="no",pad=200,plot.type="coherency",xlim=c(0,1.2), spans=c(500,500,500)) spectrum(chandler, main="smoothed Phase\nChandler Wobble Data", log="no",pad=200,plot.type="phase",xlim=c(0,1.2), spans=c(500,500,500)) 334

26 plot(chandler) spec = spectrum(chandler,plot=false, log="no",pad=200,spans=c(500,500,500)) cross.mod = sqrt(spec$coh*spec$spec[,1]*spec$spec[,2]) delf = mean(diff(spec$freq))/12 a <- rep(0,101) for( i in 1:101){ a[i] = 2*sum(cross.mod*cos(spec$phase-2*pi*(i-1) *spec$freq/12)/spec$spec[,1])*delf } b <- rep(0,101) for( i in 1:101){ b[i] = 2*sum(cross.mod*cos(spec$phase+2*pi*(i-1) *spec$freq/12)/spec$spec[,1])*delf } 335

27 > ar.fit $ar,, 1,, 2,, 3 x y x y x y x x x y y y ,, 4,, 5,, 6 x y x y x y x x x y y y ,, 7,, 8,, 9 x y x y x y x x x y y y ,, 10,, 11,, 12 x y x y x y x x x y y y ,, 13,, 14,, 15 x y x y x y x x x y y y ,, 16,, 17,, 18 x y x y x y x x x y y y ,, 19,, 20,, 21 x y x y x y x x x y y y

28 Estimates of residual variance covariance and correlation: $var.pred x y x y > sqrt(diag(ar.fit$var.pred)) x y > ar.fit$var.pred[1,2]/prod(sqrt(diag(ar.fit$var.pred))) [1]

29 Finally, the estimated impulse response function: Impulse Response Regression Coefficient lag Not enough smoothing but too much for spikes. So: not a good fitting technique for this sort of data. 338