chapter 2: fourier series (spectral)...

1. Introduction2. Basics of Fourier Series

3. Fitting a single sine wave to a time series4. Fitting a set of sine waves to a time series

5. Real data example: U.S./New Zealand exchange rate forecast6. Caution with PROC SPECTRA in SAS

CHAPTER 2: Fourier Series (Spectral) Analysis

Prof. Alan Wan

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Table of contents

1. Introduction

2. Basics of Fourier Series

3. Fitting a single sine wave to a time series

4. Fitting a set of sine waves to a time series

5. Real data example: U.S./New Zealand exchange rate forecast

6. Caution with PROC SPECTRA in SAS

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Introduction

I A Fourier series is a representation of a wave-like function asthe sum of simple sine/cosine waves.

I More formally, it decomposes any periodic function into thesum of a set of simple oscillating functions, namely sines andcosines.

I It is suitable for modeling seasonality and/or cyclicalness, andidentifying peaks and troughs;

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Introduction

I A Fourier series is a representation of a wave-like function asthe sum of simple sine/cosine waves.

I More formally, it decomposes any periodic function into thesum of a set of simple oscillating functions, namely sines andcosines.

I It is suitable for modeling seasonality and/or cyclicalness, andidentifying peaks and troughs;

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Basics of Fourier Series

I A sine wave is a repeating pattern that goes through onecycle every 2π (i.e. 2 × 3.141593 = 6.283186) units of time.

I For example,X 0 1.57 3.14 4.71 6.28

Y = sin(x) 0 1 0 -1 0

I Therefore, Y = sin(x) = sin(2π + x).

I The sine function attains a peak at π/2 and a trough at 3π/2.

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I For example,X 0 1.57 3.14 4.71 6.28

Y = sin(x) 0 1 0 -1 0

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I For example,X 0 1.57 3.14 4.71 6.28

Y = sin(x) 0 1 0 -1 0

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I One may also induce a phase shift in the spectrum of thefunction that has effect of changing the location of the peaksand the troughs.

I For example, if Y = sin(x) and Z = sin(x + π/2), then Zpeaks at x = 0 and bottoms at x = π, whereas Y peaks andbottoms at x = π/2 and x = 3π/2 respectively.

I π/2 is the amount of phase shift in this example; it has theeffect of producing a new sine wave Z that moves ahead ofthe original sine wave Y by π/2 units.

I Clearly, a phase shift of 2π will produce a sine wave identicalto Y .

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I One may also induce a phase shift in the spectrum of thefunction that has effect of changing the location of the peaksand the troughs.

I For example, if Y = sin(x) and Z = sin(x + π/2), then Zpeaks at x = 0 and bottoms at x = π, whereas Y peaks andbottoms at x = π/2 and x = 3π/2 respectively.

I π/2 is the amount of phase shift in this example; it has theeffect of producing a new sine wave Z that moves ahead ofthe original sine wave Y by π/2 units.

I Clearly, a phase shift of 2π will produce a sine wave identicalto Y .

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I One can also control the magnitude of the peaks and thetroughs by changing the amplitude of the sine function.

I For example, W = 25sin(x), then

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I One can also control the magnitude of the peaks and thetroughs by changing the amplitude of the sine function.

I For example, W = 25sin(x), then

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I Thus, a general wave like function of Y may be written as:

I Yt = Asin(2πft + P), where

Yt is the time series at time t, t = 1, · · · , n;n is the number of observations;A is the amplitude;2π is one complete cycle;f is the frequency representing the number of cycles perobservation, so that L = 1/f , the wavelength, is the numberof periods from the beginning of one cycle to the next; andP is the phase shift

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I Consider the series

Z = 300 × sin( 4t482π + 1.5708) and

Y = 200 × sin( 4t482π)

Series A f P n LZ 300 4/48 1.5708 48 12Y 200 4/48 0 48 12

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Fitting a single sine wave to a time series

I How to fit a single sine wave to a time series?I Consider the following data with n = 20 observations:

Quarter

Year 1 2 3 4

2002 1.52 0.812003 0.63 1.06 1.46 0.802004 0.71 0.98 1.50 0.852005 0.65 1.04 1.47 0.852006 0.72 0.95 1.37 0.912007 0.74 0.98

I In practice, L and f are unknown. Here, we assume L = 4,i.e., it takes four periods to complete one cycle and there are5 cycles, so f = 5/20 = 1/4.

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I Hence the model to be estimated is:Yt = µ+ A× sin(2π5/20t + P) + εt

I Note that the unknowns µ, A and P are not estimable bystandard least squares techniques (why?)

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I Hence the model to be estimated is:Yt = µ+ A× sin(2π5/20t + P) + εt

I Note that the unknowns µ, A and P are not estimable bystandard least squares techniques (why?)

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I But note that

Asin(π

2t + P) = Asin(

2t)cos(P) + Acos(

2t)sin(P)

= b1sin(π

2t) + b2cos(

where b1 = Acos(P) and b2 = Asin(P).

I Hence the model may be written as

Yt = µ+ b1sin(π2 t) + b2cos(π2 t) + εt ,

where µ, b1 and b2 can be estimated by the method of leastsquares.

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I But note that

Asin(π

2t + P) = Asin(

2t)cos(P) + Acos(

2t)sin(P)

= b1sin(π

2t) + b2cos(

where b1 = Acos(P) and b2 = Asin(P).

I Hence the model may be written as

Yt = µ+ b1sin(π2 t) + b2cos(π2 t) + εt ,

where µ, b1 and b2 can be estimated by the method of leastsquares.

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I The data for the model are:

t Yt sin(π2 t) cos(π2 t)

1 1.52 1 02 0.81 0 -1

. . . .

. . . .20 0.98 0 1

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I Note that

b21 + b22 = A2cos2(P) + A2sin2(P)

I Hence A =√b21 + b22

I Also, b2b1

= sin(P)cos(P) = tan(P)

I Hence P = tan−1(b2b1 )

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I Note that

b21 + b22 = A2cos2(P) + A2sin2(P)

I Hence A =√b21 + b22

I Also, b2b1

= sin(P)cos(P) = tan(P)

I Hence P = tan−1(b2b1 )

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data fourier1;

input y @@;

cards;

1.52 0.81 0.63 1.06 1.46 0.80 0.71 0.98 1.50 0.85

0.65 1.04 1.47 0.85 0.72 0.95 1.37 0.91 0.74 0.98;

data fourier2;

set fourier1;

pi=3.1415926;

s5=sin(pi*t/2);

c5=cos(pi*t/2);

proc reg data=fourier2;

model y = s5 c5;

output out=out1 predicted=p residual=r;

proc print data=out1;

var y p r;

run; 18 / 55

data fourier1;

input y @@;

cards;

1.52 0.81 0.63 1.06 1.46 0.80 0.71 0.98 1.50 0.85

0.65 1.04 1.47 0.85 0.72 0.95 1.37 0.91 0.74 0.98;

data fourier2;

set fourier1;

pi=3.1415926;

s5=sin(pi*t/2);

c5=cos(pi*t/2);

proc reg data=fourier2;

model y = s5 c5;

output out=out1 predicted=p residual=r;

proc print data=out1;

var y p r;

run; 19 / 55

The REG Procedure Model: MODEL1 Dependent Variable: y Number of Observations Read 20 Number of Observations Used 20 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 2 1.56010 0.78005 84.52 <.0001 Error 17 0.15690 0.00923 Corrected Total 19 1.71700 Root MSE 0.09607 R-Square 0.9086 Dependent Mean 1.00000 Adj R-Sq 0.8979 Coeff Var 9.60698 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 1.00000 0.02148 46.55 <.0001 s5 1 0.38700 0.03038 12.74 <.0001 c5 1 0.07900 0.03038 2.60 0.0187 20 / 55

The SAS System Obs y p r 1 1.52 1.38700 0.13300 2 0.81 0.92100 -0.11100 3 0.63 0.61300 0.01700 4 1.06 1.07900 -0.01900 5 1.46 1.38700 0.07300 6 0.80 0.92100 -0.12100 7 0.71 0.61300 0.09700 8 0.98 1.07900 -0.09900 9 1.50 1.38700 0.11300 10 0.85 0.92100 -0.07100 11 0.65 0.61300 0.03700 12 1.04 1.07900 -0.03900 13 1.47 1.38700 0.08300 14 0.85 0.92100 -0.07100 15 0.72 0.61300 0.10700 16 0.95 1.07900 -0.12900 17 1.37 1.38700 -0.01700 18 0.91 0.92100 -0.01100 19 0.74 0.61300 0.12700 20 0.98 1.07900 -0.09900 MAPE=8.525%

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I Therefore,

b1 = 0.387

b2 = 0.079

A =√

0.3872 + 0.0792

= 0.395

P = tan−1(0.079

0.387)

= 0.2014

I Hence the estimated model may be expressed asYt = 1 + 0.395 × sin(π2 t + 0.2014)

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Fitting a set of sine waves to a time series

I In practice, L and f are unknown. As well, amplitudes ofpeaks and troughs are rarely equal.

I The single sine wave model may be extended to the followinggeneral Fourier Series model:

Yt = µ+h∑

Ajsin(2πfj t + Pj) + εt

= µ+h∑

b1jsin(2πfj t) + b2jcos(2πfj t) + εt , (1)

I which comprises a linear combination of ”harmonic waves”,each with its own amplitude(Aj), phase shift (Pj), frequency(fj), wave length (Lj) and unique number of cycles (n/Lj).

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I In practice, L and f are unknown. As well, amplitudes ofpeaks and troughs are rarely equal.

I The single sine wave model may be extended to the followinggeneral Fourier Series model:

Yt = µ+h∑

Ajsin(2πfj t + Pj) + εt

= µ+h∑

b1jsin(2πfj t) + b2jcos(2πfj t) + εt , (1)

I which comprises a linear combination of ”harmonic waves”,each with its own amplitude(Aj), phase shift (Pj), frequency(fj), wave length (Lj) and unique number of cycles (n/Lj).

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I In order for the model to be estimable, it is assumed that theminimum number of observations for the completion of anyharmonic wave is 2, i.e., the maximum allowable number ofharmonic waves, h, is n/2.

I When n is even,

Harmonic wave, j 1 2 3 ... n2 − 1 n

Wavelength, Lj = nj n n

2n3 ... 2n

n−2 2

Frequency, fj = 1Lj

3n ... n−2

I j also represents the number of cycles for the given harmonicwave.

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I In order for the model to be estimable, it is assumed that theminimum number of observations for the completion of anyharmonic wave is 2, i.e., the maximum allowable number ofharmonic waves, h, is n/2.

I When n is even,

Harmonic wave, j 1 2 3 ... n2 − 1 n

2n3 ... 2n

n−2 2

Frequency, fj = 1Lj

3n ... n−2

I j also represents the number of cycles for the given harmonicwave.

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I Note that when j = h = n/2,fj = 1

2 and sin(2πfj t) = sin(πt) = 0

I Hence the model reduces to

Yt = µ+

n/2−1∑j=1

b1jsin(2πfj t) + b2jcos(2πfj t)

+ b2n/2cos(2πfn/2t) + εt (2)

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I Note that when j = h = n/2,fj = 1

2 and sin(2πfj t) = sin(πt) = 0

I Hence the model reduces to

Yt = µ+

n/2−1∑j=1

+ b2n/2cos(2πfn/2t) + εt (2)

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I Alternatively, when n is odd, the maximum number ofharmonic waves is (n − 1)/2

Harmonic wave, j 1 2 3 ... n−12 − 1 n−1

2n3 ... 2n

n−32nn−1

Frequency, fj = 1Lj

3n ... n−3

2nn−12n

I Hence the model is

Yt = µ+

(n−1)/2∑j=1

b1jsin(2πfj t) + b2jcos(2πfj t) + εt (3)

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I In both models (2) and (3), the number of observations equalthe number of unknowns to be estimated, resulting in zerodegree of freedom !!

I The idea is not to use the general Fourier series model forforecasting but to use it to identify significant cycles.

I Consider the previous data set with 20 observations. Supposethat a general Fourier series model is fitted to this data set.Then h = 10 and the model is

Yt = µ+9∑

+ b2,10cos(2πf10t) + εt

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Yt = µ+9∑

+ b2,10cos(2πf10t) + εt

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Yt = µ+9∑

+ b2,10cos(2πf10t) + εt

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data fourier3; set fourier1; pi=3.1415926; t+1; s1=sin (pi*t/10); c1=cos (pi*t/10); s2=sin (pi*t/5); c2=cos (pi*t/5); s3=sin (3*pi*t/10); c3=cos (3*pi*t/10); s4=sin (2*pi*t/5); c4=cos (2*pi*t/5); s5=sin (pi*t/2); c5=cos (pi*t/2); s6=sin (3*pi*t/5); c6=cos (3*pi*t/5); s7=sin (7*pi*t/10); c7=cos (7*pi*t/10); s8=sin (4*pi*t/5); c8=cos (4*pi*t/5); s9=sin (9*pi*t/10); c9=cos (9*pi*t/10); s10=sin (pi*t); c10=cos (pi*t); run; 28 / 55

proc reg data=fourier3; model y = s1 c1 s2 c2 s3 c3 s4 c4 s5 c5 s6 c6 s7 c7 s8 c8 s9 c9 c10; output out=out1 predicted=p residual=r; run; proc print data=out1; var y p r; run; proc spectra data=fourier1 out=out2; var y; run; data out2; set out2; sq=p_01; if period=. then sq=0; if round (freq, .0001)=3.1416 then sq=.5*p_01; run; proc print data=out2; sum sq; run;

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The REG Procedure Model: MODEL1 Dependent Variable: y Number of Observations Read 20 Number of Observations Used 20 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 19 1.71700 0.09037 . . Error 0 0 . Corrected Total 19 1.71700 Root MSE . R-Square 1.0000 Dependent Mean 1.00000 Adj R-Sq . Coeff Var . Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 1.00000 . . . s1 1 -0.00068100 . . . c1 1 -0.00026440 . . . s2 1 0.00339 . . . c2 1 0.01208 . . . s3 1 -0.00723 . . . c3 1 0.01673 . . . s4 1 -0.02540 . . . c4 1 0.00964 . . . s5 1 0.38700 . . . c5 1 0.07900 . . . s6 1 0.00974 . . . c6 1 -0.02258 . . . s7 1 0.03026 . . . c7 1 -0.03044 . . . s8 1 -0.00781 . . . c8 1 -0.00714 . . . s9 1 0.00672 . . . c9 1 -0.00002737 . . . c10 1 -0.07700 . . .

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Hence we have

b11 = −0.00681, b21 = −0.0002644

b12 = 0.00339 , b22 = 0.01208

b15 = 0.387 , b25 = 0.079

b2,10 = −0.077 µ = 1.

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The SAS System Obs y p r 1 1.52 1.52 2.1684E-18 2 0.81 0.81 4.944E-17 3 0.63 0.63 1.6534E-16 4 1.06 1.06 4.944E-17 5 1.46 1.46 1.4637E-16 6 0.80 0.80 2.0914E-16 7 0.71 0.71 2.6563E-17 8 0.98 0.98 6.1908E-17 9 1.50 1.50 1.03E-16 10 0.85 0.85 6.0011E-17 11 0.65 0.65 1.3623E-16 12 1.04 1.04 2.7625E-16 13 1.47 1.47 1.7445E-16 14 0.85 0.85 2.7924E-16 15 0.72 0.72 1.912E-16 16 0.95 0.95 1.1661E-16 17 1.37 1.37 2.5013E-16 18 0.91 0.91 2.2503E-16 19 0.74 0.74 1.4246E-16 20 0.98 0.98 5.8056E-16

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The SAS System

Obs FREQ PERIOD P_01 sq

1 0.00000 . 40.0000 0.00000

2 0.31416 20.0000 0.0000 0.00001

3 0.62832 10.0000 0.0016 0.00157

4 0.94248 6.6667 0.0033 0.00332

5 1.25664 5.0000 0.0074 0.00738

6 1.57080 4.0000 1.5601 1.56010

7 1.88496 3.3333 0.0060 0.00605

8 2.19911 2.8571 0.0184 0.01842

9 2.51327 2.5000 0.0011 0.00112

10 2.82743 2.2222 0.0005 0.00045

11 3.14159 2.0000 0.2372 0.11858

1.7170

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I Some explanations of the symbols are in order:Obs - 1 = j ,PERIOD = Lj = wavelength,FREQ = 2πfj ,P 01 and sq are quantities related to the line spectrum (seeslides ahead)

I Note the sum of sq = 1.717 is precisely equal to theregression’s TSS.

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I Some explanations of the symbols are in order:Obs - 1 = j ,PERIOD = Lj = wavelength,FREQ = 2πfj ,P 01 and sq are quantities related to the line spectrum (seeslides ahead)

I Note the sum of sq = 1.717 is precisely equal to theregression’s TSS.

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I Define the line spectrum as

{ n2 (b21j + b22j) if j < n/2

nb22j if j = n/2

I A plot of Sj versus Lj , the wavelength, is called theperiodogram. It measures the intensity of the specificharmonic wave.

I The P 01 column in the SAS output gives the line spectrumsfor all harmonic waves, except when j = n/2, where thecorrect line spectrum should be P 01/2.

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{ n2 (b21j + b22j) if j < n/2

nb22j if j = n/2

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{ n2 (b21j + b22j) if j < n/2

nb22j if j = n/2

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One may test for the significance of the harmonic waves by asequential testing procedure:

Source SS df F Decision

5th harmonic 1.5601 2 84.52 significantOther harmonics 0.1569 17

10th harmonic 0.11858 1 49.51 significantOther harmonics 0.03832 16

7th harmonic 0.01842 2 6.48 significant at 5%;Other harmonics 0.0199 14 not significant at 1%

4th harmonic 0.00738 2 3.54 insignificantOther harmonics 0.01252 12

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I Thus, the 5th and 10th harmonic waves are significant, and asuitable model would be:

Yt = 1 + 0.387sin(πt2 ) + 0.079cos(πt2 ) − 0.077cos(πt)

I Out-of-sample forecasts (from period 21 onwards):

Y21 = 1 + 0.387sin(π21

2) + 0.079cos(

2) − 0.077cos(π21)

= 1.464016

Y24 = 1 + 0.387sin(π24

2) + 0.079cos(

2) − 0.077cos(π24)

= 1.002016

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I Thus, the 5th and 10th harmonic waves are significant, and asuitable model would be:

Yt = 1 + 0.387sin(πt2 ) + 0.079cos(πt2 ) − 0.077cos(πt)

I Out-of-sample forecasts (from period 21 onwards):

Y21 = 1 + 0.387sin(π21

2) + 0.079cos(

2) − 0.077cos(π21)

= 1.464016

Y24 = 1 + 0.387sin(π24

2) + 0.079cos(

2) − 0.077cos(π24)

= 1.002016

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The REG Procedure Model: MODEL1 Dependent Variable: y Number of Observations Read 20 Number of Observations Used 20 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 3 1.67868 0.55956 233.64 <.0001 Error 16 0.03832 0.00239 Corrected Total 19 1.71700 Root MSE 0.04894 R-Square 0.9777 Dependent Mean 1.00000 Adj R-Sq 0.9735 Coeff Var 4.89387 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 1.00000 0.01094 91.38 <.0001 s5 1 0.38700 0.01548 25.01 <.0001 c5 1 0.07900 0.01548 5.10 0.0001 c10 1 -0.07700 0.01094 -7.04 <.0001

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Real data example: U.S./New Zealand exchange rateforecast

I Let Yt=USD equivalence of one NZD between 1986Q2 and2008Q2 (i.e., n = 89)

I The NZD was floated in March 1985 (prior to that, it waspegged to the USD) and has been in the range of 1 NZD =0.39 to 0.88 USD between 1985 and 2014;

I The NZD is one of the most widely fluctuated currencies,being affected strongly by FOREX speculative trading andcommodity prices;

I The NZD’s post-float lowest and highest daily values were0.39 USD in November 2000 and 0.88 USD in July 2011;

I Use the Fourier Series to model ∆Yt = Yt − Yt−1 (forrequirement of ”stationarity” (see Chapter 5));

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U.S. / New Zealand Exchange Rate

e actual

predict

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data nzea; set nzea; t+1; pi=3.1415926; s1= sin(pi*t/44*1); c1= cos(pi*t/44*1); . . . c43= cos(pi*t/44*43); s44= sin(pi*t/44*44); c44= cos(pi*t/44*44); run; proc reg data= nzea; model D_USNZD = s1 c1 s2 c2 s3 c3 s4 c4 s5 c5 s6 c6 s7 c7 s8 c8 s9 c9 s10 c10 s11

c11 s12 c12 s13 c13 s14 c14 s15 c15 s16 c16 s17 c17 s18 c18 s19 c19 s20 c20 s21 c21 s22 c22 s23 c23 s24 c24

s25 c25 s26 c26 s27 c27 s28 c28 s29 c29 s30 c30 s31 c31 s32 c32 s33 c33 s34 c34 s35 c35 s36 c36 s37 c37

s38 c38 s39 c39 s40 c40 s41 c41 s42 c42 s43 c43 c44 ; output out=out1 predicted=p residual=r; run;

proc spectra data= nzea out=out2; var D_USNZD; run; data out2; set out2; sq=p_01; if period=. then sq=0; if round (freq, .0001)=3.1416 then sq=.5*p_01; run; proc reg data= nzea outest=outtest3; model D_USNZD = c31 s31 c3 s3 c8 s8 ; output out=out3 predicted=p residual=r; run;

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Sequential F test results:

Source SS df F Decision

31st harmonic 0.009 2 4.8804 significantOther harmonics 0.0788 85

3rd harmonic 0.0066 2 3.7920 significantOther harmonics 0.00722 83

8th harmonic 0.0062 2 3.8210 significant;Other harmonics 0.0660 81

9th harmonic 0.0039 2 2.4908 insignificantOther harmonics 0.0620 79

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I The fitted model is:

∆Yt = 0.0025 − 0.0075sinπ31t

44− 0.0122cos

+ 0.0113sinπ3t

44+ 0.0048cos

− 0.0061sinπ8t

44− 0.0102cos

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I Out of Sample Forecasts:

∆Y2008Q3 = 0.0025 − 0.0075sin31π89

44− 0.0122cos

31π89

+ 0.0113sin3π89

44+ 0.0048cos

− 0.0061sin8π89

44− 0.0102cos

44= −0.00098

Y2008Q3 = −0.00098 + 0.7526 = 0.7516

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∆Y2008Q4 = 0.0025 − 0.0075sin31π90

44− 0.0122cos

31π90

+ 0.0113sin3π90

44+ 0.0048cos

− 0.0061sin8π90

44− 0.0102cos

44= 0.012407

Y2008Q4 = 0.012407 + 0.7516 = 0.7640

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Caution with PROC SPECTRA in SAS

I Note that PROC SPECTRA can generate the coefficientestimates of the Fourier Series automatically viaPROC SPECTRA COEF;

I However, when interpreting the results, cautions must beexercised due to the peculiar way SAS defines the FourierSeries model.

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I Note that PROC SPECTRA can generate the coefficientestimates of the Fourier Series automatically viaPROC SPECTRA COEF;

I However, when interpreting the results, cautions must beexercised due to the peculiar way SAS defines the FourierSeries model.

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I SAS defines the Fourier Series model as follows (from SASmanual):

Yt =a202

+m−1∑j=1

ζj(a1jsin(ωj t) + a2jcos(ωj t)), (4)

I wheret = 0, 1, 2, .., n − 1; ωj = 2πj

n ;m − 1 = the number of harmonics cycles; and

{12 if n is even and j = m − 11 otherwise

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Hence,

I t under (1) is equivalent to t + 1 under (4);

I µ under (1) is equivalent to a0/2 under (4);

I ωj = 2πjn under (4) is identical to 2πfj under (1);

I h under (1) is the same as m− 1 under (4), hence m = n+22 if

n is even and m = n+12 if n is odd;

I due to the introduction of ζj under PROC SPECTRA, when nis even and j = m − 1 = h = n/2, the (correct) coefficientestimate of the cosine variable is equal to 0.5 times thereported estimate of a2j ; hence b2, n

2(under (1)) = 0.5 × a2, n

(under (4)).

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I Also, PROC SPECTRA producesP 01j = n

2 (a21j + a22j) for all j ’s (j = 0, 1, · · · ,m − 1).Each P 01j is a two degrees of freedom component of thecontribution of the j th harmonic wave to the TSS, exceptwhen n is even and j = m − 1, for which sin(ωj t) = 0 is zeroand thus the corresponding P 01j is twice a one degree offreedom component (or twice the contribution of thatharmonic wave to the TSS).

I P 01j matches the definition of Sj , except whenj = m − 1 = h = n/2, for which Sj = P 01j/2.

I When j = 0, the harmonic wave reduces to a ”point” andthus corresponds to the intercept of the model. Thecorresponding P 010 = 2nY 2.

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I Consider the data set in our example. The program

data fourier1; input y @@; cards; 1.52 0.81 0.63 1.06 1.46 0.80 0.71 0.98 1.50 0.85 0.65 1.04 1.47 0.85 0.72 0.95 1.37 0.91 0.74 0.98 ; proc spectra coef data=fourier1 out=out2; var y; run; data out2; set out2; sq=p_01; if period=. then sq=0; if round (freq, .0001)=3.1416 then sq=.5*p_01; run; proc print data=out2; sum sq; run;

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I produces the following results, where cos 01 and sin 01 arethe estimates of aj and bj respectively:

Obs FREQ PERIOD COS_01 SIN_01 P_01 sq 1 0.00000 . 2.00000 0.000000 40.0000 0.00000 2 0.31416 20.0000 -0.00046 -0.000566 0.0000 0.00001 3 0.62832 10.0000 0.01176 -0.004359 0.0016 0.00157 4 0.94248 6.6667 0.00399 -0.017783 0.0033 0.00332 5 1.25664 5.0000 -0.02118 -0.017013 0.0074 0.00738 6 1.57080 4.0000 0.38700 -0.079000 1.5601 1.56010 7 1.88496 3.3333 0.01624 0.018466 0.0060 0.00605 8 2.19911 2.8571 0.04237 0.006839 0.0184 0.01842 9 2.51327 2.5000 0.00118 0.010515 0.0011 0.00112 10 2.82743 2.2222 0.00210 -0.006378 0.0005 0.00045 11 3.14159 2.0000 0.15400 0.000000 0.2372 0.11858 ======= 1.71700

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I Note that P 01 are the same as those on p.32 but thecoefficient estimates are not the same as those obtained byPROC REG on p.30;

I The difference is due to a redefinition of t, the time script byPROC SPECTRA;

I One can get around the problem by redefining the time scriptto start from 0 for the program on p.28 by replacing thecommandt + 1; bytt + 1;t = tt − 1;

I This results in the following output:

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The REG Procedure Model: MODEL1 Dependent Variable: y Number of Observations Read 20 Number of Observations Used 20 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 19 1.71700 0.09037 . . Error 0 0 . Corrected Total 19 1.71700 Root MSE . R-Square 1.0000 Dependent Mean 1.00000 Adj R-Sq . Coeff Var . Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 1.00000 . . . s1 1 -0.00056597 . . . c1 1 -0.00046190 . . . s2 1 -0.00436 . . . c2 1 0.01176 . . . s3 1 -0.01778 . . . c3 1 0.00399 . . . s4 1 -0.01701 . . . c4 1 -0.02118 . . . s5 1 -0.07900 . . . c5 1 0.38700 . . . s6 1 0.01847 . . . c6 1 0.01624 . . . s7 1 0.00684 . . . c7 1 0.04237 . . . s8 1 0.01051 . . . c8 1 0.00118 . . . s9 1 -0.00638 . . . c9 1 0.00210 . . . c10 1 0.07700 . . .

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I The results are the same as those produced by PROCSPECTRA; specifically,

I µ = 1.0 = a20/2 = 2.0/2;

I The two procedures produce the same sine and cosinecoefficient estimates; note that for j = 10, a2,10 = 0.154 andζ10 = 1/2, hence the (correct) cosine coefficient estimate isζ10a2,10=1/2 × 0.154 = 0.077;

I P 0110 = 0.2372 = 202 (0.1542). The correct line spectrum is

S10 = 0.2372/2 = 0.11858 as defined on p.35.

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chapter 2: fourier series (spectral)...

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