1 fourier series (spectral) analysis. 2 fourier series analysis suitable for modelling seasonality...

1

Fourier Series (Spectral) Analysis

2

Fourier Series Analysis

Suitable for modelling seasonality and/or cyclicalness

Identifying peaks and troughs

3

A sine wave is a repeating pattern that goes through one cycle every 2 (i.e. 2 3.141593 = 6.283186) units of time.

For example,

X 0 1.57 3.14 4.71 6.28

Y = sin (x) 0 1 0 – 1 0

4

So sin (6.283186 + x) = sin (x)

5

Since Wave Phase Shifts

Y = sin (x)

Z = sin (x + 1.5708)

0.5 (phase shift)

6

Amplitude

Y = 25 sin (x)

Amplitude

7

Combining Amplitude, Phase, Frequency and Wavelength

Yt = A*sin(2 ft + )

where

t = Time (i.e., 1, 2, 3, …, n)

Yt = Value of time series at time t

A = Amplitude

f = Number of cycle per observation

n = Number of observations in time series

2 = one complete cycle

= Phase shift

8

Amplitude determines the heights of peaks and depths of troughs

Phase shift determines where the peaks and troughs occur

Let be the wavelength, i.e. the number of

periods from the beginning of one cycle to the next.

fL

1

9

0

0 n

n

nLn

f

;1

2

;2

nL

nf

2

n

10

248

4sin200

5708.1248

4sin300

tY

tZ

Series A f P n L

Z 300 4/48 1.5708 48 12

Y 200 4/48 0 48 12

12

How to fit a single sine wave to a time series?

Consider the following data set:

YearQuarter

1 2 3 4

2002 1.52 0.81

2003 0.63 1.06 1.46 0.80

2004 0.71 0.98 1.50 0.85

2005 0.65 1.04 1.47 0.85

2006 0.72 0.95 1.37 0.91

2007 0.74 0.98

n = 20

Assume one cycle completes itself every year, then f = 5/20 and L = 4

13

So the model is :

This equation cannot be estimated by standard techniques.

t

tt

PtA

PtAY

2sin

20

52sin

14

But note that

PAb

PAb

tbtb

PtAPtA

PtA

sin

andcos

where

,2

cos2

sin

sin2

coscos2

sin

2sin

2

1

21

15

1

21

1

2

22

21

2

22222

21

tan

cos

sintan

Also,

.

sincos

Since

b

bP

b

b

P

PP

bbA

A

PPAbb

16

So the model becomes:

tt tbtbY

2cos

2sin 21

t Yt

1 1.52 1 0

2 0.81 0 – 1

20 0.98 0 1

t2

sin

t2

cos

17

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

;run;

data fourier2;

set fourier1;

pi=3.1415926;

t+1;

s5=sin(pi*t/2);

c5=cos(pi*t/2);

run;

proc reg data=fourier2;

model y = s5 c5;

output out=out1 predicted=p residual=r;

run;

proc print data=out1;

var y p r;

run;

18

The SAS System

The REG Procedure

Model: MODEL1

Dependent Variable: y

Number of Observations Read 20

Number of Observations Used 20

Analysis of Variance

Source DF Sum of Squares Mean 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

Variable DF Parameter Estimate Standard 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

19

The SAS System

Obs y p r

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

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

1.38700

0.92100

0.61300

1.07900

1.38700

0.92100

0.61300

1.07900

1.38700

0.92100

0.61300

1.07900

1.38700

0.92100

0.61300

1.07900

1.38700

0.92100

0.61300

1.07900

0.13300

–0.11100

0.01700

–0.01900

0.07300

–0.12100

0.09700

–0.09900

0.11300

–0.07100

0.03700

–0.03900

0.08300

–0.07100

0.10700

–0.12900

–0.01700

–0.01100

0.12700

–0.09900MAPE = 8.525%

20

2014.0

387.0

079.0tanˆ

395.0

07900.038700.0ˆ

07900.0ˆ

38700.0ˆ,So

1

22

2

1

P

A

b

b

2014.0

2sin395.01ˆ

is model estimated theand

tYt

21

Fitting a set of sine waves to a time series

Amplitudes of peaks and troughs are rarely equal

General Fourier series model:

So, the general model is a linear combination of cycles at the “harmonic” frequencies, each with its amplitude (Aj) and phase shift (Pj)

t

h

jjjjt PtfAY

1

2sin

t

h

jjjjj tfbtfb

121 2cos2sin

22

Assume n is even

The highest harmonic, h, is if n is even and (n-1)/2

if n is odd.

Harmonic, j

Harmonic Frequency,

Wavelength, j

j fj

nL

1

12

321 n

2

n

n

1

n

2

n

3n

n

2

2

2

1

22

2

n

nn

2

n

3

n

2

n

n

jf j

23

Note that when ,

and

So,

2

nh

0sin2sin ttf j 2

1jf

tfbtfbY jj

h

jjjt 2cos2sin 2

1

11

2 cos 2h h tb f t

24

The General Fourier series model contains n unknowns to be estimated with n observations, i.e. no degree of freedom !!

The idea is NOT to use the General Fourier Series model for forecasting, but to use it for identifying significant cycles.

25

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;

26

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=. or period=4 then sq=0;if round (freq, .0001)=3.1416 then sq=.5*p_01;run;proc print data=out2;sum sq;run;

27

The SAS System

The REG Procedure

Model: MODEL1






Model 19 1.71700 0.09037 . .

Error 0 0 .


Root MSE . R-Square 1.0000

Dependent Mean 1.00000 Adj R-Sq .

Coeff Var .

28

Parameter Estimates

Variable DF Parameter Estimate

Standard Error t Value Pr > ︳ t ︱

Intercept 1 1.00000 . . .

s1 1 -0.000681000 . . .

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 . . .

29

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 2.7934E - 16

15 0.72 0.72 1.912E - 16

16 0.95 0.95 1.165E - 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 5.8059E - 17

30

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 0.00000

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

0.15690

31

So,

0.1ˆ

387.0ˆ

00339.0ˆ

00681.0ˆ

15

12

11

b

b

b

077.0ˆ

;079.0ˆ

;01208.0ˆ

;0002644.0ˆ

210

25

22

21

b

b

b

b

32

The Line Spectrum (Periodogram)

To identify significant cyclical patterns.

The Line Spectrum is the amount of total sums of squares explained by the specific frequencies.

33

Line Spectrum can be computed as

The plot of Pj’s versus the wave length is called the periodogram. It measures the “intensity” of the specific cycles.

The “P_01” column in SAS output gives the line spectrums for all harmonics, except for the last harmonic, where the correct line spectrum is P_01/2.

jP 2 2

1 2

22

2 j j

j

nb b

nb

2

2

nh

nh

35

Source SS df F Decision

5th harmonic

Other harmonics

1.5601

0.1569

2

17

84.52 Significant

10th harmonic

Other harmonics

0.11858

0.03832

1

16

49.51 Significant

7th harmonic

Other harmonics

0.01842

0.0199

2

14

6.48 Significant

at 5% ; not significant at 1%

4th harmonic

Other harmonics

0.00738

0.01252

2

12

3.54 not significant

ANOVA test

36

Thus, the 5th & 10th harmonics (or equivalently, the

4- period & 2-period cycles) are significant which

means a suitable model would be

or equivalently,

tttYt

cos077.02

cos079.02

sin387.01ˆ

5708.1sin077.02014.02

sin395.01ˆ

t

tYt

37

Note:

since

0.077cos 0.077sin 1.5708t t

sinsin

2sincos

and

38

The SAS System

The REG Procedure

Model: MODEL1






Model 3 1.67868 0.55956 233.64 <.0001

Error 16 0.03832 0.00239


Root MSE 0.04894 R-Square 0.9777

Dependent Mean 1.00000 Adj R-Sq 0.9735

Coeff Var 4.89387

Parameter Estimates

Variable DF Parameter Estimate Standard 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

39

The SAS System

Obs y p r

1 1.52 1.46400 0.056000

2 0.81 0.84400 -0.034000

3 0.63 0.69000 -0.060000

4 1.06 1.00200 0.058000

5 1.46 1.46400 -0.004000

6 0.80 0.84400 -0.044000

7 0.71 0.69000 0.020000

8 0.98 1.00200 -0.022000

9 1.50 1.46400 0.036000

10 0.85 0.84400 0.006000

11 0.65 0.69000 -0.040000

12 1.04 1.00200 0.038000

13 1.47 1.46400 0.006000

14 0.85 0.84400 0.006000

15 0.72 0.69000 0.030000

16 0.95 1.00200 -0.052000

17 1.37 1.46400 -0.094000

18 0.91 0.84400 0.066000

19 0.74 0.69000 0.050000

20 0.98 1.00200 -0.022000

MAPE = 4.113%

40

Out-of-Sample forecasts

5708.121sin077.02014.02

21sin395.012̂1

Y

464016.1

5708.124sin077.02014.02

24sin395.012̂4

Y

002016.1

41

Example: U.S./New Zealand foreign Exchange Rate(1986 Q2 to 2008 Q2) Quarterly average

U.S. / New Zealand Exchange Rate

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Q2 19

86

Q1 19

87

Q4 19

87

Q3 19

88

Q2 19

89

Q1 19

90

Q4 19

90

Q3 19

91

Q2 19

92

Q1 19

93

Q4 19

93

Q3 19

94

Q2 19

95

Q1 19

96

Q4 19

96

Q3 19

97

Q2 19

98

Q1 19

99

Q4 19

99

Q3 20

00

Q2 20

01

Q1 20

02

Q4 20

02

Q3 20

03

Q2 20

04

Q1 20

05

Q4 20

05

Q3 20

06

Q2 20

07

Q1 20

08

Q4 20

08

Date

Rat

e actual

predict

42

data Nzea;

set Nzea;

D_USNEWZD =USNEWZD - lag(USNEWZD);

if Date = “Q2 1986” then delete;

run;

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;

43

proc reg data= Nzea;

model D_USNEWZD = 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 ;


run;

quit;

proc spectra data= Nzea out=out2;

var D_USNEWZD;

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_USNEWZD = c31 s31 c3 s3 c8 s8 ;


run;

quit;

44

ANOVA test

Source SS df F Decision

31st harmonic 0.0090 2 4.8804 Significant

Other harmonics 0.0788 85

3rd harmonic 0.0066 2 3.7920 Significant


8th harmonic 0.0062 2 3.8210 Significant


9th harmonic 0.0039 2 2.4908 Insignificant


45

Fitted model equation

Forecasts

)44

8cos(0102.0)

44

8sin(0061.0)

44

3cos(0048.0)

44

3sin(0113.0)

44

31cos(0122.0)

44

31sin(0075.00025.0ˆ tttttt

Yt

)44

898cos(0102.0)

44

898sin(0061.0)

44

893cos(0048.0)

44

893sin(0113.0)

44

8931cos(0122.0)

44

8931sin(0075.00025.0ˆ

32008

QY

)44

908cos(0102.0)

44

908sin(0061.0)

44

903cos(0048.0)

44

903sin(0113.0)

44

9031cos(0122.0)

44

9031sin(0075.00025.0ˆ

42008

QY

7526.000098.0ˆ Y

7640.0

7516.0012407.0ˆ Y

7516.0

1 fourier series (spectral) analysis. 2 fourier series analysis suitable for modelling seasonality...

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