Trigonometric functions in discrete time:
sin ( 2pπ t )
cos ( 2pπ t )
p=12
p=6
p=4
p=3
p=2.4
p=2
If one observation is sampled per unit of time, a periodic
function with period n completes one full cycle in the
observation period, a periodic function with period n/2
completes two full cycles in the observation period, etc. In
general, a periodic function with period n/k completes k full
cycles in the observation period. The frequency ωk=2πk/n
that is associated with the period n/k is called the k-th
Fourier frequency.
We consider only those Fourier frequencies ωk that are less
than or equal to the Nyquist frequency π, i.e., k≤m=[n/2].
If the improper Fourier frequency 2π 0/n=0 is also included,
there are always n non-vanishing sines and cosines.
If n is even and m=n/2, we have
1)
t ).
1, cos( 2πn⋅1 t ), …, cos( 2 π⋅n m t ), sin( 2πn⋅1 t ), ..., sin( 2 π⋅(mn
Clearly,
g(t)=cos( 2π2 t)=cos(π t)=(-1)t
is the most rapid oscillation we can observe.
The frequency π (one half a cycle per sampling interval)
is known as Nyquist frequency.
If n is odd and m=(n-1)/2, we have
1, cos( 2πn ⋅1 t ), ..., cos( 2πn⋅m t ), sin( 2πn ⋅1 t ), …, sin( 2πn⋅m t ).
Note that cos( 2 πn⋅0 t ) equals 1 and sin( 2 πn⋅0 t ) vanishes.
If n is even, then sin( 2 π⋅n m t )=sin(πt) vanishes too.
1
n
Exericse: Show that bˆ = ∑ xt yt
t =1
n
∑x
2
t
minimizes the
t =1
(Aˆ , Aˆ , Bˆ ,...)
= ((C , C , S ,...) (C , C , S ,...))
T
0
n
sum of squared errors
In the case of orthogonal regressors, the LS estimates
∑ ( yt−bxt)2.
CB
t =1
Using only Fourier frequencies ωk=2π k/n, k∈T⊆{0,…,m},
in the trigonometric regression model
yt= ∑ ( Akcos(ωkt)+Bksin(ωkt))+ut
k∈T
has the advantage that the regressors are orthogonal1, i.e.,
C S j = 0 , if 0≤j,k≤m,
T
k
1
1
1
where
1
1
−1
(C0 , C1 , S1 ,...)T y
 C0T y 
 T 
 C1 y 
 STy
 1 
 M 


= diag ( n , n / 2 , n / 2 ,...)
n
Aˆ 0 = ∑ cos(ω0t ) yt
2π j
n
n
n
∑ cos
t =1
2
(ω0t ) =
n
1
n
∑ cos(ω0t ) yt = y ,
n
t =1
n
t =1
n
t =1
n
t =1
t =1
Aˆ1 = ∑ cos(ω1t ) yt
∑ cos 2 (ω1t ) = 2n ∑ cos(ω1t ) yt ,
Bˆ1 = ∑ sin(ω1t ) yt
∑ sin 2 (ω1t ) = n2 ∑ sin(ω1t ) yt , …
t =1
n
t =1
See Appendix B.
0
 C0T C0 C0T C1 C0T S1 L
 T

 C1 C0 C1T C1 C1T S 0 L
= T

T
T
 S1 C0 S1 C1 S1 S1 L
 M
M
M
O
14444
4244444
3
t =1
 cos (
 sin (
⋅1 ) 
⋅1 ) 




Ck = 
M
M
 , Sk = 
.




2π k
2π j
 cos ( n ⋅ n) 
 sin ( n ⋅ n) 
1
1
obtained from the full model are identical to the estimates
S kT S j = CkT C j = 0 , if 0≤j≠k≤m,
2π k
n
−1
T
0
obtained from simple models with only one regressor.
CR
2
n
Exercise: Show that
∑ cos ( π
2 2 k
n
t ) =n, if k= n2 .
CN
t =1
Exercise: If n is even, then m=n/2 and ωm=π. But since
sin(π t)=0, only cos(π t) can be used as a regressor. Show
that the LS estimate of the parameter Am is given by
Aˆ m =
n
1
n
∑ y (−1) .
t
CM
t
The periodogram I(ω) of a time series y1,…,yn is for a
Fourier frequency ωk with k∈{1,2,…,[ n2−1 ]} defined by
=
((
∑y
t
t =1
n
I(ω)=
=
Rˆ k2
n
n
n
t =1
t =1
I(ω)= 2π1 n (( ∑ yt cos(ωt))2+( ∑ yt sin(ωt))2),
1
2π n
n
n
t =1
t =1
cos(ωkt)) +(
2
n
2
n
n
∑y
t
sin(ωkt)) )
t =1
2
1
2π n
2
∑ yt( cos (ωt) + i sin (ωt))
t =1
n
= 2π1 n
2
∑ yt cos (ωt) + i∑ yt sin (ωt)
n
I(ωk)= 8nπ ( Aˆ k2 + Bˆ k2 )
1
424
3
2
n
Defining the periodogram for an arbitrary frequency ω by
it can be written in complex form as
t =1
n
8π
Apart from an unimportant scaling factor, I(ωk) is just the
squared estimate of the amplitude of a sinusoid with
frequency ωk, hence its size indicates how important that
particular frequency is.
∑ yt e
2
iωt
.
CP
t =1
= 2π1 n (( ∑ yt cos(ωkt))2+( ∑ yt sin(ωkt))2).
t =1
t =1
3
Exercise: Show that it is sufficient to consider I(ω) on the
interval 0≤ω≤π by proving that I(ω) is periodic with period
2π and symmetric about the y-axis.
CS
Exercise: Revisit the not seasonally adjusted Retail and
Food Services Sales. Plot the differenced log series.
d <- y-lag(y,k=-1); d <- na.omit(d,method="r")
par(mar=c(2,2,1,1)); plot(d,type="l")
The sequence
n
∑ yt e −iωk t , k=0,…,n−1,
t =1
is called the discrete Fourier transform (DFT) of the
sequence y1,…,yn.
The R function fft uses a fast algorithm (the fast Fourier
transform) to calculate the slightly different version
n
∑ yt e −iωk (t −1) , k=0,…,n−1.
t =1
However, we have
n
∑ yt e
2
−iωk ( t −1)
=e
iωk
t =1
2
n
∑ yt e
−iωk t
t =1
=
2
eiωk
{
=1
n
∑ yt e
2
−iωk t
.
t =1
4
Exercise: Calculate the periodogram of d at the Fourier
frequencies ωk, k=1,…,[n/2] and plot it.
n <- length(d); m <- floor(n/2) # number of frequencies
f <- (2*pi/n)*(1:m) # vector of frequencies
ft <- fft(d)[2:(m+1)] # excl. k=0,m+1,m+2,...,n-1
pg <- (1/(2*pi*n))*(Mod(ft))^2 # Mod = modulus
plot(f,pg,type="l")
Exercise: Redo the last exercise, but this time omit some
observations to guarantee that the seasonal frequencies
2πj/12, j=1,…,6, are Fourier frequencies.
r12 <- n%%12 # n modulo 12
# the remainder we get when we divide n by 12
n12 <- n - r12
d12 <- d[(r12+1):n] # omit the first r12 observations
# length of new series d12 is divisible by 12
m12 <- floor(n12/2); f12 <- (2*pi/n12)*(1:m12)
ft12 <- fft(d12); ft12 <- ft12[2:(m12+1)]
pg12 <- (1/(2*pi*n12))*(Mod(ft12))^2
plot(f12,pg12,type="l")
The peaks at the seasonal frequencies are more pronounced
than the slightly displaced peaks in the last exercise.
5