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Introduction to Time Series Analysis. Lecture 15.
Spectral Analysis
1. Spectral density: Facts and examples.
2. Spectral distribution function.
3. Wold’s decomposition.
1
Spectral Analysis
Idea: decompose a stationary time series{Xt} into a combination of
sinusoids, with random (and uncorrelated) coefficients.
Just as in Fourier analysis, where we decompose (deterministic) functions
into combinations of sinusoids.
This is referred to as ‘spectral analysis’ or analysis in the‘frequency
domain,’ in contrast to the time domain approach we have considered so far.
The frequency domain approach considers regression on sinusoids; the time
domain approach considers regression on past values of the time series.
2
A periodic time series
Consider
Xt = A sin(2πνt) + B cos(2πνt)
= C sin(2πνt + φ),
whereA, B are uncorrelated, mean zero, varianceσ2 = 1, and
C2 = A2 + B2, tan φ = B/A. Then
µt = E[Xt] = 0
γ(t, t + h) = cos(2πνh).
So{Xt} is stationary.
3
An aside: Some trigonometric identities
tan θ =sin θ
cos θ,
sin2 θ + cos2 θ = 1,
sin(a + b) = sin a cos b + cos a sin b,
cos(a + b) = cos a cos b − sin a sin b.
4
A periodic time series
ForXt = A sin(2πνt) + B cos(2πνt), with uncorrelatedA, B
(mean 0, varianceσ2), γ(h) = σ2 cos(2πνh).
The autocovariance of the sum of two uncorrelated time series is the sum of
their autocovariances. Thus, the autocovariance of a sum ofrandom
sinusoids is a sum of sinusoids with the corresponding frequencies:
Xt =k∑
j=1
(Aj sin(2πνjt) + Bj cos(2πνjt)) ,
γ(h) =k∑
j=1
σ2j cos(2πνjh),
whereAj , Bj are uncorrelated, mean zero, and Var(Aj) = Var(Bj) = σ2j .
5
A periodic time series
Xt =
k∑
j=1
(Aj sin(2πνjt) + Bj cos(2πνjt)) , γ(h) =
k∑
j=1
σ2j cos(2πνjh).
Thus, we can representγ(h) using a Fourier series. The coefficients are the
variances of the sinusoidal components.
Thespectral density is the continuous analog: the Fourier transform ofγ.
(The analogousspectral representation of a stationary processXt involves
a stochastic integral—a sum of discrete components at a finite number of
frequencies is a special case. We won’t consider this representation in this
course.)
6
Spectral density
If a time series {Xt} has autocovarianceγ satisfying∑∞
h=−∞ |γ(h)| < ∞, then we define itsspectral density as
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh
for −∞ < ν < ∞.
7
Spectral density: Some facts
1. We have∑∞
h=−∞
∣
∣γ(h)e−2πiνh∣
∣ < ∞.
This is because|eiθ| = | cos θ + i sin θ| = (cos2 θ + sin2 θ)1/2 = 1,
and because of the absolute summability ofγ.
2. f is periodic, with period1.
This is true sincee−2πiνh is a periodic function ofν with period1.
Thus, we can restrict the domain off to−1/2 ≤ ν ≤ 1/2. (The text
does this.)
8
Spectral density: Some facts
3. f is even (that is,f(ν) = f(−ν)).To see this, write
f(ν) =−1∑
h=−∞
γ(h)e−2πiνh + γ(0) +∞∑
h=1
γ(h)e−2πiνh,
f(−ν) =−1∑
h=−∞
γ(h)e−2πiν(−h) + γ(0) +∞∑
h=1
γ(h)e−2πiν(−h),
=
∞∑
h=1
γ(−h)e−2πiνh + γ(0) +
−1∑
h=−∞
γ(−h)e−2πiνh
= f(ν).
4. f(ν) ≥ 0.
9
Spectral density: Some facts
5. γ(h) =
∫ 1/2
−1/2
e2πiνhf(ν) dν.
∫ 1/2
−1/2
e2πiνhf(ν) dν =
∫ 1/2
−1/2
∞∑
j=−∞
e−2πiν(j−h)γ(j) dν
=
∞∑
j=−∞
γ(j)
∫ 1/2
−1/2
e−2πiν(j−h) dν
= γ(h) +∑
j 6=h
γ(j)
2πi(j − h)
(
eπi(j−h) − e−πi(j−h))
= γ(h) +∑
j 6=h
γ(j) sin(π(j − h))
π(j − h)= γ(h).
10
Example: White noise
For white noise{Wt}, we have seen thatγ(0) = σ2w andγ(h) = 0 for
h 6= 0.
Thus,
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh
= γ(0) = σ2w.
That is, the spectral density is constant across all frequencies: each
frequency in the spectrum contributes equally to the variance. This is the
origin of the namewhite noise: it is like white light, which is a uniform
mixture of all frequencies in the visible spectrum.
11
Example: AR(1)
ForXt = φ1Xt−1 + Wt, we have seen thatγ(h) = σ2wφ
|h|1 /(1− φ2
1). Thus,
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh =σ2
w
1 − φ21
∞∑
h=−∞
φ|h|1 e−2πiνh
=σ2
w
1 − φ21
(
1 +
∞∑
h=1
φh1
(
e−2πiνh + e2πiνh)
)
=σ2
w
1 − φ21
(
1 +φ1e
−2πiν
1 − φ1e−2πiν+
φ1e2πiν
1 − φ1e2πiν
)
=σ2
w
(1 − φ21)
1 − φ1e−2πiνφ1e
2πiν
(1 − φ1e−2πiν)(1 − φ1e2πiν)
=σ2
w
1 − 2φ1 cos(2πν) + φ21
.
12
Examples
White noise:{Wt}, γ(0) = σ2w andγ(h) = 0 for h 6= 0.
f(ν) = γ(0) = σ2w.
AR(1): Xt = φ1Xt−1 + Wt, γ(h) = σ2wφ
|h|1 /(1 − φ2
1).
f(ν) =σ2
w
1−2φ1 cos(2πν)+φ2
1
.
If φ1 > 0 (positive autocorrelation), spectrum is dominated by low
frequency components—smooth in the time domain.
If φ1 < 0 (negative autocorrelation), spectrum is dominated by high
frequency components—rough in the time domain.
13
Example: AR(1)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
10
20
30
40
50
60
70
80
90
100
ν
f(ν)
Spectral density of AR(1): Xt = +0.9 X
t−1 + W
t
14
Example: AR(1)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
10
20
30
40
50
60
70
80
90
100
ν
f(ν)
Spectral density of AR(1): Xt = −0.9 X
t−1 + W
t
15
Introduction to Time Series Analysis. Lecture 16.
1. Review: Spectral density
2. Examples
3. Spectral distribution function.
4. Autocovariance generating function and spectral density.
1
Review: Spectral density
If a time series {Xt} has autocovarianceγ satisfying∑∞
h=−∞ |γ(h)| <∞, then we define itsspectral density as
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh
for −∞ < ν <∞.
2
Review: Spectral density
1. f(ν) is real.
2. f(ν) ≥ 0.
3. f is periodic, with period1. So we restrict the domain off to
−1/2 ≤ ν ≤ 1/2.
4. f is even (that is,f(ν) = f(−ν)).
5. γ(h) =∫ 1/2
−1/2
e2πiνhf(ν) dν.
3
Examples
White noise:{Wt}, γ(0) = σ2w andγ(h) = 0 for h 6= 0.
f(ν) = γ(0) = σ2w.
AR(1): Xt = φ1Xt−1 +Wt, γ(h) = σ2wφ
|h|1 /(1− φ21).
f(ν) =σ2
w
1−2φ1 cos(2πν)+φ2
1
.
If φ1 > 0 (positive autocorrelation), spectrum is dominated by low
frequency components—smooth in the time domain.
If φ1 < 0 (negative autocorrelation), spectrum is dominated by high
frequency components—rough in the time domain.
4
Example: AR(1)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
10
20
30
40
50
60
70
80
90
100
ν
f(ν)
Spectral density of AR(1): Xt = +0.9 X
t−1 + W
t
5
Example: AR(1)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
10
20
30
40
50
60
70
80
90
100
ν
f(ν)
Spectral density of AR(1): Xt = −0.9 X
t−1 + W
t
6
Example: MA(1)
Xt =Wt + θ1Wt−1.
γ(h) =
σ2w(1 + θ21) if h = 0,
σ2wθ1 if |h| = 1,
0 otherwise.
f(ν) =1
∑
h=−1
γ(h)e−2πiνh
= γ(0) + 2γ(1) cos(2πν)
= σ2w
(
1 + θ21 + 2θ1 cos(2πν))
.
7
Example: MA(1)
Xt =Wt + θ1Wt−1.
f(ν) = σ2w
(
1 + θ21 + 2θ1 cos(2πν))
.
If θ1 > 0 (positive autocorrelation), spectrum is dominated by low
frequency components—smooth in the time domain.
If θ1 < 0 (negative autocorrelation), spectrum is dominated by high
frequency components—rough in the time domain.
8
Example: MA(1)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
0.5
1
1.5
2
2.5
3
3.5
4
ν
f(ν)
Spectral density of MA(1): Xt = W
t +0.9 W
t−1
9
Example: MA(1)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
0.5
1
1.5
2
2.5
3
3.5
4
ν
f(ν)
Spectral density of MA(1): Xt = W
t −0.9 W
t−1
10
Introduction to Time Series Analysis. Lecture 16.
1. Review: Spectral density
2. Examples
3. Spectral distribution function.
4. Autocovariance generating function and spectral density.
5. Rational spectra. Poles and zeros.
11
Recall: A periodic time series
Xt =
k∑
j=1
(Aj sin(2πνjt) +Bj cos(2πνjt))
=k∑
j=1
(A2j + B2
j )1/2 sin(2πνjt+ tan−1(Bj/Aj)).
E[Xt] = 0
γ(h) =k∑
j=1
σ2j cos(2πνjh)
∑
h
|γ(h)| = ∞.
12
Discrete spectral distribution function
ForXt = A sin(2πλt) +B cos(2πλt), we haveγ(h) = σ2 cos(2πλh), and
we can write
γ(h) =
∫ 1/2
−1/2
e2πiνhdF (ν),
whereF is the discrete distribution
F (ν) =
0 if ν < −λ,σ2
2 if −λ ≤ ν < λ,
σ2 otherwise.
13
The spectral distribution function
For any stationary{Xt} with autocovarianceγ, we can write
γ(h) =
∫ 1/2
−1/2
e2πiνhdF (ν),
whereF is thespectral distribution function of {Xt}.
We can splitF into three components: discrete, continuous, and singular.
If γ is absolutely summable,F is continuous:dF (ν) = f(ν)dν.
If γ is a sum of sinusoids,F is discrete.
14
The spectral distribution function
ForXt =∑k
j=1 (Aj sin(2πνjt) +Bj cos(2πνjt)), the spectral distribution
function isF (ν) =∑k
j=1 σ2jFj(ν), where
Fj(ν) =
0 if ν < −νj ,12 if −νj ≤ ν < νj ,
1 otherwise.
15
Wold’s decomposition
Notice thatXt =∑k
j=1 (Aj sin(2πνjt) +Bj cos(2πνjt)) is deterministic
(once we’ve seen the past, we can predict the future without error).
Wold showed that every stationary process can be represented as
Xt = X(d)t +X
(n)t ,
whereX(d)t is purely deterministic andX(n)
t is purely nondeterministic.
(c.f. the decomposition of a spectral distribution function asF (d) + F (c).)
Example:Xt = A sin(2πλt) + θ(B)φ(B)Wt.
16
Introduction to Time Series Analysis. Lecture 16.
1. Review: Spectral density
2. Examples
3. Spectral distribution function.
4. Autocovariance generating function and spectral density.
17
Autocovariance generating function and spectral density
SupposeXt is a linear process, so it can be written
Xt =∑∞
i=0 ψiWt−i = ψ(B)Wt.
Consider the autocovariance sequence,
γh = Cov(Xt, Xt+h)
= E
∞∑
i=0
ψiWt−i
∞∑
j=0
ψjWt+h−j
= σ2w
∞∑
i=0
ψiψi+h.
18
Autocovariance generating function and spectral density
Define the autocovariance generating function as
γ(B) =∞∑
h=−∞
γhBh.
Then, γ(B) = σ2w
∞∑
h=−∞
∞∑
i=0
ψiψi+hBh
= σ2w
∞∑
i=0
∞∑
j=0
ψiψjBj−i
= σ2w
∞∑
i=0
ψiB−i
∞∑
j=0
ψjBj = σ2
wψ(B−1)ψ(B).
19
Autocovariance generating function and spectral density
Notice that
γ(B) =
∞∑
h=−∞
γhBh.
f(ν) =∞∑
h=−∞
γhe−2πiνh
= γ(
e−2πiν)
= σ2wψ
(
e−2πiν)
ψ(
e2πiν)
= σ2w
∣
∣ψ(
e2πiν)∣
∣
2.
20
Autocovariance generating function and spectral density
For example, for an MA(q), we haveψ(B) = θ(B), so
f(ν) = σ2wθ
(
e−2πiν)
θ(
e2πiν)
= σ2w
∣
∣θ(
e−2πiν)∣
∣
2.
For MA(1),
f(ν) = σ2w
∣
∣1 + θ1e−2πiν
∣
∣
2
= σ2w |1 + θ1 cos(−2πν) + iθ1 sin(−2πν)|
2
= σ2w
(
1 + 2θ1 cos(2πν) + θ21)
.
21
Autocovariance generating function and spectral density
For an AR(p), we haveψ(B) = 1/φ(B), so
f(ν) =σ2w
φ (e−2πiν)φ (e2πiν)
=σ2w
|φ (e−2πiν)|2 .
For AR(1),
f(ν) =σ2w
|1− φ1e−2πiν |2
=σ2w
1− 2φ1 cos(2πν) + φ21.
22
Spectral density of a linear process
If Xt is a linear process, it can be writtenXt =∑∞
i=0 ψiWt−i = ψ(B)Wt.
Then
f(ν) = σ2w
∣
∣ψ(
e−2πiν)∣
∣
2.
That is, the spectral densityf(ν) of a linear process measures the modulus
of theψ (MA(∞)) polynomial at the pointe2πiν on the unit circle.
23
Introduction to Time Series Analysis. Lecture 17.
1. Review: Spectral distribution function, spectral density.
2. Rational spectra. Poles and zeros.
3. Examples.
4. Time-invariant linear filters
5. Frequency response
1
Review: Spectral density and spectral distribution function
If a time series {Xt} has autocovarianceγ satisfying∑∞
h=−∞|γ(h)| <∞, then we define itsspectral densityas
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh
for −∞ < ν <∞. We have
γ(h) =
∫ 1/2
−1/2
e2πiνhf(ν) dν =
∫ 1/2
−1/2
e2πiνh dF (ν),
wheredF (ν) = f(ν)dν.
f measures how the variance ofXt is distributed across the spectrum.
2
Review: Spectral density of a linear process
If Xt is a linear process, it can be writtenXt =∑∞
i=0 ψiWt−i = ψ(B)Wt.
Then
f(ν) = σ2w
∣
∣ψ(
e−2πiν)∣
∣
2.
That is, the spectral densityf(ν) of a linear process measures the modulus
of theψ (MA(∞)) polynomial at the pointe2πiν on the unit circle.
3
Spectral density of a linear process
For an ARMA(p,q),ψ(B) = θ(B)/φ(B), so
f(ν) = σ2w
θ(e−2πiν)θ(e2πiν)
φ (e−2πiν)φ (e2πiν)
= σ2w
∣
∣
∣
∣
θ(e−2πiν)
φ (e−2πiν)
∣
∣
∣
∣
2
.
This is known as arational spectrum.
4
Rational spectra
Consider the factorization ofθ andφ as
θ(z) = θq(z − z1)(z − z2) · · · (z − zq)
φ(z) = φp(z − p1)(z − p2) · · · (z − pp),
wherez1, . . . , zq andp1, . . . , pp are called thezerosandpoles.
f(ν) = σ2w
∣
∣
∣
∣
∣
θq∏q
j=1(e−2πiν − zj)
φp∏p
j=1(e−2πiν − pj)
∣
∣
∣
∣
∣
2
= σ2w
θ2q∏q
j=1
∣
∣e−2πiν − zj∣
∣
2
φ2p∏p
j=1 |e−2πiν − pj |
2 .
5
Rational spectra
f(ν) = σ2w
θ2q∏q
j=1
∣
∣e−2πiν − zj∣
∣
2
φ2p∏p
j=1 |e−2πiν − pj |
2 .
As ν varies from0 to 1/2, e−2πiν moves clockwise around the unit circle
from 1 to e−πi = −1.
And the value off(ν) goes up as this point moves closer to (further from)
the polespj (zeroszj).
6
Example: ARMA
Recall AR(1):φ(z) = 1− φ1z. The pole is at1/φ1. If φ1 > 0, the pole is
to the right of1, so the spectral density decreases asν moves away from0.
If φ1 < 0, the pole is to the left of−1, so the spectral density is at its
maximum whenν = 0.5.
Recall MA(1): θ(z) = 1 + θ1z. The zero is at−1/θ1. If θ1 > 0, the zero is
to the left of−1, so the spectral density decreases asν moves towards−1.
If θ1 < 0, the zero is to the right of1, so the spectral density is at its
minimum whenν = 0.
7
Example: AR(2)
ConsiderXt = φ1Xt−1 + φ2Xt−2 +Wt. Example 4.6 in the text considers
this model withφ1 = 1, φ2 = −0.9, andσ2w = 1. In this case, the poles are
atp1, p2 ≈ 0.5555± i0.8958 ≈ 1.054e±i1.01567 ≈ 1.054e±2πi0.16165.
Thus, we have
f(ν) =σ2w
φ22|e−2πiν − p1|2|e−2πiν − p2|2
,
and this gets very peaked whene−2πiν passes near1.054e−2πi0.16165.
8
Example: AR(2)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
20
40
60
80
100
120
140
ν
f(ν)
Spectral density of AR(2): Xt = X
t−1 − 0.9 X
t−2 + W
t
9
Example: Seasonal ARMA
ConsiderXt = Φ1Xt−12 +Wt.
ψ(B) =1
1− Φ1B12,
f(ν) = σ2w
1
(1− Φ1e−2πi12ν)(1− Φ1e2πi12ν)
= σ2w
1
1− 2Φ1 cos(24πν) + Φ21
.
Notice thatf(ν) is periodic with period1/12.
10
Example: Seasonal ARMA
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
ν
f(ν)
Spectral density of AR(1)12
: Xt = +0.2 X
t−12 + W
t
11
Example: Seasonal ARMA
Another view:
1− Φ1z12 = 0 ⇔ z = reiθ,
with r = |Φ1|−1/12, ei12θ = e−i arg(Φ1).
ForΦ1 > 0, the twelve poles are at|Φ1|−1/12eikπ/6 for
k = 0,±1, . . . ,±5, 6.
So the spectral density gets peaked ase−2πiν passes near
|Φ1|−1/12 ×
{
1, e−iπ/6, e−iπ/3, e−iπ/2, e−i2π/3, e−i5π/6,−1}
.
12
Example: Multiplicative seasonal ARMA
Consider(1− Φ1B12)(1− φ1B)Xt =Wt.
f(ν) = σ2w
1
(1− 2Φ1 cos(24πν) + Φ21)(1− 2φ1 cos(2πν) + φ21)
.
This is a scaled product of the AR(1) spectrum and the (periodic) AR(1)12spectrum.
The AR(1)12 poles give peaks whene−2πiν is at one of the 12th roots of1;
the AR(1) poles give a peak neare−2πiν = 1.
13
Example: Multiplicative seasonal ARMA
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
1
2
3
4
5
6
7
ν
f(ν)
Spectral density of AR(1)AR(1)12
: (1+0.5 B)(1+0.2 B12) Xt = W
t
14
Introduction to Time Series Analysis. Lecture 17.
1. Review: Spectral distribution function, spectral density.
2. Rational spectra. Poles and zeros.
3. Examples.
4. Time-invariant linear filters
5. Frequency response
15
Time-invariant linear filters
A filter is an operator; given a time series{Xt}, it maps to a time series{Yt}. We can think of a linear processXt =
∑∞
j=0 ψjWt−j as the output ofacausal linear filterwith a white noise input.
A time series{Yt} is the output of a linear filter
A = {at,j : t, j ∈ Z} with input{Xt} if
Yt =∞∑
j=−∞
at,jXj .
If at,t−j is independent oft (at,t−j = ψj), then we say that the
filter is time-invariant.
If ψj = 0 for j < 0, we say the filterψ is causal.
We’ll see that the name ‘filter’ arises from the frequency domain viewpoint.
16
Time-invariant linear filters: Examples
1. Yt = X−t is linear, but not time-invariant.
2. Yt = 13 (Xt−1 +Xt +Xt+1) is linear, time-invariant, but not causal:
ψj =
13 if |j| ≤ 1,
0 otherwise.
3. For polynomialsφ(B), θ(B) with roots outside the unit circle,
ψ(B) = θ(B)/φ(B) is a linear, time-invariant, causal filter.
17
Time-invariant linear filters
The operation∞∑
j=−∞
ψjXt−j
is called theconvolutionof X with ψ.
18
Time-invariant linear filters
The sequenceψ is also called theimpulse response, since the output{Yt} of
the linear filter in response to aunit impulse,
Xt =
1 if t = 0,
0 otherwise,
is
Yt = ψ(B)Xt =∞∑
j=−∞
ψjXt−j = ψt.
19
Introduction to Time Series Analysis. Lecture 17.
1. Review: Spectral distribution function, spectral density.
2. Rational spectra. Poles and zeros.
3. Examples.
4. Time-invariant linear filters
5. Frequency response
20
Frequency response of a time-invariant linear filter
Suppose that{Xt} has spectral densityfx(ν) andψ is stable, that is,∑∞
j=−∞|ψj | <∞. ThenYt = ψ(B)Xt has spectral density
fy(ν) =∣
∣ψ(
e2πiν)∣
∣
2fx(ν).
The functionν 7→ ψ(e2πiν) (the polynomialψ(z) evaluated on the unit
circle) is known as thefrequency responseor transfer functionof the linear
filter.
The squared modulus,ν 7→ |ψ(e2πiν)|2 is known as thepower transfer
functionof the filter.
21
Frequency response of a time-invariant linear filter
For stableψ, Yt = ψ(B)Xt has spectral density
fy(ν) =∣
∣ψ(
e2πiν)∣
∣
2fx(ν).
We have seen that a linear process,Yt = ψ(B)Wt, is a special case, since
fy(ν) = |ψ(e2πiν)|2σ2w = |ψ(e2πiν)|2fw(ν).
When we pass a time series{Xt} through a linear filter, the spectral density
is multiplied, frequency-by-frequency, by the squared modulus of the
frequency responseν 7→ |ψ(e2πiν)|2.
This is a version of the equality Var(aX) = a2Var(X), but the equality is
true for the component of the variance at every frequency.
This is also the origin of the name ‘filter.’
22
Frequency response of a filter: Details
Why isfy(ν) =∣
∣ψ(
e2πiν)∣
∣
2fx(ν)? First,
γy(h) = E
∞∑
j=−∞
ψjXt−j
∞∑
k=−∞
ψkXt+h−k
=∞∑
j=−∞
ψj
∞∑
k=−∞
ψkE [Xt+h−kXt−j ]
=
∞∑
j=−∞
ψj
∞∑
k=−∞
ψkγx(h+ j − k) =
∞∑
j=−∞
ψj
∞∑
l=−∞
ψh+j−lγx(l).
It is easy to check that∑∞
j=−∞|ψj | <∞ and
∑∞
h=−∞|γx(h)| <∞ imply
that∑∞
h=−∞|γy(h)| <∞. Thus, the spectral density ofy is defined.
23
Frequency response of a filter: Details
fy(ν) =
∞∑
h=−∞
γ(h)e−2πiνh
=∞∑
h=−∞
∞∑
j=−∞
ψj
∞∑
l=−∞
ψh+j−lγx(l)e−2πiνh
=∞∑
j=−∞
ψje2πiνj
∞∑
l=−∞
γx(l)e−2πiνl
∞∑
h=−∞
ψh+j−le−2πiν(h+j−l)
= ψ(e2πiνj)fx(ν)
∞∑
h=−∞
ψhe−2πiνh
=∣
∣ψ(e2πiνj)∣
∣
2fx(ν).
24
Frequency response: Examples
For a linear processYt = ψ(B)Wt, fy(ν) =∣
∣ψ(
e2πiν)∣
∣
2σ2w.
For an ARMA model,ψ(B) = θ(B)/φ(B), so{Yt} has the rational
spectrum
fy(ν) = σ2w
∣
∣
∣
∣
θ(e−2πiν)
φ (e−2πiν)
∣
∣
∣
∣
2
= σ2w
θ2q∏q
j=1
∣
∣e−2πiν − zj∣
∣
2
φ2p∏p
j=1 |e−2πiν − pj |
2 ,
wherepj andzj are the poles and zeros of the rational function
z 7→ θ(z)/φ(z).
25
Frequency response: Examples
Consider the moving average
Yt =1
2k + 1
k∑
j=−k
Xt−j .
This is a time invariant linear filter (but it is not causal). Its transfer function
is the Dirichlet kernel
ψ(e−2πiν) = Dk(2πν) =1
2k + 1
k∑
j=−k
e−2πijν
=
1 if ν = 0,sin(2π(k+1/2)ν)(2k+1) sin(πν) otherwise.
26
Example: Moving average
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
ν
Transfer function of moving average (k=5)
27
Example: Moving average
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
ν
Squared modulus of transfer function of moving average (k=5)
This is alow-pass filter: It preserves low frequencies and diminishes high
frequencies. It is often used to estimate a monotonic trend component of a
series.
28
Example: Differencing
Consider the first difference
Yt = (1−B)Xt.
This is a time invariant, causal, linear filter.
Its transfer function is
ψ(e−2πiν) = 1− e−2πiν ,
so |ψ(e−2πiν)|2 = 2(1− cos(2πν)).
29
Example: Differencing
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
0.5
1
1.5
2
2.5
3
3.5
4
ν
Transfer function of first difference
This is ahigh-pass filter: It preserves high frequencies and diminishes low
frequencies. It is often used to eliminate a trend componentof a series.
30
Introduction to Time Series Analysis. Lecture 18.
1. Review: Spectral density, rational spectra, linear filters.
2. Frequency response of linear filters.
3. Spectral estimation
4. Sample autocovariance
5. Discrete Fourier transform and the periodogram
1
Review: Spectral density
If a time series {Xt} has autocovarianceγ satisfying∑∞
h=−∞ |γ(h)| <∞, then we define itsspectral densityas
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh
for −∞ < ν <∞. We have
γ(h) =
∫ 1/2
−1/2
e2πiνhf(ν) dν.
2
Review: Rational spectra
For a linear time series withMA(∞) polynomialψ,
f(ν) = σ2w
∣
∣ψ(
e2πiν)∣
∣
2.
If it is an ARMA(p,q), we have
f(ν) = σ2w
∣
∣
∣
∣
θ(e−2πiν)
φ (e−2πiν)
∣
∣
∣
∣
2
= σ2w
θ2q∏q
j=1
∣
∣e−2πiν − zj∣
∣
2
φ2p∏p
j=1 |e−2πiν − pj |2,
wherez1, . . . , zq are the zeros (roots ofθ(z))
andp1, . . . , pp are the poles (roots ofφ(z)).
3
Review: Time-invariant linear filters
A filter is an operator; given a time series{Xt}, it maps to a time series{Yt}. A linear filter satisfies
Yt =∞∑
j=−∞
at,jXj .
time-invariant: at,t−j = ψj :
Yt =
∞∑
j=−∞
ψjXt−j .
causal: j < 0 impliesψj = 0.
Yt =∞∑
j=0
ψjXt−j.
4
Time-invariant linear filters
The operation∞∑
j=−∞
ψjXt−j
is called theconvolutionof X with ψ.
5
Time-invariant linear filters
The sequenceψ is also called theimpulse response, since the output{Yt} of
the linear filter in response to aunit impulse,
Xt =
1 if t = 0,
0 otherwise,
is
Yt = ψ(B)Xt =∞∑
j=−∞
ψjXt−j = ψt.
6
Introduction to Time Series Analysis. Lecture 18.
1. Review: Spectral density, rational spectra, linear filters.
2. Frequency response of linear filters.
3. Spectral estimation
4. Sample autocovariance
5. Discrete Fourier transform and the periodogram
7
Frequency response of a time-invariant linear filter
Suppose that{Xt} has spectral densityfx(ν) andψ is stable, that is,∑∞
j=−∞ |ψj | <∞. ThenYt = ψ(B)Xt has spectral density
fy(ν) =∣
∣ψ(
e2πiν)∣
∣
2fx(ν).
The functionν 7→ ψ(e2πiν) (the polynomialψ(z) evaluated on the unit
circle) is known as thefrequency responseor transfer functionof the linear
filter.
The squared modulus,ν 7→ |ψ(e2πiν)|2 is known as thepower transfer
functionof the filter.
8
Frequency response of a time-invariant linear filter
For stableψ, Yt = ψ(B)Xt has spectral density
fy(ν) =∣
∣ψ(
e2πiν)∣
∣
2fx(ν).
We have seen that a linear process,Yt = ψ(B)Wt, is a special case, since
fy(ν) = |ψ(e2πiν)|2σ2w = |ψ(e2πiν)|2fw(ν).
When we pass a time series{Xt} through a linear filter, the spectral density
is multiplied, frequency-by-frequency, by the squared modulus of the
frequency responseν 7→ |ψ(e2πiν)|2.
This is a version of the equality Var(aX) = a2Var(X), but the equality is
true for the component of the variance at every frequency.
This is also the origin of the name ‘filter.’
9
Frequency response of a filter: Details
Why isfy(ν) =∣
∣ψ(
e2πiν)∣
∣
2fx(ν)? First,
γy(h) = E
∞∑
j=−∞
ψjXt−j
∞∑
k=−∞
ψkXt+h−k
=∞∑
j=−∞
ψj
∞∑
k=−∞
ψkE [Xt+h−kXt−j ]
=
∞∑
j=−∞
ψj
∞∑
k=−∞
ψkγx(h+ j − k) =
∞∑
j=−∞
ψj
∞∑
l=−∞
ψh+j−lγx(l).
It is easy to check that∑∞
j=−∞ |ψj | <∞ and∑∞
h=−∞ |γx(h)| <∞ imply
that∑∞
h=−∞ |γy(h)| <∞. Thus, the spectral density ofy is defined.
10
Frequency response of a filter: Details
fy(ν) =
∞∑
h=−∞
γ(h)e−2πiνh
=∞∑
h=−∞
∞∑
j=−∞
ψj
∞∑
l=−∞
ψh+j−lγx(l)e−2πiνh
=∞∑
j=−∞
ψje2πiνj
∞∑
l=−∞
γx(l)e−2πiνl
∞∑
h=−∞
ψh+j−le−2πiν(h+j−l)
= ψ(e2πiνj)fx(ν)
∞∑
h=−∞
ψhe−2πiνh
=∣
∣ψ(e2πiνj)∣
∣
2fx(ν).
11
Frequency response: Examples
For a linear processYt = ψ(B)Wt, fy(ν) =∣
∣ψ(
e2πiν)∣
∣
2σ2w.
For an ARMA model,ψ(B) = θ(B)/φ(B), so{Yt} has the rational
spectrum
fy(ν) = σ2w
∣
∣
∣
∣
θ(e−2πiν)
φ (e−2πiν)
∣
∣
∣
∣
2
= σ2w
θ2q∏q
j=1
∣
∣e−2πiν − zj∣
∣
2
φ2p∏p
j=1 |e−2πiν − pj |2,
wherepj andzj are the poles and zeros of the rational function
z 7→ θ(z)/φ(z).
12
Frequency response: Examples
Consider the moving average
Yt =1
2k + 1
k∑
j=−k
Xt−j .
This is a time invariant linear filter (but it is not causal). Its transfer function
is the Dirichlet kernel
ψ(e−2πiν) = Dk(2πν) =1
2k + 1
k∑
j=−k
e−2πijν
=
1 if ν = 0,sin(2π(k+1/2)ν)(2k+1) sin(πν) otherwise.
13
Example: Moving average
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
ν
Transfer function of moving average (k=5)
14
Example: Moving average
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
ν
Squared modulus of transfer function of moving average (k=5)
This is alow-pass filter: It preserves low frequencies and diminishes high
frequencies. It is often used to estimate a monotonic trend component of a
series.
15
Example: Differencing
Consider the first difference
Yt = (1−B)Xt.
This is a time invariant, causal, linear filter.
Its transfer function is
ψ(e−2πiν) = 1− e−2πiν ,
so |ψ(e−2πiν)|2 = 2(1− cos(2πν)).
16
Example: Differencing
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
0.5
1
1.5
2
2.5
3
3.5
4
ν
Transfer function of first difference
This is ahigh-pass filter: It preserves high frequencies and diminishes low
frequencies. It is often used to eliminate a trend componentof a series.
17
Introduction to Time Series Analysis. Lecture 18.
1. Review: Spectral density, rational spectra, linear filters.
2. Frequency response of linear filters.
3. Spectral estimation
4. Sample autocovariance
5. Discrete Fourier transform and the periodogram
18
Estimating the Spectrum: Outline
• We have seen that the spectral density gives an alternative view of
stationary time series.
• Given a realizationx1, . . . , xn of a time series, how can we estimate
the spectral density?
• One approach: replaceγ(·) in the definition
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh,
with the sample autocovarianceγ̂(·).
• Another approach, called theperiodogram: computeI(ν), the squared
modulus of the discrete Fourier transform (at frequenciesν = k/n).
19
Estimating the spectrum: Outline
• These two approaches areidenticalat the Fourier frequenciesν = k/n.
• The asymptotic expectation of the periodogramI(ν) is f(ν). We can
derive some asymptotic properties, and hence do hypothesistesting.
• Unfortunately, the asymptotic variance ofI(ν) is constant.
It is not a consistent estimator off(ν).
• We can reduce the variance by smoothing the periodogram—averaging
over adjacent frequencies. If we average over a narrower range as
n→ ∞, we can obtain a consistent estimator of the spectral density.
20
Estimating the spectrum: Sample autocovariance
Idea: use the sample autocovarianceγ̂(·), defined by
γ̂(h) =1
n
n−|h|∑
t=1
(xt+|h| − x̄)(xt − x̄), for −n < h < n,
as an estimate of the autocovarianceγ(·), and then use a sample version of
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh,
That is, for−1/2 ≤ ν ≤ 1/2, estimatef(ν) with
f̂(ν) =
n−1∑
h=−n+1
γ̂(h)e−2πiνh.
21
Estimating the spectrum: Periodogram
Another approach to estimating the spectrum is called the periodogram. It
was proposed in 1897 by Arthur Schuster (at Owens College, which later
became part of the University of Manchester), who used it to investigate
periodicity in the occurrence of earthquakes, and in sunspot activity.
Arthur Schuster, “On Lunar and Solar Periodicities of Earthquakes,”Proceedings of
the Royal Society of London, Vol. 61 (1897), pp. 455–465.
To define the periodogram, we need to introduce thediscrete Fourier
transformof a finite sequencex1, . . . , xn.
22
Introduction to Time Series Analysis. Lecture 18.
1. Review: Spectral density, rational spectra, linear filters.
2. Frequency response of linear filters.
3. Spectral estimation
4. Sample autocovariance
5. Discrete Fourier transform and the periodogram
23
Discrete Fourier transform
For a sequence(x1, . . . , xn), define thediscrete Fourier transform (DFT)as
(X(ν0), X(ν1), . . . , X(νn−1)), where
X(νk) =1√n
n∑
t=1
xte−2πiνkt,
andνk = k/n (for k = 0, 1, . . . , n− 1) are called theFourier frequencies.
(Think of {νk : k = 0, . . . , n− 1} as the discrete version of the frequency
rangeν ∈ [0, 1].)
First, let’s show that we can view the DFT as a representationof x in a
different basis, theFourier basis.
24
Discrete Fourier transform
Consider the spaceCn of vectors ofn complex numbers, with inner product〈a, b〉 = a∗b, wherea∗ is the complex conjugate transpose of the vectora ∈ Cn.
Suppose that a set{φj : j = 0, 1, . . . , n− 1} of n vectors inCn areorthonormal:
〈φj , φk〉 =
1 if j = k,
0 otherwise.
Then these{φj} span the vector spaceCn, and so for any vectorx, we canwrite x in terms of this new orthonormal basis,
x =
n−1∑
j=0
〈φj , x〉φj . (picture)
25
Discrete Fourier transform
Consider the following set ofn vectors inCn:{
ej =1√n
(
e2πiνj , e2πi2νj , . . . , e2πinνj)′
: j = 0, . . . , n− 1
}
.
It is easy to check that these vectors are orthonormal:
〈ej , ek〉 =1
n
n∑
t=1
e2πit(νk−νj) =1
n
n∑
t=1
(
e2πi(k−j)/n)t
=
1 if j = k,1ne
2πi(k−j)/n 1−(e2πi(k−j)/n)n
1−e2πi(k−j)/n otherwise
=
1 if j = k,
0 otherwise,
26
Discrete Fourier transform
where we have used the fact thatSn =∑n
t=1 αt satisfies
αSn = Sn + αn+1 − α and soSn = α(1− αn)/(1− α) for α 6= 1.
So we can represent the real vectorx = (x1, . . . , xn)′ ∈ Cn in terms of this
orthonormal basis,
x =n−1∑
j=0
〈ej , x〉ej =n−1∑
j=0
X(νj)ej .
That is, the vector of discrete Fourier transform coefficients
(X(ν0), . . . , X(νn−1)) is the representation ofx in the Fourier basis.
27
Discrete Fourier transform
An alternative way to represent the DFT is by separately considering the
real and imaginary parts,
X(νj) = 〈ej , x〉 =1√n
n∑
t=1
e−2πitνjxt
=1√n
n∑
t=1
cos(2πtνj)xt − i1√n
n∑
t=1
sin(2πtνj)xt
= Xc(νj)− iXs(νj),
where this defines the sine and cosine transforms,Xs andXc, of x.
28
Introduction to Time Series Analysis. Lecture 19.
1. Review: Spectral density estimation, sample autocovariance.
2. The periodogram and sample autocovariance.
3. Asymptotics of the periodogram.
1
Estimating the Spectrum: Outline
• We have seen that the spectral density gives an alternative view of
stationary time series.
• Given a realizationx1, . . . , xn of a time series, how can we estimate
the spectral density?
• One approach: replaceγ(·) in the definition
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh,
with the sample autocovarianceγ̂(·).
• Another approach, called theperiodogram: computeI(ν), the squared
modulus of the discrete Fourier transform (at frequenciesν = k/n).
2
Estimating the spectrum: Outline
• These two approaches areidentical at the Fourier frequenciesν = k/n.
• The asymptotic expectation of the periodogramI(ν) is f(ν). We can
derive some asymptotic properties, and hence do hypothesistesting.
• Unfortunately, the asymptotic variance ofI(ν) is constant.
It is not a consistent estimator off(ν).
3
Review: Spectral density estimation
If a time series {Xt} has autocovarianceγ satisfying∑∞
h=−∞ |γ(h)| < ∞, then we define itsspectral density as
f(ν) =∞∑
h=−∞
γ(h)e−2πiνh
for −∞ < ν < ∞.
4
Review: Sample autocovariance
Idea: use the sample autocovarianceγ̂(·), defined by
γ̂(h) =1
n
n−|h|∑
t=1
(xt+|h| − x̄)(xt − x̄), for −n < h < n,
as an estimate of the autocovarianceγ(·), and then use
f̂(ν) =
n−1∑
h=−n+1
γ̂(h)e−2πiνh
for −1/2 ≤ ν ≤ 1/2.
5
Discrete Fourier transform
For a sequence(x1, . . . , xn), define thediscrete Fourier transform (DFT) as
(X(ν0), X(ν1), . . . , X(νn−1)), where
X(νk) =1√n
n∑
t=1
xte−2πiνkt,
andνk = k/n (for k = 0, 1, . . . , n− 1) are called theFourier frequencies.
(Think of {νk : k = 0, . . . , n− 1} as the discrete version of the frequency
rangeν ∈ [0, 1].)
First, let’s show that we can view the DFT as a representationof x in a
different basis, theFourier basis.
6
Discrete Fourier transform
Consider the spaceCn of vectors ofn complex numbers, with inner product〈a, b〉 = a∗b, wherea∗ is the complex conjugate transpose of the vectora ∈ Cn.
Suppose that a set{φj : j = 0, 1, . . . , n− 1} of n vectors inCn areorthonormal:
〈φj , φk〉 =
1 if j = k,
0 otherwise.
Then these{φj} span the vector spaceCn, and so for any vectorx, we canwrite x in terms of this new orthonormal basis,
x =
n−1∑
j=0
〈φj , x〉φj . (picture)
7
Discrete Fourier transform
Consider the following set ofn vectors inCn:{
ej =1√n
(
e2πiνj , e2πi2νj , . . . , e2πinνj)′
: j = 0, . . . , n− 1
}
.
It is easy to check that these vectors are orthonormal:
〈ej , ek〉 =1
n
n∑
t=1
e2πit(νk−νj) =1
n
n∑
t=1
(
e2πi(k−j)/n)t
=
1 if j = k,1ne
2πi(k−j)/n 1−(e2πi(k−j)/n)n
1−e2πi(k−j)/n otherwise
=
1 if j = k,
0 otherwise,
8
Discrete Fourier transform
where we have used the fact thatSn =∑n
t=1 αt satisfies
αSn = Sn + αn+1 − α and soSn = α(1− αn)/(1− α) for α 6= 1.
So we can represent the real vectorx = (x1, . . . , xn)′ ∈ Cn in terms of this
orthonormal basis,
x =n−1∑
j=0
〈ej , x〉ej =n−1∑
j=0
X(νj)ej .
That is, the vector of discrete Fourier transform coefficients
(X(ν0), . . . , X(νn−1)) is the representation ofx in the Fourier basis.
9
Discrete Fourier transform
An alternative way to represent the DFT is by separately considering the
real and imaginary parts,
X(νj) = 〈ej , x〉 =1√n
n∑
t=1
e−2πitνjxt
=1√n
n∑
t=1
cos(2πtνj)xt − i1√n
n∑
t=1
sin(2πtνj)xt
= Xc(νj)− iXs(νj),
where this defines the sine and cosine transforms,Xs andXc, of x.
10
Periodogram
The periodogram is defined as
I(νj) = |X(νj)|2
=1
n
∣
∣
∣
∣
∣
n∑
t=1
e−2πitνjxt
∣
∣
∣
∣
∣
2
= X2c (νj) +X2
s (νj).
Xc(νj) =1√n
n∑
t=1
cos(2πtνj)xt,
Xs(νj) =1√n
n∑
t=1
sin(2πtνj)xt.
11
Periodogram
SinceI(νj) = |X(νj)|2 for one of the Fourier frequenciesνj = j/n (for
j = 0, 1, . . . , n− 1), the orthonormality of theej implies that we can write
x∗x =
n−1∑
j=0
X(νj)ej
∗
n−1∑
j=0
X(νj)ej
=n−1∑
j=0
|X(νj)|2 =n−1∑
j=0
I(νj).
For x̄ = 0, we can write this as
σ̂2x =
1
n
n∑
t=1
x2t =
1
n
n−1∑
j=0
I(νj).
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Periodogram
This is the discrete analog of the identity
σ2x = γx(0) =
∫ 1/2
−1/2
fx(ν) dν.
(Think of I(νj) as the discrete version off(ν) at the frequencyνj = j/n,
and think of(1/n)∑
νj· as the discrete version of
∫
ν·dν.)
13
Estimating the spectrum: Periodogram
Why is the periodogram at a Fourier frequency (that is,ν = νj) the same as
computingf(ν) from the sample autocovariance?
Almost the same—they are not the same atν0 = 0 whenx̄ 6= 0.
But if eitherx̄ = 0, or we consider a Fourier frequencyνj with
j ∈ {1, . . . , n− 1}, . . .
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Estimating the spectrum: Periodogram
I(νj) =1
n
∣
∣
∣
∣
∣
n∑
t=1
e−2πitνjxt
∣
∣
∣
∣
∣
2
=1
n
∣
∣
∣
∣
∣
n∑
t=1
e−2πitνj (xt − x̄)
∣
∣
∣
∣
∣
2
=1
n
(
n∑
t=1
e−2πitνj (xt − x̄)
)(
n∑
t=1
e2πitνj (xt − x̄)
)
=1
n
∑
s,t
e−2πi(s−t)νj (xs − x̄)(xt − x̄) =n−1∑
h=−n+1
γ̂(h)e−2πihνj ,
where the fact thatνj 6= 0 implies∑n
t=1 e−2πitνj = 0 (we showed this
when we were verifying the orthonormality of the Fourier basis) has
allowed us to subtract the sample mean in that case.
15
Asymptotic properties of the periodogram
We want to understand the asymptotic behavior of the periodogramI(ν) at
a particular frequencyν, asn increases. We’ll see that its expectation
converges tof(ν).
We’ll start with a simple example: Suppose thatX1, . . . , Xn are
i.i.d. N(0, σ2) (Gaussian white noise). From the definitions,
Xc(νj) =1√n
n∑
t=1
cos(2πtνj)xt, Xs(νj) =1√n
n∑
t=1
sin(2πtνj)xt,
we have thatXc(νj) andXs(νj) are normal, with
EXc(νj) = EXs(νj) = 0.
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Asymptotic properties of the periodogram
Also,
Var(Xc(νj)) =σ2
n
n∑
t=1
cos2(2πtνj)
=σ2
2n
n∑
t=1
(cos(4πtνj) + 1) =σ2
2.
Similarly, Var(Xs(νj)) = σ2/2.
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Asymptotic properties of the periodogram
Also,
Cov(Xc(νj), Xs(νj)) =σ2
n
n∑
t=1
cos(2πtνj) sin(2πtνj)
=σ2
2n
n∑
t=1
sin(4πtνj) = 0,
Cov(Xc(νj), Xc(νk)) = 0
Cov(Xs(νj), Xs(νk)) = 0
Cov(Xc(νj), Xs(νk)) = 0.
for anyj 6= k.
18
Asymptotic properties of the periodogram
That is, ifX1, . . . , Xn are i.i.d.N(0, σ2)
(Gaussian white noise;f(ν) = σ2), then theXc(νj) andXs(νj) are all
i.i.d. N(0, σ2/2). Thus,
2
σ2I(νj) =
2
σ2
(
X2c (νj) +X2
s (νj))
∼ χ22.
So for the case of Gaussian white noise, the periodogram has achi-squared
distribution that depends on the varianceσ2 (which, in this case, is the
spectral density).
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Asymptotic properties of the periodogram
Under more general conditions (e.g., normal{Xt}, or linear process{Xt}with rapidly decaying ACF), theXc(νj), Xs(νj) are all asymptotically
independent andN(0, f(νj)/2).
Consider a frequencyν. For a given value ofn, let ν̂(n) be the closest
Fourier frequency (that is,̂ν(n) = j/n for a value ofj that minimizes
|ν − j/n|). Asn increases,̂ν(n) → ν, and (under the same conditions that
ensure the asymptotic normality and independence of the sine/cosine
transforms),f(ν̂(n)) → f(ν). (picture)
In that case, we have
2
f(ν)I(ν̂(n)) =
2
f(ν)
(
X2c (ν̂
(n)) +X2s (ν̂
(n)))
d→ χ22.
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Asymptotic properties of the periodogram
Thus,
EI(ν̂(n)) =f(ν)
2E
(
2
f(ν)
(
X2c (ν̂
(n)) +X2s (ν̂
(n)))
)
→ f(ν)
2E(Z2
1 + Z22 ) = f(ν),
whereZ1, Z2 are independentN(0, 1). Thus, the periodogram is
asymptotically unbiased.
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