The Periodogram Recall: the discrete Fourier transform n = n 1 / 2 - - PowerPoint PPT Presentation

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Nov 19, 2023 43 likes •136 views

The Periodogram Recall: the discrete Fourier transform n = n 1 / 2 x t e 2 i j t , d j j = 0 , 1 , . . . , n 1 , t =1 and the periodogram 2 , j = 0 , 1 , . . . , n 1 , I

SLIDE 1

The Periodogram

Recall: the discrete Fourier transform

d

ωj
= n−1/2

n

xte−2πiωjt, j = 0, 1, . . . , n − 1,

and the periodogram

I

ωj
=
d
ωj
2 , j = 0, 1, . . . , n − 1,
where ωj is one of the Fourier frequencies

ωj = j n.

1

SLIDE 2

Sine and Cosine Transforms

For j = 0, 1, . . . , n − 1,

d

ωj
= n−1/2

n

xte−2πiωjt = n−1/2

n

xt cos

2πωjt
− i × n−1/2

n

xt sin

2πωjt
= dc
ωj
− i × ds
ωj
.
dc
ωj
and ds
ωj
are the cosine transform and sine trans-

form, respectively, of x1, x2, . . . , xn.

The periodogram is I
ωj
= dc
ωj

2 + ds

2.

2

SLIDE 3

Sampling Distributions

For convenience, suppose that n is odd: n = 2m + 1.
White noise:
rthogonality properties of sines and cosines

mean that dc(ω1), ds(ω1), dc(ω2), ds(ω2), . . . , dc(ωm), ds(ωm) have zero mean, variance 1

2σ2 w, and are uncorrelated.

Gaussian white noise:

dc(ω1), ds(ω1), dc(ω2), ds(ω2), . . . , dc(ωm), ds(ωm) are i.i.d. N

0, 1

2σ2 w

.
So for Gaussian white noise, I
ωj
∼ 1

2σ2 w × χ2 2.

3

SLIDE 4

General case:

dc(ω1), ds(ω1), dc(ω2), ds(ω2), . . . , dc(ωm), ds(ωm) have zero mean and are approximately uncorrelated, and var

dc
ωj
≈ var
ds
ωj
≈ 1

2fx

where fx

ωj
is the spectral density function.
If xt is Gaussian,

Ix

2fx

=

dc

2 + ds

2 1 2fx

ωj
∼ approximately χ2

2,

and Ix(ω1), Ix(ω2), . . . , Ix(ωm) are approximately indepen- dent.

4

SLIDE 5

Spectral ANOVA

For odd n = 2m + 1, the inverse transform can be written

xt − ¯ x = 2 √n

m

j=1
dc
ωj
cos
2πωjt
+ ds
ωj
sin
2πωjt
.
Square and sum over t; orthogonality of sines and cosines

implies that

n

(xt − ¯ x)2 = 2

m

2 + ds

2 = 2

m

I

5

SLIDE 6

ANOVA table: Source df SS MS ω1 2 2I(ω1) I(ω1) ω2 2 2I(ω2) I(ω2) . . . . . . . . . . . . ωm 2 2I(ωm) I(ωm) Total 2m = n − 1

(xt − ¯

x)2

6

SLIDE 7

Hypothesis Testing

Consider the model

xt = A cos

2πωjt + φ
+ wt.
Hypotheses:

– H0 : A = 0 ⇒ xt = wt, white noise; – H1 : A > 0, white noise plus a sine wave.

Note: no autocorrelation in either case.

7

SLIDE 8

Two cases:

– ωj known: use Fj = I

ωj
(m − 1)−1

j′=j I

ωj′
which is F2,2(m−1) under H0.

– ωj unknown: use max(F1, F2, . . . , Fm), or equivalently κ = max

I
ωj
, j = 1, 2, . . . , n
m−1

j I

P(κ > ξ) ≈ 1 − exp

−m exp
−ξ(m − 1 − log m)

m − ξ

8

SLIDE 9

Example: the Southern Oscillation Index

Using SAS: proc spectra program and output.
Using R:

par(mfcol = c(2, 1)) # Use fft() to calculate the periodogram directly; note that # frequencies are expressed in cycles per year, and the # periodogram values are similarly scaled by 12: freq = 12 * (0:(length(soi) - 1)) / length(soi) plotit = (freq > 0) & (freq <= 6) soifft = fft(soi) / sqrt(length(soi)) plot(freq[plotit], Mod(soifft[plotit])^2 / 12, type = "l") # Use spectrum(); override some defaults to make it match: spectrum(soi, log = "no", fast = FALSE, taper = 0, detrend = FALSE)