Complications

Question 1

Distribution of DFT for a White Noise Process

Answer 1

-if {εt} is a white noise process then:
Re[εj^] ~ N(0,σε²/2)
-and
Im[εj^] ~ N(0,σε²/2)

Question 2

I(fj) for a White Noise Process

Answer 2

I(fj) = |εj^|² = Re(εj^)² + Im(εj^)²

Question 3

Distribution of I(fj) for a White Noise Process

Answer 3

-since I(fj) is the sum of the squares of two random variables:
2/σε² I(fj) ~ χ2²
-for 1≤j∞

Question 4

Var[I(fj)] for an AR(1) Process

Answer 4

-if {Xt} is an AR(1) process then:
Var[I(fj)] = κ σε^4
-for some constant κ

Question 5

Var[I(f)] in general

Answer 5

-in general, Var[I(f)] is constant as n increases

Question 6

Inconsistency

Definition

Answer 6

• in general, Var[I(f)] is constant as n increases
• I(f) is an inconsistent estimate for the power spectrum 𝓘(f)
• as n increases we still attempt to estimate I(f) at n frequencies
• therefore, although we have more data, the amount of data per frequency remains the same

Question 7

Preferred Alternative to

Inconsistency

Answer 7

• we would prefer an estimate which gets more reliable as we gather more data
• there are a number of ways to do this e.g. truncating the autocovariance function and smoothing the raw periodogram

Question 8

Spectral Density in Terms of the Lag-k Sample Autocovariance

Answer 8

I(fj) = g0 + Σ gk cos(2πfj k)
(+ gn/2 (-1)^j if n is even)
-sum from 1≤k

Question 9

Truncating the Autocovariance Function

Definition

Answer 9

-if we do not assume {Xt} to be periodic then the number of pairs{Xt, Xt+k} decreases as k increases so gk is less reliable for larger lags
-hence we might use:
𝓘t(fj) = λ0g0 + 2 Σ λk gk cos(2πfj k)
-where subscript t denotes that this is the truncation method
-sum from k=1 to m with m

Question 10

Truncating the Autocovariance Function

Tukey Window

Answer 10

λk = 1/2 [1 + cos(πk/m)]

k=0,1,…,m

Question 11

Truncating the Autocovariance Function

Parzen Window

Answer 11

λk = 1 - 6(k/m)² + 6(k/m)³ for k=0,1,…,m/2

λk = 2 (1 - k/m)³ for k = m/2 +1, …, m

Question 12

Truncating the Autocovariance Function

Choosing m

Answer 12

• increasing m leads to the raw periodogram which is too noisy
• choosing m too small over-smooths, low variance but high bias
• a rule of thumb is to take m=√n/2

Question 13

Smoothing the Raw Periodogram

Outline

Answer 13

• this technique is called Daniel smoothing, it is the default method in R
• it is based on the assumption that the power spectrum, 𝓘(f), will be ‘similar’ for values of f ‘close’ to each other

Question 14

Smoothing the Raw Periodogram

Definition

Answer 14

-let w = 2w* + 1 be an odd, positive integer and define:
𝓘s(f) = 1/w Σ I(fj+k)
-where subscript s denotes that this is the smoothing method
-sum from k=-w* to k=w*

Question 15

Smoothing the Raw Periodogram

Var[𝓘s(f)]

Answer 15

• recall that I(f) is asymptotically unbias but has constant variance
• also, I(fj) is asymptotically independent of I(fk) if j≠k
• therefore Var[𝓘s(f)] is of order 1/w

Question 16

Smoothing the Raw Periodogram

E[𝓘s(f)]

Answer 16

E[𝓘s(f)] = 1/w Σ E[I(fj+k)]
≈ 1/w Σ 𝓘s(f)
-sum from k=-w* to k=w*

Question 17

Smoothing the Raw Periodogram

Bias

Answer 17

-the smoothing estimate 𝓘s(f) is bias since its expectation only equals 𝓘s(f) if the power spectrum is linear on the interval:
[fj-w,fj+w]
-therefore we need to balance large w (small variance) with small w (small bias)

Question 18

Smoothing the Raw Periodogram

Choosing w

Answer 18

• rule of thumb w≈√n

- or alternatively, use repeated application of the Daniel smoother with small values of w

Question 19

Equivalence of Smoothing the Periodogram and Truncating the Autocovariance Function

Answer 19

-the Daniel smoothing method is equivalent to the lag window:
λk = 1 if k=0
λk = sin(wkπ/n)/msin(kπ/n) if k=1,2,…,n-1

Question 20

Spectral Window

Definition

Answer 20

-the spectral window, K(f), is the DFT of the lag window
K(fj) = 1/√n Σ λk e^(-2πifj k)
-sum from k=0 to k=n-1
-it can be used to compare truncating the autocovariance function and smoothing the periodogram

Question 21

Spectral Window

Calculating λk

Answer 21

λk = 1/√n Σ K(fj) e^(2πifj k)

-sum from j=0 to j=n-1

Question 22

Spectral Window

Truncated Estimate of the Power Spectrum

Answer 22

-write out the definition of 𝓘t(f)
-sub in λk in terms of the spectral window
=>
𝓘t(f) = 1/n Σ K(fj’) I(fj-fj’)
-sum from j’=0 to j’=n-1

Question 23

Spectral Window

Equivalence of Procedures

Answer 23

• from the truncated estimate of the power spectrum in terms of K:
• both our estimation procedures are equivalent to smoothing the periodogram using a kernel function K
• there can be a sharp cut-off in the frequency domain (Daniel smoothing) or the time domain (truncation) but not both

Question 24

Spectral Leakage

Outline

Answer 24

• so far we have assumed that all periodic components are sine waves existing at one of the Fourier frequencies, fj=j/n
• but this need not be the case

Question 25

Spectral Leakage | Xj^ for Xt=sin(2πft) for f=fk

Answer 25

``` Xj^ = √n/2i if j=k Xj^ = -√n/2i if j=-k Xj^ = 0, else ```

Question 26

Spectral Leakage | Xj^ for Xt=e^(2πift)

Answer 26

``` Xj^ = √n if f=fj Xj^ = 0 if f=fk, k≠j Xj^ = 1/√n 1-e^[2π(f-fj)n]/1-e^[2πi(f-fj)], else ```

Question 27

Harmonics | Outline

Answer 27

-what if there is a periodic component that is not a since wave

Question 28

Harmonics | Definition

Answer 28

-let δjt be the contribution to Xj^ from time t: Xj^ = Σ δj,t -sum from t=0 to t=n-1 -from the definition of the DFT: δj,t = 1/√n Xt e^(-2πifj t) -the contribution to X2j^ from time t is: δ2j,t = 1/√n Xt e^(-2πi2fj t) = δj,t e^(-2πifj t) ... δmj,t = δj,t [e^(-2πifj t)]^(m-1) -where Xj^ is the fundamental and X2j^, X3j^, ...are harmonics

Question 29

Sine vs. Non-Sine Periodic Components

Answer 29

- if Xt contains a component, say sin(2πifk t) for some Fourier frequency fk, then it contributes only to Xk^ - if there is a non-trigonometric periodic component at frequency fj, its effects will also be felt at f2j, f3j, ... - the harmonics of the fundamental frequency fj