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Stat 153 - 13 Oct 2008 D. R. Brillinger Chapter 7 - Spectral analysis

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##### Stat 153 - 13 Oct 2008 D. R. Brillinger Chapter 7 - Spectral analysis

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**Stat 153 - 13 Oct 2008 D. R. Brillinger**Chapter 7 - Spectral analysis 7.1 Fourier analysis Xt = μ + α cos ωt + βsin ωt + Zt Cases ω known versus ω unknown, "hidden frequency" Study/fit via least squares**Fourier frequencies**ωp = 2πp/N, p = 0,...,N-1 Fourier components Σt=1N xt cos 2πpt/N Σt=1N xt sin 2πpt/N ap +i bp = Σt=1N xt exp{i2πpt/N}, p=0,...,N-1 = Σt=1N xt exp{iωpt/N}**7.3 Periodogram at frequency ωp**I(ωp ) = |Σt=1N xt exp{iωpt/N}|2 /πN = N(ap2 + bp2)/4π basis for spectral density estimates Can extend definition to I(ω) Properties I(ω) 0 I(-ω) = I(ω) I(ω+2π) = I(ω) Cp. f(ω)**Correcting for the mean**work with**Relationship - periodogram and acv**f(ω) = [γ0 + 2 Σk=1 γk cos ωk]/π I(ωp) = [c0 + 2 Σk=1N-1 ck cos ωpk]/π River height Manaus, Amazonia, Brazil daily data since 1902**Periodogram properties.**Asymptotically unbiased Approximate distribution f(ω) χ22 /2 CLT for a, b χ22 /2 : exponential variate Var {f(ω) χ22/2} = f(ω)2 Work with log I(ω)**Periodogram poor estimate of spectral density**inconsistent I(ωj), I(ωk) approximately independent Flexible estimate, χ22 + χ22 + ... + χ22 = χν2 , υ = 2 m E{ χν2} = v Var{χν2} = 2υ**Confidence interval. 100(1-α)%**work with logs Might pick m to get desired stability, e.g. 2m = 10 degrees of freedom Employ several values, compute bandwidth m = 0: periodogram**Model.**Yt = St + Nt seasonal St+s = St noise {Nt} Buys-Ballot stack years column average**General class of estimates**Example. boxcar Its bandwidth is m2π/N**Bias-variance compromise**m large, variance small, but m large, bias large (generally) Improvements prewhiten - work with residuals, e.g. of an autoregressive taper - work with {gt xt}**The partial correlation coefficient.**4-variate: (Y1 ,Y2 , Y3 , Y4 ) ε1|23 error of linear predicting Y1 using Y2 and Y3 ε4|23 error of linear predicting Y4 using Y2 and Y3 ρ14|23 = corr{ε1|23 , ε4|23} pacf(2) = 0 for AR(2) because Xt+1 , Xt+2 separate Xt+1 and Xt+4**Binary data**X(t) = 0,1 nerve cells firing biological phenomenon refractory period**Continuous time series. X(t), -< t<**Now Cov{X(t+τ),X(t)} = γ(τ), -< t, τ< and Consider the discrete time series {X(kΔt), k =0,±1,±2,...} It is symmetric and has period 2π/Δt Nyquist frequency: ωN = π/Δt One plots fd(ω) for 0 ω π/Δt, BUT ...