Time Series Exam, 2009: Solutions - STAT

So,l(σ 2 ,φ) = − 1 2 log(2πσ2 /(1−φ 2 ))− 1 2 y2 1(1−φ 2 )/σ 2which gives the desired result.−(n−1) 1 2 log(2πσ2 )− 1 2n∑(y i −φy i−1 ) 2 /σ 2 ,(c) (i)To get confidence intervals for φ and σ 2 , we need to compute ˆφ andˆσ 2 , the maximum likelihood estimates, using the log likelihood function.Then, we can compute (1−2α) confidence intervals for these parametersas ˆφ±z α v 1/211 and ˆσ2 ±z α v 1/222 , where v ii is the ith diagonal element of thematrix J((ˆφ,ˆσ 2 )) −1 .(ii) The likelihood ratio statistic for comparing the model with φ ∈ R withthe model with φ = 0 is W = 2(ˆl φ − ˆl φ=0 ). If the hypothesis φ = 0 istrue, then W ∼ χ 2 1 .3. If {γ h } h∈Z is the ACF of a stationary random sequence, then there existsa unique function F defined on [−1/2,1/2] such that F(−1/2) = 0, Fis right-continuous and non-decreasing, with symmetric increments aboutzero, and ∫γ h = e 2πihu dF(u), h ∈ Z.(−1/2,1/2]The function F is called the spectral distribution function of γ h , andits derivative f, if it exists, is called the spectral density function. If∑h |γ h| < ∞, then f exists. A function f(ω) defined on [−1/2,1/2] is thespectrum of a stationary process if and only if f(ω) = f(−ω), f(ω) ≥ 0,and ∫ f(ω)dω < ∞.We haveγ YY (h) = E[(Y t+h −E[Y t+h ])(Y t −E[Y t ])]ñ ô∑ ∑= E ψ r (ε t+h−r −E[ε t ])(ε t−s −E[ε t ])ψ sr s= ∑ ∑ψ r γ εε (h−r +s)ψ sr s= σ 2∑ ∑ψ r δ(h−r+s)ψ s ,r si=2hence,f(ω) = σ 2∑ h= σ 2∑ r∑∑ψ r δ(h−r+s)ψ s e −2πiωhr sψ r e −2πiωr∑ ψ s e 2πiωss= σ 2 |ψ(ω)| 2 .Finally, we havef(ω) = σ 2( 1+θe −2πiω)( 1+θe 2πiω) = σ 2 (1+θ 2 +2θcos(2πω)).4. A time series {Y t } is an autoregressive-moving average process of orderp,q, ARMA(p,q), model, if it is stationary and of the formY t = φ 1 Y t−1 +φ 2 Y t−2 +···+φ p Y t−p +ε t +θ 1 ε t−1 +···+θ q ε t−q ,where φ 1 ,...,φ p ,θ 1 ,...,θ q are constants with φ p ,θ q ≠ 0, and v t is whitenoise.2