Problem 1Consider a ARMA(1, 1) processYt Yt-1 +et - Oet-1=where {et}~ WN (0,0²)1. State the characteristic polynomials for the process.2. Under what condition will the process be stationary?3. Under what condition will the process be invertible?4. Obtain the Yule-Walker equations for the process and obtain the ex-pressions of 70 = Var(Y) and 1 = Cov(Yt, Yt-1).

Given the ARMA1, 1 process Yt=ϕYt-1+et-θet-1 where et~WN0, σ21. The ARMA1,1 process is Yt=ϕYt-1+et-θet-1 The model can be written as: Yt-ϕYt-1=et-θet-1 1-ϕBYt=1-θBet The characteristic polynomials for the AR part is 1-ϕλ and the characteristic polynomials for the MA part is 1-θλ. 2. The condition for the process to be stationary is the root of the characteristic equation of the AR part must be strictly greater than 1 in magnitude. 1-ϕλ=0ϕλ=1λ=1ϕ So, for stationarity, we must have ϕ<0.3. The condition for the process to be invertible is the root of the characteristic equation of the MA part must be strictly greater than 1 in magnitude. 1+θλ=0θλ=-1λ=-1θ So, for invertibility, we must have θ<0.4. Using the Yule-Walker equations, CovYt, et=CovϕYt-1+et-θet-1, et=ϕCovYt-1, et+Covet,et-θCovet-1,et=ϕ×0+Varet-θ×0=σ2 CovYt, et-1=CovϕYt-1+et-θet-1, et-1=ϕCovYt-1, et-1+Covet,et-1-θCovet-1,et-1=ϕ×σ2+0-θ×σ2=ϕ-θσ2 γ0=VarYt=CovYt, Yt=CovϕYt-1+et-θet-1, Yt=ϕCovYt-1, Yt+Covet,Yt-θCovet-1,Yt=ϕγ1+σ2-θϕ-θσ2=ϕγ1+1+θ2-θϕσ2 γ1=CovYt, Yt-1=CovϕYt-1+et-θet-1, Yt-1=ϕCovYt-1, Yt-1+Covet,Yt-1-θCovet-1,Yt-1=ϕγ0-θσ2 γ0=ϕϕγ0-θσ2+1+θ2-θϕσ2=ϕ2γ0+1+θ2-2θϕσ2γ0-ϕ2γ0=1+θ2-2θϕσ2γ0=1+θ2-2θϕ1-ϕ2σ2 γ1=ϕ1+θ2-2θϕ1-ϕ2σ2-θσ2=ϕ1+θ2-2θϕ1-ϕ2-θσ2=ϕ+θ2ϕ-2θϕ2-θ+θϕ21-ϕ2σ2=ϕ+θ2ϕ-θϕ2-θ1-ϕ2σ2 Therefore, γ0=Var(Yt)=(1+θ2-2θϕ)(1-ϕ2)σ2 and γ1=Cov(Yt,Yt-1)=(ϕ+θ2ϕ-θϕ2-θ)(1-ϕ2)σ2