TMA4285 Time Series Models Exam December

Transcription

1 Norges teknisk-naturvitenskapelige universitet Institutt for matematiske fag TMA485 Time Series Models Solution Oppgave a) A process {z t } is invertible if it can be represented as an A( ) process, z t = π j z t j +a t j= The process is invertible if and only if all the roots of θ(b) = lie outside the unit circle We have θ(b) = B + 4 B Thus, and ( θ(b) = 4 B B + = B = ± ) B = ±i 3 B = + 3 = +3 = 4 > All roots are outside the unit circle, so the model is invertible A process {z t } is covariance stationary if Ez t exists and is not a function of t, and Covz t,z t+k exists and is not a function of t, only a function of k = ± 3 4 The process is covariance stationary if and only if all roots of ϕ(b) = lie outside the unit circle We have ϕ(b) = B Thus, ϕ(b) = B = B > All roots lie outside the unit circle, so the model is covariance stationary eksdesl juni 3 Side

2 TMA485 Time Series Models b) We have Thus, we must have This gives ϕ(b)z t = θ(b)a t z t = θ(b) ϕ(b) a t ψ(b) = θ(b) ϕ(b)ψ(b) = θ(b) ϕ(b) c) ( ϕ B)( ψ B ψ B ψ 3 B ) = θ B θ B By exanding on the left hand side of this equation and setting equal coefficients in front of the same power of B, we get sequentially B : ψ ϕ = θ ψ = θ ϕ, B : ψ +ϕ ψ = θ ψ = θ +ϕ ψ = (θ ϕ )ϕ +θ, B 3 : ψ 3 +ϕ ψ = ψ 3 = ϕ ψ = ϕ ((θ ϕ )ϕ +θ ), B k : ψ k +ϕ ψ k = ψ k = ϕ ψ k = = ϕ k ((θ ϕ )ϕ +θ ) By noting that the formula ψ k = ϕ k ((θ ϕ )+θ ) is valid also for k =, we have shown what is asked for γ = Varz t = Var ψ j a t j = ψjvara t j +ψ + +(θ ϕ ) + j= ((θ ϕ )ϕ +θ ) ϕ (j ) +(θ ϕ ) +((θ ϕ )ϕ +θ ) ϕ j +(θ ϕ ) + ((θ ϕ )ϕ +θ ) j= ψ j γ = σ a ϕ γ = Covz t,z t+ = Cov ψ j a t j, ψ i a t+ i = ψ j ψ i Cova t j,a t+ i i= i= As we have { σ Cova j,a i = a if t j = t+ i i = j +, otherwise, the double sum reduces to a single sum, γ = ψ j ψ j+ σa ψ +ψ ψ + ψ j ψ j+ j= eksdesl juni 3 Side

3 γ = σ a j= TMA485 Time Series Models ψ +ψ ψ + ϕ j ϕ j+ ((θ ϕ )ϕ +θ ) ψ (+ψ )+((θ ϕ )ϕ +θ ) ϕ = σ a ψ (+ψ )+ ((θ ϕ )ϕ +θ ) ϕ ϕ ϕ j (θ ϕ )(+(θ ϕ )ϕ +θ )+ ((θ ϕ )ϕ +θ ) ϕ ϕ Oppgave a) From the plot of {z t }, the time series is clearly not stationary We can see this also from the fact that ρ k decays very slowly and φ kk goes quickly to zero The differenced time series seems to be stationary Thus, we choose d = The acf for the differenced time series decays exponentially to zero, whereas the correspondingpacfappearstocutafterlagthus,itseemsreasonabletotryanaima(,,) model It is not natural to include a deterministic trend parameter θ because the observed {z t } has no clear deterministic trend b) We first discuss the fit of each of the six proposed AIMA models in turn AIMA(,, ): the acf for the estimated residuals is significantly different from zero at lag Thus, this model does not give a good fit for the observed data AIMA(,,): the acf for the estimated residuals looks ok, and all the estimated parameters are significant This looks like a promising model AIMA(3,,): the acf for the estimated residuals looks ok, but ϕ 3 is not significantly different from zero Thus, the model seems to be overparameterised AIMA(,, ): the acf for the estimated residuals looks ok and all estimated parameters are significant This looks like a promising model AIMA(,,): the acf for the estimated residuals looks ok, but θ is not significantly different from zero Thus, the model seems to be overparameterised AIMA(,,3): the acf for the estimated residuals looks ok, but θ and θ 3 are not significantly different from zero Thus, the model seems to be overparameterised The AIMA(,,) and AIMA(,,) models both give a good fit for the observed data These models have the same number of parameters and the log-likelihood is slightly lower for the AIMA(,,) model (and thereby also the AIC is slightly lower for the AIMA(,,) model), so for this reason we prefer the AIMA(,,) model eksdesl juni 3 Side 3

4 Oppgave 3 a) ) Define x t = x t+ = zt z t zt+ z t and y t = u t We then get = 9 at x t + TMA485 Time Series Models and y t = u t = x t +b t Thus we have ) Define x t = x t+ = z t+ z t r t+ ( at Q = Cov z t z t r t = 9 Φ = ) = σ a, A = and y t = u t We then get x t + a t b t, and = Cov(b t ) = σ b and y t = u t = xt +c t Thus, we have Q = Cov Φ = a t b t = σ a σ b, A = and = Cov(c t ) = σ c b) Inserting values for Φ, A and in the Kalman recursions we get P t t+ = 4 Pt t +Q, P t+ t+ = ( K t+ )P t t+, K t+ = Pt t+ 4P t t+ + Eliminating K t+ and Pt+ t we obtain ) ( ) Pt+ ( t+ = +4Q+ 4 Pt t +Q Pt t +Q Pt t Letting t we get ( P = P +4Q )( ) P +4Q+ 4 P +Q ( ) ( ) P(P +4Q+) = 4 P +Q (P +4Q+) (P +4Q) 4 P +Q eksdesl juni 3 Side 4

5 TMA485 Time Series Models P Q Figur : Plot of P as a function of Q in Problem 3b P(P +4Q+) = 4 P +Q P +4QP +P 4 P Q = ( P + 4Q+ 3 ) Q = 4 P = ( 4Q+ 3 ) (4Q+ 4 ± 3 4) +4Q As we must have P, the root with minus in front of the square root is invalid Thus, P = ( 4Q+ 3 4 ) (4Q+ + 3 ( 4) +4Q = Q+ 3 ) ( + Q+ Q ) Plot of P as a function of Q is given in Figure When Q = the x t process is deterministic Note however, that we do not know the initial value of the x t process and our observations are noisy (since > ), so P t t > for all (finite) t When t grows we get more and more information about the (deterministic) process and in the limit when t we know exactly what it is, thus P = in this case When Q > the x t process is random and, for any t, the available information about w t is only y t Thus, the larger the w t s tend to be, the more uncertain we are about the value of x t Thus, P will grow as a function of Q In the limit when Q the x t process is a white noise process (with a very large variance) and our only source of information about x t at time t is y t As the observation noise does not change with Q, P will converge to a finite value when Q eksdesl juni 3 Side 5

6 TMA485 Time Series Models c) We have that and x t+ = Φx t +w t+ = Φ(Φx t +w t )+w t+ = Φ x t +Φw t +w t+ = = Φ t+ x +Φ t w ++Φw t +w t+ y s = Ax s +v s = AΦ s x +AΦ s w ++AΦw s +Aw s +v s Thereby the vectors x t+ y y y t T and x t+ y y y t+ T are linear combinations of independent normally distributed random variables Thus, the two vectors are multi-normally distributed Then also x t+ y,,y t and x t+ y,,y t+ are normal We have x t+ y,,y t N ( x t t+,p t t+), (3) x t+ y,,y t+ N ( x t+ t+,pt+ t+), (3) y t+ x t N(Ax t+,) (33) Using (3) and (33) we get (where proportionalities are as a function of x t+ ) { exp { exp Pt+ t f(x t y,,y t+ ) f(x t+,y t+ y,,y t ) = f(x t+ y,,y t ) f(y t+ x t+,y,,y t ) = f(x t+ y,,y t ) f(y t+ x t+ ) } { exp (x t+ x t t+ ) P t t+ (y t+ Ax t+ ) } x t+ xt t+ Pt+ t x t+ Ay t+ x t+ + A x t+ } From (3) we get (where again the proportionalities are as a function of x t+ ) { } f(x t+ y,,y t+ ) exp (x t+ x t+ t+ ) Pt+ t+ { } exp x t+ xt+ x t+ P t+ t+ Pt+ t+ Thereby we must have that the coefficients in front of x t+ in the two expressions we have found for f(x t+ y,,y t+ ) must be equal, and correspondingly for the linear terms Thus, Pt+ t+ = Pt+ t + A Pt+ = P t+ t+ = P t t+ +A P t t+ P t t+ + A and x t+ t+ P t+ t+ = xt t+ P t t+ + Ay t+ ( x x t+ t t+ = t+ Pt+ Pt+ t + Ay ) t+ eksdesl juni 3 Side 6