Examination paper for Solution: TMA4285 Time series models

Transcription

1 Department of Mathematical Sciences Examination paper for Solution: TMA4285 Time series models Academic contact during examination: Håkon Tjelmeland Phone: Examination date: December 7th 2013 Examination time (from to): 09:00 13:00 Permitted examination support material: C: Calculator CITIZEN SR-270X, CITIZEN SR-270X College or HP30S. Statistiske tabeller og formler, Tapir forlag. K. Rottman: Matematisk formelsamling. One yellow, stamped A5 sheet with own handwritten formulas and notes. Other information: Note that all answers should be justified. In your solution you can use English and/or Norwegian. Language: English Number of pages: 5 Number pages enclosed: 0 Checked by: Date Signature

2 Solution: TMA4285 Time series models, December 7th 2013 Page 1 of 5 Problem 1 a) Both ts1 and ts2 seem to be stationary, so one should have d = 0 for both time series. The acf for ts1 seems to be a damped sine wave, whereas the pacf cuts off after lag 3. This behaviour is consistent with an AR(3) model, so one should start with p = 3, d = 0 and q = 0. The acf for ts2 cuts off after lag 1, whereas the pacf seems to decay exponentially. This behaviour is consistent with an MA(1) model, so one should start with p = 0, d = 0 and q = 1. b) For the ARMA(1,1) model the estimated values for ϕ 1 and θ 1 are both significant, but the estimated acf and pacf have several lags clearly outside the confidence bands and this indicates that there are is correlation in the time series that can not be modelled by this model. For the ARMA(1,2) model the estimated value for θ 2 is (barely) not significant, which may indicate over-parameterisation. The estimated acf and pacf for the estimated residuals have at least one lag clearly outside the confidence bands, which again indicates that there is some correlation in the time series that can not be modelled by this model. For the ARMA(2,1) model the estimated values for ϕ 1, ϕ 2 and θ 1 are all significant and the estimated acf and pacf for the estimated residuals are inside the confidence bands except for a few lags where the values are slightly outside the confidence bands. Thereby this seems to be a good model for the observed time series. From the above discussion it should be clear that one should use the ARMA(2,1) model for ts3. Problem 2 a) For a time series z t to be second-order stationary one must have for all t 1, t 2, k and all x 1, x 2. F zt1,z t2 (x 1, x 2 ) = F zt1 +k,z z2 +k (x 1, x 2 ), For a time series z t to be covariance stationary all first and second order moments must exist and be time invariant.

3 Solution: TMA4285 Time series models, December 7th 2013 Page 2 of 5 A time series process z t which is covariance stationary, does not need to be second-order stationary. Even if the two first moments is time invariant, the joint distribution does not need to be time invariant. A time series process z t which is second-order stationary does not need to be covariance stationary. If all the first and second order moments exist for a second-order stationary process, then it is also covariance stationary, but the first and second order moments do not need to exist for a second-order stationary process. b) The process is invertible if and only if all the roots of θ(b) = 1 θ 1 B = 0 are outside the unit circle. The current model has only one root, which is B = 1/θ 1 so thereby the process is invertible if 1 > 1 θ ( 1, 1). θ 1 The model is covariance stationary if and only if the roots of ϕ(b) = 1 1.7B +0.72B 2 = 0 are outside the unit circle. The roots in the current model become B = 1.7 ± = 1.7 ± Thus, the model is covariance stationary. B = 1.25 or B = c) From the model we have that z t+k = ϕ 1 z t+k 1 + ϕ 2 z t+k 2 + a t+k θ 1 a t+k 1. Thereby we get for k = 0, 1, 2,..., γ k = E[z t+k z t ] = E[(ϕ 1 z t+k 1 + ϕ 2 z t+k 2 + a t+k θ 1 a t+k 1 )z t ] = ϕ 1 E[z t+k 1 z t ] + ϕ 2 E[z t+k 2 z t ] + E[a t+k z t ] θ 1 E[a t+k 1 z t ] = ϕ 1 γ k 1 + ϕ 2 γ k 2 + E[a t+k z t ] θ 1 E[a t+k 1 z t ]. For k = 2, 3,... we have that both t + k and t + k 1 is larger that t and thereby a t+k and a t+k 1 are both uncorrelated with z t, so E[a t+k z t ] = E[a t+k ]E[z t ] = 0 E[z t ] = 0 and E[a t+k 1 z t ] = E[a t+k 1 ]E[z t ] = 0 E[z t ] = 0. Thus we have found the homogeneous difference equation γ k ϕ 1 γ k 1 ϕ 2 γ k 2 = 0 for k = 2, 3,.... (1)

4 Solution: TMA4285 Time series models, December 7th 2013 Page 3 of 5 To find the initial conditions we need to consider the above expression for γ k for k < 2. We start with k = 0, γ 0 = ϕ 1 γ 1 + ϕ 2 γ 2 + E[a t z t ] θ 1 E[a t 1 z t ] = ϕ 1 γ 1 + ϕ 2 γ 2 + E[a t (ϕ 1 z t 1 + ϕ 2 z t 2 + a t θ 1 a t 1 )] θ 1 E[a t 1 (ϕ 1 z t 1 + ϕ 2 z t 2 + a t θ 1 a t 1 )] = ϕ 1 γ 1 + ϕ 2 γ 2 + E[a t a t ] θ 1 (ϕ 1 E[a t 1 z t 1 ] θ 1 E[a t 1 a t 1 ]) = ϕ 1 γ 1 + ϕ 2 γ 2 + σ 2 a θ 1 (ϕ 1 σ 2 a θ 1 σ 2 a). Thus, using that γ k is a symmetric function, the first equation becomes Doing the same for k = 1 we get γ 0 = ϕ 1 γ 1 + ϕ 2 γ 2 + σ 2 a(1 θ 1 ϕ 1 + θ 2 1). (2) γ 1 = ϕ 1 γ 0 + ϕ 2 γ 1 + E[a t+1 z t ] θ 1 E[a t z t ]. Again using what we found above, namely that E[a t+1 z t ] = 0 and E[a t z t ] = σ 2 a the second equation becomes γ 1 = ϕ 1 γ 0 + ϕ 2 γ 1 θ 1 σ 2 a. (3) As (2) and (3) includes both γ 0, γ 1 and γ 2 we need a third equation, which we get for k = 2. This, however, we get directly from (1), γ 2 = ϕ 1 γ 1 + ϕ 2 γ 0. (4) Thereby we have three equations, (2), (3) and (4), with three unknowns, γ 0, γ 1 and γ 2, which can be solved to find the initial conditions. d) We have Thereby we must have ϕ(b)z t = θ(b)a t ϕ(b) θ(b) z t = a t. ψ(b) = ϕ(b) θ(b) θ(b)ψ(b) = ϕ(b), where ψ(b) = 1 ψ 1 B ψ 2 B Thereby, (1 θ 1 B)(1 ψ 1 B ψ 2 B 2 ψ 3 B 3...) = 1 ϕ 1 B ϕ 2 B 2.

5 Solution: TMA4285 Time series models, December 7th 2013 Page 4 of 5 By expanding on the left hand side of this equation and setting equal coefficients in front of the same power of B, we sequentially get B 1 : ψ 1 θ 1 = ϕ 1 = ψ 1 = ϕ 1 θ 1, B 2 : ψ 2 + θ 1 ψ 1 = ϕ 2 ψ 2 = ϕ 2 + θ 1 ψ 1 = ϕ 2 + θ 1 (ϕ 1 θ 1 ), B 3 : ψ 3 + θ 1 ψ 2 = 0 ψ 3 = θ 1 ψ 2 = θ 1 (ϕ 2 + θ 1 (ϕ 1 θ 1 )), B k :. ψ k + θ 1 ψ k 1 = 0 ψ k = θ 1 ψ k 1 = θ k 2 1 (ϕ 2 + θ 1 (ϕ 1 θ 1 )). e) The one-step ahead forecast can be found directly from the AR( ) representation [ ] ẑ n (1) = E[z n+1..., z n 1, z n ] = E ψ k z n+1 k + a n+1..., z n 1, z n k=1 = ψ k z n+1 k, k=1 where we have used that a n+1 is uncorrelated with the observed values and that E[a n+1 ] = 0. From the definition of the model we get for k 2 ẑ n (k) = E[z n+k..., z n 1, z n ] = E[ϕ 1 z n+k 1 + ϕ 2 E[z n+k 2 + a n+k θ 1 a n+k 1..., z n 1, z n ] = ϕ 1 E[z n+k 1..., z n 1, z n ] + ϕ 2 E[z n+k 2..., z n 1, z n ] = ϕ 1 ẑ n (k 1) + ϕ 2 ẑ n (k 2), where we have used that a n+k and a n+k 1 are uncorrelated with the observed data when k 2. Thereby we have the homogeneous difference equation (inserting given parameter values) ẑ n (k) 1.7ẑ n (k 1) ẑ n (k 2) = 0 for k = 2, 3,.... To write up the general solution of this difference equation we need to find the roots of 1 1.7B B 2 = 0. These, however, we have previously found to be B = 1.25 = 1/0.8 and B = 1.11 = 1/0.9. The general solution is thereby ẑ n (k) = b k + b k. We find b 0 and b 1 from the initial conditions ẑ n (1) and ẑ n (0) = z n, b b = ẑ n (1), b b 1 1 = z n,

6 Solution: TMA4285 Time series models, December 7th 2013 Page 5 of 5 which have solution b 0 = 9z n 10ẑ n (1) and b 1 = 10ẑ n (1) 8z n. Thereby the eventual forecast function is ẑ n (k) = (9z n 10ẑ n (1)) 0.8 k + (10ẑ n (1) 8z n ) 0.9 k. Problem 3 a) We use the notation x n t = E[x t y 0,..., y n ] and Pt n = Cov[x t y 0,..., y n ]. From the model assumptions we then get for k = 1, 2,... x n n+k = E[x n+k y 0,..., y n ] = E[Φx n+k 1 + w n+k y 0,..., y n ] = ΦE[x n+k 1 y 0,..., y n ] + E[w n+k y 0,..., y n ] = Φx n n+k 1 + E[w n+k y 0,..., y n ] Using that w n+k is independent of y 0,..., y n we get E[w n+k y 0,..., y n ] = E[w n+k ] = 0. Thus, we have the recursion For P n n+k we correspondingly get x n n+k = Φx n n+k 1 for k = 1, 2,.... P n n+k = Cov[x n+k y 0,..., y n ] = Cov[Φx n+k 1 + w n+k y 0,..., y n ] = ΦCov[x n+k 1 y 0,..., y n ]Φ T + E[w n+k y 0,..., y n ] = ΦP n n+k 1Φ T + Q, where we have used that w n+k is independent of both y 0,..., y n and x n+k 1. Thus, the recursion of the covariance matrix is P n n+k = ΦP n n+k 1Φ T + Q for k = 1, 2,.... All x n+k, y 0,..., y n can be expressed as a linear combination of the independent Gaussian variables x 0, ω 1,..., ω n+k and v 0,..., v n. Thereby the joint distribution of x n+k, y 0,..., y n is Gaussian, and since the Gaussian distribution is closed under conditioning, the conditional distribution of x n+k given y 0,..., y n is also Gaussian. If the x t s are scalar quantities, a 95% prediction interval for x n+k is [ x n n+k 1.96 Pn+k n, xn n+k ] Pn+k n. If the x t s are vectors, corresponding prediction intervals can be constructed for each component of x n+k.