Norwegian University of Science and Technology Department of Mathematical Sciences Page of 5 TMA4285 December 205 Time series models, solution. Problem a) (i) The slow decay of the ACF of z t suggest that the model is non-stationary. The differenced series and its ACF appears stationary which suggests that d =. The ACF of ( B)z t cuts off after lag k = whereas the PACF exhibit exponential decay with alternating signs which suggest that ( B)z t is MA(). Overall the model is then ARIMA(0,,). (ii) The PACF cuts of after lag k = 3 whereas the ACF exhibits damped occilations aswell as possibly an exponentially decaying component consistent with an AR(3) model with two complex and one positive real root (any complex roots necessarily appear in conjungate pairs so at least one root needs to be real). So this is an ARIMA(3,0,0) model. (iii) The large autocorrelation at lag k = 2 suggest a seasonal model with period s = 2. The overall sample ACF can be seen as a convolution between the ACF of a regular AR() model (the ACF decays exponentially and the PACF at cuts off at lag k = 2) and the ACF of MA() model at multiples of the seasonal period s (the ACF at lag k = 2s = 24 is zero). Thus the observed pattern appears consistent with a seasonal ARIMA(, 0, 0) (0, 0, ) 2 model. b) Given that w t = ( B)z t is MA(), the two first autocovariances are γ 0 = σ 2 a( θ 2 ) and γ = σ 2 aθ. The theoretical autocorrelation at lag is thus ρ = θ /( θ 2 ). Equating this to the sample autocorrelation ˆρ and solving for θ yields two solutions θ =.4, θ = 0.7 out of which we choose θ = 0.7 as our estimate to satisfy the requirement of invertibility. Problem 2 a) To write in pure AR( ) form π(b) must satisfy ( φ B)Z t = ( θ B)a t π(b)z t = a t, ( π B π 2 B 2...)( θ B) = ( φ B) Expanding and equating like terms yields π = φ θ and π j = θ π j = θ j (φ θ ) for j 2. This AR( ) representation only exist when the MA part of the model is invertible for θ <.
Solution: TMA4285 Time Series Models, December 205 Page 2 of 5 b) Assuming that 0.56 is the last observation Z 5, the -step ahead forecast becomes Ẑ 5 () = E(Z 6 Z 5, Z 4,...) = π Z 5 π 2 Z 4... = (φ θ )(Z 5 θ Z 4 θ 2 Z 3...) =.5 (.56.4.93.6.5.064 2.28.0256 2.97.002.96.004.54.006 (.7)...) = 0.876. c) For lead times l 2, the forecasts become Ẑ 5 (l) = E(Z 5l Z 5,...) = E(φ Z 5l a 5l θ a 5l Z 5,...) = φ E(Z 5l Z 5,...) = φ Ẑ 5 (l ) = φ l Ẑ 5 () =.9 l 0.876. As l this tends to the stationary mean of the process, E(Z t ) = 0. d) The M A( ) polynomial ψ(b) = θ(b)/φ(b) must satisfy ( ψ B ψ 2 B 2...)( φ B) = ( θ B) Expanding and equating like terms yields ψ = φ θ and ψ j = φ ψ j for j 2. Hence ψ j = φ j (φ θ ) for j. The variance of the l-step ahead forecast error is then Var(Z tl Ẑt(l)) l = Var( ψ j a tl j ) = σ 2 a l ψj 2 l 2 (φ θ ) 2 φ 2(j ) j= l 2 2 (φ θ ) 2 (φ 2 ) j = σ 2 a As l, this tends to ( (φ θ ) 2 ) σ 2 a φ 2 the stationary variance of the process (see p. 6 in Wei). ( (φ θ ) 2 φ2(l ) φ 2 ) 2 θ 2 2φ θ φ 2 = Var(Z t ),.
Solution: TMA4285 Time Series Models, December 205 Page 3 of 5 Problem 3 a) Treating z as given, the conditional likelihood takes the form L (φ, σa) 2 = f(z 2, z 3, z 4, z 5 z ) = f(z t z t ) = = (z t φ z t ) 2 (2πσa) 2 e a 2 /2 (2πσ 2 a) 4/2 e 2 a 5 (zt φzt )2. σ Assuming that the process is stationary ( φ < ), z N(0, 2 a ). The exact likelihood is φ 2 then L(φ, σa) 2 = f(z, z 2, z 3, z 4, z 5 ) = f(z ) f(z t z t ) b) The corresponding log likelihoods are and = ( φ2 ) /2 ( φ 2 )z2 (2πσa) 2 e a 2 /2 (z t φ z t ) 2 (2πσa) 2 e a 2 /2 = ( φ2 ) /2 (2πσa) 2 5/2 e a 2 [( φ 2 )z2 5 (zt φzt )2 ]. ln L (φ, σ 2 a) = C 4 2 ln σ2 a 2 a (z t φ z t ) 2 ln L(φ, σa) 2 = C 5 2 ln σ2 a 2 ln( φ2 ) a 2 [( φ 2 )z 2 (z t φ z t ) 2 ]. For φ =.5, these are maximised for ˆσ a 2 = (z t φ z t ) 2 = 4 4 (.22.2 2.3 2.2 2 ) = 0.0525, and ˆσ 2 a = 5 [( φ2 )z 2 respectively. (z t φ z t ) 2 ] = 5 ((.52 ) 2 2.2 2.2 2.3 2.2 2 ) = 0.642, The large difference is a result of the fact that z contains information which is ignored when using the approximate conditional maximum likelihood approach. The conditional MLE of 0.0525 is not very consistent with the full data set, in particular, this value of σ 2 a would imply a stationary variance σ 2 a/( φ 2 ) = 0.07 (and a stationary standard deviation of 0.26) for which z 2 would be an extremely unlikely event with a probability of the order of 0 4.
Solution: TMA4285 Time Series Models, December 205 Page 4 of 5 Table : Results from applying the Kalman filter t Ŷ t t V t t Ŷ t t V t t Ŷ t 5 V t 5 2 0 2 0 2.5.25.765 0.9882 3.5.25.25.325 0.942.765 4.25.325.25.328.765 0.9882 5 2 0 2 0 Problem 4 a) With no observations at t = 2, 3, 4, the filtering steps at these time points become Ŷ t t = Ŷt t, V t t = V t t, () that is, the probability distribution of Y t conditional on observations up to time t is the same as the distribution conditional on observations up to time t. This can also be seen from the general formula for the filtering step by letting Ω (or rather the time dependent Ω t ) go to infinity which is equivalent to a missing observation. This leads to a kalman gain K t = 0 from which () follows. Also note that the state Y and Y 5 is observed without any error, such that Ŷt t = Z t = 2 and V t t = 0 for t = and t = 5 and so no initial values are needed. Forecasting Y 2, we get Ŷ2 = φ Ŷ = 0.5 2 = and V 2 = φ 2 V σ 2 a =. Similar calculations and () leads to the numbers in Table. b) Applying the smoothing recursions, using Ŷ5 5 = 2 and V 5 5 = 0 as initial values, we first obtain, for t = 4, J 4 = V 4 4 φ /V 5 4 = 0.5.325/.328 = 0.494, Ŷ4 5 = Ŷ4 4 J 4 (Ŷ5 5 Ŷ5 4) =.25 0.494(2 0.25) =.764, V 4 5 = V 4 4 J 2 t (V 5 5 V 5 4 ) =.3250.494 2 (0.328) = 0.9882. Similar calculations leads to the numbers in Table. Fig. shows the estimates and their associated uncertainty. The symmetry of the autocovariance function of a stationary process Y t and the pattern of the observed values translates to the estimated states being symmetric about t = 3. Additionallly, the expected value and the variances of the estimated states tends towards the stationary mean and variance of the process with increasing distance from the observations (towards the midpoint t = 3) as expected.
Solution: TMA4285 Time Series Models, December 205 Page 5 of 5 Y t 2 0 2 3 4 2 3 4 5 t Figure : Expected value and 95% probability limits of Y t Z, Z 5.