Problems of Introduction to Time Series - Exam 1 | APPM 4540, Exams of Mathematics

Download Problems of Introduction to Time Series - Exam 1 | APPM 4540 and more Exams Mathematics in PDF only on Docsity! APPM 4/5540 Sample Problems for Exam I (The actual exam will be shorter!) 1. Which of the following appears (at least visually) to be stationary? If you say that a series is not stationary, briefly justify your reasoning. (a) time 0 20 40 60 80 100 8 10 12 14 16 18 (b) Monthly Means of Daily Relative Sunspots Numbers Time su ns po ts 1750 1800 1850 1900 1950 0 50 10 0 15 0 20 0 25 0 (c) time 0 200 400 600 800 1000 -4 -2 0 2 4 6 2. An AR(1) model is selected to model a stationary series {Xt}. Suppose that we only know that the lag three correlation of {Xt} is -0.125, the mean of {Xt} is 100, and the variance of {Xt} is 16. Write down the ARMA difference equation for this model and identify all of the model parameters. (Hint: AR models all have mean 0. Fit the AR(1) model to {Xt − 100}.) 3. If I tell you that, of the following acfs, one is definitely that of an MA process and the other is definitely that of an AR process, which is which? For the MA model, give the order (q) that you would use to model the process. Lag A C F 0 5 10 15 20 25 30 -0 .5 0 .0 0 .5 1 .0 ACF Lag A C F 0 5 10 15 20 25 30 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0 ACF 4. Let {Zt} ∼ WN(0, σ 2). Is the following model stationary? Xt = ZtZt−1 If not, justify your answer. If so, specify the mean and autocovariance function. 5. For an ARMA(p,q) {Xt}, state necessary and sufficient conditions for causality. 6. List three properties that must be satisfied by an autocovariance function. 7. Explain one method for eliminating a seasonal component of period 12 from a time series {Xt}. 8. Are the following ARMA(p, q) models causal? Identify p and q. (a) Xt + 0.6Xt−1 = Zt + 1.3Zt−1 (b) Xt + 2Xt−2 = Zt (c) Xt + 1.6Xt−1 = Zt − 0.4Zt−1 + 0.04Zt−2 9. Write down the difference equation for an ARMA(1,1) and find the autocovariance function. 10. (Hint: For parts (a) and (b), do not start from scratch!) (a) Find the autocovariance function of the time series Xt = Zt + 0.3Zt−1 − 0.4Zt−2 where {Zt} ∼ WN(0, 1). (b) Find the autocovariance function of the time series Xt = Zt − 1.2Zt−1 − 1.6Zt−2 where {Zt} ∼ WN(0, 0.25) Compare your answer with part (a). 11. Consider the causal AR(1) process Xt = φXt−1 + Zt, |φ| < 1. Suppose we have observed X1 and X3 and would like to estimate the missing value X2. Find the best linear predictor of X2 given X1 and X3. 5. Write the ARMA(p,q) model as φ(B)Xt = θ(B)Zt where φ(z) = 1 − φ1z − φ2z 2 − · · · − φpzp θ(z) = 1 + θ1z + θ2z 2 + · · · + θqz q. If φ(z) and θ(z) have no common roots, then the model is causal if and only if all roots of φ(z) lie outside the unit circle. 6. (i) γ(0) ≥ 0 (ii) γ(h) = γ(−h) (iii) |γ(h)| ≤ γ(0) (There are others...) 7. Difference the series: ▽12Xt = Xt − Xt−12. 8. (a) φ(z) = 1 + 0.6z has root z = −5/3 which is outside the unit circle. This process is causal. (b) φ(z) = 1 + 2z2 has roots z = ± 1√ 2 i which are inside the unit circle. This process is not causal. (c) φ(z) = 1 + 1.6z has root z = −0.625 which is inside the unit circle. This process is not causal. 9. Xt − φXt−1 = Zt + θZt−1 Noting that we always have α0 = 1 and showing that α1 = φ+ θ, multiplying through by Xt and Xt−1 and taking expectations gives γX(0) − φγX(1) = σ 2 Z [1 + θ(φ + θ)] γX(1) − φγX(0) = σ 2 Zθ Multiplying through by Xt−k, for k ≥ 2, and taking expectations gives γX(k) − φγX(k − 1) = 0 or γX(k) = φγX(k − 1). Recursing, we get γX(k) = φ k−1γX(1) Now, solving the system of equations above for γX(0) and γX(1), we get γX(0) = σ2 Z (1 + 2φθ + θ2) 1 − φ2 and γX(1) = σ 2 Z [ φ(1 + 2φθ + θ2) 1 − φ2 + φθ ] So, for k ≥ 2, γX(k) = φ k−1σ2Z [ φ(1 + 2φθ + θ2) 1 − φ2 + φθ ] . Cool! 10. These are both MA processes. We know that the autocovariance function for an MA(q) is γX(h) = { σ2 Z ∑q−|h| i=0 θiθi+|h| , |h| ≤ q 0 , |h| > q where θ0 = 1 and the other θ’s are coefficients from the MA model. (a) MA(2): θ0 = 1, θ1 = 0.3, θ2 = −0.4, σ 2 Z = 1 γX(0) = 1 · ∑ 2−0 i=0 θiθi+|h| = θ2 0 + θ2 1 + θ2 2 = (1)2 + (0.3)2 + (−0.4)2 = 1.25 γX(1) = 1 · ∑ 2−1 i=0 θiθi+|h| = θ0θ1 + θ1θ2 = (1)(0.3) + (0.3)(−0.4) = 0.18 γX(2) = 1 · ∑ 2−2 i=0 θiθi+|h| = θ0θ2 = (1)(−0.4) = −0.4 So, γX(h) =          1.25 , h = 0 0.18 , h = ±1 −0.4 , h = ±2 0 , |h| > 2 (b) MA(2): θ0 = 1, θ1 = −1.2, θ2 = −1.6, σ 2 Z = 0.25 γX(0) = 0.25 · ∑ 2−0 i=0 θiθi+|h| = 0.25(θ2 0 + θ2 1 + θ2 2 ) = 0.25((1)2 + (−1.2)2 + (−1.6)2) = (0.25)(5) = 1.25 γX(1) = 0.25 · ∑ 2−1 i=0 θiθi+|h| = 0.25(θ0θ1 + θ1θ2) = 0.25[1(−1.2) + (−1.2)(−1.6)] = (0.25)(0.72) = 0.18 γX(2) = 0.25 · ∑ 2−2 i=0 θiθi+|h| = 0.25(θ0θ2) = 0.25(1)(−1.6) = −0.4 So, the acvf is γX(h) =          1.25 , h = 0 0.18 , h = ±1 −0.4 , h = ±2 0 , |h| > 2 which is the same as in part a! (Did you see that coming?) So, trying to recover a model from an a.c.v.f. may be tricky since there is not necessarily a unique model to recover. 11. We want to find a1 and a3 such that E[(X2 − X̂2) 2] is minimized, where X̂2 = a1X1 + a3X3. E[(X2 − X̂2) 2] = E[(X2 − a1X1 − a3X3) 2] d da1 E[(X2 − a1X1 − a3X3) 2] = E[2(X2 − a1X1 − a3X3)(−X1)] Setting this equal to zero gives E[X1(X2 − a1X1 − a3X3)] = 0 or γX(1) − a1γX(0) − a3γX(2) = 0 (1) Similarly, taking the derivative with respect to a2 leads us to the equation γX(1) − a1γX(2) − a3γX(0) = 0 (2) Plugging in γX(h) = σ 2 Z φ|h|/(1 − φ2) to (1) and (2) gives us the system φ − a1 − a3φ 2 = 0 and φ − a1φ 2 − a3 = 0