MAT 3379 (Winter 2016) FINAL EXAM (SOLUTIONS)

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1 MAT 3379 (Winter 2016) FINAL EXAM (SOLUTIONS) 15 April 2016 (180 minutes) Professor: R. Kulik Student Number: Name: This is closed book exam. You are allowed to use one double-sided A4 sheet of notes. Only non programmable calculators are permitted. There are FIFTEEN questions. Cellular phones, unauthorized electronic devices or course notes (unless an open-book exam) are not allowed during this exam. Phones and devices must be turned off and put away in your bag. Do not keep them in your possession, such as in your pockets. If caught with such a device or document, the following may occur: you will be asked to leave immediately the exam, academic fraud allegations will be filed which may result in you obtaining a 0 (zero) for the exam.by signing below, you acknowledge that you have ensured that you are complying with the above statement. Your signature: Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Total GOOD LUCK!!! Maximal no. of points Your Score 1

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3 Q1. (4 points) Consider the sequence X t = Z t Z t 1 +Z t 3, where Z t are i.i.d random variables with mean 0 and variance σ 2 Z. Compute E[X t X t+h ] for h = 0,1,3. (3 points) Is the sequence X t stationary? Why? (1 point) Solution to Q1: E[X 2 t ] = σ4 Z +σ2 Z, E[X t X t+1 ] = 0, E[X t X t+3 ] = 0, We also have E[X t ] = 0. Hence the expected value and the covariance do not depend on t. The sequence is stationary. Marking scheme for Q1: Each correct calculation in part a) - 1 point; correct conclusion in part b) - 1 point. Q2. (2 points) Consider ARMA(2, 1) model given by Is this process causal? Is this process stationary? Solution to Q2: The autoroegressive polynomial is The autoregressive polynomial is The roots are X t +0.75X t X t 2 = Z t +2.25Z t 1. φ(z) = z z , One of the solutions lies inside the unit circle. Hence, there is a stationary and a non-casual solution. Marking scheme for Q2: 1 point for correct calculation of the roots, 0.5 point for concluding stationarity, 0.5 point for concluding no-causality. Total number of points: 2. Q3. (2 points) Consider the linear process given by X t = (φ j +φ 2j )Z t j, where φ < 1 and Z t are i.i.d. random variables with mean 0 and variance σ 2 Z. Write the formula for ρ X (10), that is the correlation at lag 10. Note: You can use the formula for the autocovariance of the linear process Solution to Q3:

4 We have for all h 0, Hence, γ X (h) = σ 2 Z = σ 2 Z (φ j +φ 2j )(φ j+h +φ 2(j+h) ) { } φ h φ 2 +φ2h 1 φ 3 +φh 1 φ 3 +φ2h 1 φ 4 ρ X (10) = γ X(10) γ X (0), where γ X (0) and γ X (10) are obtained by substituting h = 0 and h = 10 in the general formula above. Marking scheme for Q3: 1 point for evaluation of γ X (10), one point for γ X (0). If you give just a general formula, you receive one point only. Total number of points: 2.. Q4. (3 points) Consider a stationary process X t = φ 1 X t 1 +φ 2 X t 2 +φ 3 X t 3 +Z t, where Z t are i.i.d random variables with mean 0 and variance σ 2 Z. For this model: Apply the Yule-Walker procedure to obtain P n X n+1. Compute the corresponding MSPE n (1). Solution to Q4: Guess the solution P n X n+3 = φ 1 X n +φ 2 X n 1 +φ 3 X n 3 and verify the Yule-Walker matrix equation. MSPE n (1) = γ X (0) φ 1 γ X (1) φ 2 γ X (2) φ 3 γ X (3). Marking scheme for Q4: Part a) - 1 point for the formula, 1 point for verification. Part b) - 1 point for the correct answer. Q5. (3 points) Consider the MA(1) model X t = Z t +θz t 1, θ R, where Z t are i.i.d. random variables with mean 0 and variance σ 2 Z. For this model: Apply thedurbin-levinsonalgorithmtogetthe coefficientsin thelinearpredictionp 2 X 3 = φ 21 X 2 +φ 22 X 1. (2 points) What are PACF at lags 1 and 2? (1 point) Solution to Q5: φ 21 = θ(1+θ2 ) 1+θ 2 +θ 4, φ 22 = 1+θ 2 +θ 4. φ 11 = θ 1+θ 2, φ 22 = 1+θ 2 +θ 4. θ 2 θ 2

5 Q6. (4 points) Consider a stationary AR(2) model X t = φ 1 X t 1 +φ 2 X t 2 +Z t, where Z t are i.i.d. normal random variables with mean zero and variance σ 2 Z. Assume that φ 1 is known. (2 points) Derive MLE for φ 2. (2 points) Derive MLE for σ 2 Z. Solution to Q6: ˆφ 2 = n t=1 X tx t 2 φ 1 n t=1 X t 1X t 2 n t=1 X2 t 2 Marking scheme for Q6: ˆσ Z 2 = 1 n (X t φ 1 X t 1 n ˆφ ) 2 X t 2 t=1. Part a) - 2 points for the correct answer, part b) - 2 points for the correct answer. Q7. (5 points) Assume that a data set X is well-described by an AR(2) model X t = φ 1 X t 1 +φ 2 X t 1 +Z t, where Z t are i.i.d. random variables with mean 0 and variance σ 2 Z. Below you can find the following output for its ACF: Autocorrelations of series TimeSeries, by lag Also, the sample variance is 1. For this model: Estimate the parameters φ 1, φ 2 and σ 2 Z. (3 points) Is the estimated model stationary? Is the estimated model causal? (2 points) Solution to Q7: Use Yule-Walker equation (φ 1,φ 2 ) T = Γ 1 2 (γ X(1),γ X (2)) T to obtain ˆφ 1 = , ˆφ2 = 0.104, ˆσ 2 Z = 0.4. Typical mistake: You used Γ 2 instead of Γ 1 2. The roots of the autoregressive polynomial are 7.87 and The model is causal and stationary. Marking scheme for Q7: Part a) - 1 point for each correct answer; part b) 1 point the roots, 1 point for correct conclusion on stationarity and causality. Q8. (6 points) We have AR(1) time series with the following characteristics: ACF:. Autocorrelations of series Data, by lag > length(data) [1] 250 > mean(data); var(data) [1] , > Data[250] [1]

6 Find the Yule-Walker estimates of φ and σ 2 Z. Compute 95% confidence interval for φ. Note: z = (c) Predict the 251st value. (d) What is the mean square error of the prediction in part (c)? (e) What is the estimated value of covariance at lag 7, i.e. ˆγ X (7)? Solution to Q8: ˆφ±z (c) If X is my time series (non-centered) then (d) (e) φ = 0.645, σ 2 Z = ˆσ Z ˆγ X (0) = (0.55,0.73) P n X n+1 = X(1 ˆφ)+ ˆφX 250 = ˆγ X (0) ˆφ ˆγ X (1) = ˆγ X (7) = Marking scheme for Q8: Part a) - 1 point for each correct answer; all other pars - 1 point. Q9. (5 points) We consider Yule-Walker estimation for AR(p) models. Consider AR(2) model X t = φ 1 X t 1 + φ 2 X t 2 + Z t, where Z t are i.i.d random variables with mean 0 and variance σ 2 Z. Write the theoretical formula for 95% confidence intervals for ˆφ 1 and ˆφ 2. Note that the appropriate normal quantile is (You do not need to derive the confidence intervals). Assume that a data set X is well-described by an AR(2) model. Below you can find the following output for its ACF: Autocorrelations of series MyTimeSeries, by lag Also, the estimated variance of the time series is Use your results from part to calculate the 95% confidence intervals for ˆφ 1 and ˆφ 2. (c) Calculate ˆσ 2 Z. Solution to Q9: ˆφ 1 ± n 1 ˆφ 2, ˆφ 2 ± n 1 ˆφ 2 (c) ˆφ 1 = 0.407, ˆφ2 = ˆσ 2 Z =

7 Marking scheme for Q9: Part a) - 1 point for each correct answer; part b) - 1 point for each correct answer; part c) - 1 point. Q10. (1 point) For a given time series X t we fitted AR(1) model. We estimated φ using the Yule-Walker estimator and computed residuals as R t = X t+1 ˆφX t. The following graph shows ACF and PACF of residuals. Is the fit appropriate? Series fit$resid[2:100] Series fit$resid[2:100] ACF Partial ACF Lag Lag Solution to Q10: Since there is no dependence left in residuals, the fit is appropriate.

8 Q11. (2 points) Consider the following sequence Y 2 t = 10 φ j Z 2 t Z2 t 1 Z2 t j, t = 1,2,..., where Z t, t = 1,2,..., are independent standard normal random variables. Compute E[Y 2 t ]. Solution to Q11: 10 E[Yt 2 ] = φ j = 1 φ11 1 φ. Q12. (6 points) Consider the following ARCH(1) process X t = σ t Z t, σ 2 t = X2 t 1, where Z t are i.i.d. random variables with mean 0 and variance σ 2 = 1. (1 point) Is the model stationary? Why? (1 points) Calculate E(X t ). (c) (2 points) Calculate Var(X t ). (d) (2 points) Assume furthermore that Z t are normal. Calculate E[X t 1 X 4 t ]. Solution to Q12: Stationary since α 1 < 1; E[X t ] = E[σ t ]E[Z t ] = 0. (c) We can use the linear representation: Xt 2 = α j 1 Z2 tzt 1 Z 2 t j 2, t = 1,2,..., (d) to get E[X t 1 X 4 t ] = E[X t 1σ 4 t ]E[Z4 t ] = 3E[X t 1σ 4 t ] Var[X t ] = E[X 2 t ] = 1 1 α 1 = = 10. = 3E[X t 1 ( X 2 t 1 )2 ] = E[X t 1 ] E[X 3 t 1 ] E[X 5 t 1 ] = 0 since for an odd q > 0 we have E[X q t 1 ] = E[σq t 1 ]E[Zq t 1 ] = 0. Q13. (2 points) To a data set we fitted an ARCH(1) model. The estimated parameters are as follows: a0 a Is the model stationary? Why? (1 point) Predict the next value of the squared volatility. Note that the last observation from the data sequence is , whereas the last fitted value of volatility is (1 point) Solution to Q13: Yes since α 1 < 1; ˆσ 2 n+1 = ( )2 =

9 Q14. (3 points) To a data set we fitted an GARCH(1,1) model. The estimated parameters are as follows: a0 a1 b e Is the model stationary? Why? (1 point) The last observation from the data sequence X is X 1000 = We also know that ˆσ 1000 = Predict the next value of the squared volatility σ (2 points) Solution to Q14: The model stationary because α 1 +β 1 < ˆσ = e Q15. (3 points) Assume that Z t = (Z t1,z t2 ) is an i.i.d. sequence of random vectors with the mean vector 0 and the covariance matrix [ ] Σ = Define the bivariate linear process by X t = Z t +ΨZ t 1, where [ ] 1 0 Ψ =. 1 1 Find Γ(0), Γ(1) and Γ( 1), where Γ(h) is the covariance matrix function of the bivariate time series X t. Solution to Q15: We have [ X t = (X t1,x t2 ) T = ] Z t1 +Z t 1,1. Z t2 +Z t 1,1 +Z t 1,2 and [ ] [ ] Var(X Γ(0) = t1 ) Cov(X t1,x t2 ) 2 2 =. Cov(X t2,x t1 ) Var(X t2 ) 2 6 [ ] [ ] Cov(Xt1,X Γ(1) = t+1,1 ) Cov(X t1,x t+1,2 ) =. Cov(X t2,x t+1,1 ) Cov(X t2,x t+1,2 ) [ ] [ ] Cov(Xt1,X Γ( 1) = t 1,1 ) Cov(X t1,x t 1,2 ) =. Cov(X t2,x t 1,1 ) Cov(X t2,x t 1,2 ) For example, how to get 6 above: Var(Z t2 +Z t 1,1 +Z t 1,2 ) = Var(Z t2 )+Var(Z t 1,1 )+Var(Z t 1,2 )+2Cov(Z t 1,1,Z t 1,2 ) = = 6. THIS IS THE LAST QUESTION