Licenciatura de ADE y Licenciatura conjunta Derecho y ADE. Hoja de ejercicios 2 PARTE A
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1 Licenciaura de ADE y Licenciaura conjuna Derecho y ADE Hoja de ejercicios PARTE A 1. Consider he following models Δy = ε ( L) Δ 1 y = ε where ε and ε are independen whie noise processes. In wha follows he raes of change are defined as he differences of he logarihms. Mark he correc answers: a) Only he process y is saionary and inverible. b) Boh processes, y and y, have sochasic seasonaliy. c) The process y is saionary, bu is rae of change is no saionary. d) Only he process y is no saionary. e) Boh processes, y and y, are no saionary, bu only he process y has a deerminisic rend. f) None of he above answers is correc.. Consider he following processes: y = ε1 + 0.ε 1 (1 0.L)(1 L)(1 L 1 )log y = (1 0.L) ε where ε and ε are independen whie noise processes. Mark he correc answers: a) Boh processes, y and logy, are saionary bu no inverible. b) The saionary ransformaion of logy does no presen regular auocorrelaion. c) The saionary ransformaion of logy is whie noise. d) The auocorrelaion of order 1 of y is negaive. e) The saionary ransformaion of logy is an AR(1) process. f) The saionary ransformaion of y is a MA(1) process. 3. Le quarerly impors of a counry o be given by he model: 4 ΔΔ log X = (1 0.4L ) ε 4 where ε is a whie noise. In wha follows he raes of change are defined as he differences of he logarihms. Mark he correc answers: a) The quarerly series of raes of change of impors is saionary. b) The annual rae of change of impors is inverible. c) The saionary ransformaion of impors has zero auocorrelaion in all he lags. d) Consider he saionary ransformaion of impors. For he orders greaer han four, he parial auocorrelaions are equal o zero.
2 e) The saionary ransformaion of impors does no presen auocorrelaion in he regular order, bu i has auocorrelaion in he seasonal order. f) None of he above answers is correc. 4. Consider he following quarerly processes: 4 (1 L )log y = 3 + ε 4 (1 L)(1 L )log y 0.4ε 1 = (1 0.6L) ε where ε 1 and ε 1 are independen whie noise processes. Mark he correc answers: 3 1 has mean equal o (b) The annual raes of change, approximaed by he annual differences of logarihmic ransformaions, are saionary in he case of y 1 bu no in he case of y. (c) The saionary ransformaion of y 1 has seasonal correlaion. (d) The auocorrelaion funcion of he saionary ransformaion of y has zero correlaions for lags greaer or equal han he lag. (e) The saionary ransformaions of y 1 and y have zero mean. (a) The series ( L 4 ) log y1
3 5. a) In he nex able are repored he resuls of a Dickey Fuller uni roo es applied o a daily ime series (including Saurdays and Sundays). The daa are daily prices of French elecriciy from 11/8/001 o 5/5/005. Inerpre he resuls of he uni roo es, indicae null and alernaive hypoheses of he es and explain why we need he regression repored in he able. Null Hypohesis: OFFPEAK1 1_8 has a uni roo Exogenous: Consan Lag Lengh: 1 (Fixed) -Saisic Prob.* Augmened Dickey-Fuller es saisic Tes criical values: 1% level % level % level *MacKinnon (1996) one-sided p-values. Augmened Dickey-Fuller Tes Equaion Dependen Variable: D(OFFPEAK1 1_8) Mehod: Leas Squares Dae: 04//08 Time: 1:39 Sample (adjused): 11/9/001 5/5/005 Included observaions: 174 afer adjusmens Variable Coefficien Sd. Error -Saisic Prob. OFFPEAK1 1_8(-1) D(OFFPEAK1 1_8(-1)) C R-squared Mean dependen var Adjused R-squared S.D. dependen var S.E. of regression Akaike info crierion Sum squared resid Schwarz crierion Log likelihood F-saisic Durbin-Wason sa.0660 Prob(F-saisic) Answer: We made a Dickey Fuller uni roo es. The null hypohesis is ha here exiss a uni roo (model I(1), no saionary) and he alernaive hypohesis is ha he model is saionary (I(0)). We es H0: ρ 1 agains H1: ρ 1 in he model y = a + ρ 1 y 1 + ε 1 = 1 0.
4 To perform he es we esimae a regression of he firs differences of he series on he firs lag of he iniial series: Δy = a + a 0 1 y 1 + ε H0: a = 1 0 H1: a 1 0 The saisic ha we ge is = , > , and consequenly we do no accep he null hypohesis. Finally, according o he es, he series has no a uni roo, and i is saionary. b) Below are repored he graphs of he series of he prices and of he esimaed auocorrelaions. Wha are he characerisics of he series of elecriciy prices ha can be derived observing boh graphs? A saionary series have auocorrelaions wih a rapid exponenial decay oward zero. Do you hink ha i is he case for he price of elecriciy?
5 Answer: The auocorrelaions of he series do no decrease exponenially, and his fac suggess ha he series is no saionary, and i is in conradicion wih he resul of he Dickey Fuller es. c) Explain, giving some reasons, if you expec ha daily elecriciy prices have a seasonal paern. Observing he graph and he auocorrelaions of he series, do you hink ha elecriciy prices have a seasonal paern? Answer: We expec ha elecriciy prices have a seasonal paern in erms of days; for example, we expec ha he firms demand for elecriciy decreases on Saurdays and Sundays, while he families demand for elecriciy increases in hese days, and his fac could affec elecriciy prices. Our expecaion is refleced in he correlogram of he series, which seems o be characerized by a cycle in he auocorrelaions (wih a period of 7 observaions, ha is a week). We can also observe ha some parial auocorrelaions are significan, in paricular he ones associaed wih he seasonal lags (7, 14, 1, 8). d) Observing he correlogram, i has been decided o make an augmened Dickey Fuller es, using informaion crieria o decide he number of lags o include in he regression. Using he resuls of quesion c), do you hink ha make sense o include 8 lags in he regression? Inerpre he resuls of he uni roo es and indicae null and alernaive hypoheses. Wha is he resul of he conras now? Commen he resuls, and make a comparison wih he resuls of quesion b).
6 Null Hypohesis: OFFPEAK1 1_8 has a uni roo Exogenous: Consan Lag Lengh: 8 (Auomaic based on SIC, MAXLAG=35) -Saisic Prob.* Augmened Dickey-Fuller es saisic Tes criical values: 1% level % level % level *MacKinnon (1996) one-sided p-values. Augmened Dickey-Fuller Tes Equaion Dependen Variable: D(OFFPEAK1 1_8) Mehod: Leas Squares Dae: 04//08 Time: 1:30 Sample (adjused): 1/6/001 5/5/005 Included observaions: 147 afer adjusmens Variable Coefficien Sd. Error -Saisic Prob. OFFPEAK1 1_8(-1) D(OFFPEAK1 1_8(-1)) D(OFFPEAK1 1_8(-)) D(OFFPEAK1 1_8(-3)) D(OFFPEAK1 1_8(-4)) D(OFFPEAK1 1_8(-5)) D(OFFPEAK1 1_8(-6)) D(OFFPEAK1 1_8(-7)) D(OFFPEAK1 1_8(-8)) D(OFFPEAK1 1_8(-9)) D(OFFPEAK1 1_8(-10)) D(OFFPEAK1 1_8(-11)) D(OFFPEAK1 1_8(-1)) D(OFFPEAK1 1_8(-13)) D(OFFPEAK1 1_8(-14)) D(OFFPEAK1 1_8(-15)) D(OFFPEAK1 1_8(-16)) D(OFFPEAK1 1_8(-17)) D(OFFPEAK1 1_8(-18)) D(OFFPEAK1 1_8(-19)) D(OFFPEAK1 1_8(-0)) D(OFFPEAK1 1_8(-1)) D(OFFPEAK1 1_8(-)) D(OFFPEAK1 1_8(-3))
7 D(OFFPEAK1 1_8(-4)) D(OFFPEAK1 1_8(-5)) D(OFFPEAK1 1_8(-6)) D(OFFPEAK1 1_8(-7)) D(OFFPEAK1 1_8(-8)) C R-squared Mean dependen var Adjused R-squared S.D. dependen var S.E. of regression Akaike info crierion Sum squared resid Schwarz crierion Log likelihood F-saisic Durbin-Wason sa Prob(F-saisic) Answer: Observing he resuls of pars b) and c), we noice ha he auocorrelaions of he series do no decrease quickly. Moreover, he series has a seasonal paern, and he model mus include more lags. As consequence, he resul of he Dickey Fuller es can be explained by he fac ha we did no include in he regression model all he necessary dynamic of he series. In fac, i has sense o include 8 lags, which corresponds o 4 weeks. Now, he resul of he Augmened Dickey Fuller es (wih 8 lags of he differences) suggess ha he series is no saionary, and also wih a 10% level we accep he null hypohesis of uni roo: he saisic is , and > The null and alernaive hypoheses of his es are he same of he Dickey Fuller es performed in par a). 6) Consider he following model y = ε σ σ = ω + αy whereε is Normal and independen wih mean 0 and variance 1. a) Assuming saionariy, ha is α<1, compue he mean and he variance of y. We have ha ε and σ are independen: i follows ha E( y ) = E( ε )E( σ ) = 0. Moreover, he process y is saionary, i follows ha i will have a consan marginal variance, sayσ, which can be compued in he following way: E( y ) = E( ε )E( σ ) = E( σ ) = E( ε ) = 1. 1 σ. I is equal o he variance of he process σ, because b) Discuss if he variable y is uncorrelaed and if i is independen. Wha are he differences beween uncorrelaion and independence? The independence beween he processes ε and σ guaranees ha he series y has no auocorrelaion and i is a whie noise process. In fac, he auocovariances of he series are given by: E( y y ) = E( k σ ε σ k k ) = E( ε ε k )E( σ σ k ) = 0, ε
8 because he process ε is independen of is pas values and of he pas and acual values of he process σ. I follows ha he process y will have null auocovariances. However, unlike he common hypoheses of ARIMA models, he series y is no independen of is pas values. In fac, y is relaed o y, and even 1 if hey are no correlaed, in general we will observe dependence in he series y. c) Compue condiional expecaion and condiional variance. Wha are he differences beween he marginal (or uncondiional) momens calculaed in par a) and he condiional momens? The condiional mean of y is E( y y 1 ) = E( ε )E( σ y 1) = 0, and i is equal o he marginal mean. The condiional variance is: Var( y y 1 ) = E( ε )E( σ y 1 ) = σ because E( ε y 1 ) = E( ε ) = 1. Therefore, unlike he marginal variance, which is consan, he condiional variance is no consan, and i changes wih ime. I follows ha he process σ has an ineresing inerpreaion: in each momen, i represens he condiional variance of he series, and i changes wih ime wih a sor of saionary dynamic. 7) Consider he following graphs which represen he observed values of a paricular variable and is sample correlaions Tasas anuales de desempleo
9 Explain how o inerpre he esimaes of he sample mean, he sample variance and he sample auocovariances of his variable. How do you inerpre ha he sample correlaions are ouside he bands of 95%? The series is abou he annual unemploymen rae, and we know ha he variable unemploymen is ransformed by a logarihmic ransformaion. Therefore, he series unemploymen has no a consan variance, and he sample variance will be no a good esimaor of he variance of he populaion (because of he nonsaionariy of he variance of he series unemploymen ). Now, analyzing he problem for he mean, he correlogram shows firs a high dependence in he observaions of he series and second non-saionariy in mean, because of he slow decrease in he FAC. Therefore, also he sample mean will be no a good esimaor of he mean of he populaion. The same hing happens for he sample auocovariances, because hey are second order momens ha depend of he firs order momen. For his reason, he fac ha he sample correlaions are ouside he bands of 95% can no be inerpreed as usually. Finally, because of he non-saionariy, sample mean, sample variance and sample auocovariances are no represenaive for his process.
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