ECON3350/7350UNIVARIATE TIME SERIES - IAlicia N. RambaldiWeek 21 / 36In this lectureReadingsIntroTime Series - Basic ConceptsCommon Time Series ModelsConcepts and PropertiesOrder of ProceedingsTheoretical FootprintThe MomentsFitting Univariate Models2 / 36Reference MaterialsAuthor Title Chapter Call NoEnders, W AppliedEconometricTime Series,3e1,2 HB139 .E552010Verbeek, M A Guide toModernEconometrics8.1,8.2,8.7.2HB139.V465 2008IEnders Chapter 1 is a background chapter. The concepts inthis chapter will be used over a number of lectures.3 / 36Australian GDP per capita (current prices)5 / 36A Time Series ProcessIA time series is a stochastic processIEach observation in a stochastic process is a random variableIThe observations evolve in time according to certainprobabilistic lawsDenitionA stochastic process may be dened as a collection of randomvariables which are ordered in time.6 / 36ExampleA rst-order moving average processyt= t+ 1t1(1)where,tis a sequence of independent random variables from adistribution with mean zero and constant variance, and1 is aparameter,t= 1, .., T.7 / 36Realisation, MomentsDenitionA realisation is one of an innite set of values ofyt(t= 1, ..., T)generated from a model, such as (1), by drawing a new set of0, 1, ..., T.The moments of a stochastic process are dened with respect tothe random variablesy1, ..., yT. We are interested in:Mean: t= E(yt),t= 1, ..., T; which can be interpreted as the averagevalue ofyttaken over all possible realisationsVariance: Var (yt) = E[(yt t)2],t= 1, ..., TCovariance: Cov(yt, ytk) = E[(yt t)(ytk tk)],t= t k + 1, ..., T8 / 36StationaryWhen a single realisation of observations is available, theaggregation of observations over time is important.DenitionA stochastic process is stationary if the data generating process issuch that the mean, variance and covariances are independent oftime.E(yt) = Var (yt) = E[(yt )2] = 2y= 0Cov(yt, ytk) = E[(yt )((ytk )] = kk= 1, 2, ...These conditions must be satised for all values oft.At this initial stage we will only consider stationary processes andwe will relax this assumption later in the course.9 / 36Common Time Series ModelsSome important models for time series processes:IDistributed Lag (DL);IAutoregressive (AR);IAutoregressive Distributed Lag (ADL);IMoving Average (MA);IAutoregressive Moving Average (ARMA).11 / 36The standard regression model is:yt= 1 + 2xt+ tE(t) = 0;V(t) = 2Here2 measures the expected response inytto a change inxt;But the eect may take time to ow through fromxttoyt.12 / 36For example, the response in the RBA cash rate to a change inination or GDP growth may take the following patterns.The Distributed lag modelyt= 1 + 2xt+ 3xt1 + 4xt2 + 5xt3 + tWith this model we can study the eect upon expectedytof achange inxtover time.13 / 36Plot the0is: E.g.,rt= 1 + 2t+ 3t1 + ... + 8t814 / 36Autoregressive (AR) ModelsTo model the dependence inytupon its own past behaviour.AR(1)yt= a0 + ayt1 + tAR(2)yt= a0 + ayt1 + a2yt2 + tAR(p)yt= a0 + ayt1 + ... + apytp + tyt= a0 +pXi =1aiyti+ t15 / 36Autoregressive Distributed Lag ModelsTo mix lags ofytandxtwe use the Autoregressive Distributed Lagmodel (ADL)E.g., an ADL(2, 2)yt= + a1yt1 + a2yt2 + 0xt+ 1xt1 + 2xt2 + tE.g., an ADL(p, l )yt= + a1yt1 + ... + apytp + 0xt+ 1xt1 + ... + lxtl+ tyt= +pXi =1aiyti+lXj =0jxtj+ tFrom this we can plot a dynamic response inytto a change in xtorto a past change inyt.16 / 36Moving Average (MA) ModelsIMA(1)yt= + t+ 1t1IMA(2)yt= + t+ 1t1 + 2t2IMA(q)yt= + t+ 1t1 + ... + qtq17 / 36The Autoregressive Moving Average Model (ARMA):IARMA(1, 1)yt= a0 + a1yt1 + t+ 1t1IARMA(2, 2)yt= a0 + a1yt1 + a2yt2 + t+ 1t1 + 2t2IARMA(3, 1)yt= a0 + a1yt1 + a2yt2 + a3yt3 + t+ 1t118 / 36Univariate ModelsIWe concentrate rst on models that only involve one stochasticprocess, e.g. yt. They are: AR(p),MA(q) andARMA(p, q):the rest of this lecture and Week 3.IWe consider DL, ADL and other extensions in Week 4IHow they behave. Theoretical footprint.IHow do we choose the model that best ts data.20 / 36Autocovariance and AutocorrelationDenitionsIVar (yt) = E[(yt )2] = 0ICov(yt, ytk) = E[(yt )(ytk )] = kk= 1, 2, ...DenitionsAutocovariance Function: k,k= 1, 2, .... If the process isstationaryk= k.Autocorrelation Function (ACF):k=k0,k= 1, 2, ...Correlogram or SACF: Plot of the sample autocorrelationfunction,rk, againstk.22 / 36IWhen a stochastic process is stationary, we can plotkagainstk. But autocovariances depend on units of measurementIThe autocorrelation is a standardised autocovariance,k (1, 1)IThe plot ofkover non-negatives values ofkis known as theautocorrelation function (ACF).IThe data generation process (DGP) ofytis unknown to thepractitioner. However,IWe know the "theoretical" form of the ACF and the (partial)ACF - PACF of Moving Average, Autoregressive and mixedtype of processes.23 / 36Partial Autocorrelation Function (PACF)DenitionThe partial autocorrelation function (kk) is given by thekthcoecients in the correspondingAR(k) system of autoregressions.AR(1)yt= a0 + a1yt1; 11= a1AR(2)yt= a0 + a1yt1 + a2yt2; 22= a2...AR(k)yt= a0 + a1yt1 + a2yt2 + ... + akytk; kk= akWe identify the DGP by plotting the sample ACF (SACF) andsample PACF (SPACF) together.24 / 36PACF (cont.)Using the Yule-Walker equations, we can work out the PACF fromthe ACF,11= 122= (221)/(1 21)...kk=k Pk1j =1k1,jkj1 Pk1j =1k1,jj, k= 3, 4, 5, . . .where,k,j= k1,j kkk1,kj, j = 1, 2, 3, . . . , k 125 / 36ACF and PACF of AR Processesyt= a0 + a1yt1 + ... + apytp + tIAutocorrelation Function (ACF) decays to zeroIPartial Autocorrelation Function (PACF) has pnon-zeropeaksACF and PACF ofAR(1)26 / 36ACF and PACF ofAR(2)27 / 36ACF and PACF of MA Processesyt= + t+ 1t1 + 2t2 + ... + qtq(2)IAutocorrelation Function (ACF) has qnon-zero peaksIPartial Autocorrelation Function (PACF) decays to zeroACF and PACF ofMA(1)28 / 36ACF and PACF ARMA Processesyt= a0+a1yt1+a2yt2+...+apytp+t+1t1+2t2+...+qtq(3)IAutocorrelation Function (ACF) decays to zeroIPartial Autocorrelation Function (PACF) decays to zeroACF and PACF ofARMA(1, 1)29 / 36The Moments and the ACFThe AR(1):yt= a0 + a1yt1 + t; 0 |a1| < 1(Stationary)E{yt} = = a0 + a1E{yt1} + E{t} =a01 a1V{yt} = 0= a21V{yt1} + V{t}0=21 a2131 / 36AR(1) cont.cov{ytytk} = k= E{ytytk}= E{(a1yt1 + t)ytk}Example1(k= 1)1= E{(a1yt1 + t)yt1}= a121 a21= a10ICan show the autocovariance forkperiods apart is ,k= ak10Iand the autocorrelation function is decaying,k=k0= ak132 / 36The MA(1)yt= + 1t1 + tE{yt} = + 1E{t1} + E{t}= V{yt} = 0= 21V{t1} + V{t}0= 2(1 + 21)33 / 36MA(1) cont.cov{ytytk} = k= E{ytytk}= E{(1t1 + t)ytk}Example1;k= 11= E{(1t1 + t)yt1}=11 + 210Then the ACF of theMA(1) have a single non-zero peak atk= 1,1=11 + 21; k= 0 k= 2, 3, ...34 / 36From Theory to PracticeNext Lecture.....IGiven a time series (e.g. prices, ination, sales), how do wechoose an ARMA(p,q) model to t the data?IVisual InspectionISelection CriteriaIDiagnostic CheckingIForecasting36 / 36