金融计量经济学双语版全套.ppt
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1、课件,5-1,Chapter 5,Univariate time series modelling and forecasting,课件,5-2,1 introduction,单变量时间序列模型只利用变量的过去信息和可能的误差项的当前和过去值来建模和预测的一类模型(设定)。与结构模型不同;通常不依赖于经济和金融理论用于描述被观测数据的经验性相关特征ARIMA(AutoRegressive Integrated Moving Average)是一类重要的时间序列模型Box-Jenkins 1976当结构模型不适用时,时间序列模型却很有用如引起因变量变化的因素中包含不可观测因素,解释变量等观测频率
2、较低。结构模型常常不适用于进行预测本章主要解决两个问题一个给定参数的时间序列模型,其变动特征是什么?给定一组具有确定性特征的数据,描述它们的合适模型是什么?,课件,5-3,A Strictly Stationary ProcessA strictly stationary process is one where For any t1,t2,tn Z,any m Z,n=1,2,A Weakly Stationary ProcessIf a series satisfies the next three equations,it is said to be weakly or covarian
3、ce stationary1.E(yt)=,t=1,2,.,2.3.t1,t2,2 Some Notation and Concepts,课件,5-4,So if the process is covariance stationary,all the variances are the same and all the covariances depend on the difference between t1 and t2.The moments,s=0,1,2,.are known as the covariance function.The covariances,s,are kno
4、wn as autocovariances.However,the value of the autocovariances depend on the units of measurement of yt.It is thus more convenient to use the autocorrelations which are the autocovariances normalised by dividing by the variance:,s=0,1,2,.If we plot s against s=0,1,2,.then we obtain the autocorrelati
5、on function(acf)or correlogram.,Some Notation and Concepts,课件,5-5,A white noise process is one with no discernible structure.Thus the autocorrelation function will be zero apart from a single peak of 1 at s=0.如果假设yt服从标准正态分布,则 approximately N(0,1/T)We can use this to do significance tests for the aut
6、ocorrelation coefficients by constructing a confidence interval.a 95%confidence interval would be given by.If the sample autocorrelation coefficient,falls outside this region for any value of s,then we reject the null hypothesis that the true value of the coefficient at lag s is zero.,A White Noise
7、Process,课件,5-6,We can also test the joint hypothesis that all m of the k correlation coefficients are simultaneously equal to zero using the Q-statistic developed by Box and Pierce:where T=sample size,m=maximum lag lengthThe Q-statistic is asymptotically distributed as a.However,the Box Pierce test
8、has poor small sample properties,so a variant has been developed,called the Ljung-Box statistic:This statistic is very useful as a portmanteau(general)test of linear dependence in time series.,Joint Hypothesis Tests,课件,5-7,Question:Suppose that we had estimated the first 5 autocorrelation coefficien
9、ts using a series of length 100 observations,and found them to be(from 1 to 5):0.207,-0.013,0.086,0.005,-0.022.Test each of the individual coefficient for significance,and use both the Box-Pierce and Ljung-Box tests to establish whether they are jointly significant.Solution:A coefficient would be si
10、gnificant if it lies outside(-0.196,+0.196)at the 5%level,so only the first autocorrelation coefficient is significant.Q=5.09 and Q*=5.26Compared with a tabulated 2(5)=11.1 at the 5%level,so the 5 coefficients are jointly insignificant.,An ACF Example(p234),课件,5-8,Let ut(t=1,2,3,.)be a sequence of i
11、ndependently and identically distributed(iid)random variables with E(ut)=0 and Var(ut)=2,then yt=+ut+1ut-1+2ut-2+.+qut-q is a qth order moving average model MA(q).Or using the lag operator notation:Lyt=yt-1 Liyt=yt-i通常,可以将常数项从方程中去掉,而并不失一般性。,3 Moving Average Processes,课件,5-9,移动平均过程的性质,Its properties
12、are E(yt)=Var(yt)=0=(1+)2Covariances自相关函数,课件,5-10,Consider the following MA(2)process:where ut is a zero mean white noise process with variance.(i)Calculate the mean and variance of Xt(ii)Derive the autocorrelation function for this process(i.e.express the autocorrelations,1,2,.as functions of the p
13、arameters 1 and 2).(iii)If 1=-0.5 and 2=0.25,sketch the acf of Xt.,Example of an MA Process,课件,5-11,(i)If E(ut)=0,then E(ut-i)=0 i.So E(Xt)=E(ut+1ut-1+2ut-2)=E(ut)+1E(ut-1)+2E(ut-2)=0Var(Xt)=EXt-E(Xt)Xt-E(Xt)Var(Xt)=E(Xt)(Xt)=E(ut+1ut-1+2ut-2)(ut+1ut-1+2ut-2)=E+cross-productsBut Ecross-products=0,si
14、nce Cov(ut,ut-s)=0 for s0.So Var(Xt)=0=E=,Solution,课件,5-12,(ii)The acf of Xt 1=EXt-E(Xt)Xt-1-E(Xt-1)=EXt Xt-1=E(ut+1ut-1+2ut-2)(ut-1+1ut-2+2ut-3)=E()=2=EXt-E(Xt)Xt-2-E(Xt-2)=EXt Xt-2=E(ut+1ut-1+2ut-2)(ut-2+1ut-3+2ut-4)=E()=,Solution(contd),课件,5-13,3=EXt Xt-3=E(ut+1ut-1+2ut-2)(ut-3+1ut-4+2ut-5)=0So s
15、=0 for s 2.now calculate the autocorrelations:,Solution(contd),课件,5-14,(iii)For 1=-0.5 and 2=0.25,substituting these into the formulae above gives 1=-0.476,2=0.190.Thus the acf plot will appear as follows:,ACF Plot,课件,5-15,An autoregressive model of order p,AR(p)can be expressed asOr using the lag o
16、perator notation:Lyt=yt-1 Liyt=yt-i or or where,4 Autoregressive Processes,课件,5-16,平稳性使AR模型具有一些很好的性质。如前期误差项对当前值的影响随时间递减。The condition for stationarity of a general AR(p)model is that the roots of 特征方程 all lie outside the unit circle.Example 1:Is yt=yt-1+ut stationary?The characteristic root is 1,so
17、it is a unit root process(so non-stationary)Example 2:p241A stationary AR(p)model is required for it to have an MA()representation.,The Stationary Condition for an AR Model,课件,5-17,States that any stationary series can be decomposed into the sum of two unrelated processes,a purely deterministic part
18、 and a purely stochastic part,which will be an MA().For the AR(p)model,ignoring the intercept,the Wold decomposition iswhere,可以证明,算子多项式R(L)的集合与代数多项式R(z)的集合是同结构的,因此可以对算子L做加、减、乘和比率运算。,Wolds Decomposition Theorem,课件,5-18,The moments of an autoregressive process are as follows.The mean is given by*The a
19、utocovariances and autocorrelation functions can be obtained by solving what are known as the Yule-Walker equations:*If the AR model is stationary,the autocorrelation function will decay exponentially to zero.,The Moments of an Autoregressive Process,课件,5-19,Consider the following simple AR(1)model(
20、i)Calculate the(unconditional)mean of yt.For the remainder of the question,set=0 for simplicity.(ii)Calculate the(unconditional)variance of yt.(iii)Derive the autocorrelation function for yt.,Sample AR Problem,课件,5-20,(i)E(yt)=E(+1yt-1)=+1E(yt-1)But alsoE(yt)=+1(+1E(yt-2)=+1+12 E(yt-2)=+1+12(+1E(yt-
21、3)=+1+12+13 E(yt-3)An infinite number of such substitutions would give E(yt)=(1+1+12+.)+1 y0So long as the model is stationary,i.e.,then 1=0.So E(yt)=(1+1+12+.)=,Solution,课件,5-21,(ii)Calculating the variance of yt:*From Wolds decomposition theorem:So long as,this will converge.,Solution(contd),课件,5-
22、22,Var(yt)=Eyt-E(yt)yt-E(yt)but E(yt)=0,since we are setting=0.Var(yt)=E(yt)(yt)*有简便方法=E=E=E=,Solution(contd),课件,5-23,(iii)Turning now to calculating the acf,first calculate the autocovariances:(*用简便方法)1=Cov(yt,yt-1)=Eyt-E(yt)yt-1-E(yt-1)1=Eytyt-1 1=E=E=,Solution(contd),课件,5-24,Solution(contd),For t
23、he second autocorrelation coefficient,2=Cov(yt,yt-2)=Eyt-E(yt)yt-2-E(yt-2)Using the same rules as applied above for the lag 1 covariance2=Eyt yt-2=E=E=,课件,5-25,Solution(contd),If these steps were repeated for 3,the following expression would be obtained3=and for any lag s,the autocovariance would be
24、 given bys=The acf can now be obtained by dividing the covariances by the variance:,课件,5-26,Solution(contd),0=1=2=3=s=,课件,5-27,Measures the correlation between an observation k periods ago and the current observation,after controlling for observations at intermediate lags(i.e.all lags k).yt-k与 yt之间的
25、偏自相关函数kk 是在给定yt-k+1,yt-k+2,yt-1 的条件下,yt-k与 yt之间的部分相关。So kk measures the correlation between yt and yt-k after removing the effects of yt-k+1,yt-k+2,yt-1.或者说,偏自相关函数kk 是对yt-k与 yt之间未被yt-k+1,yt-k+2,yt-1所解释的 相关的度量。At lag 1,the acf=pacf alwaysAt lag 2,22=(2-12)/(1-12)For lags 3+,the formulae are more comp
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