东北财经大学：《金融计量经济学导论》课程PPT教学课件（双语版）Chapter 5 Univariate time series modelling and forecasting

5-1 Chapter 5 Univariate time series modelling and forecasting

5-1 Chapter 5 Univariate time series modelling and forecasting

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(Auto Regressive Integrated Moving Average)ie- 类重要的时间序列模型 Box-enkins 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 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 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?

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 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？

5-3 2 Some Notation and concepts a Strictly Stationary Process A strictly stationary process is one where Pmv1≤b,…,yn≤bhn}=P{y+m≤b,…,yn+m≤bn} For any t1,t2,…,tn∈Z,anym∈Z,n=1,2, A Weakly stationary process If a series satisfies the next three equations, it is said to be weakly or covariance stationary 1.E0y)=p,t=1,2,…, 2.E(v2-)(v2-)=a2<0 3.E(y21-A)(y2-A)=y2-Vt,z2

5-3 • A Strictly Stationary Process A strictly stationary process is one where • For any t1 ,t2 ,…, tn∈ Z, any m ∈ Z, n=1,2,… • A Weakly Stationary Process If a series satisfies the next three equations, it is said to be weakly or covariance stationary 1. E(yt ) =  , t = 1,2,..., 2. 3.  t 1 , t 2 2 Some Notation and Concepts P{yt b ,..., yt n bn } P{yt m b ,..., yt n m bn } 1  1  = 1+  1 +  E y y t t t t ( )( ) 1 2 2 1 −  −  = −  E y y t t ( − )( − ) =    2

5-4 Some Notation and Concepts So if the process is covariance stationary, all the variances are the same and all the covariances depend on the difference between t, and t. The moments E(y-E(y)(y+s-E(y+s)=ys,s=0,1,2, are known as the covariance function The covariances, y are known as autocovariances However, the value of the autocovariances depend on the units of measurement of 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 t against S=0, 1, 2,. then we obtain the autocorrelation function(acf or correlogram

5-4 • So if the process is covariance stationary, all the variances are the same and all the covariances depend on the difference between t 1 and t 2 . The moments , s = 0,1,2, ... are known as the covariance function. • The covariances,  s , are known 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 autocorrelation function (acf) or correlogram. Some Notation and Concepts    s s = 0 E y E y y E y t t t s t s s ( − ( ))( + − ( + )) = 

5-5 A White noise process a white noise process is one with no discernible structure. E(y7)= Var(t otherwise Thus the autocorrelation function will be zero apart from a single peak of l ats=0. 如果假设服从标准正态分布,则。~ approximately n0,/n) We can use this to do significance tests for the autocorrelation coefficients by constructing a confidence interval a 95% confidence interval would be given by +196x If the sample autocorrelation coefficient, te, 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

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 autocorrelation 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 Process E y Var y if t r otherwise t t t r ( ) ( ) = = =  =   −     2 2 0   s T 1 1.96 s  ˆ

5-6 Joint Hypothesis Tests We can also test the joint hypothesis that all m of the correlation coefficients are simultaneously equal to zero using the @-statistic developed by box and pierce: Q=7x2 where T= sample size, m= maximum lag length The O-statistic is asymptotically distributed as a x m However, the Box Pierce test has poor small sample properties, so a variant has been developed, called the ljung-Box statistic O*=T(T+2 This statistic is very useful apfa portmanteau(general)test of linear dependence in time series

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 length • The Q-statistic is asymptotically distributed as a . • However, the Box Pierce test 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  m 2 = = m k Q T k 1 2  ( ) 2 1 2 2 ~ m m k k T k Q T T   =  − = +

5-7 An ACF Example (p234 · Question: Suppose that we had estimated the first 5 autocorrelation coefficients using a series of length 100 observations, and found them to be( from 1 to5):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 significant if it lies outside(0. 196, +0.196) at the 5% level, so only the first autocorrelation coefficient is significant Q5.09andQ*=5.26 Compared with a tabulated x(5=1l.1 at the 5%level, so the 5 coefficients are jointly insignificant

5-7 • Question: Suppose that we had estimated the first 5 autocorrelation coefficients 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 significant 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*=5.26 Compared with a tabulated  2 (5)=11.1 at the 5% level, so the 5 coefficients are jointly insignificant. An ACF Example (p234)

5-8 3 Moving Average processes Let u,(t1, 2,3, ..) be a sequence of independently and identically distributed (iid )random variables with E(u)=0 and Var(u =0, then y=H+l+日1n1+0u…+e4 is a gth order moving average model ma(q Or using the lag operator notation: Ly,=y L y=+∑0Ln+t1=H+(L)m (L)=1+61L+2L2+…+b 通常,可以将常数项从方程中去掉,而并不失一般性

5-8 • Let ut (t=1,2,3,...) be a sequence of independently and identically distributed (iid) random variables with E(ut )=0 and Var(ut )=  2 , then yt =  + ut +  1 ut-1 +  2 ut-2 + ... +  q ut-q is a q th order moving average model MA(q). • Or using the lag operator notation: Lyt = yt-1 L iyt = 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 t q i t t i yt i L u u (L)u 1 =  + + =  + = q  L = + L + L ++ q L 2 1 1 2 ( )

5-9 移动平均过程的性质 Its properties are E(=u var(y)=%=(1+G2+02+、+02)a2 Covariances (3+661+6。22 O or S= 0 for s>q 自相关函数

5-9 移动平均过程的性质 • Its properties are E( yt )= Var( yt ) = 0 = (1+ )2 Covariances 自相关函数 1   2 2 2 2 + +...+ q      + + + + = = + + − for s q for s q s s s q q s s 0 ( ... ) 1,2,..., 2   1 1  2  2    

5-10 Example of an Ma Process Consider the following MA(2) process A1=l1+1l1-1+62l12 where u, is a zero mean white noise process with variance o (1) Calculate the mean and variance ofX, (i) Derive the autocorrelation function for this process (i.e express the autocorrelations, Ti, t,,... as functions of the parameters 8, and 82) (iii)If 81=-0.5 and 02=0.25, sketch the acf of X

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 parameters 1 and 2 ). (iii) If 1 = -0.5 and 2 = 0.25, sketch the acf of Xt . Example of an MA Process Xt = ut + 1 ut−1 + 2 ut−2 2 