(整理)时间序列分析试题

Graduate School of Business,University of ChicagoBusiness41202,Spring Quarter2008,Mr.Ruey S.TsaySolutions to MidtermProblem A:(30pts)Answer briefly the following questions.Each question has two points.1.Describe two methods for choosing a time series model.Answer:Any two of(a)Information criteria such as AIC or BIC,(b)Out-of-sample forecasts,and(c)ACF and PACF of the series.2.Describe two applications of volatility infinance.Answer:Any two of(a)derivative(option)pricing,(b)risk management,(c)portfolio selection or asset allocation.3.Give two applications of seasonal time series models infinance.Answer:(a)Earnings forecasts and(b)weather-related derivative pricing or risk man-agement.4.Describe two weaknesses of the ARCH models in modelling stock volatility.Answer:Any two of(a)symmetric response to past positive and negative shocks,(b)restrictive,(c)Not adaptive,and(d)provides no explanation about the source ofvolatility clustering.5.Give two empirical characteristics of daily stock returns.Answer:any two of(a)heavy tails,(b)non-Gaussian distribution,(c)volatility clus-tering.6.The daily simple returns of Stock A for the last week were0.02,0.01,-0.005,-0.01,and0.025,respectively.What is the weekly log return of the stock last week?What is theweekly simple return of the stock last week?Answer:Weekly log return is0.03938;weekly simple return is0.04017.7.Suppose the closing price of Stock B for the past three trading days were$100,$120,and$100,respectively.What is the arithmetic mean of the simple return of the stock for the past three days?What is the geometric average of the simple return of the stockfor the past three days? Answer:Arithmetic mean=12120−100100+100−120120=0.017.and the geometric mean is120×100−1=0.8.Consider the AR(1)model r t=0.02+0.8r t−1+a t,where the shock a t is normally distrib-uted with mean zero and variance1.What are the variance and lag-1autocorrelation function of r t?Answer:Var(r t)=11−0.82=2.78and the lag-1ACF is0.8.19.For problems6and7,suppose the daily return r t,in percentages,of Stock A followsthe model r t=1.0+a t+0.3a t−1,where a t=σt t withσ2t =1.0+0.4a2t−1and t beingstandard normal.What is the unconditional variance of a t?What is the variance of r t?Answer:Var(a t)=11−0.4=1.67.Var(r t)=(1+0.32)σ2a=1.82.10.Suppose that a n=3.0,what is the1-step ahead forecast for r n+1at the forecast originn?What is the1-step ahead volatility forecast of r t at the forecast origin n?Answer:r n(1)=1+0.3a n=1.9,andσ2n (1)=1+0.4a2n=4.6.11.Consider the simple AR(1)model r t=100+0.8r t−1+a t,where a t is normally distributedwith mean zero and variance10.Is the r t series mean-reverting?If yes,what is the half-life of the series?Answer:Yes,r t series is mean-reverting.The half-life is ln(0.5)/ln(0.8)=3.11.12.Describe two test statistics for testing the ARCH effect of an asset return series.Writedown the associated null hypotheses.Answer:(a)The Ljung-Box statistic Q(m)of the squared shocks,i.e.a2t .The nullhypothesis is H o:ρ1=ρ2=···=ρm=0,whereρi is the lag-i ACF of a2t .(b)TheEngle F-test for the regression a2t =β0+β1a2t−1+···+βm a2t−m+e t.The null hypothesisis H o:β1=β2=···=βm=0.13.Consider the following two IGARCH(1,1)models for percentage log returns:Model A:σ2t =1.0+0.1a2t−1+0.9σ2t−1Model B:σ2t =0.1a2t−1+0.9σ2t−1.Suppose thatσ2100=20and a100=−2.0.What are the3-step ahead volatility forecastsfor Models A and B?Answer:For model A:3-step ahead volatility forecast isσ2100(3)=2+(1.0+0.1×(−2.0)2+0.9×20)=21.4.For model B,the3-step ahead volatility forecast isσ2100(3)=0.1(−2.0)2+0.9×20=18.4.14.Consider the following two models for the log price of an asset:Model A:p t=p t−1+a tModel B:p t=0.00001+p t−1+a twhere the shock a t is normally distributed with mean zero and varianceσ2>0.Suppose further that p100=5.Let p n( )be the -step ahead forecast at the forecast origin n.What are the point forecasts p100( )for both models as →∞?Answer:For model A,p100( )=5for all .For model B,p100( )converges to infinity as →∞.215.Suppose that we have T =1000daily log returns for the Decile 1portfolio.Supposefurther that the sample autocorrelation at lag-12is ˆρ12=0.15.Test the hypothesis H o :ρ12=0against the alternative hypothesis H a :ρ12=pute the test statistic and draw your conclusion.Answer :t =0.151/√1000=√1000×0.15=4.74,which is highly significant.Thus,the lag-12ACF is not zero.Problem B .(20pts)It is well-known in economics that growth rate of the domestic gross product (GDP)is negatively correlated with the change in unemployment rate.Consider the U.S.quarterly real GDP and unemployment rate from the first quarter of 1948to the first quarter of 2008.Let dgdp t be the growth rate of the GDP,i.e.dgdp t =ln(GDP t )−ln(GDP t −1),and dun t be the change in unempolyment rate,i.e.dun t =U t −U t −1with U t being the civilian unemployment rate.The data were seasonally adjusted and obtained from the Federal Reserve Bank at St.Louis.The sample size after the differencing is e the attached R output to answer the following questions.1.(5points)Write down the fitted linear regression model with dgdp t and dun t representing the dependent and independent variable,respectively,including residual standard error.What is the R 2of the linear regression?Is the fitted model adequate?Why?Answer :The fitted linear regression isdgdp t =0.017−0.017dun t +e t ,ˆσe =0.0088.The R 2is 0.37.The model is not adequate because the Q (m )statistics of the residuals show that the residuals have serial correlations,i.e.Q (12)=219.4with p-value close to zero.2.(5points)To take care of the serial correlations in the residuals,a linear regression model with time-series errors is built for the two variables.Write down the fitted model,including the residual variance.Answer :The fitted linear regression model with time-series errors is(1−0.21B −0.12B 2)(1−0.86B 4)(dgdp t −0.017+0.018dun t )=(1−0.72B 4)a t ,ˆσ2a =6.01×10−5.3.(2points)Is the model in Question 2adequate?Why?Answer :Yes,the model is adequate.The Q (m )statistics of the residuals fail to indicate the existence of any serial correlations.We have Q (12)=17.53with p-value 0.13.4.(4points)Based on the fitted model in Question 2,is the growth rate of GDP negatively correlated with the change in unempolyment rate?Why?Answer :Yes,the growth rate of GDP is negatively related to the change in unemploy-ment rate.The estimated coefficient is −0.018which is highly significant,because it standard error 0.0014is small,resulting in a large t -ratio.35.(4points)To check the predictive power of the model,it was re-estimated using thefirst236data points.This re-fitted model is used to produce1-step to4-step ahead forecasts at the forecast origin t=236.The actual value of the GDP growth rates are also given.Construct the1-step ahead95%interval forecast of the model.Is the actual growth rate in the forecasting interval?Answer:The95%interval forecast is0.012±1.96×0.0077,i.e.[−0.0031,0.027].The actual value is0.0159,which is in the interval.Problem C.(16pts)Consider the quarterly earnings per share of the Microsoft stock from thefirst quarter of1992to thefirst quarter of2008.The data were obtained from First Call. To take the log transformation,we add0.5to all data points.The R output is attached. Let x t=ln(y t+0.5)be the transformed earnings,where y t is the actual earnings per share.1.(5points)Write down thefitted model for x t,including the variance of the residuals.Answer:Thefitted model is(1−B)r t=(1−0.70B+0.39B2)(1+0.39B4)a t,ˆσ2a=0.0016, where r t=ln(x t+0.5)with x t being the earnings per share.2.(2points)Is there any significant serial correlation in the residuals of thefitted model?Why?Answer:No,the Q(m)statistics of the residuals give Q(12)=9.65with p-value0.65.3.(4points)Let T=65be the forecast origin,where T is the sample size.Based on thefitted model,and,for simplicity,use the relationship y t=exp(x t)−0.5,what are the 1-step and2-step ahead forecasts of earnings per share for the Microsoft stock?Answer:The1-step and2-step earnings forecasts are0.56and0.54,respectively.4.(2points)Test the null hypothesis H o:θ4=0vs H a:θ4=0.What is the test statistic?Draw your conclusion.Answer:The test statistic is t=0.39120.1442=2.71with two-sided p-value0.0067.Thus,the seasonal MA coefficientθ4is significantly different from zero.5.(3points)Consider the regular(i.e.,non-seasonal)part of the MA model.Is it invertible?Why?Answer:Yes,it is invertible,because the polynomial1−0.6953x+0.3889x2has roots0.89±1.33i so that the absolute value of the roots(Mod in R)is1.6,which is greaterthan1.[If you compute the roots of x2−0.6953x+0.3889,the the absolute value of the roots is less than1.]Problem D.(34pts)Consider the daily log returns of the Starbucks stock,in percentages, from January1993to December2007.The relevant R output is attached.Answer the following questions.41.(2points)Is the mean log return significant different from zero?Why?Answer:No,the basic statistics show the95%confidence interval of the mean is [−0.0103,0.1624],which contains zero.2.(2points)Is there any serial correlation in the log return series?Why?Answer:Yes,the Q(m)statistics show Q(15)=38.39with p-value0.0008.3.(2points)An MA model is used to handle the mean equation,which appears to beadequate.Is there any ARCH effect in the return series?Why?Answer:Yes,because the Q(m)statistics of the squared residuals show Q(15)=112.61 with p-value close to zero.4.(6points)A GARCH(1,1)model with Student-t distribution is used for the volatilityequation.Write down thefitted model,including the degrees of freedom of the Student-t innovations and mean equation.Answer:Thefitted model isr t=0.037+a t−0.043a t−1−0.048a t−2,a t=σt t, ∼t5.27.σ2 t =0.012+0.026a2t−1+0.973σ2t−1.5.(4points)Since the constant term of the GARCH(1,1)model is not significantly differentfrom zero at the1%level,an IGARCH(1,1)model is used.Write down thefitted IGARCH(1,1)model,including the mean equation.Answer:Thefitted IGARCH(1,1)model isr t=0.077+a t−0.029a t−1−0.044a t−2,a t=σt t, t∼N(0,1).σ2 t =0.022a2t−1+0.978σ2t−1.6.(3points)Is the IGARCH(1,1)model adequate?Why?What is the3-step aheadvolatility forecast with the last data point as the forecast origin?Answer:Yes,the Q(m)statistics for the standardized residuals give Q(10)=1.77, Q(15)=10.79,and Q(20)=18.66.The p-values of these statistics are all greater than0.05.In addition,the Q(m)statistics of the squared standardized residuals also havelarge p-values.The3-step ahead volatility forecast is √3.779=1.94.7.(5points)A GJR(or TGARCH)model with Student-t distribution is alsofitted to thelog return series.Write down thefitted model,including the mean equation and all parameters.Answer:Thwfitted GJR model isr t=0.032+a t−0.043a t−1−0.048a t−2,a t=σt t, t∼t5.31.σ2 t =0.015+(0.021+0.017N t−1)a2t−1+0.970σ2t−1,where N t−1=0if a t−1≥0and=1,otherwise.58.(2points)Is the fitted GJR (or TGARCH)model adequate?Why?Answer :Yes,the Q (m )statistics of the standardized residuals and those of the squred standardized residuals all have large p-values.9.(2points)Among the GARCH(1,1),IGARCH(1,1)and GJR(1,1)models,which one is preferred?Why?Answer :The GJR(1,1)model because it has the smallest AIC value.10.(2points)Is the leverage effect of the GJR model significant?Why?Answer :Yes,the t -ratio of the leverage parameter is 2.01,which is significant at the 5%level.11.(4points)To better understand the leverage effect,use the fitted GJR to calculate theratio σ2t (a t −1=−5.10)σ2t(a t −1=5.10),assuming σ2t −1=7.5.Answer :σ2t (a t −1=−5.10)σ2t (a t −1=5.10)=0.0154+0.0379×(−5.10)2+0.97×7.50.0154+0.0208×(5.10)2+0.97×7.5=1.057.6。

第10章 时间序列分析和预测一、单项选择题 1．已知某公司近5年经营收入的增长速度分别为6%，8.2%，9.3%，8%和10.5%，则该公司近5年的年平均增长速度为（ ）。

[浙江工商大学2017研]A ．（6%×8.2%×9.3%×8%×10.5%)/5B ．（106%×108.2%×109.3%×108%×110.5%)/5-1C ．（6%×8.2%×9.3%×8%×10.5%)1/5D ．（106%×108.2%×109.3%×108%×110.5%)1/5-1【答案】D【解析】平均增长速度也称平均增长率，它是时间序列中逐期环比值（也称环比发展速度）的几何平均数减1后的结果，其计算公式为：111n n YG Y -=⨯⨯-=-所以该商品价格的年平均增长率为：1v =-2．如果时间数列逐期增长量大体相等，则宜拟合（ ）。

[浙江工商大学2017研]A ．直线模型B．抛物线模型C．曲线模型D．众数指数曲线模型【答案】A【解析】A项，逐期增长量大体相等，说明关于时间t的曲线的斜率大体相等，应拟合直线模型；B项，抛物线模型适合于变化率逐渐减小再逐渐增大的时间序列；C项，指数曲线模型适合于呈指数增长的时间序列；D项，除直线模型意外的其他模型都属于曲线模型，包括抛物线模型和指数曲线模型。

3．定基发展速度和环比发展速度的关系是（）。

[浙江工商大学2017研]A．相邻两个定基发展速度之商＝其相应的环比发展速度B．相邻两个定基发展速度之积＝其相应的环比发展速度C．相邻两个定基发展速度之差＝其相应的环比发展速度D．相邻两个定基发展速度之和＝其相应的环比发展速度【答案】A【解析】定基发展速度是以固定一个时期为基点计算发展速度，环比增长速度是以上一个时期为基点计算发展速度，因此A项正确。

计量经济学试题时间序列分析与ARIMA模型计量经济学试题：时间序列分析与ARIMA模型1. 引言时间序列分析是计量经济学中重要的分析方法之一，能够揭示变量随时间变化的规律，并为未来趋势的预测提供依据。

ARIMA模型（差分自回归滑动平均模型）是时间序列分析中常用的模型之一，具有较强的建模和预测能力。

本文将介绍时间序列分析方法以及ARIMA模型的理论基础，并通过试题案例讲解其具体应用。

2. 时间序列分析方法概述时间序列是按时间顺序排列的一系列数据点，其特点是数据之间存在一定的时间关联性和趋势性。

时间序列分析方法可用于研究时间序列的规律，并对未来的变化进行预测。

常用的时间序列分析方法包括：平稳性检验、自相关函数（ACF）和偏自相关函数（PACF）的分析、白噪声检验、差分运算等。

3. ARIMA模型的基本原理ARIMA模型是一种广义的线性时间序列模型，它结合了自回归（AR）模型、差分（I）运算和滑动平均（MA）模型。

ARIMA模型的建立一般包括以下几个步骤：确定时间序列的平稳性、确定模型的阶数、拟合模型参数、模型检验与预测。

4. 时间序列分析与ARIMA模型的应用案例以某工业品生产量的时间序列数据为例，我们来演示时间序列分析与ARIMA模型的具体应用过程。

4.1 数据准备与描述性分析首先，我们收集了过去36个月的某工业品生产量数据，用于进行时间序列分析和ARIMA建模。

通过对数据的描述性统计分析，我们可以了解数据的分布特征、趋势以及季节性等信息。

4.2 平稳性检验为了应用ARIMA模型，首先需要检验时间序列的平稳性。

我们可以使用单位根检验（ADF检验）等方法判断时间序列是否平稳。

若时间序列不平稳，需要进行差分操作，直至得到平稳序列。

4.3 确定模型的阶数在ARIMA模型中，AR阶数表示自回归模型中的滞后阶数，MA阶数表示滑动平均模型中的滞后阶数。

通过观察自相关函数ACF和偏自相关函数PACF的图像，可以确定ARIMA模型的阶数。

Booth School of Business,University of ChicagoBusiness41202,Spring Quarter2012,Mr.Ruey S.TsaySolutions to MidtermProblem A:(34pts)Answer briefly the following questions.Each question has two points.1.Describe two improvements of the EGARCH model over the GARCHvolatility model.Answer:(1)allows for asymmetric response to past positive or negative returns,i.e.leverage effect,(2)uses log volatility to relax parameter constraint.2.Describe two methods that can be used to infer the existence of ARCHeffects in a return series,i.e.,volatility is not constant over time.Answer:(1)The sample ACF(or PACF)of the squared residuals of the mean equation,(2)use the Ljung-Box statistics on the squared residuals.3.Consider the IGARCH(1,1)volatility model:a t=σt t withσ2t =α0+β1σ2t−1+(1−β1)a2t−1.Often one pre-fixesα0=0.Why?Also,suppose thatα0=0and the1-step ahead volatility prediction at the forecast origin h is16.2%(annualized),i.e.,σh(1)=σh+1=16.2for the percentage log return.What is the10-step ahead volatility prediction?That is,what isσh(10)?Answer:(1)Fixingα0=0based on the prior knowledge that volatility is mean reverting.(2)σh(10)=16.2.4.(Questions4to8)Consider the daily log returns of Amazon stockfrom January3,2007to April27,2012.Some summary statistics of the returns are given in the attached R output.Is the expected(mean) return of the stock zero?Why?Answer:The data does not provide sufficient evidence to suggest that the mean return is not zero,because the95%confidence interval con-tains zero.5.Let k be the excess kurtosis.Test H0:k=0versus H a:k=0.Writedown the test statistic and draw the conclusion.1Answer:t-ratio =9.875√24/1340=73.79,which is highly significant com-pared with χ21distribution.6.Are there serial correlations in the log returns?Why?Answer:No,the Ljung-Box statistic Q (10)=10.69with p-value 0.38.7.Are there ARCH effects in the log return series?Why?Answer:Yes,the Ljung-Box statist of squared residuals gives Q (10)=39.24with p-value less than 0.05.8.Based on the summary statistics provided,what is the 22-step ahead point forecast of the log return at the forecast origin April 27,2012?Why?Answer:The point forecast r T (22)=0because the mean is not signif-icantly different from zero.[Give students 1point if they use sample mean.]9.Give two reasons that explain the existence of serial correlations in ob-served asset returns even if the true returns are not serially correlated.Answer:Any two of (1)bid-ask bounce,(2)nonsynchronous trading,(3)dynamic dependence of volaitlity via risk premuim.10.Give two reasons that may lead to using moving-average models inanalyzing asset returns.Answer:(1)Smoothing (or manipulation),(2)bid-ask bounce in high frequency returns.11.Describe two methods that can be used to compare different modelsfor a given time series.Answer:(1)Information criteria such as AIC or BIC,(2)backtesting or out-of-sample forecasting.12.(Questions 12to 14)Let r t be the daily log returns of Stock A.Assume that r t =0.004+a t ,where a t =σt t with t being iid N(0,1)random variates and σ2t =0.017+0.15a 2t −1.What is the unconditionalvariance of a t ?Answer:Var(a t )=0.0171−0.15=0.02.13.Suppose that the log price at t =100is 3.912.Also,at the forecastorigin t =100,we have a 100=−0.03and σ100=pute the21-step ahead forecast of the log price (not log return)and its volatility for Stock A at the forecast origin t =100.Answer:r 100(1)=0.004so that p 100(1)=3.912+0.004=3.916.Thevolatility forecast is σ2100(1)= 0.017+0.15(−0.03)2=pute the 30-step ahead forecast of the log price and its volatilityof Stock A at the forecast origin t =100.Answer:p 100(30)=3.912+0.004×30=4.032and the voaltility is the unconditional stantard error √0.02=0.141.15.Asset volatility has many applications in finance.Describe two suchapplications.Answer:Any two of (1)pricing derivative,(2)risk management,(3)asset allocation.16.Suppose the log return r t of Stock A follows the model r t =a t ,a t =σt t ,and σ2t =α0+α1a 2t −1+β1σ2t −1,where t are iid N(0,1).Under whatcondition that the kurtosis of r t is 3?That is,state the condition under which the GARCH dynamics fail to generate any additional kurtosis over that of t .Answer:α1=0.17.What is the main consequence in using a linear regression analysis whenthe serial correlations of the residuals are overlooked?Answer:The t -ratios of coefficient estimates are not reliable.Problem B .(30pts)Consider the daily log returns of Amazon stock from January 3,2007to April 27,2012.Several volatility models are fitted to the data and the relevant R output is attached.Answer the following questions.1.(2points)A volatility model,called m1in R,is entertained.Write down the fitted model,including the mean equation.Is the model adequate?Why?Answer:ARCH(1)model.r t =0.0018+a t ,a t =σt t with t being iidN(0,1)and σ2t =7.577×10−4+0.188a 2t −1.The model is inadequatebecause the normality assumption is clearly rejected.2.(3points)Another volatility model,called m2in R,is fitted to the returns.Write down the model,including all estimated parameters.3Answer:ARCH(1)model.r t=4.907×10−4+a t,a t=σt t,where t∼t∗3.56with t∗vdenoting standardized Student-t distribution with v degreesof freedom.The volatility equation isσ2t =7.463×10−4+0.203a2t−1.3.(2points)Based on thefitted model m2,test H0:ν=5versus H a:ν=5,whereνdenotes the degrees of freedom of Student-t distribution.Perform the test and draw a conclusion.Answer:t-ratio=3.562−50.366=−3.93,which compared with1.96is highlysignificant.If you compute the p-value,it is8.53×10−5.Therefore, v=5is rejected.4.(3points)A third model,called m3in R,is also entertained.Writedown the model,including the distributional parameters.Is the model adequate?Why?Answer:Another ARCH(1)model.r t=0.0012+a t,a t=σt t,where t are iid and follow a skew standardized Student-t distribution with skew parameter1.065and degrees of freedom3.591.The volatility equationisσ2t =7.418×10−4+0.208a2t−1.Ecept for the insigicant mean value,thefitted ARCH(1)model appears to be adequate based on the model checking statistics shown.5.(2points)Letξbe the skew parameter in model m3.Does the estimateofξconfirm that the distribution of the log returns is skewed?Why?Perform the test to support your answer.Answer:The t-ratio is1.065−10.039=1.67,which is smaller than1.96.Thus,the null hypothesis of symmetric innovations cannot be rejected at the 5%level.6.(3points)A fourth model,called m4in R,is alsofitted.Write downthefitted model,including the distribution of the innovations.Answer:a GARCH(1,1)model.r t=0.0017+a t,a t=σt t,where t are iid and follow a skew standardized Student-t distribution with skew parameter1.101and degrees of freedom3.71.The volatility equationisσ2t =1.066×10−5+0.0414a2t−1+0.950σ2t−1.7.(2points)Based on model m4,is the distribution of the log returnsskewed?Why?Perform a test to support your answer.Answer:The t-ratio is1.101−10.043=2.349,which is greater than1.96.Thus,the distribution is skew at the5%level.48.(2points)Among models m1,m2,m3,m4,which model is preferred?State the criterion used in your choice.Answer:Model4is preferred as it has a smaller AIC value.9.(2points)Since the estimatesˆα1+ˆβ1is very close to1,we consideran IGARCH(1,1)model.Write down thefitted IGARCH(1,1)model, called m5.Answer:r t=a t,a t=σt t,whereσ2t =3.859×10−5+0.85σ2t−1+0.15a2t−1.10.(2points)Use the IGARCH(1,1)model and the information providedto obtain1-step and2-step ahead predictions for the volatility of the log returns at the forecast origin t=1340.Answer:From the outputσ21340(1)=σ21341=3.859×10−5+0.85×(0.02108)2+0.15(.146)2=0.00361.Therefore,σ21340(2)=3.859×10−5+σ2 1340(1)=0.00365.The volatility forecasts are then0.0601and0.0604,respectively.11.(2points)A GARCH-M model is entertained for the percentage logreturns,called m6in the R output.Based on thefitted model,is the risk premium statistical significant?Why?Answer:The risk premium parameter is−0.112with t-ratio−0.560, which is less than1.96in modulus.Thus,the risk premium is not statistical significant at the5%level.12.(3points)Finally,a GJR-type model is entertained,called m7.Writedown thefitted model,including all parameters.Answer:This is an APARCH model.The model is r t=0.0014+a t,a t=σt t,where t are iid and follow a skew standardized Student-tdistribution with skew parameter1.098and degrees of freedom3.846.The volatility equation isσ2 t =7.583×10−6+0.0362(|a t−1|−0.478a t−1)2+0.953σ2t−1.13.(2points)Based on thefitted GJR-type of model,is the leverage effectsignificant?Why?Answer:Yes,the leverage parameterγ1is signfiicantly different from zero so that there is leverage effect in the log returns.5Problem C.(14pts)Consider the quarterly earnings per share of Abbott Laboratories(ABT)stock from1984.III to2011.III for110observations.We analyzed the logarithms of the earnings.That is,x t=ln(y t),where y t is the quarterly earnings per share.Two models are entertained.1.(3points)Write down the model m1in R,including residual variance.Answer:Let r t be the log earnings per share.Thefitted model is=0.00161.(1−B)(1−B4)r t=(1−0.565B)(1−0.183B4)a t,σ2a2.(2points)Is the model adequate?Why?Answer:No,the Ljung-Box statistics of the residuals give Q(12)=25.76with p-value0.012.3.(3points)Write down thefitted model m2in R,including residualvariance.Answer:Thefitted model is=0.00144.(1−B)(1−B4)r t=(1−0.470B−0.312B3)a t,σ2a4.(2points)Model checking of thefitted model m2is given in Figure1.Is the model adequate?Why?Answer:Yes,the model checking statistics look reasonable.5.(2points)Compare the twofitted model models.Which model ispreferred?Why?Answer:Model2is preferred.It passes model checking and has a smaller AIC value.6.(2points)Compute95%interval forecasts of1-step and2-step aheadlog-earnings at the forecast origin t=110.Answer:1-step ahead prediction:0.375±1.96×0.038,and2-step ahead prediction:0.0188±1.96×0.043.(Some students may use2-step ahead prediction due to the forecast origin confusion.)Problem D.(22pts)Consider the growth rate of the U.S.weekly regular gasoline price from January06,1997to September27,2010.Here growth rate is obtained by differencing the log gasoline price and denoted by gt in R output.The growth rate of weekly crude oil from January03,1997to September24,2010is also obtained and is denoted by pt in R output.Note that the crude oil price was known3days prior to the gasoline price.61.(2points)First,a pure time series model is entertained for the gasolineseries.An AR(5)model is selected.Why?Also,is the mean of the gtseries significantly different from zero?Why?Answer:An AR(5)is selected via the AIC criterion.The mean of g tis not significantly different from zero based on the one-sample t-test.The p-value is0.19.2.(2points)Write down thefitted AR(5)model,called m2,includingresidual variance.Answer:Thefitted model is=0.000326.(1−0.507B−0.079B2−0.136B3+0.036B4+0.086B5)g t=a t,σ2a3.(2points)Since not all estimates of model m2are statistically signifi-cant,we refine the model.Write down the refined model,called m3.Answer:Thefitted model is=0.000327.(1−0.504B−0.074B2−0.122B3+0.101B5)g t=a t,σ2a4.(2points)Is the refined AR(5)model adequate?Why?Answer:Yes,the Ljung-Box statistics of the residuals give Q(14)=10.27with p-value0.74,indicating that there are no serial correlationsin the residuals.5.(2points)Does the gasoline price show certain business-cycle behavior?Why?Answer:Yes,thefitted AR(5)polynomial contains compplex solutions.6.(3points)Next,consider using the information of crude oil price.Writedown the linear regression model,called m4,including R2and residualstandard error.Answer:Thefitted linear regression model isg t=0.287p t+ t,σ =0.0184,and the R2of the regression is33.66%.7.(2points)Is thefitted linear regression model adequate?Why?Answer:No,because the residuals t are serially correlated based onthe Ljung-Box test.78.(3points)A linear regression model with time series errors is enter-tained and insignificant parameters removed.Write down thefinalmodel,including allfitted parameters.Answer:The model is(1−0.404B−0.164B2−0.096B3+0.101B5)(g t−0.191p t)=a t,σ2=0.000253.a9.(2points)Model checking shows that thefittedfinal model has noresidual serial correlations.Based on the model,is crude oil pricehelpful in predicting the gasoline price?Why?Answer:Yes,because thefitted coefficient of p t is signficantly differentfrom zero.10.(2points)Compare the pure time series model and the regression modelwith time-series errors.Which model is preferred?Why?Answer:The regression model with time series error is preferred as ithas a smaller AIC criterion.8。

成都信息工程学院考试试卷2012——2013学年第2学期课程名称：《金融时间序列分析》 班级：金保111本01、02、03班一、判断题（每题1分，正确的在括号内打√，错误的在括号内打×，共15分）1．模型检验即是平稳性检验（ ）。

2．模型方程的检验实质就是残差序列检验（ ）。

3．矩法估计需要知道总体的分布（ ）。

4．ADF 检验中：原假设序列是非平稳的（ ）。

5．最优模型确定准则：AIC 值越小、SC 值越大，说明模型越优（ ）。

6．对具有曲线增长趋势的序列，一阶差分可剔除曲线趋势（ ）。

7．严平稳序列与宽平稳时序区分主要表现在定义角度不同（ ）。

8．某时序具有指数曲线增长趋势时，需做对数变换，才能剔除曲线趋势（ ）。

9．时间序列平稳性判断方法中 ADF 检验优于序时图法和自相关图检验法（ ）。

10．时间序列的随机性分析即是长期趋势分析（ ）。

11．ARMA （p,q ）模型是ARIMA(p,d,q)模型的特例（ ）。

12．若某序列的均值和方差随时间的平移而变化，则该序列是非平稳的（ ）。

13. MA(2)模型的3阶偏自相关系数等于0（ ）。

14．ARMA(p,q)模型自相关系数p 阶截尾，偏自相关系数拖尾（ ）。

15．MA(q)模型平稳的充分必要条件是关于后移算子B 的q 阶移动自回归系数多项式根的绝对值均在单位圆内（ ）。

二、填空题。

（每空2分，共20分）1．t X 满足ARMA （1,2）模型即：t X ＝0.43+0.341-t X +t ε＋0.81-t ε–0.22-t ε，则均值＝ ，1θ（即一阶移动均值项系数）＝ 。

2．设｛x t ｝为一时间序列，B 为延迟算子，则B 2X t = 。

3．在序列y 的view 数据窗，选择 功能键，可对序列y 做ADF 检验。

4．若某平稳时序的自相关图拖尾，偏相关图1阶截尾，则该拟合 模型。

5. 已知AR （1）模型：t X +0.81-t X ＝t ε，t ε服从N(0,0.36)，则一阶自相关系数＝ ，方差＝ 。

Graduate School of Business,University of ChicagoBusiness41202,Spring Quarter2005,Mr.Ruey S.TsaySolutions to MidtermAll tests are based on the5%significance level.Problem A:(30pts)Answer briefly the following questions.1.For problems1to6,consider the daily log return,in percentages,of the S&P500composite index from January1996to December31,2004for2267data points.Sum-mary statistics of the percentage log returns from SCA and Splus are attached.See Page 1of the attached output.Is the mean of percentage log returns significantly different from zero?Why?A:t-ratio=1.187<1.96.Thus,the mean return is not significantly different from zero.2.Suppose one invested$1dollar on the S&P500index at the very beginning of1996andheld the investment until the end of2004.What was the value of the investment at the market closing on December31,2004?A:total log return=0.0299*2267/100=0.67783so that the value=exp(0.67783)≈1.97.3.Test the null hypothesis that the skewness of the log returns is zero.Draw your conclu-sion.A:t-ratio=−0.0939/0.0514=−1.83<1.96so that the skewness is no signifiacnt different from zero.4.Given that the last percentage log return was−0.134(i.e.T=December31,2004),which is the corresponding simple return?A:R t=exp(−0.134/100)−1=−0.001339=.1339%.5.Are the log return serially correlated?You may use Q(10)of the series to answer thequestion.A:Q(10)=13.9with p-value0.178so that there is no serial correlation.6.Is there any ARCH effect in the log return series?You may use Q(12)of the squaredseries to answer the question.A:Yes,Q(12)of squared return is large at490(in SCA)and has a p-value of0.0from Splus.7.Give two empirical features of daily log returns of afinancial asset.A:Any two of(a)high kurtosis,(b)volatility clustering,(3)skew to left.18.What is the purpose of studying kurtosis of an asset return series?A:Understanding the tail behavior(or risk)of the return.9.Describe two applications of studying sample autocorrelation function(ACF)of an assetreturn series.A:(a)To test serial correlations in the return series,(b)to identify MA order.10.Describe two methods that can be used to identify the order of an AR model.A:(a)PACF,(b)Criterion functions such as AIC or BIC.11.Consider the AR(1)model(1−0.9B)r t=0.2+a t,where{a t}is an independent andidentically distributed random variables with mean zero and variance1.0.What is the half-life of the series?A:Half-life=ln(0.5)/ln(0.9)=6.58time units.12.Suppose that the log return r t of an asset follows the model below:r t=0.02+a t,a t=σt t,σ2t=0.116+0.42a2t−1.Let p t be the log price of the asset at time t and p T( )be the -step ahead forecast of the log price at the forecast origin T.Then,what is the value of p T( )as increases?A:0.02is the slope of time trend so that p T( )→∞as increases.13.For problems13-14,consider quarterly series of U.S.unit labor cost from1947to2004.The data were seasonally adjusted and obtained from the Federal Reserve Bank of St Louis.Let x t=(1−B)ULC t be thefirst-differenced series of the data at time t.The model(1−0.371B2)x t=0.265+(1+0.171B4)a t,σa=0.5998,fits the data reasonably well.Under thefitted model,what is the mean of x t,i.e.E(x t) =?A:E(x t)=0.265/(1−0.371)=0.421.14.Does the model imply that there exist business cycles in the unit labor cost?Why? A:No,because the two roots are real.From1−0.371x2,we have x=±1/ (0.371).15.Give two weaknesses of the GARCH-type of models for modeling asset volatility.A:Any two of(a)symmetric response to past positive and negative returns,(b)restric-tive,(c)providing no explanation of volatility clustering,(d)no adaptive in forecasting.2Problem B.(20pts)This problem is concerned with the analysis of quarterly earnings per share of the Procter&Gamble(PG)Company from1992to thefirst quarter of2003 for45data points.The data were obtained from First Call.Log transformation was taken to stabilize the variability of the puter output is attached;p.2-6of output. Due to strong seasonal pattern,which results in models that are close to being non-invertible, we analyze the seasonally differenced series in Splus.Let x t be the logarithm of quarterly earnings per share and w t=(1−B4)x t.Thus,SCA uses x t and Splus uses w t.1.(5points)Because ACF of the log earnings shows strong seasonal pattern,the seasonaldifference(1−B4)is taken.The ACF of the seasonally differenced data indicates no further differencing is necessary.Write down thefitted model for the x t series(not the differenced w t).A:(1−0.47B)(1−B4)x t=0.0508+(1−0.307B4)a t,σa=0.0502.2.(4points)Is the AR coefficient of thefitted model statistically significant?Why?A:Yes,the t-ratio is3.26,which is greater than the critical value1.96.3.(4points)Is there any serial correlation in the residuals of thefitted model?Use Q(12)of the ACF of residuals to answer the question.[Hint:for a chi-square distribution with m degrees of freedom,the expected value is m.]A:Q(12)=8.8which is less than E(χ210)=10so that p-value>0.05.4.(4points)Let T=45be the forecast origin.What are the1-step and2-step aheadforecasts of thefitted model(after taking anti-log transformation)?A:x45(1)=0.912and x45(2)=3.95(from SCA).For Splus,x45(1)=exp(0.0087−.1743)=0.847,x45(2)=exp(−.012479+1.278152)=3.55.5.(3points)Give an interpretation of the estimated constant0.0508of thefitted modelfor x t.A;Slope of time trend.3and the S&P500index from January1999to December2004with sample size T=1508.We employ the market model:r t=α+βr m,t+e t,where r t and r m,t are Wal-Mart stock return and S&P500index return,respectively.Splus output is attached;page6of output.Answer the following questions.1.(4points)Write down thefitted market model.A:r t=0.0205+0.9606r m,t+e t.2.(4points)The Ljung-Box statistics of the ACF of residuals show some minor serialcorrelations,but the ACFs are relatively small so we ignore the serial correlations and perform the ARCH effect test.Is there an ARCH effect in the residuals of thefitted market model?A:Yes,archTest gives a p-value about0.0.3.(4points)We employ a GARCH(1,1)model(called“m2”in the output).Write downthefitted ment on thefitted model.A:Let r t and r m,t be the Wal-Mart stock return and S&P500index return,respectively.The model isr t=0.9386r m,t+a t=0.9386+σt t, t∼t6.41σ2t=−4.29×10−5+0.0377a2t−1+0.963σ2t−1.The negative estimate ofα0does not make sense,but it is statistically not different from0.Also,theˆα1+ˆβ1≈1so that thefitted model indicates an IGARCH(1,1)model without the constant.4.(6points)Further analysis indicates that an EGARCH(2,1)modelfits the data better.There are two leverage effect parameters.Are these two effects statistically significant?Why?You may test the effect individually.A:Examine the t-ratio of the two leverage parameter estimates.For Lev(1),the t-atio is−2.129.For Lev(2),the t-ratio is−1.847.In both cases,the p-values are less tha0.05so that they are both significant.[It you use two-sided tests,then Lev(2)is notsignificant.]5.(2points)What are the1-step ahead forecasts for the return and its volatility of Wal-Mart stock at the forecast origin T=1508using the EGARCH model.A:zero for return and0.7082for volatility.4from January1995to December2004with2519observations.Splus output is attached;page 8of output.Answer the following questions.The ACFs of the returns are small so that the mean equation consists of a constant term only.1.(5points)Consider thefitted GARCH(1,1)model.The volatility equation isσ2t=0.042+0.049a2t−1+0.944σ2t−1.Letηt=a2t−σ2t.Rewrite the prior volatility equation in an ARMA form for the{a2t} series.A:a2t=0.042+0.993a2t−1+ηt−0.944ηt−1.2.(5points)Write down thefitted EGARCH(1,1)model with leverage effect(both meanand volatility equations and the parameter of the conditional distribution used).A:The model isr t=0.04918+a t=0.04918+σt t, t∼t6.68ln(σ2t)=−0.06316+0.1033(|a t−1|−0.5464a t−1σt−1)+0.989ln(σ2t−1).3.(4points)Test the hypotheses H o:v=5vs H a:v=5,where v is the degrees offreedom of the conditional Student t distribution.Draw your conclusion.[Hint:you may use the usual t-ratio test.]A;t-ratio=6.684−5=2.62>1.96.Thus,reject H o:v=5.4.(5points)To better understand the leverage effect,use thefitted EGARCH(1,1)modelto calculate the ratioσ2t( =−3)2t ,where t denotes the iid sequence of the innovationsdefined in class.A:From thefitetd volatility equation,we haveσ2t=exp(−0.06316)(σ2t−1).989exp(0.1033| t−1|−0.0564 t−1). Therefore,σ2t( =−3)σ2t( =3)=exp(−0.0564(−3))exp(−0.0564(3))=exp(0.0564×6)=1.403.5.(4points)Used thefitted EGARCH model and T=2519as the forecast origin.Whatare the1-step ahead forecasts of log return and volatility?A:Forecast of log return is0.0492and forecast of volatility is1.184.6.(4points)Write down the mean equation of thefitted GARCH-M model for the data.A:r t=0.058+0.00629σ2t+a t.7.(3points)Based on the GARCH-M model,is the risk premium statistically significant?Why?A:No,the t-ratio is0.453with p-value=0.65(two-sided).5。

第13章时间序列分析和预测一、单项选择题1．五月份的商品销售额为60万元，该月的季节指数为120%，则消除季节因素影响后，该月的商品销售额为（）万元。

[对外经济贸易大学2015研]A．72B．50C．60D．51.2【答案】B【解析】消除季节因素影响后的商品销售额＝该月商品实际销售额/该月季节指数＝60/120%＝50（万元）。

2．毛衣销售量时间数列分析中，如果第3季的季节指数大于100%，表明该季毛衣销售量（）。

[四川大学2013研]A．不受季节影响B．受季节因素影响C．属于旺季D．属于淡季【答案】C【解析】季节指数＝同季的平均数/历年各季总的平均数。

故若季节指数大于100%，表示该季度的销售量超过平均水平，故为销售旺季。

3．如果时间序列的逐期观察值按几何级数递增或递减，则适合的预测模型是（）。

[四川大学2013研]A．移动平均模型B．线性模型C．指数模型D．抛物线模型【答案】C【解析】时间序列的观察值按几何级数变化，说明变化幅度很大，并非线性变化情况，适合用指数模型进行拟合。

4．时间数列分析中，移动平均法只能用于修匀的数列是（）。

[四川大学2013研] A．时期数数列B．时点数数列C．空间数列D．静态数列【答案】A【解析】移动平均法适用于近期预测。

当产品需求既不快速增长也不快速下降，且不存在季节性因素时，移动平均法能有效地消除预测中的随机波动，因此可以用于修匀时期数列。

5．不存在趋势的序列称为（）。

A．平稳序列B．周期性序列C．季节性序列D．非平稳序列【答案】A【解析】时间序列可以分为平稳序列和非平稳序列两大类。

其中平稳序列是指基本上不存在趋势的序列；非平稳序列是指包含趋势、季节性或周期性的序列，它可能只含有其中一种成分，也可能是几种成分的组合。

6．时间序列在长时期内呈现出来的某种持续向上或持续下降的变动称为（）。

A．趋势B．季节性C．周期性D．随机性【答案】A【解析】趋势是指时间序列在长期内呈现出来的某种持续上升或持续下降的变动，也称长期趋势；时间序列中的趋势可以是线性的，也可以是非线性的。

2022年初级统计基础知识章节试题及答案之第五章时间序列分析含答案2022年初级统计基础学问章节试题及答案之第五章时光序列分析含答案第五章时光序列分析一、单项挑选题1.构成时光数列的两个基本要素是(C) (2022年1月)A.主词和宾词B.变量和次数C.现象所属的时光及其统计指标数值D.时光和次数2.某地区历年诞生人口数是一个(B) (2022年10月)A.时期数列B.时点数列C.分配数列D.平均数数列3.某商场销售洗衣机，2022年共销售6000台，年底库存50台，这两个指标是( C ) (2022年10)A.时期指标B.时点指标C.前者是时期指标，后者是时点指标D.前者是时点指标，后者是时期指标4.累计增长量( A ) (2022年10)A.等于逐期增长量之和B.等于逐期增长量之积C.等于逐期增长量之差D.与逐期增长量没有关系5.某企业银行存款余额4月初为80万元，5月初为150万元，6月初为210万元，7月初为160万元，则该企业其次季度的平均存款余额为( C )(2022年10)A.140万元B.150万元C.160万元D.170万元6.下列指标中属于时点指标的是( A ) (2022年10)A.商品库存量B.商品销售量C.平均每人销售额D.商品销售额7.时光数列中，各项指标数值可以相加的是( A ) (2022年10)A.时期数列B.相对数时光数列C.平均数时光数列D.时点数列8.时期数列中各项指标数值( A )(2022年1月)A.可以相加B.不行以相加C.绝大部分可以相加D.绝大部分不行以相加10.某校同学人数2022年比2022年增长了8%，2022年比2022年增长了15%，2022年比2022年增长了18%，则2022-2022年同学人数共增长了( D )(2022年10月)A.8%+15%+18%B.8%×15%×18%C.(108%+115%+118%)-1D.108%×115%×118%-1二、多项挑选题1.将不同时期的进展水平加以平均而得到的平均数称为(ABD) (2022年1月)A.序时平均数B.动态平均数C.静态平均数D.平均进展水平E.普通平均数2.定基进展速度和环比进展速度的关系是(BD) (2022年10月)A.相邻两个环比进展速度之商等于相应的定基进展速度B.环比进展速度的连乘积等于定基进展速度C.定基进展速度的连乘积等于环比进展速度D.相邻两个定基进展速度之商等于相应的环比进展速度E.以上都对3.常用的测定与分析长久趋势的办法有( ABC ) (2022年1月)A.时距扩大法B.移动平均法C.最小平办法D.几何平均法E.首末折半法4.时点数列的特点有( BCD ) (2022年10)A.数列中各个指标数值可以相加B.数列中各个指标数值不具有可加性C.指标数值是通过一次记下取得的D.指标数值的大小与时期长短没有直接的联系E.指标数值是通过延续不断的记下取得的5.增长1%的肯定值等于( AC )(2022年1)A.增强一个百分点所增强的肯定量B.增强一个百分点所增强的相对量C.前期水平除以100D.后期水平乘以1%E.环比增长量除以100再除以环比进展速度6.计算平均进展速度常用的办法有( AC )(2022年10)A.几何平均法(水平法)B.调和平均法C.方程式法(累计法)D.容易算术平均法E.加权算术平均法7.增长速度( ADE )(2022年1月)A.等于增长量与基期水平之比B.逐期增长量与报告期水平之比C.累计增长量与前一期水平之比D.等于进展速度-1E.包括环比增长速度和定基增长速度8.序时平均数是( CE )(2022年10月)A.反映总体各单位标志值的普通水平B.按照同一时期标志总量和单位总量计算C.说明某一现象的数值在不同时光上的普通水平D.由变量数列计算E.由动态数列计算三、推断题1.职工人数、产量、产值、商品库存额、工资总额指标都属于时点指标。