时间序列分析试题

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时间序列分析试题

2．设时间序列{X t}满足 ARMA(2,1)
(1 − B + 0.5B2 ) Xt = (1 + 0.4B)εt ，
（1）试分析序列{X t}的平稳性，（2）计算前 3 个 Green 函数 G0 、 G1 、 G2 。
（1）此时特征方程为： λ2 − λ + 0.5 = 0 ，特征根满足| λ1,2 |= 2 2 < 1，序列{Xt}平稳。
Xˆ t (1) = EX t +1 = E(0.6 X t + εt − 0.5εt−1) = 0.6 X t + εt − 0.5εt−1 ，
Xˆ t (2) = EX t+2 = E(0.6 X t +1 + εt+1 − 0.5εt )
= 0.6EX t+1 − 0.5εt = 0.36X t + 0.1εt − 0.3εt−1 ；
界平稳。
（4） X t − 1.1X t −1 = εt ；
此时的特征方程为 λ2 − 1.1λ = 0 ，解得 λ1 = 1.1， λ2 = 0 ；模型序列不平稳。（5） (1 − B)2 X t = εt ；
此时的特征方程为 (λ − 1)2 = 0 ，解得 λ1 = λ2 = 1 ；模型序列不平稳，但是临界平稳。
− θ1εt −2 )
=
−θ1σ
2 ε
；
ρ (1)
=
γ (1) γ (0)
=
− θ1 1 + θ12
，即
ρ (1)θ12
+ θ1
+
ρ (1)
=
0 ，根据可逆性要求，解得θ1
=
0.70

金融时间序列分析英文试题(芝加哥大学) (1)

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。

时间序列分析试题(卷)与答案解析

时间序列分析试卷1一、填空题（每小题2分，共计20分）1. ARMA(p, q)模型_________________________________，其中模型参数为____________________。

2. 设时间序列{}t X ，则其一阶差分为_________________________。

3. 设ARMA (2, 1)：1210.50.40.3t t t t t X X X εε---=++-则所对应的特征方程为_______________________。

4. 对于一阶自回归模型AR(1): 110t t t X X φε-=+＋，其特征根为_________，平稳域是_______________________。

5. 设ARMA(2, 1):1210.50.1t t t t t X X aX εε---=++-，当a 满足_________时，模型平稳。

6. 对于一阶自回归模型MA(1):10.3t t t X εε-=-，其自相关函数为______________________。

7. 对于二阶自回归模型AR(2):120.50.2t t t t X X X ε--=++则模型所满足的Yule-Walker 方程是______________________。

8. 设时间序列{}t X 为来自ARMA(p,q)模型：1111t t p t p t t q t q X X X φφεθεθε----=++++++则预测方差为___________________。

9. 对于时间序列{}t X ，如果___________________，则()~t X I d 。

10. 设时间序列{}t X 为来自GARCH(p ，q)模型，则其模型结构可写为_____________。

二、（10分）设时间序列{}t X 来自()2,1ARMA 过程，满足()()210.510.4ttB B X B ε-+=+,其中{}t ε是白噪声序列，并且()()2t t 0,E Var εεσ==。

时间序列考试试题

一，名词解释时间序列是指将某种现象某一个统计指标在不同时间上的各个数值，按时间先后顺序排列而形成的序列随机过程是一连串随机事件动态关系的定量描述平稳性决定过程特性的统计规律不随时间的变化而改变严平稳对于一切的时滞k 和时点t1，t2，。

，tn，都有Yt1，Yt2，_ _ _，Ytn 与Yt1-k，Yt2-k，。

，Ytn-k 的联合分布相同弱平稳：–均值函数在所有时间上恒为常数–对所有的时间t 和时滞k，rt,t-k=r0,k白噪声独立同分布的随机序列，属于严平稳随机趋势：在任何时间点都有零均值，方差随时间的增加而增加确定性趋势：存在周期性或季节性的趋势LS（least-square ）：最小二乘估计BLUE（best linear unbiased estimator）：最佳线性无偏估计GLS（generalized least square）：广义最小二乘QQ图（Quantile-Quantile plot）：正态得分图，显示数据的分位数和根据正态分布计算的理论分位数。

正态分布的QQ图看起来近似于一条直线。

MA：滑动平均过程AR：自回归过程ARMA：自回归滑动平均混合模型非平稳时间序列：具有时变均值的时间序列自回归滑动平均求和模型ARIMA：一个时间序列的d次差分是一个平稳的ARMA过程ACF（autocorrelation function）:自相关函数PACF（partial。

）：偏自相关函数，即预测误差之间的相关系数EACF（Extended）：扩展的自相关函数ADF单位根检验：用最小二乘回归所得估计系数的t统计量作为检验统计量，在有单位根的零假设下，该检验统计量服从某种非标准的大样本分布。

AIC（赤池信息准则）：是估计模型与真实模型的平均Kullback-Leibler偏离的估计量，定义为：AIC=-2log（极大似然估计）+2k。

AIC是有偏估计量，当参数数量相对数据容量的比值较大时，偏差很大。

统计基础知识与统计实务：时间序列试题预测(题库版)

统计基础知识与统计实务：时间序列试题预测（题库版）1、判断题发展水平是计算其他动态分析指标的基础，它只能用总量指标来表示。

（）正确答案：错2、多选时期数列的特点有（）。

A.数列中的各个指标值不能相加B.数列中的（江南博哥）各个指标值可以相加C.每个指标值的大小与时间长短无关D.每个指标值的大小与时期长短有关E.每个指标值通过连续不断登记取得正确答案：B, D, E参考解析：时期数列的特点主要有：①数列中每个指标数值可以相加，其和表示现象在更长时期内的发展总量；②数列中每个指标数值的大小与其时期长短有直接联系。

一般地，时期愈长，指标数值就愈大，反之就愈小；③数列中的每个指标数值，通常是通过连续不断地登记而取得的。

AC两项属于时点数列的特点。

3、单选?2007年到2013年某国社会经济发展基本资料如下，利用以上所给资料，完成下列题目：计算上述平均发展速度时使用的方法是（）。

A.水平法B.累计法C.叠加法D.换算法正确答案：A4、单选设（甲）代表时期数列；（乙）代表时点数列；（丙）代表加权算术平均数；（丁）代表“首尾折半法”序时平均数。

现已知2005～2009年某银行的年末存款余额，要求计算各年平均存款余额，则该数列属于____，应采用的计算方法是____。

（）A.甲；丙B.乙；丙C.甲；乙D.乙；丁正确答案：D参考解析：当时间数列中所包含的总量指标都是反映社会经济现象在某一瞬间上所达到的水平时，这种总量指标时间数列就称为时点数列。

2005～2009年某银行的年末存款余额是时点数列，可根据间隔时间相等的间断时点序列的计算方法计算各年平均存款余额，此方法为首尾折半法（简单序时平均法）。

5、单选设（甲）代表时期序列；（乙）代表时点序列；（丙）代表加权算术平均数；（丁）代表"首末折半法"序时平均数。

现已知2010~2014年某银行的年末存款余额，要求计算各年平均存款余额，则该序列属于__________，应采用的计算方法是_________。

时间序列分析试题

第九章时间序列分析一、单项选择题1、乘法模型是分析时间序列最常用的理论模型。

这种模型将时间序列按构成分解为（）等四种成分，各种成分之间( )，要测定某种成分的变动，只须从原时间序列中（）。

A. 长期趋势、季节变动、循环波动和不规则波动；保持着相互依存的关系；减去其他影响成分的变动B. 长期趋势、季节变动、循环波动和不规则波动；缺少相互作用的影响力量；减去其他影响成分的变动C. 长期趋势、季节变动、循环波动和不规则波动；保持着相互依存的关系；除去其他影响成分的变动D.长期趋势、季节变动、循环波动和不规则波动；缺少相互作用的影响力量；除去其他影响成分的变动答案：C2、加法模型是分析时间序列的一种理论模型。

这种模型将时间序列按构成分解为（）等四种成分，各种成分之间( )，要测定某种成分的变动，只须从原时间序列中（）。

A. 长期趋势、季节变动、循环波动和不规则波动；保持着相互依存的关系；减去其他影响成分的变动B. 长期趋势、季节变动、循环波动和不规则波动；缺少相互作用的影响力量；减去其他影响成分的变动C. 长期趋势、季节变动、循环波动和不规则波动；保持着相互依存的关系；除去其他影响成分的变动D.. 长期趋势、季节变动、循环波动和不规则波动；缺少相互作用的影响力量；除去其他影响成分的变动答案：B3、利用最小二乘法求解趋势方程最基本的数学要求是（）。

A.∑=-任意值2)ˆ(t Y Y B. ∑=-min )ˆ(2t Y Y C. ∑=-max )ˆ(2t Y Y D. 0)ˆ(2∑=-t Y Y 答案：B4、从下列趋势方程t Y t86.0125ˆ-=可以得出（）。

A. 时间每增加一个单位，Y 增加0.86个单位B. 时间每增加一个单位，Y 减少0.86个单位C. 时间每增加一个单位，Y 平均增加0.86个单位D. 时间每增加一个单位，Y 平均减少0.86个单位答案：D.5、时间序列中的发展水平（）。

河南大学统计学专业大二2019-2020时间序列分析测试题

河南大学统计学专业大二2019-2020时间序列分析测试题一、单选题1. 我国国内生产总值2018年为900309亿元，2019年为990865亿元，则国内生产总值2019年环比发展速度为多少? [单选题] *A、10.05%B、90.86%C、110.06(正确答案)D、9.14%答案解析：正确答案:C本题考核环比发展速度=报告期水平/基期水平=990865/900309*100%=110.06%2. 以下关于统计的说法中，错误的是（）。

[单选题] *A、统计学是关于收集、整理、分析数据和从数据中得出结论的科学B、描述统计和推断统计的作用是能分开发挥统计作用的(正确答案)C、参数估计是利用样本信息推断总体特征D、描述统计的内容包括如何用图表或数学方法对数据进行整理和展示答案解析：本题考查统计学。

描述统计和推断统计可以一起发挥作用。

3. 下列统计处理中，属于推断统计的是（）。

[单选题] *A、利用抽样调查数据估计城镇居民人均消费支出水平(正确答案)B、利用统计图表展示GDP的变化C、利用增长率描述人均可支配收入的基本走势D、利用统计表描述公司员工年龄分布答案解析：本题考查推断统计。

推断统计是研究如何利用样本数据来推断总体特征的统计学方法，其内容包括参数估计和假设检验两大类。

参数估计是利用样本信息推断总体特征；假设检验是利用样本信息判断对总体的假设是否成立。

选项A属于参数估计。

4. 某厂连续性生产电脑，为检验产品的质量，按每隔2小时取下20分钟生产的电脑，并做全部产品检验，这是属于（）。

[单选题] *A、简单随机抽样B、等距抽样(正确答案)C、分层抽样D、整群抽样、答案解析：本题考查等距抽样。

最简单的系统抽样是等距抽样，即将总体N个单位按直线排列，根据样本量n确定抽取间隔，即抽样=N/n≈k，k为最接近N/n的一个整数。

在1～k范围内随机抽取一个整数i，令位于i位置上的单位为起始单位，往后每间隔K抽取一个单位，直至抽满n。

金融时间序列分析英文试题(芝加哥大学) (3)

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。

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第九章时间序列分析一、单项选择题1、乘法模型是分析时间序列最常用的理论模型。

这种模型将时间序列按构成分解为（）等四种成分，各种成分之间( )，要测定某种成分的变动，只须从原时间序列中（）。

A.∑=-任意值2)ˆ(t Y Y B. ∑=-min )ˆ(2t Y Y C. ∑=-max )ˆ(2t Y Y D. 0)ˆ(2∑=-t Y Y 答案：B4、从下列趋势方程t Y t86.0125ˆ-=可以得出（）。

A. 只能是绝对数B. 只能是相对数C.只能是平均数D.上述三种指标均可以答案：D.6、下列时间序列中，属于时点序列的有（）。

A. 某高校“十五”期间毕业生人数B.某企业“十五”期间年末利税额C.某地区“十五”期间年末人口数D.某地区“十五”期间粮食产量答案：C7、下列时间序列中，属于时期序列的有（）。

A. 某农场“十五”期间年末奶牛存栏数B. 某企业“十五”期间年末利税额C. 某地区“十五”期间年末人口数D.某企业“十五”年末产品库存量答案：B8、对时间序列进行动态分析的基础指标是（）。

A. 发展水B. 平均发展水平C.发展速度D. 平均发展速度答案：A则该企业2005年上半年平均每个月的应收账款余额为（）。

A. 6210209688909159308502690++++++ B.71020968890915930850690++++++ C. 6968890915930850690+++++ D. 61020968890915930850+++++答案：A10、采用几何平均法计算平均发展速度时，侧重于考察（）。

A.现象的全期水平，它要求实际各期水平等于各期计算水平B.现象全期水平的总和，它要求实际各期水平之和等于各期计算水平之和C.现象全期水平的总和，它要求实际各期定基发展速度之和等于各期理论定基发展速度之和D.现象的末期水平，它要求实际末期水平等于末期计算水平答案：D.11、如果一个时间序列连续3期的环比增长速度为18%、20%和25%，则其总速度为（）。

A. 18%⨯20%⨯25%=0.9%B. 118%⨯120%⨯125%=177%C.3%25%20%18⨯⨯=20.8%D. 3%125%120%118⨯⨯=120.96%答案：B12、如果时间序列的各逐期增长量相等，则（）。

A. 各期的环比增长速度保持不变B.环比增长速度逐期上升C.环比增长速度逐期下降D.各期环比增长速度有升有降答案：C13、当时间序列的环比增长速度大体相同时，适宜拟合（）。

A. 指数曲线B. 抛物线C.直线D. 对数曲线答案：A14、元宵的销售一般在“元宵节”前后达到旺季，1月份、2月份的季节指数将（）。

A. 小于100%B. 大于100%C.等于100%D. 大于1200%答案：B15、空调的销售量一般在夏季前后最多，其主要原因是空调的供求（），可以通过计算（）来测定夏季期间空调的销售量高出平时的幅度。

A.受气候变化的影响；循环指数B.受经济政策调整的影响；循环指数C.受自然界季节变化的影响；季节指数D.受消费心理的影响；季节指数答案：C16、利用剩余法所求得的循环波动指数C%（）。

A. 包含长期趋势的影响B.包含季节变动的影响C. 消除了长期趋势、季节变动的影响，但受不规则变动的影响D.. 消除了长期趋势、季节变动和不规则变动的影响，反映循环波动的影响答案：D.17、当时间序列的二级增长量大体相同时，适宜拟合（）。

A. 抛物线B. 指数曲线C.直线D. 对数曲线答案：A18、国家统计局2005年2月28日公告，经初步核算，2004年我国的国内生产总值按可比价格计算比上年增长9.5%。

这个指标是一个（）。

A.环比发展速度B. 环比增长速度C. 定基发展速度D. 定基增长速度答案：B19、爱民制药厂2000-2004年的利润（万元）为：922，1086，1655，1982和2389。

则该企业2003年和2004年利润增长1%的绝对值为（）。

A.1982万元和2389万元B.19.82万元和23.89万元C.1655和万元1982万元D.16.55万元和19.82万元答案：D.20、移动平均法是测定（）的一种较为简单的方法。

A.长期趋势B. 循环变动C. 季节变动D. 不规则变动答案：A21、欲测定季节变动，根据时间序列乘法模型的原理需要从时间序列中（）。

A. 减去长期趋势和循环变动B.减去长期趋势、循环变动和不规则变动C.除去长期趋势和循环变动D.除去长期趋势、循环变动和不规则变动答案：D.22、欲测定循环变动，根据时间序列乘法模型的原理需要从时间序列中（）。

A. 减去长期趋势和季节变动B.减去长期趋势、季节变动和不规则变动C.除去长期趋势和季节变动D.除去长期趋势、季节变动和不规则变动答案：D.23、在时间序列分析中，一般用循环变动指数（C%）来测定循环波动，（）是计算C%的常用方法。

A. 移动平均法B. 移动平均剔除法C.指数平滑法D. 剩余法答案：D.24、本地区2000---2004年人均消费水平（元）为：2000，2090，2200，2350和2560。

则2005年的三期移动平均预测值为（）。

A. 6640235022002090=++元B.7110256023502200=++C.22133235022002090=++D.23703256023502200=++ 答案：D.25、当时间序列在长时期内呈现连续的不断增长或减少的变动趋势，其逐期增减量又大致相同时，对该时间序列未来的发展前景进行预测，应使用（）。

A. 直线趋势预测模型B.抛物线趋势预测模型C.指数曲线趋势预测模型D.对数曲线趋势预测模型答案：A26、美国劳工部于2003年11月13日公布，经季节因素调整后的第三季度非农业生产率折合成年率增长8.1%，为2002年第一季度以来的最大增幅。

计算这种“无季节性变动的年率”指标主要是为了（）。

A. 消除序列中长期趋势的影响B.消除序列中循环变动的影响C.消除序列中季节变动的影响D.消除序列中不规则变动的影响答案：C27、在（）的情况下，用时间序列预测法作（）才能收到较好的效果。

A. 序列的变化趋势比较明显且差异较大；长期预测B. 序列的变化趋势比较明显且差异较小；长期预测C. 序列的变化趋势比较明显且差异较大；短期预测D.. 序列的变化趋势比较明显且差异较小；短期预测答案：D.28、绝对误差是指（）。

A. 序列的实际观测值与预测值之间的离差B. 序列的实际观测值与预测值之间的绝对离差C. 序列的实际观测值与预测值之间的相对离差D.序列的实际观测值与预测值之间的平均离差答案：A29、彩虹冷饮厂2004年各季度的产值（万元）为：120，480，690和200，季节指数分别为44%，118%，162%和76%。

则用季节比率预测法预测的该企业2005年各季度产值为（）。

A. %44120⨯，%118480⨯，%162690⨯，%76200⨯B. %44)200690480120(⨯+++，%118)200690480120(⨯+++，%162)200690480120(⨯+++，%76)200690480120(⨯+++C.%444200690480120⨯+++，%1184200690480120⨯+++， %1624200690480120⨯+++，%764200690480120⨯+++ D. 以上说法均不正确答案：C30、平滑系数α（）。

A. 越大越好B. 越小越好C. 取值范围在-1到1之间D. 取值范围在0到1之间答案：D.二、多项选择题1.在（）时间序列中，各项指标数值不能相加。

A.绝对数时间序列B.相对数时间序列C.平均数时间序列D.时点序列E.时期序列答案：BCD.2、下列时间序列中属于时点序列的有（）。

A. 某地区高校“十五”期间招收学生人数B. 某地区高校“十五”期间毕业学生人数C. 某地区高校“十五”期间在校学生人数D.. 某企业“十五”期间年末利税额E. 某企业“十五”期间年末固定资产净值答案：CE3、下列时间序列中属于时期序列的有（）。

A. 某地区高校“十五”期间招收学生人数B. 某地区高校“十五”期间毕业学生人数C. 某地区“十五”期间国内生产总值D.某企业“十五”期间年末利税额E. 某企业“十五”期间年末固定资产净值答案：ABCD.4、已知时间序列连续5期的环比增长速度为3%，2%，4%，6%和7%，则（）。

A. 5期的定基增长速度为3%⨯2%⨯4%⨯6%⨯7%B. 5期的定基增长速度为103%⨯102%⨯104%⨯106%⨯107%—1C. 5期的平均发展速度为5%107%106%104%102%103⨯⨯⨯⨯D.5期的平均增长速度为5%7%6%4%2%3⨯⨯⨯⨯E. 5期的平均增长速度为5%107%106%104%102%103⨯⨯⨯⨯—1答案：BCE5、定基发展速度与环比发展速度的关系有（）。

A. 各环比发展速度之和等于相应的定基发展速度B. 各环比发展速度之积等于相应的定基发展速度C. 两个定基发展速度之商等于相应的环比发展速度D.相邻两个定基发展速度之商等于相应的环比发展速度E. 相邻两个定基发展速度之差等于相应的环比发展速度答案：BD.6、按几何平均法计算平均发展速度，要求现象（）。