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。

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==。

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）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_________。

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。

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。

绝密★ 启用前2019 年普通高等学校招生全国统一考试理科综合能力测试注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案标号。

回答非选择题时，将答案写在答题卡上，写在本试卷上无效。

3．考试结束后，将本试卷和答题卡一并交回。

可能用到的相对原子质量：H 1 C 12 N 14 O 16 Mg 24 S 32 Fe 56 Cu 64一、选择题：本题共13个小题，每小题6分。

共78分，在每小题给出的四个选项中，只有一项是符合题目要求的。

1．细胞凋亡是细胞死亡的一种类型。

下列关于人体中细胞凋亡的叙述，正确的是A •胎儿手的发育过程中不会发生细胞凋亡B •小肠上皮细胞的自然更新过程中存在细胞凋亡现象C.清除被病原体感染细胞的过程中不存在细胞凋亡现象D •细胞凋亡是基因决定的细胞死亡过程，属于细胞坏死2 .用体外实验的方法可合成多肽链。

已知苯丙氨酸的密码子是UUU ,若要在体外合成同位素标记的多肽链,所需的材料组合是①同位素标记的tRNA②蛋白质合成所需的酶③同位素标记的苯丙氨酸④人工合成的多聚尿嘧啶核苷酸⑤除去了DNA 和mRNA 的细胞裂解液A ．①②④B ．②③④C ．③④⑤D ．①③⑤3•将一株质量为20 g的黄瓜幼苗栽种在光照等适宜的环境中，一段时间后植株达到40 g,其增加的质量来自于A •水、矿质元素和空气B •光、矿质元素和水C.水、矿质元素和土壤D •光、矿质元素和空气4 •动物受到惊吓刺激时，兴奋经过反射弧中的传出神经作用于肾上腺髓质，使其分泌肾上腺素；兴奋还通过传出神经作用于心脏。

下列相关叙述错误的是A •兴奋是以电信号的形式在神经纤维上传导的B •惊吓刺激可以作用于视觉、听觉或触觉感受器C.神经系统可直接调节、也可通过内分泌活动间接调节心脏活动D •肾上腺素分泌增加会使动物警觉性提高、呼吸频率减慢、心率减慢5 •某种二倍体高等植物的性别决定类型为XY型。

第六章动态数列-、判斷题若将某地区社会商品库存额按时间先后顺序排列，此种动态数二、1.列属于时期数列。

（）定基发展速度反映了现象在一定时期内发展的总速度，环比发三、2.展速度反映了现象比前一期的增长程度。

（）平均增长速度不是根据各期环比增长速度直接求得的，而是根四、3.据平均发展速度计算的。

（）•用水平法计算的平均发展速度只取决于最初发展水平和最末发五、4展水平，与中间各期发展水平无关。

（）平均发展速度是环比发展速度的平均数，也是一种序时平均六、5.数。

（）1> X 2、X 3、J 4、V 5. Vo七、单项选择题•根据时期数列计算序时平均数应采用（）。

八、1几何平均法 B.加权算术平均法C.简单九、 A.算术平均法 D.首末折半法十、2•下列数列中哪一个属于动态数列（）。

十-、 A.学生按学习成绩分组形成的数列 B.工业企业按地区分组形成的数列十二、 C.职工按工资水平高低排列形成的数列 D.出口额按时间先后顺序排列形成的数列十三、3.已知某企业1月、2月、3月、4月的平均职工人数分别为190人、195人、193人和201人。

则该企业一季度的平均职工人数的计算方法为（）。

十四、心(190+195+193+201)4B.190+195 + 1933十五.(190/2)+195+193 + (201/2) 、［190/2)+195+193+(201/2)C・D・ ---------------------------------4-1 44.说明现象在较长时期内发展的总速度的指标是()。

A、环比发展速度 B.平均发展速度 C.定基发展速度 D.环比增长速度5•已知各期环比增长速度为2%、5%、8%和7%,则相应的定基增长速度的计算方法为()。

A.(102%X105%X108%X107%) -100%B.102%X105%X108%X107%C.2%X5%X8%X7%D.(2%X5%X8%X7%) -100%6•定基增长速度与环比增长速度的关系是()。