金融时间序列分析_期中试卷

一、单项选择题（每题2分，共20分） P61关于严平稳与（宽）平稳的关系；弱平稳的定义：对于随机时间序列y t ，如果其期望值、方差以及自协方差均不随时间t 的变化而变化，则称y t 为弱平稳随机变量，即y t 必须满足以下条件： 对于所有时间t ，有 （i ）E （yt ）=μ为不变的常数；（ii ） Var （yt ）=σ²为不变的常数；（iii ） γj =E[y t -μ][y t-j -μ]，j=0，±1，,2，… （j 为相隔的阶数）（μ=0，cov （y t ，y t-j ）=0，Var （yt ）=σ²时为白噪音过程，常用的平稳过程。

） 从以上定义可以看到，凡是弱平稳变量，都会有一个恒定不变的均值和方差，并且自协方差只与y t 和y t-j 之间的之后期数j 有关，而与时间t 没有任何关系。

严平稳过程的定义：如果对于任何j 1,，j 2，...,j k ,随机变量的集合（y t ，y t+j1,，y t+j2，…，y t+jk ）只依赖于不同期之间的间隔距离（j 1，j 2，…，j k ），而不依赖于时间t ，那么这样的集合称为严格平稳过程或简称为严平稳过程，对应的随机变量称为严平稳随机变量。

P46 t X 的k 阶差分是；△ｋX t =△ｋ-1X t -△ｋ-1X t-1，△ 表示差分符号。

滞后算子；P54对于AR ： L p y t =y t-p ，对于MA ：L ｐεt =εt-ｐAR （p ）模型即自回归部分的特征根—平稳性；确定好差分方程的阶数，则其特征方程为：λp-α1λp-1-α2λp-2-…-αp =0，若所有的特征根的│λ│<1则平稳补充：逆特征方程为：1-α1ｚ1-α2ｚ²-…-αp ｚp=0，若所有的逆特征根│ｚ│>1，则平稳。

注意：特征根和逆特征方程的根互为倒数。

如：p57作业3： y t =1.2y t-1-0.2y t-2+εt ，为二阶差分，其特征方程为：λ2-1.2λ+0.2=0，解得λ1=1，λ2=0.2，由于λ1=1，所以不平稳。

芜湖广播电视大学2 0 11 -2 0 1 2学年度第二学期"开放本科"期末考试货币银行学试题一、名词解释{每小题5分，共1 5分)1. 金融结构:是指构成金融总体的各个组成部分的分布、存在、相对规模、相互关系与配合的状态。

2. 格雷欣法则:金银复本位制条件下出现劣币驱逐良币现象。

指当两种实际价值不同而名义价值相同的铸币同时流通时，实际价值较高的通货即良币，被熔化、收藏或输出国外，退出流通，而市价价值较低的劣币则充斥市场的现象。

3. 货币乘数:指货币供应量对基础货币的倍数关系，亦即基础货币每增加或减少一个单位所引起的货币供给量增加或减少的倍数。

不同口径的货币供给量有各自不同的货币乘数。

二、判断正确与错误{正确的打.J .错误的打×。

每小题1分，共1 0分)1.银行信用与商业信用是对立的，银行信用发展起来以后，逐步取代商业信用。

(X )2. 直接标价法指以一定单位的外国货币为基准来计算应付多少单位的本国货币。

(V )3. 升水是远期汇率低于即期汇率，贴水是远期汇率高于即期汇率。

( X)4. 黄金是金属货币制度下货币发行的准备，不兑现的信用货币制度中货币发行不需要黄金准备。

(V )5. 凯恩斯的流动偏好论认为利率是由借贷资金的供求关系决定的。

(X )6. 公开市场业务是通过增减商业银行借款成本来调控基础货币的。

(X )7. 通货膨胀得以发生的前提是现代货币供给的形成机制。

( V)8. 中央银行充当最后贷款人是作为银行的银行的表现。

( V)9. 关于技资的收支都应该记入国际收支平衡表的资本和金融项目中。

(X )10. 国际货市制度创新的其中一个重要表现是区域性货币一体化趋势，它与国际金融监管创新同属于金融组织结构创新。

(X )三、单项选择题{每小题l分，共1 0分，每小题有-项答案正确，请将正确答案的序号填写在括号内}1.货币执行支付手段职能的特点是(C )。

A. 货币是商品交换的媒介B. 货币运动伴随商品运动C. 货币作为价值的独立形式进行单方面转移D 货币是一般等价物2. 目前人民币汇率实行的是(A )。

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。

金融市场中的时间序列分析随着现代经济的发展和供求关系的变化，金融市场日益成为世界经济的核心。

在这个动态的市场中，各种金融工具交易的价格、利率和汇率等变量都在时刻发生着变化，这些变化背后隐藏着丰富的信息和规律。

时间序列分析是研究金融时间序列波动的统计方法，通过对历史数据的分析，可以为金融市场提供有效的预测和决策依据。

一、时间序列分析简介时间序列是指按时间顺序排列的一系列随机变量的观察值。

时间序列分析是对这些观察值的统计分析、模型构建和预测，其基本假设是序列的常见值或趋势改变具有一定的稳定性。

在金融市场中，时间序列分析通常用于对金融变量如股票价格、利率、汇率、价格指数进行分析和预测。

时间序列分析的主要方法包括平稳性检验、白噪声检验、自相关函数和偏自相关函数的绘制、时间序列模型选择和估计等。

常用的时间序列模型包括随机游走模型、自回归移动平均模型（ARMA）、自回归条件异方差模型（ARCH）和广义自回归条件异方差模型（GARCH）等。

二、平稳性检验平稳性是时间序列分析的基本假设，它的意义在于序列的均值、方差和自相关系数等统计量不随时间变化而发生显著变化。

若序列是非平稳的，则需要对其进行差分或变换，使其变为平稳序列。

常见的平稳性检验方法包括ADF检验、KPSS检验、PP检验等。

ADF检验的假设是序列有单位根，即序列不平稳。

检验统计量的值越小，拒绝序列有单位根的假设越强，即序列越平稳。

KPSS检验的假设是序列具有趋势性，即序列不平稳。

检验统计量的值越大，拒绝序列无趋势的假设越强，即序列越不平稳。

PP检验是另一种检测序列平稳性的方法，其假设是序列有单位根。

检验统计量和ADF检验类似，其值越小，拒绝序列有单位根的假设越强。

三、自相关函数和偏自相关函数的绘制自相关函数（ACF）和偏自相关函数（PACF）是判断时间序列是否平稳，以及确定合适的时间序列模型的重要工具。

自相关函数是指对平稳序列按照时间先后顺序计算的各个时刻之间的相关系数。

内蒙古财经学院2011——2012学年第1学期《金融时间序列分析》试卷答案一、填空题（1分*15空二15分）1.Xt =0心_1+0X^2 ---------- 0pXt・p+$t 一厲斫-1 ------ %、朝、…帖厲、…、%。

2.描述性：3.兀=斗』+耳，0,1,0；4平稳性检验，纯随机性检验；5. VPx t = （1 - B）Pxt, V k x t = （1 - B k）x t；6 宽平稳，严平稳，宽平稳：7.自回归二、不定项选择题（2分*5题=10分）1、AC2、ABD3、AB4、ABCD5、ABD三、判断并说明理由（2题*5分=10分）1、如果一个时间序列宽平稳，则它肓定不是严平稳；如果一个时间序列严平稳，则它一定是宽平稳。

答：说法是错误的。

（1分）严平稳是一种条件比较苛刻的平稳性定义，该定义表明，一个序列的所有统计均平稳时，该序列才是平稳的。

而宽平稳则是条件宽松的平稳性定义，即只要求序列的二阶矩平稳，则序列就是平稳的。

由定义町知，在一般情况下，如果一个时间序列是宽平稳的，则它肯定不是严平稳的；如果一个时间序列是严平稳的，则它一定是宽平稳的。

（2分）但两种情况各有例外，如多元正态分布，二阶矩包插所有•统计性质，所以对于服从多元正态分布的序列，宽平稳也是严平稳：再比如柯酋分布不存在二阶矩，因此如果一个序列服从柯西分布，且为严平稳, 但却推不出其为宽平稳。

确切的说应该是对于存在二阶矩的序列，严平稳才能推出宽平稳。

（2分）2、差分运算的实质是使用自回归的方式提取确定性信息答：说法是正确的。

（5分〉四、简答题：（25分）1、简述平稳序列的建模步骤（7分）答：（1）时间序列分析的第-步是获得观察值序列，然后对这个序列进行平稳性检验，对平稳的序列进行纯随机性检验，如果是纯随机序列，分析结束：如果不是纯随机序列，选择模型拟合该序列：（2）求出该观察值序列的样本自相关系数（ACF）和样本偏自相关系数（PACF）的值。

一．名词解释（每小题5分，共20分） 1. 零均值白噪声序列 2. 严平稳序列 3. 自协方差函数 4. 均方意义下收敛二．简答题（每小题10分，共20分）1.请简述ARMA(p,q)序列的之平稳域的定义。

2.请简要描述Ljung-Box检验的过程（包括原假设、备择假设、统计量的定义和抽样分布）三．计算分析题（每小题15分，共45分）1. 设cos()sin()t t t X ξη+=，其中ξ，η独立且服从N(0,1)。

完成下列问题： 1a ： 给出{}t X 的自协方差函数(,)t t k γ+和自相关函数(,)t t k ρ+的表达式；（10分）1b ：{}t X 是二阶宽平稳序列吗，为什么？（5分）2.考虑时间序列120.25t t t t X X X ε--=+-其中t ε为一零均值白噪声序列，方差为2。

2a: 判断该序列的种类，并说明该序列是否是二阶宽平稳序列（5分）；2b ： 给出该序列的自协方差函数（10分）。

3. 判定下列各序列的阶数，然后判断平稳性（给出详细的理由和判断步骤） 3a. 12100.30.20.3t t t t t X X X εε----+=-（5分） 3b. 21.2t t t X εε-=-（5分） 3c. 1t t t X X ε--=（5分）四．证明题（15分）若序列0t k k k t X c ε∞-==∑，其中{},0,1,2,...t t ε=是零均值白噪声序列，方差为2τ，02k k c =∞∞<∑。

试证明： 1. 0t EX = （5分）； 2. ()20cov ,t s j s t j j X X c c σ∞-+==∑ （10分）。