Financial Time Series Prediction Using Exogenous Series and

garch模型外生变量英文回答：GARCH (Generalized Autoregressive Conditional Heteroskedasticity) models are commonly used in finance and econometrics to model and forecast volatility in financial time series data. These models are able to capture thetime-varying nature of volatility, which is an important characteristic of financial markets.In the context of GARCH models, an exogenous variable refers to a variable that is not directly influenced by the volatility of the financial time series being modeled. Including exogenous variables in a GARCH model allows for the incorporation of additional information that may have an impact on the volatility of the series.There are several reasons why one might want to include exogenous variables in a GARCH model. Firstly, exogenous variables can provide additional explanatory power inmodeling volatility. For example, in the case of stock returns, including macroeconomic variables such as GDP growth or interest rates as exogenous variables can help explain changes in volatility that are not captured by the lagged volatility terms in the GARCH model.Secondly, exogenous variables can help improve the forecast accuracy of the GARCH model. By incorporating additional information into the model, the forecasts can be more accurate and reliable. For example, if there is a known event or announcement that is expected to have an impact on the volatility of a financial time series, including this information as an exogenous variable can help the model capture the effect of the event on volatility.Lastly, including exogenous variables in a GARCH model can help in understanding the underlying dynamics and relationships between the financial time series and the exogenous variables. This can provide valuable insightsinto the factors that drive volatility and help in making informed decisions.To illustrate the use of exogenous variables in a GARCH model, let's consider the example of modeling thevolatility of stock returns. In addition to the lagged volatility terms in the GARCH model, we can include exogenous variables such as interest rates, GDP growth, and company-specific variables like earnings announcements or news events. By including these exogenous variables, we can capture the impact of macroeconomic factors and company-specific events on the volatility of the stock returns.中文回答：GARCH（广义自回归条件异方差）模型是金融和计量经济学中常用的模型，用于建模和预测金融时间序列数据的波动率。

西南财经大学Southwestern University of Finance and Economics2014届本科毕业论文（设计）论文题目：基于小波分解的时间序列预测及应用学生姓名：魏相育所在学院：经济信息工程学院专业：信息系统与信息管理（金融智能与信息管理）学号：41032103指导教师：叶淋宁成绩：2014 年 5 月本科毕业论文（设计）原创性及知识产权声明本人郑重声明：所呈交的毕业论文（毕业设计）是本人在导师的指导下取得的成果。

对本论文（设计）的研究做出重要贡献的个人和集体，均已在文中以明确方式标明。

因本毕业论文引起的法律结果完全由本人承担。

本毕业论文（毕业设计）成果归西南财经大学所有。

特此声明毕业论文作者签名：作者专业：信息系统与信息管理（金融智能与信息管理）作者学号：410321032014年5月10日西南财经大学本科学生毕业论文（设计）开题报告表论文（设计）类型：A—理论研究；B—应用研究；C—软件设计等。

摘要摘要：黄金作为重要的储值工具，其价格对货币市场有相当大的影响。

从2002年起，国际金价持续上涨，其时间序列具有明显的不平稳性。

2008年爆发次贷危机，使全球经济陷入发展疲软状态，但国际金价具有相当好的回调力度，所以其数据具有一定的可研究性。

受全球经济一体化等因素的影响，金融市场波动性加剧，用传统时间序列分析进行预测，会带来一定的误差。

用小波变换对时间序列进行分解和去噪，然后再建立相应的时间序列模型进行预测，可以改变预测的效果。

本文探讨了小波分解与时间序列的理论知识和相关应用，并用国际金价进行了相应的实证分析：先将原始数据进行小波分解处理，对分解后的各低频信号和高频信号进行时间序列的建模及预测，然后用原始的数据进行时间序列的建模和预测，最后比较二者的预测误差和百分比，得到在小波分解的基础上进行的时间序列预测结果更好的结论。

同时，在对各个时间序列进行分析的过程中，本文用到了单位根检验、AIC 准则定阶法、相关函数确定模型以及残差序列等统计检验，为模型的建立和预测提供支持和验证。

时间序列分析在金融领域的应用随着社会经济的发展，金融业逐渐成为社会经济的发展动力。

金融业对经济的发展具有非常重要的作用，能够有效地推动国家经济的发展，提高人们的生活水平。

因此，金融领域的研究一直备受关注，而时间序列分析对于金融领域的研究具有重大意义。

时间序列分析是一种用于对观测数据进行分析和预测的科学方法。

它被广泛应用于金融领域。

时间序列分析包括时间序列模型、时间序列预测和时间序列模型相关性研究等。

通过对时间序列分析的研究，可以确定未来市场方向、预测商品价格走势、制定投资策略等。

时间序列分析在金融领域的应用可以分为以下几个方面：一、股票价格预测投资者通过股票市场买卖股票来赚取收益，对股票价格的预测成为投资者决策的重要依据。

时间序列分析可用来预测股票价格的变化。

例如，通过时间序列分析，可以确定未来市场方向，进而选择合适的投资产品和策略，减小投资风险，提高投资收益。

二、货币政策的制定货币政策是央行通过调整货币供应量、利率等手段，用于稳定经济增长和物价稳定的政策。

而时间序列分析则是央行制定货币政策的一种重要方法。

通过对货币领域的时间序列数据的预测和分析，央行可以有效地调整各种金融政策，实现货币政策的稳定和有效实施。

三、汇率预测汇率波动影响了国内外经济贸易的发展，因而对于汇率波动的预测成为了金融领域研究的一个重要方向。

时间序列分析在汇率预测中能够发挥重要作用。

通过对汇率领域的时间序列数据的分析，可以有效预测未来汇率走势，提供对外经济决策的依据。

四、债券价格预测在金融市场中，债券是一种重要的投资产品。

在债券交易中，需要对债券价格做出预测。

时间序列分析能够对债券价格进行有效的预测，给投资者提供科学的决策依据，从而减小投资风险，提高投资收益。

总之，时间序列分析在金融领域的应用已经成为了一种重要的研究方法。

随着数据交换和处理的发展，时间序列分析将会更加便捷和高效，为金融领域提供更加准确和科学的决策依据。

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。