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1、Booth School of Business,University of ChicagoBusiness 41202,Spring Quarter 2012,Mr.Ruey S.TsaySolutions to MidtermProblem A:(34 pts)Answer briefly the following questions.Each questionhas two points.1.Describe two improvements of the EGARCH model over the GARCHvolatility model.Answer:(1)allows for
2、asymmetric response to past positive or negativereturns,i.e.leverage effect,(2)uses log volatility to relax parameterconstraint.2.Describe two methods that can be used to inferthe existenceof ARCHeffects in a return series,i.e.,volatility is not constant over time.Answer:(1)The sample ACF(or PACF)of
3、 the squared residuals ofthe mean equation,(2)use the Ljung-Box statistics on the squaredresiduals.3.Consider the IGARCH(1,1)volatility model:at=ttwith 2t=0+12t1+(1 1)a2t1.Often one pre-fixes 0=0.Why?Also,suppose that 0=0 and the 1-step ahead volatility prediction at theforecast origin h is 16.2%(an
4、nualized),i.e.,h(1)=h+1=16.2 for thepercentage log return.What is the 10-step ahead volatility prediction?That is,what is h(10)?Answer:(1)Fixing 0=0 based on the prior knowledge that volatilityis mean reverting.(2)h(10)=16.2.4.(Questions 4 to 8)Consider the daily log returns of Amazon stockfrom Janu
5、ary 3,2007 to April 27,2012.Some summary statistics ofthe 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 thatthe mean return is not zero,because the 95%confidence interval con-tains zero.5.L
6、et k be the excesskurtosis.Test H0:k=0 versus Ha:k=0.Writedown the test statistic and draw the conclusion.1Answer:t-ratio=9.87524/1340=73.79,which is highly significant com-pared with 21distribution.6.Are there serial correlations in the log returns?Why?Answer:No,the Ljung-Box statistic Q(10)=10.69
7、with 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)=39.24 with p-value less than 0.05.8.Based on the summary statistics provided,what is the 22-step aheadpoint forecast of the log return at the forecast origin Apri
8、l 27,2012?Why?Answer:The point forecast rT(22)=0 because the mean is not signif-icantly different from zero.Give students 1 point if they use samplemean.9.Give two reasons that explain the existence of serial correlations in ob-served asset returns even if the true returns are not serially correlate
9、d.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 highfrequency returns.11.Describe t
10、wo methods that can be used to compare different modelsfor a given time series.Answer:(1)Information criteria such as AIC or BIC,(2)backtestingor out-of-sample forecasting.12.(Questions 12 to 14)Let rtbe the daily log returns of Stock A.Assume that rt=0.004+at,where at=ttwith tbeing iid N(0,1)random
11、 variates and 2t=0.017+0.15a2t1.What is the unconditionalvariance of at?Answer:Var(at)=0.01710.15=0.02.13.Suppose that the log price at t=100 is 3.912.Also,at the forecastorigin t=100,we have a100=0.03 and 100=0.025.Compute the21-step ahead forecast of the log price(not log return)and its volatility
12、for Stock A at the forecast origin t=100.Answer:r100(1)=0.004 so that p100(1)=3.912+0.004=3.916.Thevolatility forecast is2100(1)=0.017+0.15(0.03)2=0.131.14.Compute the 30-step ahead forecast of the log price and its volatilityof Stock A at the forecast origin t=100.Answer:p100(30)=3.912+0.00430=4.03
13、2 and the voaltility is theunconditional stantard error0.02=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 rtof Stock A follows the model rt=at,at=tt,and
14、2t=0+1a2t1+12t1,where tare iid N(0,1).Under whatcondition that the kurtosis of rtis 3?That is,state the condition underwhich the GARCH dynamics fail to generate any additional kurtosisover that of t.Answer:1=0.17.What is the main consequence in using a linear regression analysis whenthe serial corre
15、lations of the residuals are overlooked?Answer:The t-ratios of coefficient estimates are not reliable.Problem B.(30 pts)Consider the daily log returns of Amazon stock fromJanuary 3,2007 to April 27,2012.Several volatility models are fitted to thedata and the relevant R output is attached.Answer the
16、following questions.1.(2 points)A volatility model,called m1 in R,is entertained.Writedown the fitted model,including the mean equation.Is the modeladequate?Why?Answer:ARCH(1)model.rt=0.0018+at,at=ttwith tbeing iidN(0,1)and 2t=7.577 104+0.188a2t1.The model is inadequatebecause the normality assumpti
17、on is clearly rejected.2.(3 points)Another volatility model,called m2 in R,is fitted to thereturns.Write down the model,including all estimated parameters.3Answer:ARCH(1)model.rt=4.907104+at,at=tt,where tt3.56with tvdenotingstandardizedStudent-t distribution with v degreesof freedom.The volatility e
18、quation is 2t=7.463 104+0.203a2t1.3.(2 points)Based on the fitted model m2,test H0:=5 versus Ha:=5,where denotes the degrees of freedom of Student-t distribution.Perform the test and draw a conclusion.Answer:t-ratio=3.56250.366=3.93,which compared with 1.96 is highlysignificant.If you compute the p-
19、value,it is 8.53 105.Therefore,v=5 is rejected.4.(3 points)A third model,called m3 in R,is also entertained.Writedown the model,including the distributional parameters.Is the modeladequate?Why?Answer:Another ARCH(1)model.rt=0.0012+at,at=tt,wheretare iid and follow a skew standardized Student-t distr
20、ibution with skewparameter 1.065 and degrees of freedom 3.591.The volatility equationis 2t=7.418 104+0.208a2t1.Ecept for the insigicant mean value,the fitted ARCH(1)model appears to be adequatebasedon the modelchecking statistics shown.5.(2 points)Let be theskew parameter in model m3.Does the estima
21、teof confirm that the distribution of the log returns is skewed?Why?Perform the test to support your answer.Answer:The t-ratio is1.06510.039=1.67,which is smaller than 1.96.Thus,the null hypothesis of symmetric innovations cannot be rejectedat the5%level.6.(3 points)A fourth model,called m4 in R,is
22、also fitted.Write downthe fitted model,including the distribution of the innovations.Answer:a GARCH(1,1)model.rt=0.0017+at,at=tt,where tare iid and follow a skew standardized Student-t distribution with skewparameter 1.101 and degrees of freedom 3.71.The volatility equationis 2t=1.066 105+0.0414a2t1
23、+0.9502t1.7.(2 points)Based on model m4,is the distribution of the log returnsskewed?Why?Perform a test to support your answer.Answer:The t-ratio is1.10110.043=2.349,which is greater than 1.96.Thus,the distribution is skew at the 5%level.48.(2 points)Among models m1,m2,m3,m4,which model is preferred
24、?State the criterion used in your choice.Answer:Model 4 is preferred as it has a smaller AIC value.9.(2 points)Since the estimates 1+1is very close to 1,we consideran IGARCH(1,1)model.Write down the fitted IGARCH(1,1)model,called m5.Answer:rt=at,at=tt,where 2t=3.859 105+0.852t1+0.15a2t1.10.(2 points
25、)Use the IGARCH(1,1)model and the information providedto obtain 1-step and 2-step ahead predictions for the volatility of thelog returns at the forecast origin t=1340.Answer:From the output 21340(1)=21341=3.859 105+0.85(0.02108)2+0.15(.146)2=0.00361.Therefore,21340(2)=3.859105+21340(1)=0.00365.Thevo
26、latility forecasts are then 0.0601 and 0.0604,respectively.11.(2 points)A GARCH-M model is entertained for the percentage logreturns,called m6 in the R output.Based on the fitted model,is therisk premium statistical significant?Why?Answer:The risk premium parameter is 0.112 with t-ratio 0.560,which
27、is less than 1.96 in modulus.Thus,the risk premium is notstatistical significant at the 5%level.12.(3 points)Finally,a GJR-type model is entertained,called m7.Writedown the fitted model,including all parameters.Answer:This is an APARCH model.The model is rt=0.0014+at,at=tt,where tare iid and follow
28、a skew standardized Student-tdistribution with skew parameter 1.098 and degrees of freedom 3.846.The volatility equation is2t=7.583 106+0.0362(|at1|0.478at1)2+0.9532t1.13.(2 points)Based on the fitted GJR-type of model,is the leverage effectsignificant?Why?Answer:Yes,the leverage parameter 1is signf
29、iicantly different fromzero so that there is leverage effect in the log returns.5Problem C.(14 pts)Consider the quarterly earnings per share of AbbottLaboratories(ABT)stock from 1984.III to 2011.III for 110 observations.Weanalyzedthe logarithmsof the earnings.That is,xt=ln(yt),where ytis thequarterl
30、yearningsper share.Two models are entertained.1.(3 points)Write down the model m1 in R,including residual variance.Answer:Let rtbe the log earningsper share.The fitted model is(1 B)(1 B4)rt=(1 0.565B)(1 0.183B4)at,2a=0.00161.2.(2 points)Is the model adequate?Why?Answer:No,the Ljung-Box statistics of
31、 the residuals give Q(12)=25.76 with p-value 0.012.3.(3 points)Write down the fitted model m2 in R,including residualvariance.Answer:The fitted model is(1 B)(1 B4)rt=(1 0.470B 0.312B3)at,2a=0.00144.4.(2 points)Model checking of the fitted model m2 is given in Figure 1.Is the model adequate?Why?Answe
32、r:Yes,the model checking statistics look reasonable.5.(2 points)Compare the two fitted model models.Which model ispreferred?Why?Answer:Model 2 is preferred.It passes model checking and has asmaller AIC value.6.(2 points)Compute 95%interval forecasts of 1-step and 2-step aheadlog-earnings at the fore
33、cast origin t=110.Answer:1-step ahead prediction:0.3751.960.038,and 2-step aheadprediction:0.01881.960.043.(Some students may use 2-step aheadprediction due to the forecast origin confusion.)Problem D.(22 pts)Consider the growth rate of the U.S.weekly regulargasoline price from January 06,1997 to Se
34、ptember 27,2010.Here growthrate is obtained by differencing the log gasoline price and denoted by gt inR output.The growth rate of weekly crude oil from January 03,1997 toSeptember 24,2010 is also obtained and is denoted by pt in R output.Notethat the crude oil price was known 3 days prior to the ga
35、soline price.61.(2 points)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 gtis not significantly different from zero
36、based on the one-sample t-test.The p-value is 0.19.2.(2 points)Write down the fitted AR(5)model,called m2,includingresidual variance.Answer:The fitted model is(10.507B0.079B20.136B3+0.036B4+0.086B5)gt=at,2a=0.000326.3.(2 points)Since not all estimates of model m2 are statistically signifi-cant,we re
37、fine the model.Write down the refined model,called m3.Answer:The fitted model is(1 0.504B 0.074B2 0.122B3+0.101B5)gt=at,2a=0.000327.4.(2 points)Is the refined AR(5)model adequate?Why?Answer:Yes,the Ljung-Box statistics of the residuals give Q(14)=10.27 with p-value 0.74,indicating that there are no
38、serial correlationsin the residuals.5.(2 points)Does the gasolineprice show certain business-cycle behavior?Why?Answer:Yes,the fitted AR(5)polynomial contains compplex solutions.6.(3 points)Next,consider using the information of crude oil price.Writedown the linear regression model,called m4,includi
39、ng R2and residualstandard error.Answer:The fitted linear regression model isgt=0.287pt+t,=0.0184,and the R2of the regression is 33.66%.7.(2 points)Is the fitted linear regression model adequate?Why?Answer:No,because the residuals tare serially correlated based onthe Ljung-Box test.78.(3 points)A lin
40、ear regression model with time series errors is enter-tained and insignificant parameters removed.Write down the finalmodel,including all fitted parameters.Answer:The model is(10.404B0.164B20.096B3+0.101B5)(gt0.191pt)=at,2a=0.000253.9.(2 points)Model checking shows that the fitted final model has no
41、residual serial correlations.Based on the model,is crude oil pricehelpful in predicting the gasoline price?Why?Answer:Yes,because the fitted coefficient of ptis signficantly differentfrom zero.10.(2 points)Compare the pure time series model and the regressionmodelwith 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