2020Semester TwoAssignment 2(September2020)STUDENT ID:__ __ __ __ __ __ __ __SURNAME: __ __ __ __ __ __ __ __ __ __ __ __ ___ GIVEN NAME: __ __ __ __ __ __ __ ____ __ __ __ __ __ __ UNIT CODE:ETC4460/ETC5460UNIT TITLE: Financial Econometrics IITASK DURATION:Monday 14th9:00 hrs–Friday 18th16:00 hrsUpon completion of this task please upload to Moodle as a Word or PDF document through Moodle submission. This submission must occurby 4pm onFriday 18thSeptember2020.(Australian Eastern Daylight Time).Please attach a copy of this page as a cover page to your submission with your personal detailscompleted as asked for above. ETC4460/ETC5460Financial Econometrics IIAssignment 2, 2020InstructionsThe due date for this assignment is4pmonFriday 18th September 2020. Your completedassignments must be uploaded to theAssignment 2link on Moodle no later than this time anddate. The assignment is worth 15% of your final grade.TheRpackages that you may need to complete this assignment have already been encountered inthe tutorial exercises (Exercises 1-6). You can use other “non-R”software if you wish.Eviewsisalso available on MoVE (see Moodle page) for example.All of the data is available on Moodle in the Excel spreadsheet namedCokePepsi.csvNote (as with Assignment 1) that while we are happy for you all to discuss your individual dataanalyses with each other (and indeed you are encouraged to do so!), your submitted written as-signment must be your own work.2 Question 1(Abstract 15 marks)Assume that an assetRhas a return densityf(r) that has a polynomial left hand tail, orequivalently a loss density with a polynomial right hand tail, butf(r) is otherwise unspecified.Because the return density is assumed to have a polynomial left tail, the return densityf(r)satisfiesf(r)∼Ar−(a+1)asr→−∞,whereA >0 is a constant anda >0. The parameterais referred to as the tail index and thenotation∼means that the ratio of the left-hand side to the right-hand side converges to 1 asr→−∞. Suppose we wish to estimate a Value at Risk (VaR), where ifRis revenue thenV aR(α)is the negative of theαquantile ofR, so there is only a 100α% chance of a loss exceedingV aR(α).(a) Show thatPr(R≤r)∼Aaraasr→−∞and hence thatPr(R≤−r)Pr(R≤−r0)∼(rr0)a.for any pair of valuesr >0 andr0>0.(b) IfVis the value of the current position, thenV aR(α) =−V×q(α) whereq(α) is theαquantile ofRand the minus sign converts revenue (return times initial investment) into aloss. Verify that ifr=V aR(α) andr0=V aR(α0), where 0< α < α0, thenα1/aV aR(α)∼α1/a0V aR(α0).(c) How might the previous expression be used to construct estimates ofV aR(α) for smallvalues ofαfrom an estimate ofV aR(α0) for a large value ofα0?Question 2(Practical 15 marks)This question uses the data file namedCokePepsi.csv, containing daily prices of Coca-Colastock. The following code computes the daily returns for Coca-Cola stock from January 2007 toNovember 2012 and provides a plot of the returns.1. CokePepsi = read.table("CokePepsi.csv", header=T)2. price = CokePepsi[,1]3. returns = diff(price)/lag(price)[-1]4. ts.plot(returns)5. alpha = 0.056. qalpha = as.numeric(quantile(returns, alpha))a. What is being calculated in the last line of this code?b. Use the above calculations, together with any others that are needed, to evaluate anestimate ofV aR(0.05) for an investment of A$50,000 in Coca-Cola stock.3 Now assume that the returns are i.i.d. and follow a Student’st-distribution. At-distribution ischaracterised by it’s meanμ, a scale parameterλ, and it’s degrees-of-freedomν.1The standarddeviation of at-distributionσ=λ√ν/(ν−2). Run the followingRcommands to fit at-distribution to the data and get corresponding parameter estimates.1. library(MASS)2. fitt = fitdistr(returns, "t")3. param = as.numeric(fitt$estimate)4. mean = param[1]5. nu = param[3]6. sd = param[2] * sqrt((nu) / (nu - 2))7. lambda = param[2]8. qalpha = qt(alpha, df = nu)c. What quantity is being computed in the last line of this code?d. Use the above calculations, together with any others that are needed, to evaluate anestimate ofV aR(0.05) for a A$50,000 investment.Now suppose that we have data on returns forntime periods,r1,...,rn, and we want to estimateVaR for the next return in periodn+ 1. Letμn+1|nandσn+1|ndenote the conditional mean andvariance of tomorrows returnRn+1, conditional on the current information set, which we take assimplyR1,...,Rn. If we assume thatRn+1has a conditionalt-distribution with tail indexνandthatRis anARMA+GARCHprocess, then we can estimatesμn+1|n,σn+1|nandν. Thefollowing code will fit such anARMA+GARCHmodel to the Coca-Cola returns and calculateone step ahead forecasts.1. library(fGarch) # for qstd() function2. library(rugarch)3. garch.t = ugarchspec(mean.model=list(armaOrder=c(0,0)),4. variance.model=list(garchOrder=c(1,1)),5. distribution.model = "std")6. CC.garch.t = ugarchfit(data=returns, spec=garch.t)7. show(CC.garch.t)8. plot(CC.garch.t, which = 2)9. pred = ugarchforecast(CC.garch.t, data=returns, n.ahead=1) ; pred10. fitted(pred) ; sigma(pred)11. nu = as.numeric(coef(CC.garch.t)[5])1At-distribution has polynomial tails with a tail index equal to it’s degrees-of-freedom.4 12. q = qstd(alpha, mean = fitted(pred), sd = sigma(pred), nu = nu) ; q13. sigma(pred)/sqrt( (nu)/(nu-2) )14. qt(alpha, df=nu)e. Carefully express the fittedARMA+GARCHmodel that has been fitted to the data inmathematical notation. What is being plotted in line 8?f. What are the one-step ahead predictions of the conditional mean and conditional standarddeviation?g. What is the difference between the calculations being conducted in lines 12 and 14?h. Use the above calculations, together with any others that are needed, to construct an predic-tion ofV aR(0.05) for a A$50,000 investment.Now suppose that we wish to estimateV aR(0.01) for a A$50,000 investment.i. Provide an econometric/statistical explanation of why the three values ofV aR(0.05) for aA$50,000 investment that you have calculated in questions 2b. 2d. and 2h. are different.j. If you were going to use one of these estimates to estimateV aR(0.01) for a A$50,000 invest-ment, which one would you choose and why.Question 3(Abstract 15 marks)LetFBtbe generated by{Bs:s∈[0,t]}. Show that the following process:I(Bt) =12(B2t−t)isa continuous–time martingale with respect toFBt.Question 4(Practical 15 marks)Consider a discretized model of the form: fori= 0,1,2,···,10∆St+i∆tSt+i∆t=μ∆t+σ√∆tε,(4.1)whereε∼N(0,1).Suppose that the expected return from the stock is 14% per annum and that the standard deviationof the return (i.e., the volatility) is 20% per annum. If time is measured in years, it follows thatμ= 0.14 andσ= 0.20.Suppose that ∆t= 0.01 so that we are considering changes in the stock price in time intervals oflength 0.01 years (or 3.65 days). It follows that∆StSt= 0.14×0.01 + 0.2×√0.01ε.(4.2)A path for the stock price can be simulated by sampling repeatedly from model (4.1). One procedurefor doing this is to sample values fromε∼N(0,1) and then use (4.2) to calculate ∆St.The initial stock price is assumed to be $20. For the first period of the random numbers,ν1,sampled fromN(0,1) is 0.52. Using equation (4.2), this gives a random sample of 0.0118 for∆StSt.Thus,∆St= 20×0.0118 = 0.236.5 At the beginning of the next period the stock price is thereforeSt+∆t=St+ ∆St= $20 + $0.236 = $20.236.Equation (4.1) then provides ∆St+∆t. By repeatedly simulating movements in the stock price asfollows:St+2∆t=St+∆t+ ∆St+∆t,St+3∆t=St+2∆t+ ∆St+2∆t,···so on,St+10∆t=St+9∆t+ ∆St+9∆t,produce a tablefor a complete probability distribution of the stock prices:St+∆t,St+2∆t,···,St+10∆tfor the final step at the end of one–tenth of a year, 10 ∆t= 10×0.01 = 0.1.Note: You should regenerateν1in each step, rather than using the first value of 0.52. If you dogenerate more than one value fromε∼N(0,1) in each step, you may use the simple mean for thatstep.6